General Relativity and Gravitational Waves: Session 2. General Coordinates Thomas A. Moore — Les Houches — July 4, 2018 Overview of this session:

2.2 Definition of a Coordinate Basis 2.3 Tensors in a Coordinate Basis 2.4 The Tensor Gradient 2.5 The Geodesic Equation 2.6 Schwarzschild Geodesics 2.7 LOFs and LIFs u=–1 u=0 u=1 u=2

ew Q w=5 w=4+dw ds

eu dw ew P w=4 P due w=4 u

w=3 u=1 u=1+du

Figure 1: This drawing shows an arbitrary coordinate system in a possibly curved space, a point , the basis P vectors eu, ew at that point, and a close-up view of how we define the basis vectors so that an infinitesimal displacement ds is the vector sum of the basis vectors times simply du and dw respectively.

I am not going to emphasize rigor in what follows, but rather an intuitive approach that I think will help you quickly develop a working understanding of the mathematics. Some might complain about all of the assumptions I am leaving unstated and the hands that I am waving, but my time with you is too short. No matter how we construct our coordinate system, the di↵erential distance ds between two infinitesimally separated points in a two-dimensional space (or the spacetime separation between two events in a spacetime) is a coordinate-independent quantity, because we can measure it directly with a ruler (or a clock in spacetime) without having to define a coordinate system at all. The fundamental way that we connect arbitrary coordinates to physical reality is by specifying how the spacetime separation between two infinitesimally- separated points (or events) depends on their coordinate separations. Let’s first consider a two-dimensional flat space. In such a space, a cartesian xy coordinate system is one in which the distance ds between two infinitesimally-separated points is given by ds2 = dx2 + dy2 at all points in the space. A curvilinear coordinate system is any non-cartesian coordinate system where this simple pythagorean relationship is not true. How can we connect the coordinate-independent distance between two points with their coordinate separations in such a case? Consider arbitrary coordinatesCoordinateu, w for a 2D space. Basis When using index notation, we will interpret dxu as being equivalent to du, and dxu as being equivalent to dw, and we will assume that Greek indices have two possible values u and w. (In the last session, in the context of cartesian coordinates in non-curved spacetime, u=–1 u=0 u=1 u=2 I stated that indices could represent either t, x, y, or z, but when we use arbitrary coordinates, the indices take on values corresponding to whatever namesew the coordinate have.) I willQ also represent vectors in this two-dimensional space with the same bold-face notation as wew=5 usedw=4+dw for four-vectorsds last time. This will keep eu dw ew the notation from changing when we generalizeP to 4D spacetimes.w=4 P due Now, no matter how our u, w coordinate system is defined,w at=4 each pointu in the space, we can define P a pair of basis vectors eu, ew such that w=3 u=1 u=1+du 1. eu points tangent to the w = constant curve toward increasing values of u. Define basis vectors eu, ew such that

2. ew points tangent to the 1.u =eu constant points tangent curve toward to the increasing w = constant values curve of w. in the direction of increasing u 3. Their lengths are defined so that the displacement vector ds between a point at coordinates u, w 2. e points tangent to the u = constant curve and an infinitesimally separatedw neighboring point at coordinates u + du, w +Pdw can be written in the direction of increasingQ w µ 3. ddss= =du dueeuu ++ dwdweww = dx eµ (See figure 1.) (2.1)

(Note: the lower index on eµ tells us which basis vector, not which component, we are talking about.) Now, this only works for di↵erential separations, because the directions of the basis vectors change as we move significant distances. Moreover, vector addition as I have illustrated it in Figure 1 is really only defined in a flat space. In a curved space, the separation between and must be so small that we can treat the space as flat, just as we treat a city map as flat on the curvedP surfaceQ of the earth. (Technically, we are doing this vector addition in a flat space that is tangent to the curved space at point .) P

2 Now, ifNow, we define if we define basis vectors basis vectors this way, this then way,du thenanddudwandbecomedw become the components the components of ds in of thatds in basis that and basis and we callwe the call set the of set basis of vectors basis vectorseu, eweau,coordinateew a coordinate basis. basis.A coordinateA coordinate basis is basis generally is generally di↵erent di↵ thanerent than Now, if we define basis vectors this way, then du and dw become the components of ds in that basis and the cartesianthe cartesian coordinate coordinate basis vectors basis vectorsex, ey einx, thatey in (1) thateu (1)andeuewandmayew notmay be not perpendicular, be perpendicular, (2) eu (2)andeuewand ew we call the set of basis vectors eu, ew a coordinate basis. A coordinate basis is generally di↵erent than may notmay have not unit have length, unit length, and (3) andeu (3)and/oreu and/orew mayew changemay change in magnitude in magnitude and/or and/or direction direction as we move as we from move from the cartesianpoint coordinate topoint point. to basis point. It is vectors also It is importante alsox, ey importantin that to note (1) toe thatu noteand defining thatew may defining a not coordinate be a perpendicular, coordinate basis is basis not (2) is theeu notandonly theewwayonly toway define to define e e may not havea set unit ofa length, set basis of vectors basis and (3) vectors oru aand/or coordinate or a coordinatew may system. change system. For in magnitudeexample, For example, standard and/or standard direction polar coordinates polar as we coordinates move in from two-dimensional in two-dimensional point to point. It is also important to note that defining a coordinate basis is not the only way to define flat spaceflat and space spherical and spherical coordinates coordinates in a three-dimensional in a three-dimensional flat space flat do spacenot dousenot a coordinateuse a coordinate basis, becausebasis, because a set of basis vectors or a coordinate system. For example, standard polar coordinates in two-dimensional the basisthe vectors basis vectors in those in coordinate those coordinate systems systems always always have unit have magnitude. unit magnitude. Using Using a coordinate a coordinate basis makes basis makes flat space and spherical coordinates in a three-dimensional flat space do not use a coordinate basis, because componentscomponentsof theof displacement the displacement vector vectords simplerds simpler (at the (at expense the expense of complexity of complexity in the in basis the vectors), basis vectors), and and the basis vectors in those coordinate systems always have unit magnitude. Using a coordinate basis makes this turnsthis out turns to out be crucial to be crucial in what in follows. what follows. components of the displacement vector ds simpler (at the expense of complexity in the basis vectors), and Once weOnce have we defined have definedCoordinate a coordinate a coordinate basis for basis coordinates for Basis coordinatesu and uwand,thenwew,thenwedefine definethe componentsthe components of a of a this turns out to be crucial in what follows. four-vectorfour-vectorA at anyA at point any pointto be Atou,A bewAsuchu,Aw thatsuch that Once we have defined a coordinateP basisP for coordinates u and w,thenwedefine the components of a u w four-vector A at any point to be A ,A such that u u w w µ µ P Define a four-vector’sA = AAe=u components+AAeue+w A= Aew soe=µ thatA eµ (2.2) (2.2) A = Aue + Awe = Aµe (2.2) This ensuresThis ensures that the that components the components of dus and of dsAwandtransformA transformµ alike. alike. Now, the scalar product of ds with itself is the square of the physical distance between the endpoints of This ensures thatNow, the the components scalar product of ds ofandds Awithtransform itself is alike. the square of the physical distance between the endpoints of Now, thethe scalar displacementthe product displacement of itd represents:s with it represents:The itself metric is the in square a coordinate of the physical basis distanceis between the endpoints of the displacement it represents: 2 2 ds = dsds ⇧=dsd=(s ⇧ duds =(eu +dudweue+w)dw⇧ (eduw)e⇧u(+dudweue+w)dw ew) 2 2 2 2 2 ds = ds ⇧ ds =(=dudueue=u+du⇧dweuee+uw⇧du)e⇧u( dwdu+ dueeuu dw+⇧ edwwe+ue⇧wdwe)w dw+ dwew dw⇧ eue+w ⇧dweu +ewdw⇧ ewew ⇧ ew 2 µ ⌫ µ ⌫ µ ⌫ µ ⌫ 2 = du eu ⇧ eu=+dxdudx= dwdxeeuµdx⇧⇧ eew⌫e+µ dw⇧geµ⌫⌫ dwdx gedxwµ⌫⇧dxeu dx+ dw ew ⇧ ew (2.3) (2.3) µ ⌫ µ ⌘⌫ ⌘ = dx dx eµ ⇧ e⌫ gµ⌫ dx dx (2.3) The setThe of fourset of components four componentsgµ⌫ ⌘= geµµ⌫⇧=e⌫eµrepresents⇧ e⌫ represents the metric the metric tensor tensorof ourof two-dimensional our two-dimensional coordinate coordinate The set ofbasis. four componentsbasis. Note that Noteg becauseµ that⌫ = e becauseµ both⇧ e⌫ represents the both magnitudes the the magnitudesmetric and directions tensor and directionsof of our the two-dimensional of basis the vectors basis vectors depend coordinate depend on position, on position, the the basis. Notevalue that ofvalue becausegµ⌫ ofalsog bothµ⌫ generallyalso the generally magnitudes depends depends on and position. directions on position. But of the since But basis the since vectors dot-product the dot-product depend of on vectors position, of vectors is commutative, the is commutative, this this value of gµdefinition⌫ alsodefinition generally implies depends implies that the that on metric position. the metric is always But is sincealways symmetric: the symmetric: dot-productg↵ =gg↵ of↵ vectors.= Thisg↵. is means This commutative, means that the that this metric the metric of a two- of a two- definition impliesdimensionaldimensional that the space metric spacehas onlyis hasalways 3 only independent symmetric: 3 independent components,g↵ = components,g↵. and This in means and spacetime, in that spacetime, the the metric metric the of metric has a two- 10 has independent 10 independent dimensionalcomponents spacecomponents has only(of the 3 (of independent 16 the possible 16 possible combinations components, combinations of and index in of spacetime, values, index values, 4 involve the 4 metric involve the same has the 10 value same independent repeated value repeated and the and half the half componentsof (of the theof remaining the 16 possible remaining 12 arecombinations 12 the are same the asof same index the as other values, the half, other 4 involve for half, a total for the a ofsame total 4 + value of 6 4 = + repeated 10). 6 = 10). and the half of the remainingYou 12 can areYou easily the can same easily see as how the see we other how can we half, generalize can for generalize a total this of to 4 this four + 6 to = dimensions four 10). dimensions in spacetime. in spacetime. In spacetime, In spacetime,gµ⌫ =gµ⌫ = You caneµ easily⇧ e⌫e(whereµ see⇧ e⌫ how(where the we indices can the indices generalize now range now this overrange to four four over values) dimensions four values) represents in represents spacetime. the generalization the In generalization spacetime, of thegµ of metric⌫ the= metric tensor tensor⌘µ⌫ ⌘µ⌫ eµ ⇧ e⌫ (whereintroduced theintroduced indices in the now in last range the session. last over session. four In flat values) Inspacetime,flat representsspacetime, we can the we generalization always can always find a find cartesian of the a cartesian metric coordinate tensor coordinate⌘ basisµ⌫ where basis where the the introducedbasis in the vectorsbasis last session. vectors are everywhere are In flat everywherespacetime, orthogonal orthogonal we (e can⇧ e always (e=µ ⇧ 0e for⌫ find=µ 0 a= for cartesian⌫),µ and= ⌫), have coordinate and unit have magnitude unit basis magnitude where (e the⇧ e (e=µ ⇧ e1⌫ when= 1 when µ ⌫ 6 6 µ ⌫ ± ± basis vectorsµ = are⌫,with everywhereµ = ⌫,with1 indicating orthogonal1 indicating a ( timeeµ ⇧ ae coordinate)⌫ time= 0 for coordinate)µ = at⌫), all and events at allhave events and unit still magnitude and have still the have (e componentsµ ⇧ thee⌫ = components1 when of the of di↵ theerential di↵erential 6 6 ± µ = ⌫,withfour-displacement6 1 indicatingfour-displacement a timeds be coordinate)d simplys be simplydt, at dx, alldt, dy, events dx,and dy,dz andand. This stilldz. have is This not the is generally notcomponents generally possible of possible the in di curved↵erential in curved spacetime. spacetime. The The 6 2 2 µ ⌫ µ ⌫ four-displacementquantityquantitydsdsbe= simplydsgµ⌫ dx=dt,gdxµ dx,⌫ dxgeneralizes dy,dxandgeneralizesdz. this This idea thisis not for idea generally an forarbitrary an possiblearbitrarycoordinate incoordinate curved system spacetime. system using a using The coordinate a coordinate basis basis 2 µ ⌫ quantity dsin= angarbitraryµin⌫ dx an dxarbitraryspacetime.generalizesspacetime. The this idea quantity The for quantity andsarbitraryrepresentsds representscoordinate the spacetime the system spacetime separation using separation a coordinate if the events if the basis events at ends at of ends the of the in an arbitrarydi↵erentialspacetime.di↵erential displacement The displacement quantity haveds a have spacelikerepresents a spacelike separation, the spacetime separation, and d separation⌧ and= pd⌧ ds= if2p theis theds events2 infinitesimalis the at infinitesimal ends of spacetime the spacetime interval interval di↵erential displacement(which is have the a same spacelike as the separation, proper time and ind⌧ the= p infinitesimalds2 is the infinitesimal limit) between spacetime the events interval if they have a timelike (which is the same as the proper time in the infinitesimal limit) between the events if they have a timelike (which is theseparation. sameseparation. as the The proper point The time is point that in is the the that infinitesimal metric the metric connects limit) connects the between arbitrary the arbitrary the coordinates events coordinates if they with have the with a physical timelike the physical distances distances or or separation.intervals Theintervals point in the is that in physical the the physical metric universe connects universe behind the behind that arbitrary coordinate that coordinate coordinates system. system. with the physical distances or intervals in the physical universe behind that coordinate system. 2.2.12.2.1 Exercise: Exercise: The Metric The Metric for Spherical for Spherical Coordinates Coordinates 2.2.1 Exercise: The Metric for Spherical Coordinates ConsiderConsider✓- coordinates✓- coordinates on the on surface the surface of a sphere of a sphere of radius of radiusR, whereR, where curves curves of constant of constant✓ and ✓ andare lines are of lines of Consider ✓- coordinates on the surface of a sphere of radius R, where curves of constant ✓ and are lines of latitudelatitude and longitude, and longitude, respectively respectively (but assume (but assume that ✓ that= 0 at✓ = the 0 at north the pole, north as pole, is normal as is normal in physics, in physics, rather rather latitude and longitude, respectively (but assume that ✓ = 0 at the north pole, as is normal in physics, rather than atthan the at equator). the equator). Note that Note these that curves these curves are perpendicular are perpendicular everywhere everywhere but the but poles. the poles. By considering By considering than at the equator). Note that these curves are perpendicular everywhere but the poles. By considering what thewhat formula the formula for ds2 forbetweends2 between infinitesimally-separated infinitesimally-separated points points must be must in be terms in terms of d✓ ofandd✓dand,findd,find what the formula for ds2 between infinitesimally-separated points must be in terms of d✓ and d,find the metricthe metric components componentsg✓✓,g✓g,g✓✓,g✓✓, and,g✓g, andasg a functionas a function of position of position for a coordinate for a coordinate basis based basis based on the on the the metric components g✓✓,g✓,g✓, and g as a function of position for a coordinate basis based on the coordinatescoordinates✓ and ✓.and Also,. what Also, are what the are lengths the lengths of the ofe✓ theandee✓ andbasise vectorsbasis vectors as a function as a function of position? of position? coordinates ✓ and . Also, what are the lengths of the e✓ and e basis vectors as a function of position? 2.3 Tensors in a Coordinate Basis 2.3 Tensors2.3 Tensors in a Coordinate in a Coordinate Basis Basis Again, for the moment, let’s go back to considering a two-dimensional, possibly curved surface in space. Again, forAgain, the moment, for the let’s moment, go back let’s to go considering back to considering a two-dimensional, a two-dimensional, possibly curved possibly surface curved in space. surface in space. Consider a general transformation between our original coordinates u, w to new coordinates p(u, w) and Consider aConsider general transformation a general transformation between our between original our coordinates original coordinatesu, w to newu, coordinates w to newp coordinates(u, w) and p(u, w) and q(u, w). Theq(u, chain w).q(u, Therule w). for chain The partial rule chain derivatives for rule partial for partial impliesderivatives derivatives that implies infinitesimal implies that infinitesimal changes that infinitesimal in the changes new changes coordinates in the in new the are coordinates new coordinates are are

3 3 3 Now, if we define basis vectors this way, then du and dw become the components of ds in that basis and we call the set of basis vectors eu, ew a coordinate basis. A coordinate basis is generally di↵erent than the cartesian coordinate basis vectors ex, ey in that (1) eu and ew may not be perpendicular, (2) eu and ew may not have unit length, and (3) eu and/or ew may change in magnitude and/or direction as we move from point to point. It is also important to note that defining a coordinate basis is not the only way to define a set of basis vectors or a coordinate system. For example, standard polar coordinates in two-dimensional flat space and spherical coordinates in a three-dimensional flat space do not use a coordinate basis, because the basis vectors in those coordinate systems always have unit magnitude. Using a coordinate basis makes components of the displacement vector ds simpler (at the expense of complexity in the basis vectors), and this turns out to be crucial in what follows. Once we have defined a coordinate basis for coordinates u and w,thenwedefine the components of a four-vector A at any point to be Au,Aw such that P u w µ A = A eu + A ew = A eµ (2.2)

This ensures that the components of ds and A transform alike. Now, the scalar product of ds with itself is the square of the physical distance between the endpoints of the displacement it represents:

2 ds = ds ⇧ ds =(du eu + dw ew) ⇧ (du eu + dw ew) 2 2 = du eu ⇧ eu + du dw eu ⇧ ew + dw dw ew ⇧ eu + dw ew ⇧ ew = dxµdx⌫ e ⇧ e g dxµdx⌫ (2.3) µ ⌫ ⌘ µ⌫

The set of four components gµ⌫ = eµ ⇧ e⌫ represents the metric tensor of our two-dimensional coordinate basis. Note that because both the magnitudes and directions of the basis vectors depend on position, the value of gµ⌫ also generally depends on position. But since the dot-product of vectors is commutative, this definition implies that the metric is always symmetric: g↵ = g↵. This means that the metric of a two- dimensional space has only 3 independent components, and in spacetime, the metric has 10 independent components (of the 16 possible combinations of index values, 4 involve the same value repeated and the half of the remaining 12 are the same as the other half, for a total of 4 + 6 = 10). You can easily see how we can generalize this to four dimensions in spacetime. In spacetime, gµ⌫ = eµ ⇧ e⌫ (where the indices now range over four values) represents the generalization of the metric tensor ⌘µ⌫ introduced in the last session. In flat spacetime, we can always find a cartesian coordinate basis where the basis vectors are everywhere orthogonal (eµ ⇧ e⌫ = 0 for µ = ⌫), and have unit magnitude (eµ ⇧ e⌫ = 1 when µ = ⌫,with 1 indicating a time coordinate) at all events6 and still have the components of the di↵±erential four-displacement6 ds be simply dt, dx, dy, and dz. This is not generally possible in curved spacetime. The 2 µ ⌫ quantity ds = gµ⌫ dx dx generalizes this idea for an arbitrary coordinate system using a coordinate basis in an arbitrary spacetime. The quantity ds represents the spacetime separation if the events at ends of the di↵erential displacement have a spacelike separation, and d⌧ = p ds2 is the infinitesimal spacetime interval Exercise(which is the same as the proper time in the infinitesimal limit) between the events if they have a timelike separation. The point is that the metric connects the arbitrary coordinates with the physical distances or intervals in the physical universe behind that coordinate system.

2.2.1 Exercise: The Metric for Spherical Coordinates Consider ✓- coordinates on the surface of a sphere of radius R, where curves of constant ✓ and are lines of latitude and longitude, respectively (but assume that ✓ = 0 at the north pole, as is normal in physics, rather than at the equator). Note that these curves are perpendicular everywhere but the poles. By considering what the formula for ds2 between infinitesimally-separated points must be in terms of d✓ and d,find the metric components g✓✓,g✓,g✓, and g as a function of position for a coordinate basis based on the coordinates ✓ and . Also, what are the lengths of the e✓ and e basis vectors as a function of position?

2.3 Tensors in a Coordinate Basis Again, for the moment, let’s go back to considering a two-dimensional, possibly curved surface in space. Consider a general transformation between our original coordinates u, w to new coordinates p(u, w) and q(u, w). The chain rule for partial derivatives implies that infinitesimal changes in the new coordinates are

3 related to changes in the old coordinates as follows: @p @p @q @q dp = du + dw and dq = du + dw (2.4) @u @w @u @w related to changes in the old coordinates as follows: If we consider the p, q coordinates the primed coordinate system and u, w coordinates the unprimed system, then we can write this transformation@p rule compactly@p as @q @q dp = du +Tensorsdw and dq = indu + adw coordinate(2.4) basis @u @w µ @u @w µ @x0 ⌫ dx0 = dx (2.5) If we consider the p, q coordinates the primed coordinate@x⌫ system and u, w coordinates the unprimed system, relatedthen we to can changes write this in the transformation old coordinates rule as compactly follows: as (with an implicitrelated sum over to changes the ⌫ subscript: in the oldTransformation we coordinates will consider as a follows: superscript from u, win to the p, denominator q coordinates of a partial @p @p µ @q @q derivative to equivalent to a subscript).dp = du + dwµ and@x0 dq =⌫ du + dw (2.4) dx0 =@p dx@p @q @q (2.5) Now note that in a coordinate basis,@u the values@wdp of=dp and@dux⌫ dq+ are@dw theu actualand@w componentsdq = duof+ the infinitesimaldw (2.4) displacement vector ds in the primed system and du @andu dw are@w the same in the unprimed@u system.@w Since If(with we consider an implicit the sump, q coordinates over the ⌫ subscript: the primed we coordinate will consider system a superscript and u, w coordinatesin the denominator the unprimed of a system, partial bythen definition, we can the writeIf components we this consider transformation of the anp, arbitrary q coordinates ruleor compactly vector abstractly:A thetransform as primed in coordinate the same ways system as the and componentsu, w coordinates of the the unprimed system, displacementderivative to vector, equivalent The transformation to a subscript). law for the components of A is Now note thatthen in we a coordinate can write basis, this transformation the values of dp and ruleµ dq compactlyare the actual as components of the infinitesimal µ @x0 ⌫ dx0 = µ dx (2.5) displacement vector ds in the primed system andµ du@x0and⌫ ⌫dw are the sameµ in the unprimed system. Since A = @xA @x0 (2.6) by definition, the components of an arbitrary vector0 A ⌫transform inµ the same ways⌫ as the components of the @x dx0 = ⌫ dx (2.5) (withdisplacement an implicit vector, sum The over transformation the ⌫ subscript: lawVector we for will the transformation considercomponents a superscript of A is @law:x in the denominator of a partial Again,derivative the extensionto equivalent to four-dimensional to a subscript). spacetime is straightforward. But this approach is only straight- forward if we are(with using an a coordinate implicit sum basis over, something the ⌫ subscript: that weµ will we simply will consider assume from a superscript now on. in the denominator of a partial Now note that in a coordinate basis, the valuesµ of dp@xand0 dq⌫ are the actual components of the infinitesimal derivative to equivalent to a subscript).A0 = A (2.6) displacementNote in that vector contextds thatin the equation primed 2.6 system looks and justdu like@andx⌫ thedw transformationare the same in law the for unprimed the components system. of Since a ⌫ µ µ four-vector in spacetimeNow with note@ thatx0 /@ inx areplacing coordinate⇤ basis,. Indeed, the values if you take of dp partialand dq derivativesare the actual of the components Lorentz of the infinitesimal byAgain, definition, the extension the components to four-dimensional of an arbitrary spacetime vector⌫ A istransform straightforward. in the same But this ways approach as the components is only straight- of the transformation functionsdisplacementt0(t, x, vector y, z), xd0(st,in x,Generalizes y,the z), primedy0(t, x, y, system z ),Lorentz and andz0(t,du transformation, x, y,and z),dw youare will the see same thatbecause in in fact the for unprimed a LT: system. Since displacementforward if we vector, are using The a transformationcoordinate basis law, something for the components that we will of simplyA is assume from now on. by definition, the components of anµ arbitrary vector A transform in the same ways as the components of the Note in that context that equation 2.6 looks@x0 just likeµ the transformation law for the components of a displacement vector, The transformationµ =@⇤xµ0 law⌫ for the components of A is (2.7) four-vector in spacetime with @x ⌫ /@xµ replacingA0 ⌫⇤=µ . Indeed,⌫ A if you take partial derivatives of the Lorentz(2.6) 0 @x ⌫ @x⌫ transformation functions t0(t, x, y, z), x0(t, x, y, z), y0(t, x, y, z), and z0(t, x, y,µ z), you will see that in fact µ @x0 ⌫ forAgain, that particular the extension coordinate to four-dimensional transformation. spacetime We see, is therefore, straightforward.A that0 = expressing ButA this the approach transformation is only straight- coe- (2.6) cients in terms of partial derivatives is consistent with@x µ but generalizes the@ transformationx⌫ between cartesian forward if we are using a coordinate basis, something0 = that⇤µ we will simply assume from now on. (2.7) coordinatesNote in in that inertialAgain, context frames the that extension that equation we considered to 2.6 four-dimensional looks earlier.@x just⌫ like⌫ the spacetime transformation is straightforward. law for the components But this approach of a is only straight- Now, basic partial di↵erential calculus⌫ µ implies thatµ four-vectorfor that particular inforward spacetime coordinateif with we are@ transformation.x0 using/@x areplacing coordinate We⇤ see,⌫ . basis Indeed, therefore,, something if you that take expressing that partial we will thederivatives simply transformation of assume the Lorentz coe from- now on. transformation functions t0(t, x, y, z), x0(t, x, y, zµ), y0⌫(t, x, y, z), and z0(t, x, y, z), you will see that in fact cients in terms ofNote partial in derivatives that context is consistent that@x0 equation@ withx but 2.6µ generalizes looks just the like transformation the transformation between cartesian law for the components of a ⌫ =µ µ (2.8) coordinates infour-vector inertial frames in spacetime that we considered with @⌫ x earlier./µ@↵x replacing↵ ⇤ . Indeed, if you take partial derivatives of the Lorentz @x @@0xx0 0 ⌫ Now, basic partial di↵erential calculus implies that= ⇤µ (2.7) For example, if wetransformation write this out functions for our u,t w0(t, x,p,@ y,x q z⌫),transformations,x0(t,⌫ x, y, z), y this0(t, x, says y, z that), and z0(t, x, y, z), you will see that in fact ! µ ⌫ @x0 @x µ for that particular coordinate@p @u transformation.@p @w dp We see,@ therefore,p=@uµ @p@ thatx@0w expressingdpµ the transformation coe(2.8)- + = =1@x⌫, @x ↵ ↵+ == ⇤ ⌫=0, (2.7) cients in terms of partial@u derivatives@p @w @ isp consistentdp with0 @ butu @ generalizesq @w@x@⌫q thedq transformation between cartesian coordinates in inertial frames that we considered earlier. For example, if we write@q this@u out@q for@w our u,dq w p, q transformations,@q @u @q @w thisdq says that Now, basicfor partial that di particular↵erential+ calculus coordinate= implies=0! transformation., that + We see,= therefore,= 1 that expressing (2.9) the transformation coe- cients in@ termsu@@pp@u of@w partial@p@p@w derivativesdpdp is@ consistentu@@pq@u @w@ withp@q@w butdqdp generalizes the transformation between cartesian + = @x=1µ ,@x⌫ + = =0, Equation 2.8 basicallycoordinates says@u that in@p inertial@x@wµ/@ framesxp⌫ anddp that@0xµ/ we@x considered⌫=@representu@µq @w earlier. inverse@q transformations,dq analogous(2.8) to 0 @x⌫ @x ↵0 ↵ ⇤µ and (⇤ 1)µ inNow, flat spacetime. basic@q @u partial@q @ diw↵erentialdq calculus0 @q implies@u @q that@w dq ⌫ ⌫ + = =0, + = = 1 (2.9) ForNow example, we can if easily we write state@ this generalizedu @p out for@w our transformation@p u, wdp p, q lawstransformations,@u for@q arbitrary@µw @q tensors: this⌫ saysdq an that nth-rank tensor’s com- ! ↵ @x0 @x n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index= objectµ (with 4 components) (2.8) @p @u @p µ@w ⌫ dp µ @⌫p @u @⌫ p @w↵ dp ↵ Equation 2.8 basically says that @x0 /@x and @x /@···x0 represent@x @ inversex0 transformations, analogous to thatµ transforms1 accordingµ to + = =1, + = =0, ⇤ ⌫ and (⇤ ) ⌫ in flat spacetime.@u @p @w @p dp @u @q @w @q dq For example, if we write this↵ out for⌫ our u, w p, q transformations, this says that Now we can easily state@q @ generalized↵u @q @w transformation@x0dq @x laws@q@x@ for0u arbitrary@qµ@w tensors: dq an nth-rank tensor’s com- 0 ··· ··· ! T + = = µ =0, ↵ + T ···⌫ =··· = 1n (2.10) (2.9) ponents transform: we define@u @p an n···@th-rankw @p @@px tensor@dpu···@@xTp0 ···@···w@u@···@xqdpto··· be@w an@nq···-index@dqp @ objectu @ (withp @w 4 components)dp that transforms according to + ··· = =1, + = =0, that is, a partial-derivative factor with theµ @ primedu⌫ @p coordinate@µw @p ⌫ indp the numerator@u for@q every@ upperw @q (superscript)dq Equation 2.8 basically says that @x0 /@x and @x /@x0 represent inverse transformations, analogous to µ 1 µ ↵ ⌫ index⇤ andand a (⇤ a partial-derivative) in flat spacetime. factor↵ with@ theq@@x0u primed@@qx coordinate@w @dqx0 in theµ denominator@q @u for@q every@w lowerdq (sub- ⌫ ⌫ T 0 ··· ··· = T (2.10) script) index. Using methods similar to what weµ did+ in the last= session, =0, you···⌫ can··· prove+ in particular= that= 1 the (2.9) Now we can easily state generalized··· transformation@u@@xp ···@@wx0@ lawsp··· for@dpx arbitrary··· tensors:···@u @q an nth-rank@w @q tensor’sdq com- metric is a tensor with two lower indices: ↵ n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) that is, a partial-derivative factor with the primed coordinateµ ··· ⌫ in the numeratorµ ⌫ for every upper (superscript) that transformsEquation according 2.8 to basically says that@x@x↵0 @/x@⌫ x and @x /@x0 represent inverse transformations, analogous to index and a a partial-derivativeµ 1 µ factor withg the= primed0 coordinateg in the denominator for every lower(2.11) (sub- ⇤ ⌫ and (⇤ ) ⌫ in flat spacetime.µ0 ⌫ µ ↵ script) index. Using methods similar to what we↵ @ didx in@⌫x the0 last session, you can prove in particular that the ↵ @x0 @x @x0 µ Now we canT 0 easily··· ··· state= generalized transformationT ··· laws··· for arbitrary tensors:(2.10) an nth-rank tensor’s com- metric is a tensor with two lower indices: µµ⌫ µ⌫ ↵ ⌫ ⌫ that the Kronecker delta is still a tensor,··· that@gx ···defined@x0 such··· @ thatx ···g gµ↵ =··· ↵ (the matrix inverse of the n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) metric tensor), is still a tensor, that multiplying by g ↵and g⌫µ⌫ and summing··· over µ still lowers or raises a that is, a partial-derivativethat transforms factor according with the primedto @ coordinateµx⌫0 @x in the numerator for every upper (superscript) gµ0 ⌫ = µ g↵ (2.11) @x @x0 index and a a partial-derivative factor with the primed coordinate↵ in the⌫ denominator for every lower (sub- µ↵⌫ @x0 µ@⌫ x @⌫ x0 µ script)that the index. Kronecker Using delta methods is still similar a tensor, to what that weg0 ··· diddefined4 in··· the such last that session,g g you= can prove(the in matrix particular inverse that of the T = µ µ↵ ↵ T ···⌫ ··· (2.10) ··· @µx⌫ ···@x ··· @x ··· ··· metric istensor), a tensor is still with a two tensor, lower that indices: multiplying by gµ⌫ and g and summing0 over µ still lowers or raises a ↵ ⌫ that is, a partial-derivative factor@ withx0 the@x primed coordinate in the numerator for every upper (superscript) gµ0 ⌫ = µ g↵ (2.11) index and a a partial-derivative factor@x4 @ withx0 the primed coordinate in the denominator for every lower (sub- script) index. Using methods similarµ⌫ to what we didµ⌫ in the last⌫ session, you can prove in particular that the that the Kronecker delta is still a tensor, that g defined such that g gµ↵ = ↵ (the matrix inverse of the metric is a tensor with two lower indices: µ⌫ metric tensor), is still a tensor, that multiplying by gµ⌫ and g and summing over µ still lowers or raises a ↵ ⌫ @x0 @x gµ0 ⌫ = µ g↵ (2.11) 4 @x @x0 µ⌫ µ⌫ ⌫ that the Kronecker delta is still a tensor, that g defined such that g gµ↵ = ↵ (the matrix inverse of the µ⌫ metric tensor), is still a tensor, that multiplying by gµ⌫ and g and summing over µ still lowers or raises a

4 related to changes in the old coordinates as follows: related to changes in the old coordinates as follows: @p @p @q @q @p @p dp = du +@q dw @qand dq = du + dw (2.4) dp = du + dw and @udq = @wdu + dw @u @w (2.4) @u @w @u @w Ifrelated we consider to changes the in thep, q oldcoordinates coordinates the as follows: primed coordinate system and u, w coordinates the unprimed system, If we consider the p, q coordinates the primed coordinate system and u, w coordinates the unprimed system, then we can write this transformation@p @p rule compactly@q as @q then we can write this transformation rule compactlydp = du as+ dw and dq = du + dw (2.4) @u @w @u @w µ µ µ @x0 ⌫ If we consider the p, q coordinatesµ the primed@x0 coordinate⌫ dx system0 = and u,dx w coordinates the unprimed system, (2.5) dx0 = dx ⌫ (2.5) then we can write this transformation rule@x⌫ compactly as @x µ (with an implicit sum over the ⌫ subscript:µ @ wex0 will⌫ consider a superscript in the denominator of a partial (with an implicit sum over the ⌫ subscript: we will considerdx0 a= superscriptdx in the denominator of a partial (2.5) derivative to equivalentderivative to a subscript).to equivalent to a subscript). @x⌫ Now note that in(with a coordinateNow an implicit note basis, that sum in the over a coordinate values the ⌫ subscript: of dp basis,and wedq the willare values consider the actual of adp superscript componentsand dq arein the theof denominator theactual infinitesimal components of a partialof the infinitesimal displacement vectordisplacementderivativeds in theto primed equivalent vector system tods ain subscript). and thedu primedand dw systemare the and samedu and in thedw unprimedare the same system. in the Since unprimed system. Since by definition, the componentsby definition,Now note of that an the inarbitrary a components coordinate vector basis, ofA an thetransform arbitrary values of indp vector theand samedq areA transform ways the actual as the components in components the sameof the ways of infinitesimal the as the components of the displacement vector ds in the primed system and du and dw are the same in the unprimed system. Since displacement vector,displacement The transformation vector, law The for transformation the components law of forA is the components of A is by definition, the components of an arbitrary vector A transform in the same ways as the components of the displacement vector, The transformation lawµ for the components of Aµ is µ @x0 ⌫ µ @x0 ⌫ A0 = A (2.6) A0 = ⌫ A µ ⌫ (2.6) @x µ @x0 ⌫ @x A0 = A (2.6) @x⌫ Again, the extensionAgain, to four-dimensional the extension spacetime to four-dimensional is straightforward. spacetime But is thisstraightforward. approach is only But straight- this approach is only straight- forward if we are usingforwardAgain, a coordinate theif extension we are basis using to four-dimensional, something a coordinate that spacetime basis we will, something is simply straightforward. assume that we from But will this now simply approach on. assume is only from straight- now on. forward if we are using a coordinate basis, something that we will simply assume from now on. Note in that contextNote that in equation that context 2.6 looks that just equation like the 2.6 transformation looks just like law the for transformation the components law of for a the components of a Note in that⌫ contextµ that equationµ 2.6 looks just like the transformation law for the components of a four-vector in spacetime with @x0 /@x replacing⌫ ⇤ µ.⌫ Indeed,µ ifµ you takeµ partial derivatives of the Lorentz four-vectorfour-vector in in spacetime spacetime with with@x0 /@@xx⌫ 0 replacing/@x replacing⇤ ⌫ . Indeed,⇤ ⌫ if. you Indeed, take partial if you derivatives take partial of the derivatives Lorentz of the Lorentz transformation functionstransformationtransformationt0(t, x, y, functions z), functionsx0(t,t x,0(t, y, x,t z0 y,(),t, zy), x,0(xt, y,0( x,t, z), x, y, y,x z0 z),(),t, andy x,0(t, y,z x,0 z( y,),t, z x,y),0( y, andt, z x,),z0 y,( yout, z x,), y, will and z), see youz0( thatt, will x, see y, in z that fact), you in fact will see that in fact µ µ @x0 µ @x0 µ µ = ⇤@⌫x0 µ (2.7) ⌫ = ⇤ ⌫ @x⌫ = ⇤ (2.7) (2.7) @x @x⌫ ⌫ for that particular coordinatefor that particular transformation. coordinate transformation. We see, therefore, We see, that therefore, expressing that expressing the transformation the transformation coe- coe- forcients that in terms particular of partial coordinateTensors derivatives transformation. is consistent in with buta We generalizescoordinate see, therefore, the transformation that expressing between basis thecartesian transformation coe- cients in terms of partialcientscoordinates derivatives in terms in inertial of is partial consistent frames that derivatives with we considered but is generalizes consistent earlier. the with transformation but generalizes between the transformation cartesian between cartesian coordinates in inertialcoordinates framesNow, basic that in partial we inertial considered di↵erential frames calculus earlier. that we implies considered that earlier. Now, basic partial di↵erential calculus implies that Now, basic partialAn di↵ importanterential calculus identity:@x µ implies@x⌫ that 0 = µ (2.8) µ ⌫ @x⌫ @x ↵ ↵ @x0 @x µ 0 µ ⌫ = @x0 @x (2.8) For example, if we write this out⌫ for our↵u, w ↵ p, q transformations, thisµ says that @x @x0 ⌫ ↵ = ↵ (2.8) ! @x @x0 For example, if we write this out for our u,@p w@u @p,p q@wtransformations,dp @p this@u says@p @ thatw dp For example, if we write this!+ out for= our=1u,, w p, q+transformations,= =0, this says that @Example:u @p @w @p dp !@u @q @w @q dq @p @u @p @w @q dp@u @q @w @pdq@u @p @@qw@u dp@q @w dq + = @p+=1@u, @=p @w=0+, dp =+ @=0p @,=u =@p 1@w dp (2.9) @u @p @w @p @u dp@p @w +@p @udp@q = @w=1@@uq@,q dq@w @q +dq = =0, @u @p @w @p dp @u @q @w @q dq @q @u @q @w dq µ ⌫@q @u µ @q @⌫ w dq Equation 2.8+ basically says= that=0@x,0 /@x and @+x /@x0 represent= = inverse 1 transformations, analogous (2.9) to µ 1 µ @q @u @q @w dq @q @u @q @w dq ⇤ ⌫ and@u @ (⇤p ) @⌫ win@ flatp spacetime.dp + @u @q = @w=0@q , dq + = = 1 (2.9) @u @p @w @p dp @u @q @w @q dq Now we can easilyµ state⌫ generalizedµ transformation⌫ laws for arbitrary tensors: an nth-rank tensor’s com- Equation 2.8 basically says that @x0 /@x and @x /@x0 represent↵ inverse transformations, analogousn to ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) µ 1 µ µ ⌫ ··· µ ⌫ ⇤ ⌫ and (⇤ ) ⌫ inEquationthat flat transforms spacetime. 2.8 accordingbasically to says that @x0 /@x and @x /@x0 represent inverse transformations, analogous to µ 1 µ General tensor transformation law: Now we can easily⇤ state⌫ and generalized (⇤ ) ⌫ in transformation flat spacetime. laws↵ for arbitrary⌫ tensors: an nth-rank tensor’s com- ↵ ↵ @x0 @x @x0 µ n 0 ··· ··· ponents transform: we defineNow we an cannth-rank easily state tensorT generalized T ···= ···µ transformationto be an n-index laws objectT ··· for⌫ (with arbitrary··· 4 components) tensors: an(2.10) nth-rank tensor’s com- ··· ···@x ···@x0 ··· @x↵ ··· ··· n that transforms accordingponents to transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) that is, a partial-derivative factor with the primed coordinate in the··· numerator for every upper (superscript) that transforms according↵ to ⌫ index and a a↵ partial-derivative @x0 factor@x with the@ primedx0 coordinateµ in the denominator for every lower (sub- 0 ··· ··· script) index.T Using methods= µ similar to what we did in↵ theT last···⌫ session,⌫··· you can prove in particular(2.10) that the ··· @x ···@↵x0 ··· @x @x···0 ···@x @x0 µ metric is a tensor with two lower indices:T 0 ··· ··· = T ···⌫ ··· (2.10) @xµ ···@x ··· @x ··· ··· that is, a partial-derivative factor with the primed coordinate··· in the↵ numerator⌫ 0 for every upper (superscript) @x0 @x index and a a partial-derivative factor with the primed coordinategµ0 ⌫ = µ in the g↵ denominator for every lower (sub-(2.11) that is, a partial-derivative factor with the@ primedx @x0 coordinate in the numerator for every upper (superscript) script) index. Using methods similar to what we did in the last session, you can prove in particular that the index and a a partial-derivative factor withµ⌫ the primed coordinateµ⌫ ⌫ in the denominator for every lower (sub- that the Kronecker delta is still a tensor, that g defined such that g gµ↵ = ↵ (the matrix inverse of the metric is a tensor with two lower indices: µ⌫ script)metric tensor), index. is Using still a tensor, methods that similar multiplying to what by gµ⌫ weand didg inand the summing last session, over µ still you lowers can prove or raises in a particular that the ↵ ⌫ metric is a tensor with two lower@x0 indices:@x gµ0 ⌫ = g↵ (2.11) @xµ @x 4 ↵ ⌫ 0 @x0 @x µ⌫ gµ0 ⌫ =µ⌫ µ ⌫ g↵ (2.11) that the Kronecker delta is still a tensor, that g defined such that g g@µ↵x =@x↵0 (the matrix inverse of the µ⌫ metric tensor), is still a tensor, that multiplying by gµ⌫ and g andµ summing⌫ over µ still lowersµ⌫ or raises⌫ a that the Kronecker delta is still a tensor, that g defined such that g gµ↵ = ↵ (the matrix inverse of the µ⌫ metric tensor), is still a tensor, that multiplying by gµ⌫ and g and summing over µ still lowers or raises a 4 4 related to changes in the old coordinates as follows: @p @p @q @q dp = du + dw and dq = du + dw (2.4) @u @w @u @w If we consider the p, q coordinates the primed coordinate system and u, w coordinates the unprimed system, then we can write this transformation rule compactly as

µ µ @x0 ⌫ dx0 = dx (2.5) @x⌫ (with an implicit sum over the ⌫ subscript: we will consider a superscript in the denominator of a partial derivative to equivalent to a subscript). Note inNow that note context that that in aequation coordinate 2.6 looks basis, just the like values the transformation of dp and dq laware for the theactual components components of a of the infinitesimal ⌫ µ µ four-vectordisplacement in spacetime vector with @dxs0 in/@x thereplacing primed⇤ system⌫ . Indeed, and if youdu takeand partialdw are derivatives the same of the in the Lorentz unprimed system. Since transformationby definition, functions thet0 components(t, x, y, z), x0(t, of x, y,an z), arbitraryy0(t, x, y, z vector), and zA0(t,transform x, y, z), you in will the see same that in ways fact as the components of the displacement vector, The transformationµ law for the components of A is @x0 µ ⌫ = ⇤ ⌫ (2.7) Note in that context that equation 2.6 looks just@x like the transformationµ law for the components of a µ @x0 ⌫ four-vector in spacetime with @x ⌫ /@xµ replacing ⇤µ . Indeed,A if0 you= take partialA derivatives of the Lorentz (2.6) for that particular coordinate0 transformation. We⌫ see, therefore, that@ expressingx⌫ the transformation coe- transformationcients in terms functions of partialt0(t, derivatives x, y, z), x0 is(t, consistent x, y, z), y0( witht, x, y, but z), generalizes and z0(t, x, the y, z transformation), you will see that between in fact cartesian coordinatesAgain, in the inertial extension frames that to wefour-dimensional considered earlier. spacetime is straightforward. But this approach is only straight- @x µ Now,forward basic partialif we di are↵erential using calculus a coordinate implies0 basis that= ⇤µ, something that we will simply assume(2.7) from now on. @x⌫ ⌫ µ ⌫ Note in that context that equation@x0 @x 2.6 looks just like the transformation law for the components of a for that particular coordinate transformation. We⌫ see,µ therefore,= µ thatµ expressing the transformation coe(2.8)- four-vector in spacetime with @x0 /@⌫x replacing↵ ↵ ⇤ . Indeed, if you take partial derivatives of the Lorentz cients in terms of partial derivatives is consistent@ withx @ butx0 generalizes the⌫ transformation between cartesian transformation functions t (t, x, y, z), x (t, x, y, z), y (t, x, y, z), and z (t, x, y, z), you will see that in fact coordinatesFor example, in inertial if we write frames this that out we for considered0 our u, w earlier.p,0 q transformations,0 this says that 0 ! Now, basic partial di↵erential calculus implies that µ @p @u @p @w dp @p @u@x0@p @w µ dp + = µ=1,⌫ + ⌫ = ⇤=⌫ =0, (2.7) @u @p @w @p @xdp0 @x @µu @q @x@w @q dq ⌫ ↵ = ↵ (2.8) @q @u @q @w @dqx @x0 @q @u @q @w dq for that particular coordinate+ = transformation.=0, We+ see, therefore,= = 1 that expressing (2.9)the transformation coe- For example, if we write this@u out@p for@ ourw @u,p w dp p, q transformations,@u @q @w this@q saysdq that cients in terms of partial derivatives! is consistent with but generalizes the transformation between cartesian µ ⌫ µ ⌫ Equationcoordinates 2.8 basically in@p inertialsays@u that@p frames@@xw0 /@xdp thatand we@x considered/@@xp0@urepresent@ earlier.p @w inversedp transformations, analogous to µ 1 µ + = =1, + = =0, ⇤ ⌫ and (⇤Now, ) ⌫ basicin flat@u partial spacetime.@p @ diw ↵@erentialp dp calculus@u implies@q @w that@q dq Now we can easily state generalized transformation laws for arbitrary tensors: an nth-rank tensor’s com- @q @u @q @w dq ↵@q @u @q @w dq n ponents transform: we define+ an nth-rank= tensor=0, T ··· ···+toµ be an⌫=n-index= 1 object (with 2 components (2.9) @x0 @x µ n @u @p @w @p dp @u ···@q @w @q =dq (2.8) in a 2D space and 4 components in spacetime) that transforms@x⌫ according@x ↵ to ↵ µ ⌫ µ ⌫ 0 Equation 2.8 basically says that @x0 /@x and @↵x /@x0 ⌫ represent inverse transformations, analogous to µ 1 µ ↵ @x0 @x @x0 µ ⇤ andFor (⇤ example,) in flat spacetime. if we write0 ··· this··· out for our u, w p, q transformations, this says that ⌫ ⌫ T = µ T ···⌫ ··· (2.10) Now we can easily state generalized··· transformation@x ···@ lawsx0 ··· for arbitrary@!x ··· tensors:··· an nth-rank tensor’s com- @p @u @p @↵w dp @p @u @p @wn dp ponentsthat is, transform: a partial-derivative we define an factornth-rank with the tensor primedT coordinate··· ··· to in be the an numeratorn-index object for every (with upper 2 components (superscript) n + ···= =1, + = =0, in aindex 2D space and one and with 4 components the primed coordinatein spacetime)@u @p in the that@w denominator@ transformsp dp according for every lower to@u @ (subscript)q @w @ index.q dq In particular, we can prove that@q the@u metric↵ @q correctly@w⌫ dq transforms as a@q tensor@u with@q two@w lowerdq indices as ↵ @x0 @x @x0 µ follows. The coordinate-independence0 ··· ··· of the+ spacetime= separation=0, implies that + = = 1 (2.9) T = µ T ···⌫ ··· (2.10) ··· @u @@px ···@w@x@0p ··· dp@x ··· ···@u @q @w @q dq µ ⌫ ↵ Tensorsgµ0 ⌫ dx0 dx0 =ing↵dx µadx coordinate⌫ µ ⌫ basis that is, aEquation partial-derivative 2.8 basically factor with says the that primed@x coordinate0 /@x and in the@ numeratorx /@x0 forrepresent every upper inverse (superscript) transformations, analogous to ↵ index andµ one with the primed1 µ coordinate in the denominator@x ↵ for every@x lower (subscript) index. ⇤ ⌫ and (⇤ ) ⌫ in flat spacetime.= g dx0 dx In particular, we can prove that the metric correctly↵ @x transforms@x as a tensor with two lower indices as Now we can easily state generalized✓ transformation0 ◆✓ 0 laws◆ for arbitrary tensors: an nth-rank tensor’s com- follows. The coordinate-independence of the spacetime↵ separation implies that Example: the metric tensor@x @x is a tensor ↵ n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) = g↵ dx0 dx0 µ ⌫ @↵x0 @x0 ··· that transforms accordinggµ0 ⌫ dx0 dx to0 = g↵✓dx dx ◆ @x↵ @x @x↵ @x µ ⌫ = µ ⌫↵g↵↵ dx0 dx0⌫ = g↵↵ @x0 @dxx00@x0 dx@x @x0 µ T 0 ···✓ @x···0 = ◆@x0 T ··· ··· (2.10) ✓ ↵ ◆✓µ ◆ ⌫ ···↵@x @x @x ···@x0 µ···⌫@x ··· ··· 0=@x @x g↵ gµ0 ⌫ dx0 dx0 (2.11) ) = @x µ @xg↵⌫ dx0 dx0 @✓x @0 x 0 ◆ that is, a partial-derivative factor✓ with0 the0 primed◆ coordinate in the numerator for every upper (superscript) µ ↵⌫ Now, weindex can’t and just a divide a partial-derivative both sides by dx0 factordx@x0 because@x with the a sum primedµ can⌫ be coordinate zero even if the in the individual denominator terms in for every lower (sub- = g dx0 dx0 the sum are not. But this relation must be@ truex µ for@x arbitrary⌫ ↵ di↵erential displacements. So if I choose the script) index. Using methods similar✓ 0 to0 what◆ we did in the last session, you can prove in particular that the displacement to be entirely in the p direction (↵dw = 0), then the only nonzero term in the sum is the one metric is a tensor with two lower@ indices:x @x µ ⌫ where µ = ⌫ = p, so the quantity in parentheses0= µ must⌫ g be↵ 0 whengµ0 ⌫ µdx=0 ⌫dx=0 p. In a similar way, I can(2.11) choose ) @x0 @x0 ↵ ⌫ displacement components to show that all✓ of the other components◆@x of0 that@x quantity must be zero as well: µ ⌫ gµ0 ⌫ = µ g↵ (2.11) Now, we can’t just divide both sides by↵ dx0 dx0 because a sum can@ bex zero↵@x even⌫ if the individual terms in @x @x @x0 @0x the sum are not. But this relation0= must be trueg forg0 arbitrary dig↵0 erential= displacements.g So if I choose the(2.12) @x µ @x ⌫ ↵ µ⌫ ) µµ⌫⌫ @xµ @x ↵ µ⌫ ⌫ displacementthat to the be Kronecker entirely in the delta✓ p direction0 is still0 a (dw tensor,= 0),◆ thenthat theg onlydefined nonzero such0 term that in theg sumgµ↵ is= the ↵ one(the matrix inverse of the µ⌫ wherewhichµ metric= is⌫ the= correctp, tensor), so the transformation quantity is still in a parentheses tensor, law for a that tensor. must multiplying Using be 0 when similarµ by= methods,g⌫µ=⌫ pand. In youg a can similarand show way, summing that I the can Kronecker choose over µ still lowers or raises a µ⌫ µ⌫ ⌫ displacementdelta is still components a tensor, that to showg defined that all such of the that otherg componentsgµ↵ = ↵ (the of that matrix quantity inverse must of the be metric zero as tensor), well: is still a tensor, that multiplying by g and gµ⌫ and summing over µ still lowers or raises a tensor index, that ↵ µ⌫ ↵ ⌫ @x @x @x0 4 @x contracting over an upper0= and lower indexg ofg a0 tensor quantityg0 = still yields ag tensor with rank n 2,(2.12) etc. @x µ @x ⌫ ↵ µ⌫ ) µ⌫ @xµ @x ↵ ✓ 0 0 ◆ 0 which is the correct transformation law for a tensor. Using4 similar methods, you can show that the Kronecker µ⌫ µ⌫ ⌫ delta is still a tensor, that g defined such that g gµ↵ = ↵ (the matrix inverse of the metric tensor), is µ⌫ still a tensor, that multiplying by gµ⌫ and g and summing over µ still lowers or raises a tensor index, that contracting over an upper and lower index of a tensor quantity still yields a tensor with rank n 2, etc.

4 related to changes in the old coordinates as follows: @p @p @q @q related to changes indp the= olddu coordinates+ dw asand follows:dq = du + dw (2.4) @u @w @u @w @p @p @q @q dp = du + dw and dq = du + dw (2.4) If we consider the p, q coordinates the primed@u coordinate@w system and u, w@coordinatesu @w the unprimed system, then we can write this transformation rule compactly as If we consider the p, q coordinates the primed coordinate system and u, w coordinates the unprimed system, µ then we can write this transformation ruleµ compactly@x0 ⌫ as dx0 = ⌫ dx (2.5) @x µ µ @x0 ⌫ dx0 = dx (2.5) (with an implicit sum over the ⌫ subscript: we will consider a@ superscriptx⌫ in the denominator of a partial derivative to equivalent to a subscript). Now note(with that an in implicit a coordinate sum over basis, the the⌫ valuessubscript: of dp weand willdq considerare the actual a superscript componentsin theof the denominator infinitesimal of a partial displacementderivative vectortods equivalentin the primed to a subscript). system and du and dw are the same in the unprimed system. Since by definition,Now the components note that in a of coordinate an arbitrary basis, vector theA valuestransform of dp and in thedq sameare the waysactual as the components componentsof the of the infinitesimal displacementdisplacement vector, The vector transformationds in the primed law for system the components and du and of dwA isare the same in the unprimed system. Since by definition, the components of an arbitrary vector A transform in the same ways as the components of the µ displacement vector, The transformationµ law@x for0 the⌫ components of A is A0 = ⌫ A (2.6) @x µ µ @x0 ⌫ A0 = A (2.6) Again, the extension to four-dimensional spacetime is straightforward.@x⌫ But this approach is only straight- forward if we are using a coordinate basis, something that we will simply assume from now on. Note inAgain, that the context extension that equation to four-dimensional 2.6 looks just spacetime like the is transformation straightforward. law But for this the components approach is only of a straight- forward if we are using⌫ a coordinateµ basis,µ something that we will simply assume from now on. four-vector in spacetime with @x0 /@x replacing ⇤ ⌫ . Indeed, if you take partial derivatives of the Lorentz Note in that context that equation 2.6 looks just like the transformation law for the components of a transformation functions t0(t, x, y, z), x0(t, x, y, z), y0(t, x, y, z), and z0(t, x, y, z), you will see that in fact ⌫ µ µ four-vector in spacetime with @x0 /@x replacing ⇤ ⌫ . Indeed, if you take partial derivatives of the Lorentz µ transformation functions t (t, x, y, z), x@(xt,0 x, y, z), y (t, x, y, z), and z (t, x, y, z), you will see that in fact 0 0 = ⇤µ 0 0 (2.7) @x⌫ ⌫ @x µ 0 = ⇤µ (2.7) for that particular coordinate transformation. We see, therefore,@x⌫ that⌫ expressing the transformation coe- cients in terms of partial derivatives is consistent with but generalizes the transformation between cartesian coordinatesfor in that inertial particular frames coordinate that we considered transformation. earlier. We see, therefore, that expressing the transformation coe- Now, basiccients partial in terms di↵ oferential partial calculus derivatives implies is consistent that with but generalizes the transformation between cartesian coordinates in inertial frames that we considered earlier. µ ⌫ Now, basic partial di↵erential calculus@x0 @ impliesx thatµ ⌫ ↵ = ↵ (2.8) @x @x0 µ ⌫ @x0 @x µ For example, if we write this out for our u, w p, q transformations,⌫ ↵ = ↵ this says that (2.8) ! @x @x0 @p @u @p @w dp @p @u @p @w dp For example, if we write+ this out= for our=1u,, w p, q +transformations,= =0 this, says that @u @p @w @p dp !@u @q @w @q dq @p @u @p @w dp @p @u @p @w dp @q @u @q @w+ dq = =1@q @, u @q @w+ dq = =0, + @u @p =@w @=0p , dp + @u @q =@w @=q 1dq (2.9) @u @p @w @p dp @u @q @w @q dq @q @u @q @w dq @q @u @q @w dq µ +⌫ =µ =0⌫ , + = = 1 (2.9) Equation 2.8 basically says that @@xu0 @/p@x @andw @p@x /dp@x0 represent@u @ inverseq @w transformations,@q dq analogous to µ 1 µ ⇤ and (⇤ ) in flat spacetime. ⌫ ⌫ µ ⌫ µ ⌫ Now weEquation can easily 2.8 state basically generalized says that transformation@x0 /@x and laws@x for/@ arbitraryx0 represent tensors: inverse an nth-rank transformations, tensor’s com- analogous to µ 1 µ ↵ ⇤ ⌫ and (⇤ ) ⌫ in flat spacetime. n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) that transformsNow according we can easily to state generalized transformation··· laws for arbitrary tensors: an nth-rank tensor’s com- ↵ n ponents transform: we define an nth-rank tensor T ··· ··· to be an n-index object (with 4 components) ↵ ⌫ ··· that transforms according↵ to @x0 @x @x0 µ 0 ··· ··· T = µ T ···⌫ ··· (2.10) ··· @x ···@x0 ↵ ··· @x ⌫ ··· ··· ↵ @x0 @x @x0 µ 0 ··· ··· that is, a partial-derivative factor withT the primed = coordinateµ in the numerator T for···⌫ every··· upper (superscript) (2.10) ··· @x ···@x0 ··· @x ··· ··· index and a a partial-derivative factor with the primed coordinate in the denominator for every lower (sub- script) index.that is,Using a partial-derivative methods similar factor to what with we the didTensors primed in the last coordinate session, in in you the cana numerator provecoordinate in for particular every upper that (superscript) the basis metric isindex a tensor and with a a two partial-derivative lower indices: factor with the primed coordinate in the denominator for every lower (sub- script) index. Using methods similar to what we did in the last session, you can prove in particular that the Other↵ tensors⌫ and tensor operations (which you can metric is a tensor with two lower indices:@x0 @x gµ0 ⌫ = proveµ in ga↵ similar way): (2.11) @x @x0 ↵ ⌫ @x0 @x µ⌫ g0 = µg⌫↵ ⌫ (2.11) that the Kronecker delta is still a tensor, that g definedµThe⌫ such Kroneckerµ that g delta:gµ↵ = (the matrix inverse of the @x @x0 ↵ metric tensor), is still a tensor, that multiplying by g and gµ⌫ and summing over µ still lowers or raises a µ⌫ µ⌫ µ⌫ ⌫ that the Kronecker delta is still a tensor, thatTheg inversedefined metric: such that g gµ↵ = ↵ (the matrix inverse of the µ⌫ metric tensor), is still a tensor, that multiplying by gµ⌫ and g and summing over µ still lowers or raises a ν 4Lowering an index: Aμ = gμνA μ μν Raising 4an index: B = g Bν Gradients tensor index, that contracting over an upper and lower index of a tensor quantity still yields a tensor with tensor index, that contracting over an upper and lower index of a tensor quantity still yields a tensor with rank n 2, and so on. The gradient of a scalar also still yields a tensor (covector): by the chain rule rank n 2, and so on. The gradientThe gradient of a scalar of a alsoscalar still is yields a covector: a tensor (covector): by the chain rule @ @x⌫ @ @x⌫ ⌫ ⌫ @@µ0 @xµ =@ µ @⌫x= µ (@⌫ ) (2.12) ⌘ @x0 @x0 @x @x0 @µ0 µ = µ ⌫ = µ (@⌫ ) (2.12) which is the correct generalized⌘ transformation@x0 @x0 @ lawx for@ ax tensor0 with one lower index. However (unlike in which is thethe correct Lorentz generalized transformation transformation case), the gradient law for of a vector tensor (or with any one larger-rank lower index. tensor) However is not a tensor: (unlike by in the the Lorentzchain transformation and product case), rulesThe the gradient gradient of of a a vector vector (or is NOT any larger-rank a tensor: tensor) is not a tensor: by the chain and product rules ↵ ↵ ↵ ↵ @A @ @x0 ⌫ @x @ @x0 ⌫ @µ0 A µ = µ ⌫ A = µ ⌫ A @A↵ ⌘ @x@0 @@xx0 ↵ @x @x @@x0 @@xx ↵ @x @ A↵ = 0 A✓⌫ = ◆ 0✓ A⌫ ◆ µ0 µ @xµ @2x⌫ ↵ @xµ @x↵ @A⌫ ⌘ @x0 @x0 @x 0 ⌫ @x0 @x0 @x = µ ✓ ⌫ A◆ + µ ⌫ ✓ ◆ (2.13) @2x0 ↵ @x @x @x0↵ @x ⌫ @x @x @ x0 ✓ ⌫ @x◆ @x0 @A ✓ ◆ The last term looks like= the transformationµ ⌫ A law+ for a tensorµ ⌫ quantity with one upper and one lower index,(2.13) but @x0 @x @x @x0 @x @x the first term in the last line does✓ not. This◆ term did not arise✓ in the◆ Lorentz transformation case because The last termthe looks Lorentz like transformation the transformation coecients law arefor constant,a tensor quantity implying that with the one double upper partial and one derivatives lower index, are all but zero. the first term inThis the is last a serious line does problem, not. because This term many did physics not arise equations in the involve Lorentz calculating transformation derivatives case of because quantities the Lorentzthat transformation will generalize coe to vectorscients are or tensors constant, of higher implying rank. that We the will double address partial this problem derivatives in the are next all section. zero. This is a serious problem, because many physics equations involve calculating derivatives of quantities that will generalize2.3.1 Exercise: to vectors Parabolic or tensors coordinates. of higher rank. We will address this problem in the next section. Consider the parabolic coordinate system p, q shown in figure 2. The transformation functions from ordinary 2.3.1 Exercise: Parabolic coordinates. Consider the parabolic coordinate system p, q shown in figure 2. The transformationq = 3 m functions from ordinary q = 2 m q = 3 mq = 1 m eq q ep q = 2 mq = 0 A q = –1 m qA =y 1 m eq eq q ep Ax q = 0 q = –2 m O ep q = –3 m A A q = –1 m e y q A q = –2 m p = –4 m –2 m 0 x 2 m 4 m O ep q = –3 m Figure 2: Parabolic coordinates for a flat two-dimensional plane. Adapted from Moore, A General Relativity Workbook, University Science Books, 2013, p. 60.) p = –4 m –2 m 0 2 m 4 m cartesian x, y coordinates are p(x, y)=x and q(x, y)=y cx2 (2.14) Figure 2: Parabolic coordinates for a flat two-dimensional plane. Adapted from Moore, A General Relativity Workbook, Universitywhere c is a Science constant Books, with units 2013, of p.inverse 60.) meters. (a) Show that the inverse transformation functions are cartesian x, y coordinates are 2 p(x, y)=x x(andp, q)=qp(x,and y)=yy(p, qcx)=2 cp + q (2.14)(2.15) µ ⌫ µ ⌫ where c is a(b) constantEvaluate with all units eight of partial inverse derivatives meters. @x0 /@x and @x /@x0 .

(a) Show that(c) theThe inverse metric tensor transformation components functions for cartesian are coordinates in space are gxx = gyy =1,gxy = gyx = 0. Use the general tensor transformation rule to show that the metric tensor for p, q coordinates is x(p, q)=p and y(p, q)=cp2 + q (2.15) 1+4c2p2 2cp g0 = (2.16) µ ⌫ µ⌫ µ2cp ⌫ 1 (b) Evaluate all eight partial derivatives @x0 /@x and @x /@x0 . p q (c) The metric(d) Let tensor a vector componentsA have components for cartesianA coordinates=1,A =0inthe in spacep, q arecoordinategxx = gyy system.=1,g Findxy = thisgyx vector’s= 0. Use the generalcomponents tensor in transformation the x, y coordinate rule system to show (as that a function the metric of x tensorand y). for Butp, show q coordinates that A2 = isA ⇧ A has the same value in both coordinate systems at every position. 1+4c2p2 2cp g = (2.16) µ0 ⌫ 2cp 1  5 (d) Let a vector A have components Ap =1,Aq =0inthep, q coordinate system. Find this vector’s components in the x, y coordinate system (as a function of x and y). But show that A2 = A ⇧ A has the same value in both coordinate systems at every position.

5 tensor index, that contracting over an upper and lower index of a tensor quantity still yields a tensor with rank n 2, and so on. The gradient of a scalar also still yields a tensor (covector): by the chain rule @ @x⌫ @ @x⌫ @µ0 µ = µ ⌫ = µ (@⌫ ) (2.12) ⌘ @x0 @x0 @x @x0 which is the correct generalized transformation law for a tensor with one lower index. However (unlike in the Lorentz transformation case), the gradient of a vector (or any larger-rank tensor) is not a tensor: by the chain and product rules ↵ ↵ ↵ ↵ @A @ @x0 ⌫ @x @ @x0 ⌫ @0 A = A = A µ ⌘ @x µ @x µ @x⌫ @x µ @x @x⌫ 0 0 ✓ ◆ 0 ✓ ◆ @x @2x ↵ @x @x ↵ @A⌫ = 0 A⌫ + 0 (2.13) @x µ @x@x⌫ @x µ @x⌫ @x 0 ✓ ◆ 0 ✓ ◆ The last term looks like the transformation law for a tensor quantity with one upper and one lower index, but the first term in the last line does not. This term did not arise in the Lorentz transformation case because the Lorentz transformation coecients are constant, implying that the double partial derivatives are all zero. This is a serious problem, because many physics equations involve calculating derivatives of quantities that will generalize to vectors or tensors of higher rank. We will address this problem in the next section.

2.3.1 Exercise: ParabolicExercise: coordinates. Parabolic coordinates Consider the parabolic coordinate system p, q shown in figure 2. The transformation functions from ordinary

q = 3 m p(x, y) = x q = 2 m 2 q = 1 m q(x, y) = y − cx eq q ep q = 0 A q = –1 m Ay eq Ax q = –2 m O ep q = –3 m

p = –4 m –2 m 0 2 m 4 m

Figure 2: Parabolic coordinates for a flat two-dimensional plane. Adapted from Moore, A General Relativity Workbook, University Science Books, 2013, p. 60.) cartesian x, y coordinates are p(x, y)=x and q(x, y)=y cx2 (2.14) where c is a constant with units of inverse meters. (a) Show that the inverse transformation functions are x(p, q)=p and y(p, q)=cp2 + q (2.15)

µ ⌫ µ ⌫ (b) Evaluate all eight partial derivatives @x0 /@x and @x /@x0 .

(c) The metric tensor components for cartesian coordinates in space are gxx = gyy =1,gxy = gyx = 0. Use the general tensor transformation rule to show that the metric tensor for p, q coordinates is 1+4c2p2 2cp g = (2.16) µ0 ⌫ 2cp 1  (d) Let a vector A have components Ap =1,Aq =0inthep, q coordinate system. Find this vector’s components in the x, y coordinate system (as a function of x and y). But show that A2 = A ⇧ A has the same value in both coordinate systems at every position.

5 tensor index, that contracting over an upper and lower index of a tensor quantity still yields a tensor with rank n 2, and so on. The gradient of a scalar also still yields a tensor (covector): by the chain rule @ @x⌫ @ @x⌫ @µ0 µ = µ ⌫ = µ (@⌫ ) (2.12) ⌘ @x0 @x0 @x @x0 which is the correct generalized transformation law for a tensor with one lower index. However (unlike in the Lorentz transformation case), the gradient of a vector (or any larger-rank tensor) is not a tensor: by the chain and product rules ↵ ↵ ↵ ↵ @A @ @x0 ⌫ @x @ @x0 ⌫ @0 A = A = A µ ⌘ @x µ @x µ @x⌫ @x µ @x @x⌫ 0 0 ✓ ◆ 0 ✓ ◆ @x @2x ↵ @x @x ↵ @A⌫ = 0 A⌫ + 0 (2.13) @x µ @x@x⌫ @x µ @x⌫ @x 0 ✓ ◆ 0 ✓ ◆ The last term looks like the transformation law for a tensor quantity with one upper and one lower index, but the first term in the last line does not. This term did not arise in the Lorentz transformation case because the Lorentz transformation coecients are constant, implying that the double partial derivatives are all zero. This is a serious problem, because many physics equations involve calculating derivatives of quantities that will generalize to vectors or tensors of higher rank. We will address this problem in the next section.

2.3.1 Exercise: Parabolic coordinates. Consider the parabolic coordinate system p, q shown in figure 2. The transformation functions from ordinary

q = 3 m q = 2 m q = 1 m eq q ep q = 0 A q = –1 m Ay eq Ax q = –2 m O ep q = –3 m Exercise:p = –4 mParabolic –2 m 0 2 m 4 m coordinates Figure 2: Parabolic coordinates for a flat two-dimensional plane. Adapted from Moore, A General Relativity Workbook, University Science Books, 2013, p. 60.) cartesian x, y coordinates are 2 p(x, y) =p(x,x y)=x andq(xq,(x,y) y)== yy −cx2cx (2.14) where c is a constant with units of inverse meters. (a) Show that the inverse transformation functions are x(p, q)=p and y(p, q)=cp2 + q (2.15)

µ ⌫ µ ⌫ (b) Evaluate all eight partial derivatives @x0 /@x and @x /@x0 .

(c) The metric tensor components for cartesian coordinates in space are gxx = gyy =1,gxy = gyx = 0. Use the general tensor transformation rule to show that the metric tensor for p, q coordinates is 1+4c2p2 2cp g = (2.16) µ0 ⌫ 2cp 1  (d) Let a vector A have components Ap =1,Aq =0inthep, q coordinate system. Find this vector’s components in the x, y coordinate system (as a function of x and y). But show that A2 = A ⇧ A has the same value in both coordinate systems at every position.

5 2.4 The Tensor Gradient To see more clearly why the simple gradient of a vector field is not a tensor consider the simple case of a constant vector field A(xµ) in a flat two-dimensional space. In flat space at least, we can define “constant” in a coordinate-independent way by saying that at all points, the vector points in the same direction and 2.4 The Tensor Gradient has2.4 the same The magnitude. Tensor Gradient Therefore the physical gradient of such a field should be zero in any coordinate To see more clearly why the simple gradient of a vector field is not a tensor consider the simple case of a systemTo see (because more clearly the vector why the does simple not gradient change as of a we vector change field positions). is not a tensor In a consider cartesian the coordinate simple case system, of a the constant vector field Aµ(xµ) in a flat two-dimensional space. In flatµ space at least, we can define “constant” components A of suchA a fieldµ are constant, so @↵A = 0 as expected. But in a curvilinear coordinate system, in a coordinate-independentconstant vector field way(x by) saying in a flat that two-dimensional at all points, the space. vector In points flat space in the at same least, direction we can and define “constant” thein components a coordinate-independent of even a truly way constant by saying vector that field at all may points,not thebe vector constant, points because in the the same basis direction vectors and used to has the same magnitude. Therefore the physical gradient of such a field should be zero in any coordinate µ systemdefine (becausehas the the the same components vector magnitude. does change not Therefore change as one as we the goes change physical from positions). gradient point to In of point. a such cartesian a We field coordinate need should to find be system, zero a way the in any to correct coordinate@↵A in µ µ componentssuchsystem aA coordinateof (because such a field system the are vector constant, so does as to so not remove@↵ changeA = the 0 as as part expected. we change due But to positions). changes in a curvilinear in In the a cartesian coordinate basis vectors coordinate system, if we system, hope to the find the µ µ µ the componentstruecomponents change of even inA the aof truly vector such constant a fieldfunction are vector constant,A( fieldx ). may so @not↵Abe= constant, 0 as expected. because But the basis in a curvilinear vectors used coordinate to system, ⌫ µ define thetheTensor components components Gradient change of even as of one a a truly Vector. goes constant fromLet point vector us to define point. field a may We set need ofnot coe tobe find constant,cients a way because to(which correct the@ physicists↵ basisA in vectors call usedChristo to ↵el ⌫↵ µ suchsymbols a coordinatedefine the) at system components a given so as point to change remove or event as the one part goessuch due from to that changes point in to the point. basis We vectors need if to we find hope a to way find to the correct @↵A in true changesuch in a the coordinate vector function systemA so(xµ as). to removeP the part due to changes in the basis vectors if we hope to find the 2.4 The Tensor Gradient µ ⌫ Tensortrue Gradient change in of the a Vector. vector functionLet us defineA(x a). set of coe@e↵cients ⌫⌫↵ (which physicists call Christo↵el To seesymbols more clearly) at a given why the point simple or event gradientsuch of a that vector field is not a tensorµ↵ considere⌫ the⌫ simple case of a (2.17) Tensor Gradient of a Vector. Let us define@x aµ set⌘ of coecients ⌫↵ (which physicists call Christo↵el constant vector field A(xµ) in a flat two-dimensionalP space. In flat space at least, we can define “constant” symbols) at a given point or event @esuch that in a coordinate-independentThe numerator here way by is saying the di that↵erential at allP ↵ points, change⌫ the in vector the basis points vector in the samee as direction we move and from point to a point µ µ↵e⌫ ↵ (2.17) µ @x ⌘ P has the sameadi magnitude.↵erential displacement Therefore the physicaldx along gradient a curve of such@e where a↵ field should⌫ the other be zero coordinates in any coordinate are constant, divided by that µ↵e⌫ (2.17) systemThe (because numerator the here vector is does the di not↵erential changeµ change as we change in the positions). basis@ vectorxµ ⌘ Ine a↵ cartesianas we move coordinate from point system,to the a point e di↵erentialµ displacementµ dx (whereµ ↵ and µ have some specific values here).P Since the change @ ↵ is a componentsadi↵erentialA of displacement such a field aredx constant,along so a@ curve↵A = where 0 as expected. the other But coordinates in a curvilinear are constant, coordinate divided system, by that vectorThe (an numerator arrow here withµ is a the certain di↵erential magnitude change pointing in the basis in a vector certaine↵ direction),as we move we from can point write thatto a change point as a the componentsdi↵erential displacement of even a trulydx constant(where vector↵ and fieldµ have may not somebe specific constant, values because here). the basis Since vectors the change used to@e↵ is a µ µ P definevector thesum components (anadi over arrow↵erential the with change basis a displacement certain as vectors one magnitude goese⌫dx fromevaluatedalong point pointing to a at curvepoint. in point a certain where We need: this direction), the to is other find the a we point coordinates way can to of write correct the that are sum@↵A constant, change overin ⌫ as. aIndivided a two-dimensional by that µ P such a coordinatedi↵erential system displacement so as to removedx the(where part due↵ toand changesµ have in the some basis specific vectors values if we hope here). to find Since the the change @e↵ is a sum overspace, the there basis vectors are 8 suche⌫ evaluated coeTensorcients; at point in four-dimensional: this gradient is the point of spacetime, the sum of over a there⌫. Invector a are two-dimensional 64 such coecients. µ truespace, change there invectorNow the are vector consider (an 8 such arrow function coe a withcients; vectorA(x a). certain in field four-dimensionalA magnitudethatP is a pointing spacetime, function in there aof certain position. are 64 direction), such The coe productcients. we can write rule thatimplies change that as the a true ⌫ TensorNowamount Gradientsum consider overdA a of thethat vector a basis Vector.A changes field vectorsLetA that use when⌫ defineevaluated is a we functiona set move of at coe point of an position.cients arbitrary: this⌫↵ The(which is infinitesimal the product point physicists of rule the displacement call implies sumChristo over that↵⌫el the.d Ins true(whose a two-dimensional components are symbols) at a given point or event such that P amount dspace,A that thereA changes are 8 when suchP we coe moveDefinecients; an arbitrary ↵Christoffel in four-dimensional infinitesimal symbols: displacement spacetime, thereds (whose are 64 components such coe arecients. some specific set of components↵ dx )is some specificNow set of consider components a vectordx )is field@Ae that is a function of position. The product rule implies that the true ↵ ⌫ e A A µ µ↵ ⌫ µ (2.17)s amount d that changes when@x we⌘ moveµµ an arbitrary@A infinitesimal displacementµ @eµ ↵ d (whose components are µdA =↵ d(@AA e )= µ dx@eµ ↵e + A dx (2.18) some specific set of componentsdA = d(A eµdx)=)is dxµ eµ + A dx µ ↵ (2.18) The numerator here is the di↵erential changeThe in physical the basisdx change vector e in↵dx asa vector we@x move↵ is then: from point@x to a point ✓ ◆ P adi↵erential displacement dxµ along a curve where✓ the other◆ coordinatesµ are constant, divided by that µ @A µ @eµ ↵ di↵erentialOne canOne displacement use can equation use equationdx 2.17µ (where and 2.17 rename↵ and and someµd renamehaveA = indices somed(A some toe specificµ)= rewrite indices values thisdx to as here). rewriteeµ Since+ A this the as changedx @e is a (2.18) dx @x↵ ↵ vector (an arrow with a certain magnitude pointingµ in a certain✓ direction),◆ we can write that change as a @A µ @⌫Aµ ↵ µ ↵ sum over the basis vectors e⌫ evaluateddA = at point +: this↵⌫ A is theeµdx pointµ of( ⌫ the↵A sum)eµdx over↵ ⌫. In aµ two-dimensional↵ (2.19) One can use equation 2.17 and@x↵d renameA = some+ indices↵⌫⌘ Ar toe rewriteµdx this( as↵A )eµdx (2.19) space, there are 8 such coecients; in four-dimensional P @ spacetime,x↵ there are 64 such⌘ coercients.  µ NowThe consider quantities a ( vectorAµ) fielddx↵ Aforthat di↵erent is a function values of of@Aµ position.are by definition The product components rule implies (in whatever that the true coordinate ↵ µ ↵ dA = + µ A⌫ e dx↵ ( Aµ)e dx↵ (2.19) amount dTheA that quantitiesA changesr when ( ↵A we) movedx anfor arbitrary di↵erent infinitesimal↵ values↵⌫ of displacementµ µare byd definitions (whose↵ µ components components are (in whatever coordinate system we are using) of ther true changeSoin thisA for tensor that@x digradient↵erential is displacement. a tensor⌘ r Therefore, we define some specificsystem set of we components are using)dx↵ of)is the true change in A for that di↵erential displacement. Therefore, we define µ ↵ @Aµ The quantities ( ↵A )dx for di↵erentµ values ofµ µ⌫are by definition components (in whatever coordinate ↵A µ + A (2.20) r @A ↵ ↵⌫ @eµ µ system we are using) of theµ truer change⌘ @inx A forµ thatµ @ diA↵↵erentialµ displacement.⌫ Therefore, we define dA = d(A eµ)= dx eµ + A ↵ dx (2.18) dx ↵A @x ↵ + ↵⌫ A (2.20) to be the tensor gradient of the vector✓ field A: the◆r term involving⌘ @µx the Christo↵el symbols corrects the µ @A µ ⌫ µ Onepartial can use derivative equation 2.17 of the and field rename for the some variations indices to in rewrite the basis this vectors. as The tensor gradient ↵A must be a ↵A ↵ + ↵⌫ A (2.20) (second-rank)to be the tensor,tensor because gradient we see thatof equation the vector 2.19r field implies⌘A:@ thatx the ( termAµ)dx involving↵ yields the ther components Christo↵el of symbolsa corrects the @Aµ r↵ µ vector.partial However,to be derivative the neithertensor thed ofA gradient the= partial field derivative+of for µ theA the⌫ vector alonee variationsdx↵ nor field( theA inA: Christoµ the) thee dx term↵ basis↵el symbol involving vectors. alone the The are Christo tensors: tensor(2.19)↵el only gradient symbols the corrects↵A must the be a ↵ ↵⌫ µ ↵ µ µ ↵ r (second-rank) tensor, because@x we see that equation⌘ r 2.19 implies that ( A )dx yields theµ components of a combinationpartial given derivative by equation of the 2.20 field is a for tensor. the variations in the basis vectors. The tensor↵ gradient ↵A must be a The Tensor Gradientµ ↵ of a Covector. To determine a covector’s tensor gradient, werµ can take↵ advantager The quantitiesvector.(second-rank) ( ↵ However,A )dx for tensor, neither di↵erent because the values partial we of seeµ are that derivative by equation definition alone 2.19 components nor implies the that Christo (in whatever ( ↵Ael) symboldx coordinateyields alone the componentsare tensors: of only a the r µ ↵ systemof the wecombination fact arevector. using) that A However, ofB the givenµ istrue a scalar by neither change equation, and thein A the partialfor 2.20 tensor that is derivative di gradient a↵erential tensor. of alone displacement. a scalar nor is the the Therefore,Christo same as↵rel its we symbol ordinary define alone gradient: are tensors: only the combinationThe Tensor given Gradient by equation of a@ 2.20 Covector.Aµ isµ a tensor.@BTo determine a covector’s tensor gradient, we can take advantage µ µ @A µ µ µ⌫ µ @↵(A Bµ)=A Bµ ++A A (2.20) of theThe fact Tensor that A GradientBµ is a scalarr of↵ a@ Covector.x,⌘↵ and@x↵ the tensor↵⌫@Tox↵ determine gradient a covector’s of a scalar tensor is the gradient, same as we its can ordinary take advantage gradient: µ of the fact that A Bµ is a scalar= ,( andAµB the)=( tensorAµ gradient)B + Aµ of( a scalarB ) is the same as its (2.21) ordinary gradient: to be the tensor gradient of the vector fieldrA↵: theµ term involvingrµ↵ µ the Christor↵ ↵µel symbols corrects the µ @A µ @Bµ µ partial derivative of the field for the variations in the basis vectors.µ The tensorµ gradient A must be a Subtracting one side from the other and@↵( substitutingAµBµ)=@ whatA ↵ B weµ know+ Aµ @BµA↵ to be yields↵ @x µ@↵x ↵ r (second-rank) tensor, because we see that equation@↵(A B 2.19µ)= implies↵ B thatµ + (A↵Ar)↵dx yields the components of a µ @x µ r @x µ µ vector. However, neither the partial@A derivativeµ @ aloneBµ nor= theµ↵ Christo(Aµ B↵µµel)=( symbol↵µA alone)B areµ +µ tensors:A ( ↵ onlyBµ) the (2.21) 0= Bµ + A (=↵A )(BAµ B A)=(( ↵BµA) )B + A ( B ) (2.21) combination given by equation 2.20@x↵ is a tensor.@x↵ r r↵ µ r ↵r µ ↵ rµ r r r µ Subtracting one side fromµ ⇢ the other andµ substituting⇢ what we know A to be yields The Tensor Gradient of a Covector.@A ⇢ To@ determineBµ @A a⇢ covector’s tensor gradient, we can takeµ↵ advantage Subtractingµ one side from theµ other and substitutingµ ⌫ what weµ know ↵rA to be yields of the fact that A B is a scalar= , and⇢ B theµ + tensorA gradient⇢ ofB aµ scalar(↵⌫ isA the)Bµ sameA as( its↵B ordinaryµ) gradient: µ @x↵ µ@x↵ @x↵ r r ⇢ @A µ ⇢ µ @Bµ µ µ @0=B @µA B + Aµ @Bµ ( µA )B µA ( B ) µµ 0=µ@A ↵⌫ Bµ +µA@Bµ ↵ ( ↵A↵ )Bµ µ A ( ↵Bµ↵) µ @ =(AAB )= @xB↵µB+⌫ A ↵B@µx↵ r r (2.22) ↵ µ @x↵ @x@↵x µ r@x↵@x r r  µµµ ⇢ µ µµµ⇢⇢ = @↵@A(AA ⇢⇢Bµ)=(µµ↵@@ABBµ)µBµ @+A@AA⇢(⇢↵Bµ)µ µ⌫ ⌫ µ µ (2.21) ==r ⇢⇢↵↵ BBµ ++AAr ↵↵ ⇢↵⇢↵BrµBµ (↵⌫(A↵⌫ A)Bµ)BµA (A ↵(Bµ↵) Bµ) ⇢⇢@@xx @6@xx ⇢@⇢x@x µ r r Subtracting one side from the other and substituting what we know ↵A to be yields @@Bµ r = Aµµ µ ⌫⌫ B B (2.22) @Aµ =@AB ↵ ↵↵µµB⌫⌫ ↵ ↵Bµ µ (2.22) 0= B + Aµ µ @(x↵Aµ)B Aµr(rB ) @x↵ µ @x↵  r↵ µ r↵ µ µ ⇢ µ ⇢ @A ⇢ µ @Bµ @A ⇢ µ ⌫ µ = ⇢ Bµ + A ⇢ Bµ (↵⌫ A )Bµ 6 A ( ↵Bµ) ⇢@x↵ @x↵ ⇢@x↵ 6 r @B = Aµ µ ⌫ B B (2.22) @x↵ ↵µ ⌫ r↵ µ 

6 2.4 The Tensor Gradient To see2.4 more The clearly Tensor why the Gradient simple gradient of a vector field is not a tensor consider the simple case of a constantTo see vector more field clearlyA(xµ why) in athe flat simple two-dimensional gradient of a space. vector In field flat is space not a at tensor least, consider we can define the simple “constant” case of a in a coordinate-independentconstant vector field A(x wayµ) in by a saying flat two-dimensional that at all points, space. the In vector flat space points at least, in the we same can define direction “constant” and has thein samea coordinate-independent magnitude. Therefore way the by physical saying that gradient at all of points, such a the field vector should points be zero in the in same any coordinate direction and systemhas (because the same the magnitude. vector does Therefore not change the as physical we change gradient positions). of such In a a field cartesian should coordinate be zero in system, any coordinate the µ µ componentssystemA (becauseof such the a field vector are does constant, not change so @↵A as= we 0 change as expected. positions). But in In a a curvilinear cartesian coordinate coordinate system, system, the µ µ the componentscomponents ofA evenof such a truly a field constant are constant, vector field so @↵ mayA =not 0 asbe expected. constant, But because in a curvilinear the basis vectors coordinate used system, to µ definethe the components components of change even a astruly one constant goes from vector point field to may point.not Webe need constant, to find because a way the to basis correct vectors@↵A usedin to µ such adefine coordinate the components system so changeas to remove as one the goes part from due point to changes to point. in the We basis need vectors to find a if way we hope to correct to find@↵ theA in true changesuch a incoordinate the vector system function so asA to(xµ remove). the part due to changes in the basis vectors if we hope to find the µ ⌫ Tensortrue change Gradient in the of vector a Vector. functionLetA us(x define). a set of coecients ⌫↵ (which physicists call Christo↵el ⌫ symbolsTensor) at a given Gradient point or of event a Vector.suchLet that us define a set of coecients ⌫↵ (which physicists call Christo↵el symbols) at a given point or eventP such that P @e↵ ⌫ µ @eµ↵e⌫ (2.17) @x ⌘ ↵ ⌫ e (2.17) @xµ ⌘ µ↵ ⌫ The numerator here is the di↵erential change in the basis vector e↵ as we move from point to a point The numerator here is theµ di↵erential change in the basis vector e as we move from pointP to a point adi↵erential displacement dx along a curve where the other coordinates↵ are constant, dividedP by that adi↵erential displacementµ dxµ along a curve where the other coordinates are constant, divided by that di↵erential displacement dx (where ↵ and µ have some specific values here). Since the change @e↵ is a µ vectordi (an↵erential arrow displacement with a certaindx magnitude(where ↵ pointingand µ have in a some certain specific direction), values we here). can write Since that the change change@ ase↵ ais a vector (an arrow with a certain magnitude pointing in a certain direction), we can write that change as a sum over the basis vectors e⌫ evaluated at point : this is the point of the sum over ⌫. In a two-dimensional sum over the basis vectors e evaluated at pointP : this is the point of the sum over ⌫. In a two-dimensional space, there are 8 such coecients;⌫ in four-dimensionalP spacetime, there are 64 such coecients. Nowspace, consider there are a vector 8 such field coeAcients;that is in a four-dimensional function of position. spacetime, The there product are 64 rule such implies coecients. that the true amount dNowA that considerA changes a vector when field we moveA that an is arbitrary a function infinitesimal of position. displacement The productds (whose rule implies components that the aretrue someamount specificd setA that of componentsA changes whendx↵)is we move an arbitrary infinitesimal displacement ds (whose components are some specific set of components dx↵)is µ @A @eµ dA = d(Aµe )= dx@Aµ e + Aµ dx@e↵ (2.18) dA = d(µAµe )=dx dxµ e +@Ax↵µ µ dx↵ (2.18) µ✓ dx◆ µ @x↵ ✓ ◆ One can use equation 2.17 and rename some indices to rewrite this as One can use equation 2.17 and rename some indices to rewrite this as @Aµ @Aµµ ⌫ ↵ µ ↵ dA = ↵ + ↵⌫ A µ eµ⌫dx (↵ ↵A )eµµdx ↵ (2.19) dA@=x + A eµ⌘dx r ( ↵A )eµdx (2.19)  @x↵ ↵⌫ ⌘ r  The quantities ( Aµ)dx↵ for di↵erent values of µ are by definition components (in whatever coordinate The quantities↵ ( Aµ)dx↵ for di↵erent values of µ are by definition components (in whatever coordinate system we are using)r of↵ the true change in A for that di↵erential displacement. Therefore, we define system we are using)r of the true change in A for that di↵erential displacement. Therefore, we define µ µ @A µ µ ⌫ A µ @+A A µ ⌫ (2.20) ↵ A ↵ ↵⌫+ A (2.20) r r↵⌘ @x⌘ @x↵ ↵⌫ to be the tensor gradient of the vector field A: theA term involving the Christo↵el symbols corrects the to be the tensor gradient of the vector field : the term involving the Christo↵el symbolsµ corrects the partialpartial derivative derivative of the of field the forfield the for variations the variations in the in basis the basis vectors. vectors. The The tensor tensor gradient gradient↵A Amustµ must be a be a µ ↵ r ↵ (second-rank) tensor, because we see that equation 2.19 implies that ( ↵A )dxµ yields↵ the componentsr of a (second-rank) tensor, because we see that equation 2.19 implies thatr ( ↵A )dx yields the components of a vector.vector. However, However, neither neitherTensor the partial the partial derivative derivativegradient alone alone nor the nor Christo the Christoof↵el symbol↵rael symbol covector alone alone are tensors:are tensors: only only the the combinationcombination given given by equation by equation 2.20 is 2.20 a tensor. is a tensor. The TensorThe Tensor Gradient Gradient of a Covector. of a Covector.To determineTo determine a covector’s a covector’s tensor tensor gradient, gradient, we can we can take take advantage advantage of the fact that AµB isµ a scalar, and the tensor gradient of a scalar is the same as its ordinary gradient: of the fact thatµA TensorBµ is a scalar gradient, and the of tensora scalar gradient = ordinary of a scalar gradient, is the same so as its ordinary gradient: µ µ µ @A @A µ @Bµ @Bµ @↵(A@B(µA)=µB )=Bµ +BA + Aµ ↵ µ@x↵ @x↵ µ @x↵ @x↵ µ µ µ = ↵=(A B(µA)=(µB )=(↵A )BAµµ+)BA+( A↵µB( µ)B ) (2.21) (2.21) r r↵ µ r r↵ µ r r↵ µ µ µ μ SubtractingSubtracting one side one from sideSubtracting from the other the other and and substituting and substituting substituting what what we yields, know we know for↵A arbitrary↵toA beto yields be yields r r A µ µ @A @A µ @Bµµ @Bµ µ µ µ µ 0= 0=Bµ +BA + A ( ↵A( )BAµ )BA ( A↵B( µ) B ) @x↵ @x↵ µ @x↵ @x↵r r↵ µ r r↵ µ µ ⇢ µ ⇢ µ ⇢µ ⇢ @A ⇢@A @Bµ @B@A ⇢@A = B +⇢Aµ µ µ B ⇢( µ A⌫µ)B ⌫ Aµ( µB ) where in the last step,⇢=↵ I exchangedµ⇢↵ Bµ + A↵ the µ,↵ ⇢⌫↵indexµ⇢↵ B namesµ ↵⌫(↵⌫ inA theµ )B thirdµ A term↵( µ↵ ofBµ the) line above so that I could ⇢@x ⇢@x @x @x⇢@x⇢@x r r pull out the common factor of Aµ. Since this must be true for arbitrary four-vectors A, the quantity in µ @Bµµ @Bµ⌫ ⌫ = A = A B⌫ B ↵Bµ B (2.22)(2.22) brackets must be zero for@x all↵ @ indexx↵↵µ values↵rµ ⌫µr, implying↵ µ that   @B B = µ ⌫ B (2.23) So we must have: r↵6 µ 6 @x↵ ↵µ ⌫ µ⌫ The Tensor Gradient of a Tensor. The generalization to arbitrary tensors (e.g., T ) is not hard: @T µ⌫ T µ⌫ = + µ T ⌫ + µ T µ + T µ⌫ (2.24) r↵ @x↵ ↵ ↵ ↵ The general rule is that the tensor gradient of a tensor is the sum of its ordinary gradient + plus a positive Christo↵el symbol (times the tensor) for each upper index (summing over that index and the second lower Christo↵el symbol index) and a negative Christo↵el symbol term (times the tensor) for each lower index (summing over that index and the upper Christo↵el symbol index. Calculating Christo↵el Symbols. We will assume in what follows that the Christo↵el symbols are symmetric in their lower indices (which amounts to assuming that spacetime is free of “torsion”1):

↵ ↵ µ⌫ = ⌫µ (2.25)

Given this assumption, one can calculate Christo↵el symbols in terms of the metric tensor as follows (the proof is one of your homework problems):

↵ = 1 g↵ [ @ g + @ g @ g ] (2.26) µ⌫ 2 µ ⌫ ⌫ µ µ⌫ This is easier to remember than one might think. A factor of the inverse metric generates the Christo↵el symbol’s superscript index. The final negative term has the symbol’s lower indices as the indices of the metric, and one generates the two positive terms by rotating the last term’s indices either right or left.

2.4.1 Exercise: The Tensor Gradient in Parabolic Coordinates. Consider the parabolic coordinate system described in the previous exercise, where p(x, y)=x and q(x, y)= y cx2. Consider a truly constant vector field in Cartesian coordinates defined so that Hx = 1 and Hy = 0. (a) Find the components of H in the p, q coordinate system. (b) In the previous exercise, we found the metric tensor for p, q coordinates to be

1+4c2p2 2cp g = (2.27) µ0 ⌫ 2cp 1  Verify (by matrix multiplication) that the inverse metric is given by

1 2cp g µ⌫ = (2.28) 0 2cp 1+4 c2p2  (c) From equations 2.26 through 2.28, one can show that the Christo↵el symbols for p, q coordinates are

q ↵ pp =2c, all other µ⌫ = 0 (2.29)

(This makes sense, because we can see from Figure 2 that only the ep unit vector changes with position, and then only in the q direction as we vary p.) Use equations 2.26 through 2.28 to verify that these q p are the correct values for pp and pp. (d) Calculate all four components of H⌫ in p, q coordinates. Are the results what you expect? rµ

7 where in the last step, I exchanged the µ, ⌫ index names in the third term of the line above so that I could pull out the common factor of Aµ. Since this must be true for arbitrary four-vectors A, the quantity in brackets must be zero for all index values µ, implying that @B Tensor gradientB = µ of ⌫ Ba tensor (2.23) r↵ µ @x↵ ↵µ ⌫ µ⌫ The Tensor GradientOne ofChristoffel a Tensor. termThe for generalizationeach index: to arbitrary tensors (e.g., T ) is not hard: @T µ⌫ T µ⌫ = + µ T ⌫ + µ T µ T µ⌫ (2.24) r↵ @x↵ ↵ ↵ ↵ The general rule is that the tensor gradient of a tensor is the sum of its ordinary gradient + plus a positive Christo↵el symbol (times the tensor) for each upper index (summing over that index and the second lower Christo↵el symbol index) and a negative Christo↵el symbol term (times the tensor) for each lower index (summing over that index and the upper Christo↵el symbol index. Calculating Christo↵el Symbols. We will assume in what follows that the Christo↵el symbols are symmetric in their lower indices (which amounts to assuming that spacetime is free of “torsion”1):

↵ ↵ µ⌫ = ⌫µ (2.25)

Given this assumption, one can calculate Christo↵el symbols in terms of the metric tensor as follows (the proof is one of your homework problems):

↵ = 1 g↵ [ @ g + @ g @ g ] (2.26) µ⌫ 2 µ ⌫ ⌫ µ µ⌫ This is easier to remember than one might think. A factor of the inverse metric generates the Christo↵el symbol’s superscript index. The final negative term has the symbol’s lower indices as the indices of the metric, and one generates the two positive terms by rotating the last term’s indices either right or left.

2.4.1 Exercise: The Tensor Gradient in Parabolic Coordinates. Consider the parabolic coordinate system described in the previous exercise, where p(x, y)=x and q(x, y)= y cx2. Consider a truly constant vector field in Cartesian coordinates defined so that Hx = 1 and Hy = 0. (a) Find the components of H in the p, q coordinate system. (b) In the previous exercise, we found the metric tensor for p, q coordinates to be

1+4c2p2 2cp g = (2.27) µ0 ⌫ 2cp 1  Verify (by matrix multiplication) that the inverse metric is given by

1 2cp g µ⌫ = (2.28) 0 2cp 1+4 c2p2  (c) From equations 2.26 through 2.28, one can show that the Christo↵el symbols for p, q coordinates are

q ↵ pp =2c, all other µ⌫ = 0 (2.29)

(This makes sense, because we can see from Figure 2 that only the ep unit vector changes with position, and then only in the q direction as we vary p.) Use equations 2.26 through 2.28 to verify that these q p are the correct values for pp and pp. (d) Calculate all four components of H⌫ in p, q coordinates. Are the results what you expect? rµ

7 where in the last step, I exchanged the µ, ⌫ index names in the third term of the line above so that I could where in the last step, I exchanged the µ, ⌫ index names in the third term of the line above so that I could pull out the common factor of Aµ. Since this must be true for arbitrary four-vectors A, the quantity in pull out the common factor of Aµ. Since this must be true for arbitrary four-vectors A, the quantity in brackets must be zero for all index values µ, implying that brackets must be zero for all index values µ, implying that @B @B B = µ ⌫ B (2.23) B = µr↵ µ⌫ B@x↵ ↵µ ⌫ (2.23) r↵ µ @x↵ ↵µ ⌫ The Tensor Gradient of a Tensor. The generalization to arbitrary tensors (e.g., T µ⌫ ) is not hard: µ⌫ The Tensor Gradient of a Tensor. The generalization to arbitrary tensors (e.g., T ) is not hard: @T µ⌫ µ⌫ µ⌫ µ ⌫ µ µ µ⌫ @T ↵T = + ↵T + ↵T ↵T (2.24) T µ⌫ = r + µ T@x⌫↵ + µ T µ T µ⌫ (2.24) r↵ @x↵ ↵ ↵ ↵ The general rule is that the tensor gradient of a tensor is the sum of its ordinary gradient + plus a positive The general ruleChristo is that↵el the symbol tensor (times gradient the tensor) of a tensor for each is the upper sum index of its (summing ordinary gradient over that + index plus and a positive the second lower Christo↵el symbolChristo (times↵el symbol the tensor) index) for and each a negative upper indexChristo (summing↵el symbol over term that (times index theand tensor) the second for each lower lower index Christo↵el symbol(summing index) over and that a negative index andChristo the upper↵el symbol Christo↵ termel symbol (times index. the tensor) for each lower index (summing over thatCalculating index and the Christo upper↵el Christo Symbols.Calculating↵el symbolWe will index.assume Christoffelin what follows symbols: that the Christo↵el symbols are Calculatingsymmetric Christo in↵el their Symbols. lower indicesWe (which will assume amountsin what to assuming follows that that spacetime the Christo is free↵el ofsymbols “torsion” are1): symmetric in their lower indices (which amounts to assuming that spacetime is free of “torsion”1): We assume this symmetry↵ (spacetime↵ is “torsion-free”): µ⌫ = ⌫µ (2.25) ↵ = ↵ (2.25) Given this assumption, one can calculateµ⌫ Christo⌫µ ↵el symbols in terms of the metric tensor as follows (the Given this assumption,proof is one one of can your calculate homework Christo problems):Then↵ oneel symbols can prove in that: terms of the metric tensor as follows (the proof is one of your homework problems): ↵ = 1 g↵ [ @ g + @ g @ g ] (2.26) µ⌫ 2 µ ⌫ ⌫ µ µ⌫ ↵ = 1 g↵ [ @ g + @ g @ g ] (2.26) This is easier to rememberµ⌫ than2 oneµ might⌫ think.⌫ µ A factorµ⌫ of the inverse metric generates the Christo↵el This is easier tosymbol’s remember superscript than one index. might The think. final A negative factor of term the has inverse the symbol’s metric generates lower indices the Christo as the↵ indicesel of the metric, and one generates the two positive terms by rotating the last term’s indices either right or left. symbol’s superscript index. The final negative term has the symbol’s lower indices as the indices of the metric, and one generates the two positive terms by rotating the last term’s indices either right or left. 2.4.1 Exercise: The Tensor Gradient in Parabolic Coordinates. 2.4.1 Exercise:Consider The the Tensor parabolic Gradient coordinate in system Parabolic described Coordinates. in the previous exercise, where p(x, y)=x and q(x, y)= y cx2. Consider a truly constant vector field in Cartesian coordinates defined so that Hx = 1 and Hy = 0. Consider the parabolic coordinate system described in the previous exercise, where p(x, y)=x and q(x, y)= y cx2. Consider(a) aFind truly the constant components vector of fieldH in in the Cartesianp, q coordinate coordinates system. defined so that Hx = 1 and Hy = 0. (a) Find the components(b) In the previous of H in exercise, the p, q coordinate we found the system. metric tensor for p, q coordinates to be

(b) In the previous exercise, we found the metric tensor for p, q1+4coordinatesc2p2 2cp to be g = (2.27) µ0 ⌫ 2cp 1 1+4c2p2 2cp g = (2.27) Verify (by matrix multiplication)µ0 ⌫ that2cp the inverse1 metric is given by  1 2cp Verify (by matrix multiplication) that the inverse metricg µ⌫ = is given by (2.28) 0 2cp 1+4 c2p2  µ⌫ 1 2cp g 0 = (2.28) (c) From equations 2.26 through 2.28,2 onecp can1+4 showc2p that2 the Christo↵el symbols for p, q coordinates are  q ↵ (c) From equations 2.26 through 2.28, one can show thatpp =2 thec, Christoall other↵el symbolsµ⌫ = 0 for p, q coordinates are (2.29) (This makes sense, because we can see from Figure 2 that only the e unit vector changes with position, q =2c, all other ↵ = 0p (2.29) and then only in the q directionpp as we vary p.)µ⌫ Use equations 2.26 through 2.28 to verify that these q p are the correct values for pp and pp. (This makes sense, because we can see from Figure 2 that only the ep unit vector changes with position, and then only in the q direction as we vary p.) Use⌫ equations 2.26 through 2.28 to verify that these (d) Calculate all four components of µH in p, q coordinates. Are the results what you expect? q p r are the correct values for pp and pp. (d) Calculate all four components of H⌫ in p, q coordinates. Are the results what you expect? rµ

7

7 Subtracting one side from the other and substituting what we know Aµ to be yields r↵ @Aµ @B 0= B + Aµ µ ( Aµ)B Aµ( B ) @x↵ µ @x↵ r↵ µ r↵ µ µ ⇢ µ ⇢ @A ⇢ µ @Bµ @A ⇢ µ ⌫ µ = ⇢ Bµ + A ⇢ Bµ (↵⌫ A )Bµ A ( ↵Bµ) ⇢@x↵ @x↵ ⇢@x↵ r @B = Aµ µ ⌫ B B (2.25) @x↵ ↵µ ⌫ r↵ µ  where in the last step, I exchanged the µ, ⌫ index names in the third term of the line above so that I could pull out the common factor of Aµ. Since this must be true for arbitrary four-vectors A, the quantity in brackets must be zero for all index values µ, implying that @B B = µ ⌫ B (2.26) r↵ µ @x↵ ↵µ ⌫ µ⌫ The Tensor Gradient of a Tensor. The generalization to arbitrary tensors (e.g., T ) is not hard: @T µ⌫ T µ⌫ = + µ T ⌫ + µ T µ T µ⌫ (2.27) r↵ @x↵ ↵ ↵ ↵ The general rule is that the tensor gradient of a tensor is the sum of its ordinary gradient + plus a positive Christo↵el symbol (times the tensor) for each upper index (summing over that index and the second lower Christo↵el symbol index) and a negative Christo↵el symbol term (times the tensor) for each lower index (summing over that index and the upper Christo↵el symbol index. Calculating Christo↵el Symbols. We will assume in what follows that the Christo↵el symbols are symmetric in their lower indices (which amounts to assuming that spacetime is free of “torsion”1):

↵ ↵ µ⌫ = ⌫µ (2.28) Given this assumption, one can calculate Christo↵el symbols in terms of the metric tensor as follows (the proof is one of your homework problems): ↵ = 1 g↵ [ @ g + @ g @ g ] (2.29) µ⌫ 2 µ ⌫ ⌫ µ µ⌫ ThisExercise: is easier to remember than one might think. A factor of the inverse metric generates the Christo↵el symbol’s superscript index. The final negative term has the symbol’s lower indices as the indices of the metric, and one generates the two positive terms by rotating the last term’s indices either right or left.

2.4.1 Exercise: The Tensor Gradient in Parabolic Coordinates. Consider the parabolic coordinate system described in the previous exercise, where p(x, y)=x and q(x, y)= y cx2. Consider a truly constant vector field in Cartesian coordinates defined so that Hx = 1 and Hy = 0. (a) Find the components of H in the p, q coordinate system. (b) In the previous exercise, we found the metric tensor for p, q coordinates to be 1+4c2p2 2cp g = (2.30) µ0 ⌫ 2cp 1  Verify (by matrix multiplication) that the inverse metric is given by 1 2cp g µ⌫ = (2.31) 0 2cp 1+4 c2p2  (c) The Christo↵el symbols for p, q coordinates are

q ↵ pp =2c, all other µ⌫ = 0 (2.32)

(This makes sense, because we can see from Figure 2 that only the ep unit vector changes with position, ↵ 1 ↵ and then only in the q direction as we vary p.) Use µ⌫ = 2 g [ @µg⌫ + @⌫ gµ @gµ⌫ ]toverify q p that these are the correct values for pp and pp. (d) Calculate all four components of H⌫ in p, q coordinates. Are the results what you expect? rµ

7 2.5 The Geodesic Equation. 2.5 The Geodesic Equation.2.5 The GeodesicThe four-velocity Equation.u = ds/d⌧ at every event along a particle’s worldline is a vector parallel to the di↵erential displacement ds that the particle moves during a di↵erential proper time d⌧ (as measured by a clock traveling The four-velocity u = ds/d⌧ at everyThe four-velocityevent along au particle’s=withds/d the⌧ worldlineat particle). every event is The a vector along four-velocity parallel a particle’s to is therefore the worldline di↵erential always is a vector tangent parallel to the to particle’s the di↵erential worldline. Now a geodesic displacement ds that the particledisplacement moves duringd as dithat↵erential theis2.5 by particle definitionproper The moves time Geodesic “thed during⌧ straightest(as measured a Equation. di↵erential possible” by a proper clock path traveling time at everyd⌧ (as point. measured A mathematical by a clock traveling way to define such a path is with the particle). The four-velocitywith is the therefore particle). always Theto tangent four-velocity say that to at the all is eventsparticle’s therefore along worldline. always the particle’s tangent Now a worldline,to geodesic the particle’s the particle’s worldline. four-velocity Now a geodesic is constant: du/d⌧ = 0. u s is by definition “the straightest possible”is by definition path at “the every straightestThe point.In four-velocity a curvedA possible” mathematical space= pathd or/d spacetime, way at⌧ at every to every define point. what event such A we along mathematical a mean path a particle’s by is this way is worldline that to define the is vector a such vectoru a pathdoes parallel notis to change the di direction↵erential displacement ds that the particle moves during a di↵erential proper time d⌧ (as measured by a clock traveling to say that at all events along theto particle’s say that worldline, at all eventsduring the along particle’s an the infinitesimal particle’s four-velocity stepworldline, is of constant: proper the particle’stimedud/d⌧⌧, as= four-velocity we 0. would determine is constant: in adu cartesian/d⌧ = 0. coordinate system set with the particle). The four-velocity is therefore always tangent to the particle’s worldline. Now a geodesic In a curved space or spacetime,In what a curved we mean space byup or this in spacetime, a is patch that ofthe what area vector around we meanu does the by particle not this change is that that direction is the small vector enoughu does to be not considered change locally direction flat. For example, we is by definition “the straightest possible” path at every point. A mathematical way to define such a path is during an infinitesimal step of properduring time an infinitesimald⌧, as we wouldmight step determine map of proper a great timein circle a cartesiand⌧, on as the we coordinate earth’s would determine surface system by in laying set a cartesian out a cartesian coordinate coordinate system system set on a city-sized to say that at all events along the particle’s worldline, the particle’s four-velocity is constant: du/d⌧ = 0. up in a patch of area around theup particle in a patch that is of small area aroundpatch,enough draw the to be particle a considered straight that line is locally smallin that flat. enough coordinate For to example, be system considered we a couple locally of miles flat. For long, example, then set we up a new coordinate In a curved space or spacetime, what we mean by this is that the vector u does not change direction might map a great circle on the earth’smight map surface a great by laying circlesystem out on centeredthe a cartesian earth’s at thesurface coordinate endpoint, by laying system draw out a on new a a cartesian city-sized straight coordinate line segment, system and repeat. on a city-sized during anThe infinitesimal Geodesic step of proper time d⌧ ,Equation as we would determine in a cartesian coordinate system set patch, draw a straight lineIn in an that arbitrary coordinate coordinate system system a couple in a of curved miles or long, flat then space set or up spacetime, a new coordinate we can calculate this path patch, draw a straight line in that coordinate systemup a couple in a patch of miles of area long, around then the set particle up a new that coordinate is small enough to be considered locally flat. For example, we system centered at the endpoint,system draw a centered new straight at the linemathematically endpoint, segment, draw and as a repeat. follows: new straight line segment, and repeat. might map a great circle on the earth’sdu surfaced by layingdu outu a cartesiande coordinate system on a city-sized In an arbitrary coordinate systemIn in an a arbitrary curved or coordinate flat space system or spacetime, in a curved we can or0= flat calculate space= this or(u spacetime,µ pathe )= wee can+ uµ calculateµ this path (2.30) patch, drawA a geodesic straight line is locally in that straight: coordinated⌧ d⌧ systemµ a coupled⌧ µ of milesd⌧ long, then set up a new coordinate mathematically as follows: mathematically as follows: Now,systemeu centereddepends at on the⌧ because endpoint, the draw particle’s au new position straight in line spacetime segment, depends and repeat. on ⌧ and e depends on position. du d µ du µ µ dedµu d µ du µ deµ µ 0= = (u eµ)=So usingIn aneµ the arbitrary+ u chain0= rule coordinate= and(u substitutinge systemµ)= inueµ a+ curvedudx(2.30)µ/d or⌧ into flat the space above, or spacetime, we find that we(2.30) can calculate this path d⌧ d⌧ mathematicallyd⌧ d as⌧d⌧ follows:d⌧ d⌧ ⌘ d⌧ u Now, eµ depends on ⌧ because theNow, particle’seµ depends position on in⌧ because spacetime the depends particle’s2 u on position⌧ and µeµ independs⌫ spacetimedu on2d depends position.u µ onµdu⌧ and⌫ eµ dependsµ deµ 2 u on position.↵ d x µ dxµ dx0=deµ =d x (u eµ)=dx dx eµ↵+ u d x µ dx dx (2.30) So using theµ chainµ rule and substituting0= u eµ +dx /d⌧ into the= above,eµ we+ find that e↵ = + eµ So using the chain rule and substituting u dx /d⌧ into the above, wed⌧ find2 thatd⌧ d⌧ dx⌫d⌧ dd⌧⌧2 d⌧ dd⌧⌧ µ⌫ d⌧ d⌧ 2 ↵ d⌧ d⌧ ⌘ ⌘  2 uNow, eµ dependsµ ⌫ on2 ⌧u because2 u the↵ particle’sµ ⌫ position in2 spacetimeu depends↵ on ⌧ and eµ depends on position. 2 u µ ⌫ 2 u d x µ ⌫dx dx ded x2 ud x µ dx dx↵dx dx d x dx dx d x dx dx deµ d x dxSo usingdx the↵ chainµd rulex and substitutingµ dx dx uµ↵ dxµ/d⌧ into theµ above, we find that 0= e + = 0=e + eµ + e0== =2 ++↵eµ + e e↵ = + eµ (2.31) d⌧ 2 µ d⌧ d⌧ dx⌫ d⌧ 2 µ d⌧ 2d⌧ d⌧ d)⌧µ⌫d⌧↵ dx⌫d⌧d⌧ 2 d⌧ 2 ↵d⌧d⌧dd⌧⌧d⌧d⌧ µµ⌘⌫ d⌧ 2 ↵ d⌧ d⌧ 2 u µ ⌫ 2 u  µ ⌫ 2 u ↵ 2 u ↵ 2 u ↵ d x dx dx deµ d x dx dx d x µ dx dx d x µ dx dx d x whereµ indx goingdx to the last step on the first line, I renamed the bound↵ µ, ⌫, ↵ indices in the last term to ↵, ,µ, 0= + 0= eµ + = eµ + µ⌫ e↵ = + ↵(2.31) eµ 0= + ↵ 2 ↵ Therefore,d⌧ 2 the geodesicd⌧ d⌧ dx ⌫equationd⌧ 2 (2.31) is: d⌧ d⌧ d⌧ 2 d⌧ d⌧ ) d⌧ 2 d⌧ d⌧ ) d⌧ respectively.d⌧ d The⌧ last line follows because a vector will only be zero if each of its components are zero. The 2 u ↵ last equation is thed xgeodesicµ dx equationdx : it represents four second-order di↵erential equations that one can where in going to the last step onwhere the first in line, going I to renamed the last the step bound on the0=µ, first⌫, ↵ line,indices+ I renamed inµ the last the term bound to ↵µ,, ⌫,,µ↵,indices in the last term to ↵, ,µ, (2.31) solve to yield the equationsd⌧ 2 ↵x d(⌧⌧)d that⌧ parametrically describe a geodesic in an arbitrary coordinate system respectively. The last line follows) because a vector will only be zero if each of its components are zero. The respectively. The last line follows because a vector willin only an arbitrary be zero if curved each of or its flat components space or spacetime. are zero. Note The that in flat spacetime, where the metric gµ⌫ = ⌘µ⌫ = last equation is the geodesic equationlast equation: it represents is the geodesicconstant,where four second-order in going equation all the to theChristo: di it↵ lasterential represents↵ stepel symbols on equations the four first (which second-order thatline, involve I one renamed can di derivatives↵erential the bound equations ofµ, the⌫, metric)↵ indices that one are in the zero, can last and term the to geodesic↵, ,µ, µ solve to yield the equations xµ(⌧)solve that to parametrically yield the equations describeequationrespectively.x a( yields⌧ geodesic) thatThedx last parametricallyµ/d in⌧ line an2 = arbitrary follows 0, whose describe because coordinate solutions a a vector geodesic are system straight will in only an worldlines, arbitrary be zero if as coordinate each we of would its system components expect. are zero. The in an arbitrary curved or flat spacein oran spacetime. arbitrary curved Notelast that orThe flat equation in geodesicspace flat spacetime, is or the equation spacetime.geodesic where is one Note equation the of that metric the intwo: itg flatµ core represents⌫ = spacetime,⌘ equationsµ⌫ = four where of second-order general the metric relativity. di↵gerentialµ⌫ = It⌘ expressesµ equations⌫ = the that first one clause can µ constant, all the Christo↵el symbolsconstant, (which all involve the Christo derivativesofsolve Wheeler’s↵el to symbols yield of aphorism: the the (which equations metric) “Spacetime involve arex ( zero,⌧) derivatives that tells and parametrically matter the of geodesic the how metric) to describe move.” are a If zero, geodesic we know and in the anthe arbitrary geodesic metric of coordinate spacetime system in any µ 2 equation yields dxµ/d⌧ 2 = 0, whoseequation solutions yields aredx straight/dcoordinatein⌧ an= worldlines, arbitrary0, whose system, solutionscurved as we we would orcan are flat use expect. straight space this equation or worldlines, spacetime. to calculate as Note we that would the in worldlines expect.flat spacetime, of free where particles the in metric thatg spacetime.µ⌫ = ⌘µ⌫ = The geodesic equation is one of theThe two geodesic core equations equationconstant, of is one general all of the the relativity. Christo two core↵el It equations symbols expresses (which of the general first involve clause relativity. derivatives It expresses of the metric) the first are clause zero, and the geodesic equation yields dxµ/d⌧ 2 = 0, whose solutions are straight worldlines, as we would expect. of Wheeler’s aphorism: “Spacetimeof Wheeler’s tells matter aphorism: how to2.5.1 move.” “Spacetime Exercise: If we tells know Geodesics matter the metric how in to of Parabolic move.” spacetime If Coordinates. we in know any the metric of spacetime in any coordinate system, we can use thiscoordinate equation system, to calculate we can theThe use worldlines geodesic this equation equation of free to particles is calculate one of in the the that two worldlines spacetime. core equations of free particles of general in relativity. that spacetime. It expresses the first clause Considerof Wheeler’s the aphorism: parabolic coordinate “Spacetime system tells matter for two-dimensional how to move.” flat If space we know described the metric in the of previous spacetime two in exer- any 2 2.5.1 Exercise: Geodesicscises,coordinate where in system,p Parabolic(x, y)= wex canand Coordinates. useq(x, this y)= equationy cx to. calculateIn a two dimensional the worldlines space of free like particles this, we in parameterize that spacetime. paths 2.5.1 Exercise: Geodesics in Parabolic Coordinates.using the arclength s along the path instead of the proper time, so the geodesic equation becomes Consider the parabolic coordinateConsider system for the two-dimensional parabolic2.5.1 coordinate Exercise: flat space system described Geodesics for two-dimensional in in the Parabolic previous flat two space Coordinates. exer- described in the previous two exer- 2 d2xu dx↵ dx cises, where p(x, y)=x and q(x,cises, y)=y wherecx2p.(x, In y a)= twox dimensionaland q(x, y)= spacey cx like. this, In a we two parameterize dimensional0= paths space+ µ like this, we parameterize paths (2.32) Consider the parabolic coordinate system for two-dimensional2 ↵ flat space described in the previous two exer- using the arclength s along the pathusing instead the arclength of the propers along time, the pathso the instead geodesic of equationthe proper becomes time, sods the geodesicds equationds becomes cises, where p(x, y)=x and q(x, y)=y cx2. In a two dimensional space like this, we parameterize paths µ 2 u andusing we the↵ look arclength for solutionss along of2 theu the path form insteadx↵(s). of We the found proper in the time, last so exercise the geodesic that equation the Christo becomes↵el symbols for d x µ dx dx d x µ dx dxq 0= + this coordinate system0= all zero+ except for pp =2(2.32)c. (2.32) 2 ↵ ds2 ↵ ds ds ds ds ds d2xu dx↵ dx (a) Find the p and q components of the0= geodesic+ equation. µ (2.32) and we look for solutions of the formand wexµ( looks). We for solutions found in the of the last form exercisexµ(s). that We the found Christo in the↵el last symbolsds exercise2 for ↵ thatds theds Christo↵el symbols for this coordinateq system(b) all zeroThe except solution for to theq =2p-componentc. equation is easy: p = as,wherea is a constant of integration, if we this coordinate system all zero except for pp =2c. and we look for solutionspp of the form xµ(s). We found in the last exercise that the Christo↵el symbols for this coordinatedefine s to system be zero all where zero exceptp is zero. for Use q this=2c to. show that the solution to the q-component equation is (a) Find the p and q components(a) of theFind geodesic the p and equation.q componentsq = ca2 ofs2 + thebs geodesic+ q ,where equation.b and q ppare constants of integration. 0 0 (a) Find the p and q components of the geodesic equation. (b) The solution to the p-component(b) The equation solution is easy: to the(c)p p=The-componentas,where transformationsa equationis a constant back is easy: to of cartesianp integration,= as,where coordinates if wea is a are constantx(p, q)= of integration,p and y(p, q if)= wecp2 + q. Use these define s to be zero where p is zero.define Uses thisto be to zero show(b) where thattransformationsThe thep solutionis solution zero. to Use the to to this convert thep-componentq to-component show the solutions that equation the equation for solution isp easy:(s is) and top the=q(ass)q,where-component to solutionsa is a equation for constantx(s) and is of integration,y(s). Argue if that we 2 2 q = ca2s2 + bs + q ,whereb and q are constants of integration. q = ca s + bs + q0,whereb and q0are constants of0 thedefine integration. resultings to be solutions zero0 where arep straightis zero. lines Use this (Hint: to showExpress thaty theas a solution function to of thex.)q-component equation is 2 2 q = ca s + bs + q ,whereb and q are constants of integration.2 (c) The transformations back to(c) cartesianThe transformations coordinates are backx(p, to q)= cartesianp and coordinatesy0(p, q)=cp2 are+ qx0.(p, Use q)= thesep and y(p, q)=cp + q. Use these transformations to convert the solutionstransformations for p(s) to and(c) convertTheq(s) transformations to the solutions solutions for for backx(sp)(s and to) and cartesiany(sq).(s) Argue to coordinates solutions that for arexx((sp,) andq)=yp(sand). Arguey(p, q)= thatcp2 + q. Use these the resulting solutions are straightthe lines resulting (Hint: solutionsExpresstransformations arey as straight a function lines to of convert (Hint:x.) Express the solutionsy as a for functionp(s) and of qx(.)s) to solutions for x(s) and y(s). Argue that the resulting solutions are straight lines (Hint: Express y as a function of x.) 8

8 8 8 2.5 The Geodesic Equation. The four-velocity u = ds/d⌧ at every event along a particle’s worldline is a vector parallel to the di↵erential displacement ds that the particle moves during a di↵erential proper time d⌧ (as measured by a clock traveling with the particle). The four-velocity is therefore always tangent to the particle’s worldline. Now a geodesic is by definition “the straightest possible” path at every point. A mathematical way to define such a path is to say that at all events along the particle’s worldline, the particle’s four-velocity is constant: du/d⌧ = 0. In a curved space or spacetime, what we mean by this is that the vector u does not change direction during an infinitesimal step of proper time d⌧, as we would determine in a cartesian coordinate system set up in a patch of area around the particle that is small enough to be considered locally flat. For example, we might map a great circle on the earth’s surface by laying out a cartesian coordinate system on a city-sized patch, draw a straight line in that coordinate system a couple of miles long, then set up a new coordinate system centered at the endpoint, draw a new straight line segment, and repeat. In an arbitrary coordinate system in a curved or flat space or spacetime, we can calculate this path mathematically as follows: du d duu de 0= = (uµe )= e + uµ µ (2.30) d⌧ d⌧ µ d⌧ µ d⌧

Now, eµ depends on ⌧ because the particle’s position in spacetime depends on ⌧ and eµ depends on position. So using the chain rule and substituting uµ dxµ/d⌧ into the above, we find that ⌘ d2xu dxµ dx⌫ de d2xu dxµ dx⌫ d2xu dx↵ dx 0= e + µ = e + ↵ e = + µ e d⌧ 2 µ d⌧ d⌧ dx⌫ d⌧ 2 µ d⌧ d⌧ µ⌫ ↵ d⌧ 2 ↵ d⌧ d⌧ µ  d2xu dx↵ dx 0= + µ (2.31) ) d⌧ 2 ↵ d⌧ d⌧ where in going to the last step on the first line, I renamed the bound µ, ⌫, ↵ indices in the last term to ↵, ,µ, respectively. The last line follows because a vector will only be zero if each of its components are zero. The last equation is the geodesic equation: it represents four second-order di↵erential equations that one can solve to yield the equations xµ(⌧) that parametrically describe a geodesic in an arbitrary coordinate system in an arbitrary curved or flat space or spacetime. Note that in flat spacetime, where the metric gµ⌫ = ⌘µ⌫ = constant, all the Christo↵el symbols (which involve derivatives of the metric) are zero, and the geodesic equation yields dxµ/d⌧ 2 = 0, whose solutions are straight worldlines, as we would expect. The geodesic equation is one of the two core equations of general relativity. It expresses the first clause ofExercise Wheeler’s aphorism: “Spacetime tells matter how to move.” If we know the metric of spacetime in any coordinate system, we can use this equation to calculate the worldlines of free particles in that spacetime.

2.5.1 Exercise: Geodesics in Parabolic Coordinates. Consider the parabolic coordinate system for two-dimensional flat space described in the previous two exer- cises, where p(x, y)=x and q(x, y)=y cx2. In a two dimensional space like this, we parameterize paths using the arclength s along the path instead of the proper time, so the geodesic equation becomes

d2xu dx↵ dx 0= + µ (2.32) ds2 ↵ ds ds and we look for solutions of the form xµ(s). We found in the last exercise that the Christo↵el symbols for q this coordinate system all zero except for pp =2c. (a) Find the p and q components of the geodesic equation. (b) The solution to the p-component equation is easy: p = as,wherea is a constant of integration, if we define s to be zero where p is zero. Use this to show that the solution to the q-component equation is q = ca2s2 + bs + q ,whereb and q are constants of integration. 0 0 (c) The transformations back to cartesian coordinates are x(p, q)=p and y(p, q)=cp2 + q. Use these transformations to convert the solutions for p(s) and q(s) to solutions for x(s) and y(s). Argue that the resulting solutions are straight lines (Hint: Express y as a function of x.)

8 2.6 Schwarzschild Geodesics. As we will see in a subsequentThe lecture,Schwarzschild one can solve the Einstein equationmetric in the vacuum around a spherical 2.6 Schwarzschild Geodesics. star to get the Schwarzschild metric: As we will see in a subsequent lecture, one can solve the Einstein equation in the vacuum around a spherical star to get the Schwarzschild2 metricGM : dr2 ds2 = 1 dt2 + + r2d✓2 + r2 sin2 ✓ d2 (2.33) r 2GM 1 2GM/rdr2 ds✓2 = 1 ◆ dt2 + + r2d✓2 + r2 sin2 ✓ d2 (2.33) r 1 2GM/r ✓ ◆ 1 2 2 2 (meaning that gtt = meaning(1 2GM/r that),grr =(1 2GM/r) ,g✓✓ = r and g = r sin ✓ and the o↵-diagonal 1 2 2 2 metric components(meaning that areg zero).tt = (1 This2GM/r is by),g norr means=(1 2 theGM/r only) ,g solution✓✓ = r and eveng for= ther sin vacuum✓ and the spacetime o↵-diagonal around a sphericalmetric star, components since we can are redefine zero). This coordinates is by no means in completely the only solution arbitrary even for ways the vacuum and so spacetime generate around an infinite number coordinatea spherical systems star,and since other that we canmetric describe redefine components the coordinates same physical inare completely zero. reality. arbitrary But this ways coordinate and so generate system an infinite does have number coordinate systems that describe the same physical reality. But this coordinate system does have some advantages:some advantages: it is independent it is independent of the of time the time coordinate coordinatet (int (in a a diagonal diagonal metric, metric, the the time time coordinate coordinate is the is the coordinatecoordinate whose metric whose component metric component is negative) is negative) and and this this metric metric reduces reduces to to the the metric metric for for flat flat spacetime spacetime (in (in a sphericala coordinate-basis) spherical coordinate-basis) as r as r . I should. I should note note that that though though I I am am using using units units where where time time and space and space are both measuredare both measured in meters in meters(so that!1 (soc that=!1 1),c = I 1), am I amnotnotusingusing units units where where the the universal universal gravitational gravitational constant constant G = 1 as well,G = 1 as as is well, often as is done. often I done. am doing I am doing this this so that so that connections connections to to Newtonian mechanics mechanics will will be clearer. be clearer. But it is goodBut it to is keep good in to mindkeep in that mind in that this in unit this unit system systemGMGMhashas units units of meters, meters, and and in in fact factGMGM= 1477= 1477 m m for a star with the mass of the sun (and is 4.45 mm for the Earth). for a star withIt the is not mass my of purpose the sun here (and to discuss is 4.45 the mm many for fascinating the Earth). features of this metric: it looks like others It is notmay my talk purpose about that, here and to the discuss topic is the somewhat many tangential fascinating to the features main line of thisof my metric: argument. it But looks I want like to others may talk aboutintroduce that, this and metric the so topic that we is somewhat can explore the tangential implications to the of the main geodesic line equation of my argument. in a realistic But situation I want to introduce this(one metric that even so applies that we to our can daily explore lives!). the implications of the geodesic equation in a realistic situation (one that evenCalculating applies to the our Christo daily↵el lives!). symbols in this case is simply an example of applying equation 2.26 repeatedly Calculatinguntil you the have Christo evaluated↵el symbols all 40 that in this might case be independent. is simply an There example are various of applying labor-saving equation ways 2.26 to do repeatedly this (including using computer tools), but a straightforward and low-tech way to do this is to use something until you havethat I evaluated call the Diagonal all 40 that Metric might Worksheet be independent.. This worksheet There are assumes various a diagonal labor-saving metric of ways the to form do this (includingds using2 = computerAx0 + Bx1 + tools),Cx2 + Dx but3 (where a straightforward the numbers are and placeholders low-tech for way coordinate to do names,this is not to exponents) use something that I calland the thatDiagonal the metric Metric components WorksheetA, B, C, D can. Thisdepend worksheet on any or all assumes of the coordinates. a diagonal The metric worksheet of then the form ds2 = Axsimply0 + Bx lists1 + allCx of2 the+ Dx Christo3 (where↵el symbols the numbers in terms of are the placeholders metric components for coordinate and their derivatives names, not (the exponents) latter 0 and that thewritten metric in a components shorthand formA, where B, C,A D0 can@A/ depend@x and on so any on. On or all a copy of the of the coordinates. worksheet, one The merely worksheet writes then above each term what the Christo↵el symbol⌘ evaluates to for whatever the functions for A, B, C, D actually simply lists all of the Christo↵el symbols in terms of the metric components and their derivatives (the latter are in a given particular circumstance. 0 written in a shorthandIn the Schwarzschild form where case,A0 we identify@A/@xx0and= t, so x1 on.= r, On x2 a= copy✓,x3 of= the, and worksheet,A =1 one2GM/r, merely B = writes 1 2 ⌘ 2 2 above each(1 term2GM/r what) the,C Christo= r , and↵elD symbol= r sin evaluates✓. First note to for that whatever no metric the component functions depends for A, on B,t or C, D, andactually are in a givenonly particularD depends on circumstance.✓, so a lot of the derivative terms that might appear in general will be zero. So we look 0 1 2 0 3t 1 In theon Schwarzschild the Diagonal Metric case, Worksheet, we identify andx see,= fort, examplex = r, that x =00 ✓=,xtt == 2A,A and0 = 0,A because=1 A2doesGM/r, not B = depend1 on x0 =2 t. The next listing2 on2 the worksheet tells us that (1 2GM/r) ,C = r , and D = r sin ✓. First note that no metric component depends on t or , and only D depends on0 ✓, so0 a lot1 of the derivativet t terms that1 might@ appear2GM in generalGM will be zero. So we look 01 = 10 = A1 tr = rt = 01 t =1 (2.34) on the Diagonal Metric Worksheet,2A ) and see, for2(1 example2GM/r that) @r = r = rA2(10 =2 0,GM/r because) A does not ✓00 tt ◆ 2A depend on x0 = t. The next listing on the worksheet tells us that and the remaining entries on that line are zero because A does not depend on ✓ or . One can go through the worksheet in a similar way to show that the complete1 set of nonzero@ Christo2GM↵el symbols forGM the Schwarzschild 0 =coordinate 0 = 1 systemA are (in additiont = tot = the pair evaluated above):1 = (2.34) 01 10 2A 1 ) tr rt 2(1 2GM/r) @r r r2(1 2GM/r) GM 2GM ✓ GM◆ r = r = 1 , r = , and the remaining entries on thattt linett are zeror2 because r A does notrr dependr2(1 on2GM/r✓ or ). One can go through the ✓ ◆ worksheet in a similar way to showr that the complete set of nonzeror Christo↵el symbols2 for the Schwarzschild ✓✓ = (r 2GM), = (r 2GM)sin ✓ (2.35) coordinate system are (in addition to the pair evaluated above): ✓ ✓ 1 ✓ r✓ = ✓r = , = cos ✓ sin ✓ (2.36) r r GM 2GM r r GM tt = tt = 2 1 1 , rr = 2 , r = r = , = cotr (1✓ 2GM/r) (2.37) ✓ r r ◆r ✓ ✓ r = (r 2GM), r = (r 2GM)sin2 ✓ (2.35) As an application✓✓ of the geodesic equation, first consider a particle at least momentarily rest at some radial coordinate r. The spatial components of such a particle’s four-velocity u are zero by definition (dr/d⌧ = d✓/d⌧ = d/d⌧ = 0),✓ but in✓ an arbitrary1 ✓ coordinate system we cannot conclude that the four- r✓ = ✓r = , = cos ✓ sin ✓ (2.36) velocity’s time component ut = dt/d⌧, as we couldr in special relativity. Rather, we must go back to the basic 1 = = , = cot ✓ (2.37) r r r ✓9 ✓ As an application of the geodesic equation, first consider a particle at least momentarily rest at some radial coordinate r. The spatial components of such a particle’s four-velocity u are zero by definition (dr/d⌧ = d✓/d⌧ = d/d⌧ = 0), but in an arbitrary coordinate system we cannot conclude that the four- velocity’s time component ut = dt/d⌧, as we could in special relativity. Rather, we must go back to the basic

9 Example: four-velocity of a particle at rest result for the magnitude of the four-velocity, which (by definition of the di↵erential proper time) still says: result for the magnitudeRemember of the the four-velocity, magnitude which of (by a four-vector definition of theis –1: di↵erential proper time) still says: µ ⌫ µ ⌫ 2 2 µ dx⌫ dx µ ⌫g dx2 dx 2ds d⌧ result for the magnitude of the four-velocity, which (by definition ofµ thedx di⌫↵erentialdx propergµ⌫ dx dx time)µ⌫ ds still says:d⌧ u ⇧ u uu⇧µgu u⌫u=gµ⌫ ug = =gµ⌫ = = = = = =1 = 1 (2.38) (2.38) ⌘ µ⌫⌘ d⌧ µ⌫ d⌧d⌧ dd⌧⌧ 2 dd⌧⌧22 d⌧ 2 d⌧ 2 d⌧ 2 dxµ dx⌫ g dxµdx⌫ ds2 d⌧ 2 u ⇧ u uµg uFor⌫ = aFor particle ag particle at rest,= at thisµ rest,⌫ equation this equation= reduces= to reduces = to 1 (2.38) ⌘ µ⌫ d⌧ µ⌫ d⌧ This allowsd⌧ 2 us tod find⌧ 2 utd for⌧ 2 a particle at rest: dt 1 1 For a particle at rest, this equation reduces to 1=uµg u⌫µ= g (⌫ut)2 +0+0+0t 2 ut = = t dt= 1 1 (2.39) 1=µ⌫ u gµ⌫ttu = gtt(u ) +0+0+0) d⌧ u =g = = (2.39) ) r ttd⌧ 1 2gGM/rtt 1 2GM/r r dt 1 1 µ ⌫ Thist 2 equation says that a clockt attached to a particle at rest registers a properp time between events along 1=u gµ⌫ u = gtt(u ) +0+0+0This equation saysu that= a clock= attached= to a particle at rest(2.39) registers a properp time between events along ) d⌧ gtt 1 2GM/r its worldline that is somewhat smallerr than the Schwarzschild coordinate time t between those events. This one consequenceits worldline of that the curvature is somewhat of spacetime smaller in than the the Schwarzschild Schwarzschild geometry. coordinate We see time that tt,between though it those is a events. This This equation says that a clock attached“timeone coordinate,” to consequence a particle is not at of restthe the time registers curvature that just a of proper any spacetime clockp time at any between in the location Schwarzschild events would along read, geometry. but rather Wea kind see of that globalt, though it is a its worldline that is somewhat smallerbookkeeper’s than“time the coordinate,” time Schwarzschild stamp. is It not only coordinate the coincides time that time with justt thebetween any time clock that those a at clock events. any would location This directly would read read, if the but clock rather is at a kind of global one consequence of the curvature ofrest spacetime atbookkeeper’s infinity. in the time Schwarzschild stamp. It geometry. only coincides We see with that thet time, though that it a is clock a would directly read if the clock is at “time coordinate,” is not the time thatThe justrestr any-componentat infinity. clock at of any the location geodesic would equation read, in thisbut rather situation a kind says ofthat global the particle’s radial coordinate bookkeeper’s time stamp. It only coincidesproper accelerationThe withr-component the is time given that by of a the clock geodesic would equationdirectly read in this if the situation clock is says at that the particle’s radial coordinate rest at infinity. proper2 accelerationµ is⌫ given by d r r dx dx r t t GM 2GM 1 2GM The r-component of the geodesic equation= inµ this⌫ situation= saysttu u that+ zeros the = particle’s1 radial coordinate = (2.40) d⌧ 2 d2r d⌧ d⌧dxµ dx⌫ r2 GMr 1 22GMGM/r 1r2 2GM proper acceleration is given by = r = r utut + zeros✓ = ◆1 = (2.40) d⌧ 2 µ⌫ d⌧ d⌧ tt r2 r 1 2GM/r r2 This seems say us that the radial acceleration of a particle initially at✓ rest (according◆ to the particle’s own d2r dxµ dx⌫ clock time) is exactlyGM what we would2GM expect from1 Newtonian2GM mechanics. However, this is a bit misleading, = r = r utut + zeros = 1 = (2.40) d⌧ 2 µ⌫ d⌧ d⌧ ttbecauseThis radial seems distance sayr us2 a that particle the movesr radial1 between acceleration2GM/r two events of ar as particle2 it falls initially radially is at not restquite (accordingthe same to as the the particle’s own di↵erenceclockdr time)in the is radial exactly✓ coordinate what we◆r. would In fact, expect according from to Newtonian the metric, mechanics. the spatial distance However, between this is two a bit misleading, This seems say us that the radial accelerationeventsbecause that occur of radial a at particle the distance same initially time a particle but at restat moves points (according with between a purely to two the radial events particle’s coordinate as it own falls separation radially ofisdr notisquite the same as the clock time) is exactly what we would expectdi↵erence fromdr Newtonianin the radial mechanics. coordinate However,r. In fact, this is according a bit misleading, to the metric, the spatial distance between two 2 µ ⌫ 2 dr because radial distance a particle movesevents between that two occur eventsds at= p the asds sameit= fallsg timeµ radially⌫ dx butdx at is= notpointsgrrquitedr with+the zeros a same purely = as radial the coordinate separation(2.41) of dr is 1 2GM/r di↵erence dr in the radial coordinate r. In fact, according to the metric,p the spatialp distance between two 2 µ ⌫ 2 dr events that occur at the same timeHowever, but at points as long with as r> a purely2GMds, radial this= p distinctionds coordinate= gµ is⌫ dx not separationdx important.= ofgrr Fordrdris example,p+ zeros at = the surface of the earth, (2.41) where r = 6380 km and 2GM 0.9 cm, we’d have to keep track better than 8 decimal1 2GM/r places in our ⇡ p p 2 measurementsµ ⌫ to notice that2 the Newtonian resultdr for falling objects is not correct. ds = pds = gµ⌫ dx dx = grrdr + zeros = (2.41) p Still,However, there is as an long important as r> lesson2GM, here this1 that distinction2GM/r I am going is not to repeat important. over and For over example, in the daysat the to surface come. of the earth, where r = 6380 km and 2GM 0.9 cm, we’d have to keep track better than 8 decimal places in our Coordinatesp by themselvesp have no physical⇡ meaning. Calling coordinates t, r, ✓, may cause you to However, as long as r>2GM, thisthink distinctionmeasurementsthat you is know not what to important. notice they mean, that For the but example,p Newtonian we could at have the result called surface for these falling of coordinates the objects earth, anything. is not correct.The only thing where r = 6380 km and 2GM that0.9 cm,givesStill, we’d coordinates there have is to an physical keep important track meaning better lesson is than here the 8that metric decimal I am, which placesgoing is what to in repeat our anchors over human and coordinates over in the to days to come. measurements to notice that the⇡ NewtonianphysicalCoordinates reality. result One for by must falling themselves always objects resort is have not to the correct. no metric physical to learn meaning. what yourCalling coordinates coordinates actually mean.t, r, ✓, may cause you to Still, there is an important lessonThis herethink is that athat tricky I you am idea, know going because what to repeat we they are mean, overso used and but to overworkingwe could in with the have days coordinates called to come. these in flat coordinates spacetime anything. (for example,The only thing Coordinates by themselves havespherical nothat physical coordinates gives meaning. coordinates with anCalling orthonormal physical coordinates basis) meaning thatt, r, have is✓, themay a reasonably metric cause, you which direct to intuitiveis what anchors meaning. human But in coordinates to general relativity, we have to give this up and use the metric exclusively to learn about what coordinates think that you know what they mean, butphysical we could reality. have One called must these always coordinates resort to anything. the metricThe to only learn thing what your coordinates actually mean. mean. But if you find this hard to grasp or remember, be comforted. Even Einstein struggled with this in This is a tricky idea, because we are so used to working with coordinates in flat spacetime (for example, that gives coordinates physicalthe meaning months leading is the up metric to his, final which November is what 1915 anchors paper human describing coordinates the Einstein to equation. 2 physical reality. One must always resortNowspherical to let’s the metric consider coordinates to a learn particle with what in an orbit your orthonormal around coordinates an basis) object actually that whose have mean. exterior a reasonably geometry direct is described intuitive by the meaning. But in This is a tricky idea, because weSchwarzschild aregeneral so used relativity, metric. to working Since we with the have star coordinates to and give its this surrounding in up flat and spacetime use spacetime the (for metric are example, sphericallyexclusively symmetric,to learn we about can take what coordinates spherical coordinates with an orthonormalany planemean. basis) through But that if the you have star’s find a center this reasonably hard to be to the direct grasp equatorial or intuitive remember, (✓ = ⇡ meaning./2) plane, be comforted. But and if in the Even particle’s Einstein initial struggled velocity with this in general relativity, we have to givelies this inthe up that monthsand plane use then leading the it metric must up stay toexclusively his on final that plane Novemberto learn by symmetry. about 1915 paper what So without coordinates describing loss of the generality, Einstein we equation. can consider2 mean. But if you find this hard toonly grasp orbits orNow remember, in the let’s equatorial consider be comforted. plane, a particle where Even insin orbit2 Einstein✓ = 1 around and struggledd✓/d an⌧ = object 0. with Now, whose this note in exterior under these geometry circumstances is described by the the months leading up to his final NovemberthatSchwarzschild the component 1915 paper metric. of describing the geodesic Since the equation Einsteinstar and implies equation. its surrounding that 2 spacetime are spherically symmetric, we can take any plane through the star’s center to be the equatorial (✓ = ⇡/2) plane, and if the particle’s initial velocity Now let’s consider a particle in orbit around an objectd whose2 exteriordxµ dx⌫ geometryd2 is describeddr d by thed✓ d lies in that plane then0= it must+ stay on that= plane+2 by symmetry.+2 So without loss of generality,(2.42) we can consider Schwarzschild metric. Since the star and its surrounding spacetimed⌧ 2 areµ⌫ d spherically⌧ d⌧ d⌧ symmetric,2 r d⌧ d we⌧ can✓ taked⌧ d⌧ any plane through the star’s center to beonly the orbits equatorial in the (✓ equatorial= ⇡/2) plane, plane, and where if thesin particle’s2✓ = 1 and initiald✓/d velocity⌧ = 0. Now, note under these circumstances lies in that plane then it must stay onsince thatthat only plane the these bycomponent Christo symmetry.↵el symbols of So the without geodesic with lossin equation the of generality, upper implies position we that can are consider nonzero. But the last term is zero because d✓/d2⌧ = 0, and substituting in the value of the remaining Christo↵el symbol yields only orbits in the equatorial plane, where sin ✓ = 1 and d✓/d⌧ = 0.2 Now, note underµ ⌫ these2 circumstances d dx dx d dr d d✓ d that the component of the geodesic equation implies thatd2 0=2 dr d +1 d d = +2d +2 (2.42) 0= + d⌧ 2= µ⌫ dr⌧2 d⌧ d⌧ 2 r2 =r constantd⌧ d⌧ ` ✓ d⌧ d⌧ (2.43) d⌧ 2 r d⌧ d⌧ r2 d⌧ d⌧ ) d⌧ ⌘ d2 dxµ dx⌫ d2 dr d d✓ ✓d ◆ since only these Christo ↵el symbols with in the upper position are nonzero. But the last term is zero 0= 2 + µ⌫ = 2 +2r +2✓ (2.42) d⌧ becaused⌧ d⌧ d✓/dd⌧⌧ = 0, andd substituting⌧ d⌧ ind⌧ thed⌧ value of the remaining Christo↵el symbol yields since only these Christo↵el symbols with in the upper position are nonzero. But10 the last term is zero d2 2 dr d 1 d d d because d✓/d⌧ = 0, and substituting in the value of the remaining0= Christo+ ↵el symbol= yieldsr2 r2 = constant ` (2.43) d⌧ 2 r d⌧ d⌧ r2 d⌧ d⌧ ) d⌧ ⌘ ✓ ◆ d2 2 dr d 1 d d d 0= + = r2 r2 = constant ` (2.43) d⌧ 2 r d⌧ d⌧ r2 d⌧ d⌧ ) d⌧ ⌘ ✓ ◆ 10

10 2.6 Schwarzschild Geodesics. 2.6 Schwarzschild Geodesics. As we will see in a subsequent lecture, one can solve the Einstein equation in the vacuum around a spherical starAs to we get will the seeSchwarzschild in a subsequent metric lecture,: one can solve the Einstein equation in the vacuum around a spherical star to get the Schwarzschild metric: 2GM dr2 ds2 = 1 dt2 + + r2d✓2 + r2 sin2 ✓ d2 (2.36) r2GM 1 2GM/rdr2 ds2 =✓ 1 ◆ dt2 + + r2d✓2 + r2 sin2 ✓ d2 (2.36) r 1 2GM/r 1 2 2 2 (meaning that gtt = (1 2GM/r✓),grr =(1◆ 2GM/r) ,g✓✓ = r and g = r sin ✓ and the o↵-diagonal 1 2 2 2 metric(meaning components that gtt are= zero).(1 2 ThisGM/r is) by,grr no=(1 means2 theGM/r only) solution,g✓✓ = r evenand forg the= vacuumr sin ✓ spacetimeand the o↵ around-diagonal a sphericalmetric components star, since we are can zero). redefine This is coordinates by no means in completely the only solution arbitrary even ways for the and vacuum so generate spacetime an infinite around numbera spherical coordinate star, sincesystems we that can redefine describe coordinates the same physical in completely reality. arbitrary But this ways coordinate and so system generate does an have infinite somenumber advantages: coordinate it is independentsystems that of describe the time the coordinate same physicalt (in a reality. diagonal But metric, this coordinate the time coordinate system does is the have coordinatesome advantages: whose metric it is component independent is of negative) the time and coordinate this metrict (in reduces a diagonal to the metric, metric the for time flat coordinate spacetime (in is the a sphericalcoordinate coordinate-basis) whose metric component as r . is I negative) should note and that this though metric reduces I am using to the units metric where for time flat spacetime and space (in !1 area both spherical measured coordinate-basis) in meters (so as thatr c = 1),. I I should am not noteusing that units though where I the am universal using units gravitational where time constant and space G =are 1 both as well, measured as is often in meters done. I(so am that!1 doingc = this 1), Iso am thatnot connectionsusing units to where Newtonian the universal mechanics gravitational will be clearer. constant ButG it= is 1 good as well, to keep as is oftenin mind done. that I in am this doing unit this system so thatGM connectionshas units of to meters, Newtonian and mechanics in fact GM will= be 1477 clearer. m forBut a star it is with good the to mass keep of in the mind sun that (and in is this 4.45 unit mm system for theGM Earth).has units of meters, and in fact GM = 1477 m forIt is a star not mywith purpose the mass here of the to discuss sun (and the is many 4.45 mm fascinating for the Earth). features of this metric: it looks like others may talkIt is about not my that, purpose and the here topic to is discuss somewhat the many tangential fascinating to the main features line of of this my argument. metric: it But looks I wantlike others to introducemay talk this about metric that, so that and the we can topic explore is somewhat the implications tangential of to the the geodesic main line equation of my argument. in a realistic But situation I want to (oneintroduce that even this applies metric to so our that daily we lives!).can explore the implications of the geodesic equation in a realistic situation (oneConsider that even a particle applies at to least our momentarily daily lives!). rest at some radial coordinate r. The spatial components of such aConsider particle’s a four-velocity particle at leastu are momentarily zero by definition rest at ( somedr/d⌧ radial= d✓/d coordinate⌧ = d/d⌧r.= The 0), but spatial in an components arbitrary of coordinatesuch a particle’s system we four-velocity cannot concludeu are that zero theby definition four-velocity’s (dr/d time⌧ = d component✓/d⌧ = d/dut ⌧==dt/d 0),⌧ but, as in we an could arbitrary in t specialcoordinate relativity. systemresult Rather, we for cannot we the must magnitude conclude go back that to of the the the basic four-velocity, four-velocity’s result for which the time magnitude component (by definition ofu the= of four-velocity,dt/d the⌧, di as↵erential we which could proper in time) still says: (byspecial definition relativity. of the di Rather,↵erential we proper must go time) back still to the says: basic result for the magnitude of the four-velocity, which (by definition of the di↵erential proper time) still says: µ ⌫ µ ⌫ 2 2 µ µ⌫ ⌫ dxµ ⌫ dx 2 gµ⌫ dx2 dx ds d⌧ µ ⌫ udx u dx gµ⌫ dx dx ds d⌧ u u ⇧ u gµ⌫ u = gµ⌫ = 2 = 2 = 2 = 1 (2.38) ⇧ u gµ⌫ u = gµ⌘µ⌫ =⌫ d2 ⌧ µ =⌫ d⌧ 2 =2 2d⌧=2 1d⌧ d⌧ (2.37) ⌘ µ ⌫ d⌧dx d⌧dx gdµ⌧⌫ dx dx d⌧ ds d⌧ d⌧ u ⇧ u u gµ⌫ u = gµ⌫ = 2 = 2 = 2 = 1 (2.37) For a particle at rest,For athis particle equation⌘ at rest, reduces thisd to⌧ equationd⌧ reducesd⌧ to d⌧ d⌧ For a particle at rest, this equation reduces to dt 1 dt1 1 1 1=uµg u⌫ = g (ut)2 +0+0+0µ ⌫ t 2ut = = = t (2.38) µ⌫ tt 1=u gµ⌫ u = gtt(u ) +0+0+0dt 1 u = =1 = (2.39) µ ⌫ t 2 ) t d⌧ g)tt 1 d2⌧GM/r gtt 1 2GM/r 1=u gµ⌫ u = gtt(u ) +0+0+0 u = r= = r (2.38) ) d⌧ gtt 1 2GM/r r p This equation saysThis that equation a clock attached says that to a a particle clock attached at rest registers to a particle a proper at time rest between registers events a proper alongp time its between events along p worldlineThis equation that is smaller saysits worldline that than a clock the that Schwarzschild attached is somewhat to a coordinate particle smaller at time thanrestt registersbetween the Schwarzschild a those proper events. time coordinate betweenThis one events consequence time t alongbetween its those events. This of the curvature of spacetime in the Schwarzschild geometry. We see that t, though it is a “time coordinate,” worldline that isone smaller consequence than the Schwarzschild of the curvature coordinate of spacetime time t between in the Schwarzschild those events. This geometry. one consequence We see that t, though it is a is notof the the curvature time that of just spacetime any clock in the at any Schwarzschild finite r would geometry. read, but We rather see that a kindt, though of global it is bookkeeper’s a “time coordinate,” time “time coordinate,” is not the time that just any clock at any location would read, but rather a kind of global stamp.is not It the only time coincides thatExample: just with any the clock time at that any a finite clockr wouldwould directly read, but read rather if the a kindclock of is global at rest bookkeeper’s at infinity. time bookkeeper’s time stamp. It only coincides with the time that a clock would directly read if the clock is at stamp.The r-component It only coincides of the with geodesic the time equation that a in clock this would situation directly says read that if the the particle’s clock is at radial rest at coordinate infinity. properThe accelerationr-componentrest isgeodesic given at of infinity. the by geodesic for equation a particle in this situation initially says that the particle’sat rest radial coordinate The r-component of the geodesic equation in this situation says that the particle’s radial coordinate proper acceleration is given by 2 µ ⌫ properSo the acceleration r-componentd r is givenr ofdx bythedx geodesicr equationt t is: 2 2= µ⌫ µ =⌫ ttu u + zeros (2.39) d⌧ d r dr ⌧dxd⌧dx r t t 2 =µ ⌫ = u u + zeros (2.39) d r dx2 dx µ⌫ tt GM 2GM 1 2GM So to evaluate this radial acceleration,= dr⌧ we only need=d⌧ todr⌧ evaluateutut+ zeros = 1 = (2.40) d⌧ 2 µ⌫ d⌧ d⌧ tt r2 r 1 2GM/r r2 So to evaluate this radialr 1 acceleration,r↵ we only need to1 evaluaterr ✓ ◆ tt = 2 g (@tgt↵ + @tg↵t @↵gtt)= 2 g (0 + 0 @rgtt) This seemsr say1 r↵ us that the radial acceleration1 rr of a particle initially at rest (according to the particle’s own tt =12 g (@2tgGMt↵ + @tdg↵t @↵gtt2GM)= 2 g (0 + 0 @2rGMgtt) GM clock time)= is1 exactly what we would1+ expect=+ from1 Newtonian mechanics. However,(2.40) this is a bit misleading, 2 1 r2GMdr d r2GM r2GM r2GM because radial= ✓ distance1 ◆ a particle✓ moves1+ between◆ =+✓ two1 events◆ as it falls radially is(2.40) not quite the same as the 2 r dr r r r2 Substituting this intodi↵erence the geodesicdr in✓ the equation radial yields◆ coordinate✓ r. In◆ fact, according✓ to◆ the metric, the spatial distance between two Substituting thisevents into thethat geodesic occur at equation the same yields time but at points with a purely radial coordinate separation of dr is d2r GM 2GM 1 GM = r utut = 1 = (2.41) d⌧ 2d2r tt r2GM r2GM1 2GM/r1 r2GM dr = r ututds== pds✓ 2 =1 g ◆dxµdx⌫ = g dr= 2 + zeros = (2.41) (2.41) d⌧ 2 tt r2 µr⌫ 1 2GM/rrr r2 This seems say us that the radial acceleration of a particle✓ initially◆ at rest (according to the particle’s1 2GM/r own p p clockThis time) seems is exactly say us that what the we radial would acceleration expect from of Newtonian a particle mechanics. initially at rest However, (according this is to a thebit misleading, particle’s own However, as long as r>2GM, this distinction is not important. For example,p at the surface of the earth, clock time) is exactly what we would expect from Newtonian mechanics. However, this is a bit misleading, where r = 6380 km and 2GM 0.9 cm, we’d have to keep track better than 8 decimal places in our measurements to notice that the⇡ Newtonian9 result for falling objects is not correct. Still, there is an important lesson9 here that I am going to repeat over and over in the days to come. Coordinates by themselves have no physical meaning. Calling coordinates t, r, ✓, may cause you to think that you know what they mean, but we could have called these coordinates anything. The only thing that gives coordinates physical meaning is the metric, which is what anchors human coordinates to physical reality. One must always resort to the metric to learn what your coordinates actually mean. This is a tricky idea, because we are so used to working with coordinates in flat spacetime (for example, spherical coordinates with an orthonormal basis) that have a reasonably direct intuitive meaning. But in general relativity, we have to give this up and use the metric exclusively to learn about what coordinates mean. But if you find this hard to grasp or remember, be comforted. Even Einstein struggled with this in the months leading up to his final November 1915 paper describing the Einstein equation. 2 Now let’s consider a particle in orbit around an object whose exterior geometry is described by the Schwarzschild metric. Since the star and its surrounding spacetime are spherically symmetric, we can take any plane through the star’s center to be the equatorial (✓ = ⇡/2) plane, and if the particle’s initial velocity lies in that plane then it must stay on that plane by symmetry. So without loss of generality, we can consider only orbits in the equatorial plane, where sin2✓ = 1 and d✓/d⌧ = 0. Now, note under these circumstances that the component of the geodesic equation implies that

d2 dxµ dx⌫ d2 dr d d✓ d 0= + = +2 +2 (2.42) d⌧ 2 µ⌫ d⌧ d⌧ d⌧ 2 r d⌧ d⌧ ✓ d⌧ d⌧ since only these Christo↵el symbols with in the upper position are nonzero. But the last term is zero because d✓/d⌧ = 0, and substituting in the value of the remaining Christo↵el symbol yields

d2 2 dr d 1 d d d 0= + = r2 r2 = constant ` (2.43) d⌧ 2 r d⌧ d⌧ r2 d⌧ d⌧ ) d⌧ ⌘ ✓ ◆

10 Exercise

It turns out one solution to the full geodesic equation for a radially falling particle is dr/dτ = − 2 G M / r . At what value of r is the particle at rest? At what value of r is dr/dτ = 1? Does that necessarily mean that it is traveling at the speed of light? 2.7 Locally Orthonormal and Locally Inertial Frames. The problem with restricting ourselves to coordinate bases is that the kinds of coordinate systems that we use every day in the laboratory employ orthonormal basis vectors. If we want to know what a particle’s energy or velocity or some other physical quantity would be “according to (some specified) observer,” we are implicitly assuming that the observer is doing his or her measurements or calculations using a frame having a standard orthonormal basis. The purpose of this section is to explore a method for transforming physical quantities from their values determined in a large-scale coordinate-basis coordinate system like Schwarzschild coordinatesLocal and a local reference Flatness frame that uses an orthonormalTheorem basis. The local flatness theorem is important in this context (and will be important to us later as well). A statement of the theorem follows. At any given event in spacetime, our freedom to choose coordinates allows us to construct a coordinate system whereP (1) g ( )=⌘ (with six degrees of freedom left over that correspond µ⌫ P µ⌫ to arbitrary rotations and Lorentz boosts), (2) all 40 of the independent values of @↵gµ⌫ = 0, |P and (3) all but 20 of the 100 independent values of @↵@gµ⌫ equal to zero. |P The proof of this theorem is too involved to go into here: if you are interested, look at pp. 207-209 in my textbook.3 But the result is very important. It tells us that at every event in spacetime, we can (in principle) construct a reference frame that at least locally (if we don’t get too far from event )behaveslike laboratories we are used to. P We call a reference frame where g ( )=⌘ a locally orthonormal frame (LOF), because ⌘ = µ⌫ P µ⌫ µ⌫ gµ⌫ eµ ⇧ e⌫ means that the basis vectors for this coordinate system are orthogonal and have unit length, at least⌘ at event . If our coordinate system only satisfies this first criterion but not the second (meaning that we don’t botherP to set the first derivatives of the metric equal to zero), then at least some Christo↵el symbols will not be zero, and geodesics will not be straight lines in our coordinate system. This would be analogous to a laboratory coordinate system at rest on the surface of the Earth. We can definitely set up such a coordinate system (we do it all the time), but geodesics (the worldlines of freely falling particles) are not straight lines in such a coordinate system. We call a LOF that also satisfies the second criterion (all the first derivatives of the metric are zero) a locally inertial frame (LIF). In this frame, the Christo↵el symbols are all zero, and the worldlines of geodesics are (locally) straight lines. Near the earth’s surface, such a frame would be a freely-falling frame, such as the reference frame in a falling elevator or the International Space Station. In such a frame, we would need to look at geodesics over fairly large distances (or for fairly long times) to see the tidal e↵ects that distinguish a freely-falling frame from a globally inertial frame in deep space. But no matter how good our coordinate system is, we will eventually see such e↵ects, because we cannot set all of the second metric derivatives to zero (and thus first derivatives of the Christo↵el symbols) to zero. This means that as we move far enough away from in space and/or time, we will eventually see the geodesics begin to converge or diverge as the Christo↵el symbolsP begin to become appreciably nonzero. But how can we transform quantities expressed in Schwarzschild coordinates into values we would measure in either a LOF or LIF? In principle, if we knew the coordinate transformation equations from Schwarzschild coordinates to the new coordinates, then we could use the normal tensor transformation rule. But it is usually very dicult to derive those transformation equations. Fortunately, there is another way. Consider for the sake of argument that we want to know the components of a four-vector A in a certain LOF, and suppose we know the orthonormal basis vectors eµ of the LOF, ⌫ which we will take to be the primed frame. Since in any coordinate basis A = A e⌫0 and gµ0 ⌫ eµ0 ⇧ e⌫0 ,we have ⌘ ⌫ ⌫ eµ0 ⇧ A = eµ0 ⇧ e⌫0 A0 = gµ0 ⌫ A0 = Aµ0 (2.47)

In our particular LOF, gµ0 ⌫ = ⌘µ⌫ by definition, so we can say that

↵µ ↵µ ↵ ⌘ (e ⇧ A)=⌘ A0 = A0 (2.48) 0µ µ So, we can evaluate the components of the vector A in any LOF if we can evaluate the dot product of A with the LOF’s basis vectors eµ0 . The clever trick is that since the dot product must have a coordinate-independent value, we can evaluate the dot product in whatever global coordinate system (such as the Schwarzschild coordinate system) that describes our spacetime. All that we need to is to evaluate the Schwarzschild components of both the LOF’s basis vectors eµ0 and the four-vector A of interest.

12 2.7 Locally Orthonormal and Locally Inertial Frames. 2.7The Locally problem with Orthonormal restricting ourselves and Locally to coordinate Inertial bases Frames. is that the kinds of coordinate systems that we 2.7 Locally OrthonormalThe anduse problem every Locally with day in restricting Inertial the laboratory Frames. ourselves employ to coordinate orthonormal bases basis is that vectors. the kinds If we of coordinate want to know systems what that a particle’s we energy or velocity or some other physical quantity would be “according to (some specified) observer,” we are The problem with restricting ourselvesuse every to coordinate day in the bases laboratory is that the employ kinds orthonormal of coordinate basis systems vectors. that we If we want to know what a particle’s implicitly assuming that the observer is doing his or her measurements or calculations using a frame having use every day in the laboratoryenergy employ or orthonormal velocity or basis some vectors. other physical If we want quantity to know would what be “according a particle’s to (some specified) observer,” we are energy or velocity or some other physicalimplicitlya standard quantity assuming orthonormal would that be the “according basis. observer The to is (some purpose doing specified) his of or this her observer,” section measurements is we to are explore or calculations a method using for transforming a frame having physical implicitly assuming that the observera standardquantities is doing orthonormal his from or her their measurements basis. values The determined purpose or calculations in of a this large-scale usingsection a coordinate-basisframe is to explore having a method coordinate for transforming system like Schwarzschild physical a standard orthonormal basis. Thequantities purposecoordinates from of this their and section a values local is to determined reference explore framea inmethod a large-scale that for uses transforming an coordinate-basis orthonormal physical basis. coordinate system like Schwarzschild quantities from their values determinedcoordinates inThe a large-scalelocal and a flatness local coordinate-basis reference theorem frame coordinateis that important uses system an in orthonormal this like Schwarzschild context basis. (and will be important to us later as well). A coordinates and a local reference frameThestatement thatlocal uses offlatness an the orthonormal theorem theorem follows. basis.is important in this context (and will be important to us later as well). A The local flatness theoremstatementis important of the in this theorem context follows. (and will be important to us later as well). A At any given event in spacetime, our freedom to choose coordinates allows us to construct a statement of the theorem follows. P At anycoordinate given event system wherein spacetime, (1) gµ⌫ ( our)= freedom⌘µ⌫ (with to choose six degrees coordinates of freedom allows left us over to construct that correspond a P P At any given event in spacetime,coordinate ourto arbitrary freedom system to rotations choose where (1) coordinates andg Lorentz( )= allows⌘ boosts),(with us to (2) six construct all degrees 40 of a of the freedom independent left over values that correspond of @↵gµ⌫ = 0, P µ⌫ P µ⌫ |P coordinate system where (1) gµ⌫ (to)= arbitraryand⌘µ⌫ (3)(with all rotations butsix degrees 20 andof the of Lorentz freedom 100 independent boosts), left over (2) that values all correspond 40 of of@ the↵@ independentgµ⌫ equal valuesto zero. of @ g = 0, P P ↵ µ⌫ to arbitrary rotations and Lorentz boosts), (2) all 40 of the independent values of @↵gµ⌫ = 0, | |P and (3) all but 20 of the 100 independent values of @↵|P@gµ⌫ equal to zero. and (3) all but 20 of the 100 independentThe proof values of this of theorem@↵@gµ⌫ isequal too involved to zero. to go into here:| ifP you are interested, look at pp. 207-209 in |P Themy proof textbook. of this3 theoremBut the is result too involved is very important. to go into here: It tells if us you that are at interested, every event look in at spacetime, pp. 207-209 we can in (in The proof of this theorem is toomy involvedprinciple) textbook. to go3 constructBut into the here: a result reference if you is arevery frame interested, important. that at look least It at tells locally pp. us 207-209 that (if we at don’t in every get event too far in spacetime, from event we)behaveslike can (in my textbook.3 But the result is very important. It tells us that at every event in spacetime, we can (in P principle)laboratories construct we are a reference used to. frame that at least locally (if we don’t get too far from event )behaveslike principle) construct a reference frame that at least locally (if we don’t get too far from event )behaveslike P laboratoriesWe call we a are reference used to. frame where gµ⌫ ( )=⌘µ⌫ aPlocally orthonormal frame (LOF), because ⌘µ⌫ = laboratories we are used to. P Wegµ⌫ calleµ a⇧ referencee⌫ means frame that wherethe basisgµ⌫ vectors( )=⌘ forµ⌫ a thislocally coordinate orthonormal system are frame orthogonal (LOF) and, because have unit⌘µ⌫ = length, We call a reference frame where gµ⌫ (⌘)=⌘µ⌫ a locally orthonormalP frame (LOF), because ⌘µ⌫ = gµ⌫ at leastePµ ⇧ e at⌫ means event that. If the our basis coordinate vectors system for this only coordinate satisfies system this first are criterion orthogonal but and not have the unitsecond length, (meaning gµ⌫ eµ ⇧ e⌫ means that the basis vectors⌘ for this coordinate system are orthogonal and have unit length, ⌘ at leastthat at we event don’t bother. IfP our to coordinate set the first system derivatives only satisfies of the metric this first equal criterion to zero), but then not the at secondleast some (meaning Christo↵el at least at event . If our coordinate system onlyP satisfies this first criterion but not the second (meaning that we don’t botherP to set the firstthatsymbols derivatives we don’t will bother of not the be metric to zero, set equal the and first to geodesics zero), derivatives then will at of not least the be metric some straight Christo equal lines to↵el in zero), our coordinatethen at least system. some Christo This would↵el be symbols will not be zero, and geodesicssymbolsanalogous will will not not to be a be straight laboratory zero, and lines geodesics coordinate in our coordinate will system not be system. at straight rest This on lines the would surface in our be ofcoordinate the Earth. system. We can This definitely would be set up analogous to a laboratory coordinateanalogoussuch system a coordinate to at a rest laboratory on system the surface coordinate (we doof the it system all Earth. the at time), We rest can on but definitely the geodesics surface set (the of up the worldlines Earth. We of freely can definitely falling particles) set up are such a coordinate system (we dosuch it allnot a the coordinate straight time), but lines system geodesics in such (we (the a do coordinate it worldlines all the time), system. of freely but falling geodesics particles) (the worldlines are of freely falling particles) are not straight lines in such a coordinatenot straight system.We call lines a LOFin such that a coordinate also satisfies system. the second criterion (all the first derivatives of the metric are zero) We call a LOF that also satisfiesWea thelocally call second a inertial LOF criterion that frame also (all the satisfies (LIF) first. derivatives the In this second frame, of criterion the the metric Christo (all are the↵ zero)el first symbols derivatives are all of zero, the andmetric the are worldlines zero) of a locally inertial frame (LIF)a. Inlocallygeodesics this frame, inertial are the (locally) frame Christo straight↵ (LIF)el symbols. In lines. this are Near frame, all zero, the the andearth’s Christo the surface, worldlines↵el symbols such of a are frame all zero, would and be the a freely-falling worldlines of frame, geodesics are (locally) straight lines.geodesicssuch Near as the are the earth’s (locally) reference surface, straight frame such lines. in a frame a Near falling would the elevator earth’s be a freely-falling or surface, the International such frame, a frame Space would Station. be a freely-falling In such a frame, frame, we such as the reference frame in asuch fallingwould as elevator the need reference to or look the frame International at geodesics in a falling Space over elevator fairly Station. large or In the distances such International a frame, (or for we Space fairly Station. long times) In tosuch see a the frame, tidal we e↵ects would need to look at geodesics overwouldthat fairly need distinguish large to look distances at a freely-falling geodesics (or for fairly over frame fairly long from times) large a globally distances to see the inertial (or tidal for e frame↵ fairlyects in long deep times) space. to But see the no matter tidal e↵ howects good that distinguish a freely-falling framethatour from distinguish coordinate a globally a freely-falling system inertial is, frame we frame will in deep eventually from space. a globally But see no such inertial matter e↵ects, frame how because good in deep we space. cannot But set no all matter of the howsecond goodmetric our coordinate system is, we willour eventuallyderivatives coordinate see such to system zero e↵ects, is, (and we because thus will eventually first we cannot derivatives see set such all of of e the the↵ects, Christosecond becausemetric↵el we symbols) cannot to set zero. all of This the second meansmetric that as we derivatives to zero (and thus first derivatives of the Christo↵el symbols) to zero. This means that as we derivativesmove far to enough zero (and away thus from firstin derivatives space and/or of the time, Christo we↵ willel symbols) eventually to see zero. the This geodesics means begin that to as converge we move far enough away from in space and/or time, we will eventually see the geodesics begin to converge moveor far diverge enough as the away Christo from ↵elin symbolsP space and/or begin totime, become we will appreciably eventually nonzero. see the geodesics begin to converge or diverge as the Christo↵el symbolsP begin to become appreciablyP nonzero. or divergeBut as how the can Christo we transform↵el symbols quantities begin to expressed become inappreciably Schwarzschild nonzero. coordinates into values we would measure But how can we transform quantitiesBut expressed how can we in Schwarzschild transform quantities coordinates expressed into values in Schwarzschild we would measure coordinates into values we would measure in either a LOF or LIF? In principle,in if we either knew a theLOF coordinate or LIF? In transformation principle, if we equations knew the from coordinate Schwarzschild transformation equations from Schwarzschild in eithercoordinates a LOF toor LIF? the new In principle, coordinates, ifFinding we then knew we the could coordinate Components use the transformation normal tensor equations transformationin a from LOF Schwarzschild rule. But it is coordinates to the new coordinates,coordinates then we to could the use new the coordinates, normal tensor then transformation we could use rule. the normal But it tensor is transformation rule. But it is usually very dicult to derive thoseusually transformation very di equations.cult to derive those transformation equations. usually very dicult to derive those transformation equations. Fortunately, there is another way. ConsiderFortunately, for the sake there of is argument another way.that we Consider want to for know the the sake components of argument that we want to know the components Fortunately, there is another way. ConsiderAssume that for the we sakeknow of basis argument vectors that of wea LOF. want Since: to know the components of a four-vector A in a certain LOF,of and a four-vector suppose weA knowin a the certain orthonormal LOF, and basis suppose vectors weeµ knowof the the LOF, orthonormal basis vectors eµ of the LOF, of a four-vector A in a certain LOF, and suppose⌫ we know the orthonormal basis vectors⌫ eµ of the LOF, which we will take to be the primedwhich frame. we Since will in takeany tocoordinate be the primed basis A frame.= A e⌫0 Sinceand g inµ0 ⌫anyeµ0coordinate⇧ e⌫0 ,we basis A⌫= A e⌫0 and gµ0 ⌫ eµ0 ⇧ e⌫0 ,we which we will take to be the primed frame. Since in any coordinate⌘ basis A = A e0 and g0 e0 ⌘⇧ e0 ,we have have ⌫ µ⌫ ⌘ µ ⌫ have ⌫ ⌫ ⌫ ⌫ eµ0 ⇧ A = eµ0 ⇧ e⌫0 A0 = gµ0 ⌫ A0 = Aµ0 eµ0 ⇧ A = eµ0 ⇧ e⌫⌫0 A0 = gµ0⌫⌫(2.47)A0 = Aµ0 (2.47) eµ0 ⇧ A = eµ0 ⇧ e⌫0 A0 = gµ0 ⌫ A0 = Aµ0 (2.47) In our particular LOF, g = ⌘ by definition, so we can say that µ0 ⌫ µ⌫ In our particular LOF, gµ0 ⌫ = ⌘µ⌫ by definition, so we can say that In our particular LOF, gµ0 ⌫ = ⌘µ⌫ by definition,implying that so we can say that ↵µ ↵µ ↵ ⌘ (e µ ⇧ A)=⌘ A0 = A0 ↵µ ↵µ (2.48)↵ 0 µ ↵µ⌘ (e µ ⇧ A)=↵µ ⌘ Aµ0 =↵ A0 (2.48) ⌘ (eµ0 ⇧ A0)=⌘ Aµ0 = A0 (2.48) So, we can evaluate the componentsSo, of the we canvector evaluateA in any the LOF components if we can of evaluate the vector the dotA in product any LOF of A if we can evaluate the dot product of A with the LOF’s basis vectors e . So, we can evaluate the components of the vector A in any LOF if we can evaluate the dot product of A µ0 with the LOF’s basis vectors e . The clever trick is that since thewith dot the product LOF’s must basis have vectors a coordinate-independente . µ0 value, we can evaluate The clever trick is thatµ since0 the dot product must have a coordinate-independent value, we can evaluate the dot product in whatever global coordinateThe clever system trick is (such that since as the the Schwarzschild dot product coordinate must have system) a coordinate-independent that value, we can evaluate the dot product in whatever global coordinate system (such as the Schwarzschild coordinate system) that describes our spacetime. All thatthe we need dot product to is to evaluate in whatever the Schwarzschild global coordinate components system of (such both theas the LOF’s Schwarzschild coordinate system) that describes our spacetime. All that we need to is to evaluate the Schwarzschild components of both the LOF’s basis vectors eµ0 and the four-vectordescribesA of interest. our spacetime. All that we need to is to evaluate the Schwarzschild components of both the LOF’s basis vectors eµ0 and the four-vector A of interest. basis vectors eµ0 and the four-vector A of interest. 12 12 12 would need to look at geodesics over fairly large distances (or for fairly long times) to see the tidal e↵ects that distinguish a freely-falling frame from a globally inertial frame in deep space. But no matter how good would need to look at geodesics over fairly largeour distances coordinate (or system for fairly is, we long will times) eventually to see see the such tidal e↵ eects,↵ects because we cannot set all of the second metric that distinguish a freely-falling frame from a globallyderivatives inertial to zero frame (and in deep thus space. first derivatives But no matter of the how Christo good↵el symbols) to zero. This means that as we our coordinate system is, we will eventually seemove such far e↵ enoughects, because away from we cannotin space set all and/or of the time,second wemetric will eventually see the geodesics begin to converge derivatives to zero (and thus first derivativesor of diverge the Christo as the↵el Christo symbols)↵el symbolsP to zero. begin This to means become that appreciably as we nonzero. move far enough away from in space and/or time,But we how will can eventually we transform see quantities the geodesics expressed begin into Schwarzschild converge coordinates into values we would measure P or diverge as the Christo↵el symbols begin toin become either appreciably a LOF or LIF? nonzero. In principle, if we knew the coordinate transformation equations from Schwarzschild But how can we transform quantities expressedcoordinates in Schwarzschild to the new coordinates coordinates, into values then we we could would use measure the normal tensor transformation rule. But it is in either a LOF or LIF? In principle, if we knewusually the coordinate very dicult transformation to derive those equations transformation from Schwarzschild equations. coordinates to the new coordinates, then we couldFortunately, use the normal there is tensor another transformation way. Suppose rule. we want But it to is know the components of a four-vector A in a usually very dicult to derive those transformationcertain equations. LOF, and suppose we know the orthonormal basis vectors eµ of the LOF, which we will take to be ⌫ Fortunately, there is another way. Supposethe we primed want frame. to know Since the in componentsany coordinate of a four-vector basis A = AAein0 and a g0 e0 ⇧ e0 ,wehave ⌫ µ⌫ ⌘ µ ⌫ certain LOF, and suppose we know the orthonormal basis vectors eµ of the LOF, which we will take to be ⌫ ⌫ ⌫ the primed frame. Since in any coordinate basis A = A e0 and g0 e0 ⇧ e0 ,wehaveeµ0 ⇧ A = eµ0 ⇧ e⌫0 A0 = gµ0 ⌫ A0 = Aµ0 (2.43) ⌫ µ⌫ ⌘ µ ⌫ ⌫ ⌫ eµ0 ⇧ A = eInµ0 ⇧ oure⌫0 A particular0 = gµ0 ⌫ A0 LOF,= Agµ0µ0 ⌫ = ⌘µ⌫ by definition, so we can(2.43) say that ↵µ e A ↵µ ↵ In our particular LOF, gµ0 ⌫ = ⌘µ⌫ by definition, so we can say that ⌘ ( µ0 ⇧ )=⌘ Aµ0 = A0 (2.44) ↵µ ↵µ ↵ A A ⌘ (eµ0So,⇧ A we)= can⌘ A evaluateµ0 = A0 the components of the vector in(2.44) any LOF if we can evaluate the dot product of with the LOF’s basis vectors eµ0 . So, we can evaluate the components of the vectorTheA cleverin any trick LOF is if that we can since evaluate the dot theproduct dot product must have of aA coordinate-independent value, we can evaluate with the LOF’s basis vectors eµ0 . the dot product in whatever global coordinate system (such as the Schwarzschild coordinate system) that The clever trick is that since the dot productdescribes must have our a spacetime. coordinate-independent All that we need value, to is we to can evaluate evaluate the Schwarzschild components of both the LOF’s the dot product in whatever global coordinatebasis system vectors (sucheµ0 asand the the Schwarzschild four-vector A coordinateof interest. system) that describes our spacetime. All that we need to is toConsider evaluate the the Schwarzschild following application. components Suppose of both we the have LOF’s a particle that is falling from rest at infinity in basis vectors eµ0 and the four-vector A of interest.Schwarzschild spacetime, and we would like to know the particle’s ordinary velocity according to an ob- Consider the following application. Supposeserver we at have rest a at particle a given that Schwarzschild is falling from coordinate rest atr infinity. In order in to do this calculation, we must first find the Schwarzschild spacetime, and we would like toSchwarzschild know the particle’s coordinates ordinary of the velocity observer’s according LOF basis to an vectors, ob- which we will call oT , ox, oy, and oz, corre- server at rest at a given Schwarzschild coordinatespondingr. In to order the observer’s to do this time calculation,T and spatial we mustx, y, first z coordinates, find the respectively. (Note that the observer’s time Schwarzschild coordinates of the observer’s LOFcoordinate basis vectors,T is not whichthe Schwarzschild we will call oT time, ox, o coordinatey, and oz,t corre-: this is why I am giving them di↵erent names. In sponding to the observer’s time T and spatial x,what y, z follows,coordinates, we will respectively. assume that (NoteAT that,Ax,A they,A observer’sz refer to time the components of A evaluated in the LOF, and t r ✓ coordinate T is not the Schwarzschild timeConsider coordinateA ,A the,A followingt,A: thisto is application. its why components I am givingSuppose in the them we Schwarzschild have di↵erent a particle names. system. that In is This falling will from avoid rest all at the infinity primes.) in T x y z what follows, we will assume that A Schwarzschild,A ,A ,A Werefer spacetime, begin to the by and components noting we would that of the likeA observer’s toevaluated know the four-velocity in particle’s the LOF, ordinary anduobs is velocity appropriately according normalized to an ob- for a time basis At,Ar,A✓,A to its components in theserver Schwarzschild atvector rest at ( au system. given⇧ u Schwarzschild This= will1). avoid Also, coordinate all the the observerr primes.). In order is notto do moving this calculation, in his or we her must reference first find frame, the so this vector obs obs We begin by noting that the observer’sSchwarzschild four-velocitypoints coordinates purelyuobs in ofis the the appropriately observer’s observer’s time LOF normalized direction. basis vectors, for Therefore, a time which basis weo willT must call o beT , ouxobs, oy., Whatand oz, are corre- the Schwarzschild vector (u ⇧ u = 1). Also, thesponding observercomponents to is the not observer’s moving of this time in four-vector? hisT and or her spatial reference Thesex, y, z componentscoordinates, frame, so this are respectively. [ vectorut,ur,u✓ (Note,u ] that=[ thedt/d observer’s⌧, dr/d⌧ time,d✓/d⌧,d/d⌧ ] obs obs obs obs points purely in the observer’s timecoordinate direction.byT Therefore, definition.is not theo Schwarzschild IfT themust observer be uobs time is. at coordinateWhat rest are relativet the: this Schwarzschild to is the why Schwarzschild I am giving them coordinates, di↵erent names. then we In know that dr = t r ✓ r T ✓ x y z A t components of this four-vector? Thesewhat components follows,d✓ = we ared will=0 [u ,u assume,u ,u thatu ]obsA==[,Au dt/d,A=⌧,Au, dr/drefer=⌧ 0.,d to We✓ the/d can⌧,d components then/d⌧ compute]obs of evaluatedu using inuobs the⇧ LOF,uobs = and 1: t r ✓ obs obs obs obs by definition. If the observer is at restA ,A relative,A ,A to theto its Schwarzschild components) in coordinates, the Schwarzschild then we system. know This that willdr avoid= all the primes.) r ✓ We begin by noting that thet observer’sµ four-velocity⌫ t uobs2 is appropriatelyt normalized1 for a1 time basis d✓ = d =0 uobs = uobs = u = 0. We can then compute uobs using uobs ⇧ uobs = 1: ) obsvector (u ⇧ u = 1). Also,1= the observergµ⌫ uobsu isobs not= movinggtt(uobs in) his or heru referenceobs = frame,= so this vector points (2.45) obs obs ) p gtt 1 2GM/r purely in the observer’s time direction. Therefore, o must be u . The Schwarzschild components of this µ ⌫ t 2 t 1 1 T obs t r ✓ 1=gµ⌫ uobsuobs =four-vector?gtt(uobs) TheseExample: componentsuobs = are [=u an,u ,u observer,u ]obs =[dt/d⌧ ,at dr/d (2.45)rest⌧,d✓/d⌧,d/d⌧ ]obsp by definition. If We) know that thep g Schwarzschildtt 1 2GM/r basis vectors are orthogonal at every point (because the Schwarzschild the observer is at rest relative to the Schwarzschild coordinates, then we know that dr = d✓ = d =0 r ✓metric is diagonal), so we can convenientlyt choose the observer’s spatial basis vectors to) align with the uobs = uobs = u in= 0. Schwarzschild We can then computep uobs using spacetimeuobs ⇧ uobs = 1: We know that the Schwarzschild basis vectorsSchwarzchild areobs orthogonal basis vectors. at every Let’s point choose (becauseoz parallel the Schwarzschild to the Schwarzschild +er (that is, “upward”), ox parallel metric is diagonal), so we can conveniently chooseto +e, the and observer’soy to be parallel spatial tobasise✓ vectors(the minus to align sign with is necessary the to make the observer’s coordinate system µ ⌫ t 2 t 1 1 Schwarzchild basis vectors. Let’s choose o parallelright-handed, to theoT1= Schwarzschild= u asgobsµ you⌫.u Weobs can uknowobs check += efromgtt(that( withu obsbefore) is, your “upward”),that fingers).uobs = Thiso parallel means= that the only nonzero Schwarzschild(2.49) compo- z r ) px gtt 1 2GM/r to +e, and oy to be parallel to e✓ (the minusnents sign of isox necessary, oy and o toz are make their the observer’s, ✓, and r coordinatecomponents, system respectively. The only thing we then need to do is We can choose the spatial basis vectors to be normalized versions of the right-handed, as you can check with yourWe fingers). knowensure that This that the means the Schwarzschild observer’s that the basis only basis nonzero vectors vectors are Schwarzschild normalized. orthogonal For at compo- every the op pointz vector, (because we have the Schwarzschild spatial Schwarzschild basis vectors. In particular: nents of ox, oy and oz are their , ✓, andmetricr components, is diagonal), respectively. so we can conveniently The only thing choose we the then observer’s need to spatial do is basis vectors to align with the Schwarzchild basis vectors. Let’s choose o parallelµ ⌫ to the Schwarzschildr r +e (thatr is, “upward”),1 o parallel ensure that the observer’s basis vectors normalized. For the1=oz ovector,z ⇧ oz = wegµ have⌫ (zoz) (oz) = grr(oz) (oz) r (oz) = = 1x 2GM/r (2.46) to +e , and o to be parallel to e (the minus sign is necessary to make) the observer’spg coordinaterr system y ✓ µ right-handed,⌫ r as your can check withr your1 fingers). This means that the only nonzero Schwarzschildp compo- 1=oz ⇧ oz = gµ⌫ (oz) (oz) = grr(oz) (oz) (oz) = = 1 2GM/r (2.46) nents of oandx, oy alland otheroz are) Schwarzschild their , ✓, andp componentsgrrr components, of the respectively. basis vectors The are only zero. thing One we then can need find theto do components is of the ensure thatother the basis observer’s vectors basis similarly. vectors normalized:p and all other Schwarzschild components of the basis vectors are zero. One can find the components of the µ ⌫ 1 1 other basis vectors similarly. 1=ox ⇧ ox = gµ⌫ (ox) (ox) = g(ox) (ox) (ox) = = ) pg r sin ✓

µ ⌫ ✓ ✓ ✓ 1 1 1=oy ⇧ oy = gµ⌫ (oy) (oy) = g✓✓(oy) (oy) (11oy) = = ) pg✓✓ r

µ ⌫ r r r 1 1=11oz ⇧ oz = gµ⌫ (oz) (oz) = grr(oz) (oz) (oz) = = 1 2GM/r (2.50) ) pgrr p and all other Schwarzschild components of the LOF basis vectors are zero. Now consider the four-velocity of the falling particle. We saw in exercise 2.6.1 that we have e = 1 and ur = dr/d⌧ = 2GM/r, u✓ = d✓/d⌧ = 0 and u = d/d⌧ = 0 for this particle. Equation 2.44 says that p 2GM dt dt e 1 e = 1 ut = = (2.51) r d⌧ ) ⌘ d⌧ 1 2GM/r 1 2GM/r ✓ ◆ in this case. This gives us all of the Schwarzschild components of the particle’s four-velocity. So, to summarize, the Schwarzschild coordinates of all the vectors of interest are:

t r ✓ 1 [(oT ) , (oT ) , (oT ) , (oT ) ]= , 0, 0, 0 (2.52a) " 1 2GM/r # 1 [(o )t, (o )r, (o )✓, (o ) ]= p0, 0, 0, (2.52b) x x x x r sin ✓  1 [(o )t, (o )r, (o )✓, (o ) ]= 0, 0, , 0 (2.52c) y y y T r  [(o )t, (o )r, (o )✓, (o ) ]= 0, 1 2GM/r, 0, 0 (2.52d) z T T T h p 1 2GM i [ ut,ur,u✓,u ]= , , 0, 0 (2.52e) 1 2GM/r r 

13 Example: an observer at rest

Now, as we sawin in exerciseSchwarzschild 2.6.1, a radially fallingspacetime particle yields ur = dr/d⌧ = 2GM/r if we assume the particle started at rest at infinity. Then 1=u ⇧ u requires that r r Now, as we sawNow, in as exercise we saw 2.6.1, in exercise a radially 2.6.1,A particle a radially falling falling falling particleat rest particlefrom yields infinity: yieldsu u= =dr/ddr/d⌧⌧== 22GM/rGM/rifif we we assume assumer p Now, as we saw in exercise 2.6.1, a radially falling particle yields u = dr/d⌧ = 2GM/r if we assume the particle started at rest at infinity. Then 1=u ⇧ u requires that the particle started at rest at infinity. Then 1=u⇧ ut requires2 thatr 2 2GM p t 2 1 2GM the particle1= startedgtt(u at) rest+ g atrr( infinity.u ) = Then1 1=up⇧(uu requires) + that 2GM 1r 2GM 1 2GM/r r p 1=g (ut)2 + g r(ur)2 = 21GM (ut)2 ✓+ 1 ◆ 2GM✓ ◆ Now, as we saw in exercise 2.6.1, a radially falling particlet 2 tt yieldsr 2 rru = dr/d⌧ = 22GM/rt 2 if we assume 1=gtt(u) + grr(u ) = 1 r (ut 2) + 1r 22GM/r r2GM t 2 1 2GM 2GM⇢ ✓1=2GMg (u ◆) t +2 g 2(✓GMu ⇢) = ◆1 t 1(u ) + the particle started at rest at infinity. Then 1=u ⇧ u 1requires⇢ that= 2 1 r tt (u ) +rr 1⇢ 2GM/r u =r (2.47) 2GM⇢ 2GM ✓t 2 2GM⇢◆ t ✓ 1 ◆ r 1 2GM/r r 1 )⇢ =⇢1r 2 (u )+ r⇢ p u =⇢ r )✓ 1 ◆2GM/r(2.47) ✓ ◆ 2GM⇢ ⇢ ✓ 2GM ◆ ✓2GM⇢⇢ ◆ 21 ⇢) r rt 2 2GM⇢⇢ r ) t 2GM1 2GM/r 2GM⇢ 1 t 12 r ✓2= 1 ◆ 2GM✓ (ut )2◆+ ⇢ 1 u2GM= t 2 ⇢ (2.47)t 1=)gtt(u ) +⇢grrr(inu this) = case.1 Thisr gives( usu1) all+ of⇢ ther Schwarzschild= ) 1 components1 2GM/r(u ) + of the particle’su four-velocity.= (2.47) ✓ in this case.◆ This gives✓ us allr of) the◆ Schwarzschild⇢ r1 components2GM/r of ther r particle’s four-velocity.⇢ r ) 1 2GM/r So, to summarize,So, to✓ summarize, the Schwarzschild◆ the✓ Schwarzschild coordinates✓ ◆ of all coordinates the✓ ◆ vectors of of interest◆ all the are: vectors of interest are: in this case. This gives2 us all of the Schwarzschild components of the particle’s four-velocity. 2GM⇢ 2GM t 2 2inGM⇢ this case. Thist gives1 us all of the Schwarzschild components of the particle’s four-velocity. 1 ⇢ So,= to summarize,1 the(u Schwarzschild) + ⇢ coordinatesu = of all the vectors of interest(2.47) are: ) ⇢ r r ⇢ r So,t to) summarize,r ✓ 1 2 theGM/r Schwarzschild1 coordinates1 of all the vectors of interest are: ✓ ◆ ✓ ◆ [(oT ) , (oT ) ,[((ooT ) ),t(,o(To) )]=r, (o )✓, (o ) ]=, 0, 0, 0 , 0, 0, (2.48a)0 (2.48a) T T " T 1 2GM/rT # " 1 2GM/r # in this case. This gives us all of the Schwarzschildt componentsr t ✓ ofr the✓ particle’s four-velocity.1 [(oT ) , (oT [() o,z()o,T()oT,)(,o(ToT)) ]=, (oT ) ]= p0,t 1 2rGM/r,, 0, 0,✓0,00 1 (2.48b)(2.48a) So, to summarize, the Schwarzschild coordinates of all the vectors[(o ) oft, interest(o )r[(1, (ooT are:2))GM/r✓,,((ooT )), (]=oT ) ,p0(o, T )1 ]=2GM/r, 0, 0 , 0, 0, 0 (2.48b) (2.48a) z "T h T T # i 1 2GM/r t r ✓ p 1 2GM " # t r ✓ [ u ,u ,u ,u ]= , h , 0, 0 (2.48c)i [(o ) , (o ) , (o ) , (o ) ]= p0, t11 tr 22GM/rGM/r,✓ r 0,✓r 0 p1 2GM (2.48b) t r ✓ z T T1 T [[(uo,u),,u(o ,u) , (]=o ) , (o ) ]= ,p0, 1 2,GM/r,0, 0 0, 0 (2.48c) (2.48b) [(oT ) , (oT ) , (oT ) , (oT ) ]= , 0, 0, 0 z T T 1 T(2.48a)2GM/r r We can now calculate" the1 components2GM/r of the particle’sh #p 1 four-velocity2GM in the observer’si frame by calculating t r ✓ h p 1 2GM i the necessary dot products[ uin,u the,u global,u Schwarzschild]= coordinate, [ systemut,ur: ,u, ✓0,u, 0 ]= ,(2.48c) , 0, 0 (2.48c) o t o r o ✓Weo can nowp calculate the components1 2GM/r of the particle’s r four-velocity in the observer’s frame by calculating [( z) , ( T ) , ( T ) , ( T ) ]= T 0, Tµ1 2GM/r,TT 0, 0 ↵ t(2.48b)t 1 2GM/r r the necessaryu = dot⌘ productsoµ⇧ u = ⌘ inoT the⇧ u global=( 1)( SchwarzschildoT ) g↵u = ( coordinateoT ) gttu  system : We can now calculate the componentsh of the particle’s four-velocityi in the observer’s frame by calculating t r ✓ Wep can1 now1 calculate2GM2 theGM components1 of the particle’s1 four-velocity in the observer’s frame by calculating the necessary dot[ u products,u ,u ,uin]= the global=+ SchwarzschildT , Tµ1 coordinate, 0, 0 TT system=: (2.48c)↵ (2.49a)t t the1 necessary2GM/r1 u2GM/r= dot⌘ productsorµ ⇧ ur = ⌘in1 theo2TGM/r global⇧ u =( Schwarzschild1)(1 o2TGM/r) g↵u coordinate= (oT ) systemgttu :  ✓ ◆ z zµ zz ↵ r r T Tµo u =u ⌘ poµTT⇧ uo = ⌘u oz ⇧ u =1 (+1)(o o↵z) g↵u 2=+(GMpooz) tgrru t1 1 We can now calculate the components ofu the= particle’s⌘ µ ⇧ four-velocity= ⌘ T=+⇧ in=( theu1)( observer’sT =T⌘)Tµg↵o frame1u⇧ u== by⌘(TTT calculatingo) gtt⇧ u =( 1)(=o )↵g u = (o )tg ut (2.49a) 2GM 1 1 2GM/r µ 2rGM/rT1 2GM/r T 1 ↵ 2GM/r T tt the necessary dot products in the global Schwarzschild coordinate=1 system2GM: 1 2GM/r1 ✓ = ◆ 1 (2.49b) =+ r z11 2zµGM/r zz 1=1 2GM/r ↵ 2GM 1r (2.49a)r 1 r u =⌘ oµ ⇧ u ==+⌘ oz ⇧ u = (+1)(o1z) g↵u =+(oz) grru = (2.49a) T Tµ TT 1 2GM/r↵ pr p t1 2tGM/r 1 2GM/r p u = ⌘ oµ ⇧ u = ⌘ oT ⇧ u =(1)(oT ) g↵✓ u = (oT◆) gttu 1 2pGM/r r 1 2GM/r 1 2GM/r Finally,z we canzµ evaluate thezz components of the particle’s↵ ordinary speedr in✓ ther observer’s◆ frame as follows: u = ⌘ o u = ⌘ o u = (+1)(2GMzo ) gzµ1u =+(ozz) g u 2GM/r↵ r r 1 2pGMµ ⇧ 1 z ⇧= u z1= ⌘ ↵o ⇧ u = p1⌘z o2GM/rrr⇧ u = (+1)(= o ) g u =+(o ) g u (2.49b) ux uy r uz 1 2pGM/r2µGM/r z 2GMz ↵ pz rr =+ 1v = =0,v= =0,v=r = = 1 2GM/r(2.49a)= 1 2GM/r(2.50) 1 2GM/r x rT2GM1 y 2GM/r1 T z 1 T2GM/r 2GM/r ✓ = u ◆ u 1 2uGM/r 1=2GM2GM/rp 1 r (2.49b)2GM/r = p 1 p2GM/r = (2.49b) uz = ⌘zµo u = ⌘zzo Finally,u=r (+1)( wer o can1)↵ evaluateg2GM/ru =+( theo components)rg ur ofr the1 1 particle’s2GM/r2GM/r ordinary speed in the observer’s frame as follows: p µ ⇧ The particle’sz ⇧ measured speedz approaches↵ thatpz of lightrr aspr the observer’s radial coordinate r 2GM. (An1 2GM/r p ! observer will not measure a speedx larger than thaty of light atp a positionz r<2GMp because an observer Finally,2 weGM can evaluate1 the componentsFinally,u of we the can2 particle’sGM/r evaluateu ordinary the components speedu in theof the observer’s2GM/r particle’s frame ordinary asp follows: speed in the2GM observer’s frame as follows: = cannot be at rest when1 vr<2xGM/r=2GMT .)==0,v y = T =0,vz = T = (2.49b) 1 2GM/r = (2.50) r r 1Sox the2GM/r general approach y tou calculating1 quantities2zGM/ru in an observer’su LOF or LIF1 is2 asGM/r follows: r u p u uux 2GM/ruy uz 2GMp2GM/r 2GM vx = T =0,vy = T =0,vz = T = 1 2GM/r = (2.50) 1.u Use the observer’su four-velocitypuvobsx =asu the=0 the observer’s,vy = time-directed=0,vp basisz = vector= oT r. 1 2GM/r = (2.50) Finally, we can evaluate the componentsThe of the particle’s particle’s measured ordinary speed speedu approachesT in the1 observer’s2GM/r thatuT of frame light as as follows: theu observer’sT 1 radial2GM/r coordinate r 2GM . (Anr p ! 2. Constructobserver a set will of spatial not measure basis vectors a speedo , o , largero such than that o that⇧ o = of⌘ light. at a position r<2GMp because an observer ux uy uz 2GM/r x ypz 2µGM⌫ µ⌫ The particle’s measuredcannot speed be approaches atThe rest particle’s when thatr< measuredof2GM light.) as speed the observer’s approaches radial that coordinate of light aspr the2 observer’sGM. (An radial coordinate r 2GM. (An vx = =0,vy = 3.=0 Find,v thez = components= ofo in whatever global1 2GM/r coordinate= system describes spacetime(2.50) on the! large scale. uT observer willu notT measureSo a theu speedT general larger1 approachµ 2 thanGM/r that to calculating of light at quantities a positionr inr< an observer’s2GM because LOF an or LIF observer is as follows: ! observer will not measure a speed larger than that of light at a position r<2GM because an observer cannot be at rest4. when Also determiner<2GM the.)cannot components be at of rest thep when four-vectorr

12 Now, as we saw in exercise 2.6.1, a radially falling particle yields ur = dr/d⌧ = 2GM/r if we assume the particle started at rest at infinity. Then 1=u ⇧ u requires that p 2GM 1 2GM 1=g (ut)2 + g (ur)2 = 1 (ut)2 + tt rr r 1 2GM/r r ✓ ◆ ✓ ◆ 2GMExample:⇢ 2GM an2 observer2GM⇢ at rest1 1 ⇢ = 1 (ut)2 + ⇢ ut = (2.47) ) ⇢ r r ⇢ r ) 1 2GM/r ✓ in◆ Schwarzschild✓ ◆ spacetime in this case. This gives us all of the Schwarzschild components of the particle’s four-velocity. So, to summarize,Summary: the Schwarzschild coordinates of all the vectors of interest are:

t r ✓ 1 [(oT ) , (oT ) , (oT ) , (oT ) ]= , 0, 0, 0 (2.48a) " 1 2GM/r # [(o )t, (o )r, (o )✓, (o ) ]= p0, 1 2GM/r, 0, 0 (2.48b) z z z z h p 1 2GM i [ ut,ur,u✓,u ]= , , 0, 0 (2.48c) 1 2GM/r r  We can now calculate the components of the particle’s four-velocity in the observer’s frame by calculating the necessary dot products in the global Schwarzschild coordinate system:

uT = ⌘Tµo ⇧ u = ⌘TTo ⇧ u =( 1)(o )↵g u = (o )tg ut µ T T ↵ T tt 1 2GM 1 1 =+ 1 = (2.49a) 1 2GM/r r 1 2GM/r 1 2GM/r ✓ ◆ z zµ zz ↵ r r u = ⌘ poµ ⇧ u = ⌘ oz ⇧ u = (+1)(oz) g↵u =+(poz) grru 2GM 1 2GM/r = 1 2GM/r = (2.49b) r r 1 2GM/r 1 2GM/r p Finally, we can evaluate the components of the particle’s ordinaryp speed in the observer’s frame as follows:

ux uy uz 2GM/r 2GM vx = =0,vy = =0,vz = = 1 2GM/r = (2.50) uT uT uT 1 2GM/r r p The particle’s measured speed approaches that of light asp the observer’s radial coordinate r 2GM. (An observer will not measure a speed larger than that of light at a position r<2GM because! an observer cannot be at rest when r<2GM.) So the general approach to calculating quantities in an observer’s LOF or LIF is as follows:

1. Use the observer’s four-velocity uobs as the the observer’s time-directed basis vector oT .

2. Construct a set of spatial basis vectors ox, oy, oz such that oµ ⇧ o⌫ = ⌘µ⌫ .

3. Find the components of oµ in whatever global coordinate system describes spacetime on the large scale. 4. Also determine the components of the four-vector A of interest in that global coordinate system.

µ µ⌫ 5. The components of the four-vector in the observer’s system are A0 = ⌘ o⌫ ⇧ A, where one evaluates ↵ the dot product oµ ⇧ A = g↵(oµ) A in the global coordinate system. One can generalize this scheme to tensor quantities as well (see the homework). This is a very powerful scheme for translating abstract tensor quantities expressed in arcane coordinates into quantities that one can more easily interpret physically.

Homework Problems 2.1 What is the metric for an r, ✓ polar-coordinate system whose basis vectors point in the same directions as those in the usual orthonormal polar coordinates, but whose magnitudes are appropriate for a coordinate basis? What are the magnitudes of the basis vectors in this case?

12 Now, as we saw in exercise 2.6.1, a radially falling particle yields ur = dr/d⌧ = 2GM/r if we assume the particle started at rest at infinity. Then 1=u ⇧ u requires that p 2GM 1 2GM 1=g (ut)2 + g (ur)2 = 1 (ut)2 + tt rr r 1 2GM/r r ✓ ◆ ✓ ◆ 2GM⇢ 2GM 2 2GM⇢ 1 1 ⇢ = 1 (ut)2 + ⇢ ut = (2.47) ) ⇢ r r ⇢ r ) 1 2GM/r ✓ ◆ ✓ ◆ in this case. This gives us all of the Schwarzschild components of the particle’s four-velocity. So, to summarize, the Schwarzschild coordinates of all the vectors of interest are:

t r ✓ 1 [(oT ) , (oT ) , (oT ) , (oT ) ]= , 0, 0, 0 (2.48a) " 1 2GM/r # [(o )t, (o )r, (o )✓, (o ) ]= p0, 1 2GM/r, 0, 0 (2.48b) z z z z Example: an observerh p 1 at rest2GM i [ ut,ur,u✓,u ]= , , 0, 0 (2.48c) 1 2GM/r r in Schwarzschild spacetime We can now calculate the components of the particle’s four-velocity in the observer’s frame by calculating the necessary dot products in the global Schwarzschild coordinate system: Calculate the LOF components in Schwarzschild coordinates: uT = ⌘Tµo ⇧ u = ⌘TTo ⇧ u =( 1)(o )↵g u = (o )tg ut µ T T ↵ T tt 1 2GM 1 1 =+ 1 = (2.49a) 1 2GM/r r 1 2GM/r 1 2GM/r ✓ ◆ z zµ zz ↵ r r u = ⌘ poµ ⇧ u = ⌘ oz ⇧ u = (+1)(oz) g↵u =+(poz) grru 2GM 1 2GM/r = 1 2GM/r = (2.49b) r r 1 2GM/r 1 2GM/r p Finally, we can evaluate the components of the particle’s ordinaryp speed in the observer’s frame as follows:

ux uy uz 2GM/r 2GM vx = =0,vy = =0,vz = = 1 2GM/r = (2.50) uT uT uT 1 2GM/r r p The particle’s measured speed approaches that of light asp the observer’s radial coordinate r 2GM. (An observer will not measure a speed larger than that of light at a position r<2GM because! an observer cannot be at rest when r<2GM.) So the general approach to calculating quantities in an observer’s LOF or LIF is as follows:

1. Use the observer’s four-velocity uobs as the the observer’s time-directed basis vector oT .

2. Construct a set of spatial basis vectors ox, oy, oz such that oµ ⇧ o⌫ = ⌘µ⌫ .

3. Find the components of oµ in whatever global coordinate system describes spacetime on the large scale. 4. Also determine the components of the four-vector A of interest in that global coordinate system.

µ µ⌫ 5. The components of the four-vector in the observer’s system are A0 = ⌘ o⌫ ⇧ A, where one evaluates ↵ the dot product oµ ⇧ A = g↵(oµ) A in the global coordinate system. One can generalize this scheme to tensor quantities as well (see the homework). This is a very powerful scheme for translating abstract tensor quantities expressed in arcane coordinates into quantities that one can more easily interpret physically.

Homework Problems 2.1 What is the metric for an r, ✓ polar-coordinate system whose basis vectors point in the same directions as those in the usual orthonormal polar coordinates, but whose magnitudes are appropriate for a coordinate basis? What are the magnitudes of the basis vectors in this case?

12 We can now calculate the components of the particle’s four-velocity in the observer’s frame by calculating the necessary dot products in the global Schwarzschild coordinate system:

We can now calculate the components of the particle’sT four-velocityTµ inTT the observer’s frame↵ by calculating t t u = ⌘ oµ ⇧ u = ⌘ oT ⇧ u =( 1)(oT ) g↵u = (oT ) gttu the necessary dot products in the global Schwarzschild coordinate system: 1 2GM 1 1 T Tµ TT =+↵ 1t t = (2.53a) u = ⌘ oµ ⇧ u = ⌘ oT ⇧ u =( 1)(oT ) g1↵u2GM/r= (oT ) gttu r 1 2GM/r 1 2GM/r ✓ ◆ 1 2GM z zµ1 zz 1 ↵ r r =+ 1 u = ⌘ poµ ⇧ u ==⌘ oz ⇧ u = (+1)(oz) g↵u =+((2.53a)poz) grru 1 2GM/r r 1 2GM/r 1 2GM/r ✓ ◆ 2GM 1 2GM/r z zµ zz = ↵ r r 1 2GM/r = (2.53b) u = ⌘ poµ ⇧ u = ⌘ oz ⇧ u = (+1)(oz)g↵ur =+(1 po2zGM/r) grru 1 2GM/r Example: anr observer at rest 2GM 1 x xµ 2GM/rxx p ↵ = 1 u2GM/r= ⌘ o=µ ⇧ u= ⌘ ox ⇧ u = (+1)(ox) g↵up=+((2.53b)ox) gu = 0 (2.53c) r 1 2GM/r y yµ 1 2yyGM/r ↵ ✓ ✓ r in Schwarzschildu = ⌘ oµ ⇧ u =⌘ ospacetimey ⇧ u = (+1)(oy) g↵u =+(oy) g✓✓u = 0 (2.53d) x xµ xx p ↵ u = ⌘ oµ ⇧ u = ⌘ ox ⇧ u = (+1)(ox) g↵up=+(ox) gu = 0 (2.53c) y yµ Finally,yy we can evaluate↵ the components ✓ of the✓ particle’s ordinary speed in the observer’s frame as follows: u = ⌘ oµ ⇧ u = ⌘ Nowoy ⇧ weu = can (+1)( calculateoy) g ↵theu falling=+( oparticle’sy) g✓✓u speed= 0 in the LOF: (2.53d) ux uy uz 2GM/r 2GM Finally, we can evaluate the components ofvx the= particle’s=0,v ordinaryy = speed=0,v in thez = observer’s= frame as follows:1 2GM/r = (2.54) uT uT uT 1 2GM/r r ux uy uz 2GM/r 2GM p v = =0,v=The particle’s=0,v measured= = speed approaches1 that2GM/r of light= asp the observer’s(2.54) radial coordinate r 2GM. (An x uT y uT z uT r observer will not measure1 a speed2GM/r larger than that of light at a position r<2GM because! an observer p The particle’s measured speed approachescannot be at that rest of when lightr< asp the2GM observer’s.) radial coordinate r 2GM. (An observer will not measure a speedSo larger the than general that approach of light to at calculating a position quantitiesr<2GM inbecause an observer’s! an observer LOF or LIF is as follows: cannot be at rest when r<2GM.) 1. Use the observer’s four-velocity uobs as the the observer’s time-directed basis vector oT . So the general approach to calculating quantities in an observer’s LOF or LIF is as follows: 2. Construct a set of spatial basis vectors ox, oy, oz such that oµ ⇧ o⌫ = ⌘µ⌫ . 1. Use the observer’s four-velocity uobs as the the observer’s time-directed basis vector oT . 3. Find the components of oµ in whatever global coordinate system describes spacetime on the large scale. 2. Construct a set of spatial basis vectors ox, oy, oz such that oµ ⇧ o⌫ = ⌘µ⌫ . 4. Also determine the components of the four-vector A of interest in that global coordinate system. 3. Find the components of oµ in whatever global coordinate system describes spacetime on the large scale. µ µ⌫ 5. The components of the four-vector in the observer’s system are A0 = ⌘ o⌫ ⇧ A, where one evaluates 4. Also determine the components of the four-vector A of interest in↵ that global coordinate system. the dot product oµ ⇧ A = g↵(oµ) A in the global coordinate system. µ µ⌫ 5. The components of the four-vectorOne can in the generalize observer’s this system scheme are toA tensor0 = ⌘ quantitieso⌫ ⇧ A, where as well one (see evaluates the homework). This is a very powerful ↵ the dot product oµ ⇧ A = gscheme↵(oµ) forA translatingin the global abstractcoordinate tensor system. quantities expressed in arcane coordinates into quantities that one One can generalize this schemecan to more tensor easily quantities interpret as physically. well (see the homework). This is a very powerful scheme for translating abstract tensor quantities expressed in arcane coordinates into quantities that one can more easily interpret physically.Homework Problems 2.1 What is the metric for an r, ✓ polar-coordinate system whose basis vectors point in the same directions Homework Problems as those in the usual orthonormal polar coordinates, but whose magnitudes are appropriate for a 2.1 What is the metric for an r, ✓ polar-coordinatecoordinate basis? system What whose are the basis magnitudes vectors point of the in the basis same vectors directions in this case? as those in the usual orthonormal2.2 We polarcan derive coordinates, equation but 2.26 whose as follows. magnitudes Calculate are appropriate the partial forderivative a of the metric and use the coordinate basis? What are the magnitudes of the basis vectors in this case? definition of the metric gµ⌫ = eµ ⇧ e⌫ and equation 2.17 that defines the Christo↵el symbol: 2.2 We can derive equation 2.26 as follows. Calculate the partial derivative of the metric and use the @gµ⌫ @(eµ ⇧ e⌫ ) @eµ @e⌫ definition of the metric g = e ⇧ e and equation= 2.17= that defines⇧ e⌫ + theeµ ⇧ Christo=↵el symbol:e ⇧ e⌫ + e ⇧ eµ = g⌫ + gµ (2.55) µ⌫ µ ⌫ @x↵ @x↵ @x↵ @x↵ µ↵ ⌫↵ µ↵ ⌫↵ @g @(eµ ⇧ e ) @e @e µ⌫ = ⌫ = µ ⇧ eThe+ followinge ⇧ ⌫ = equations e ⇧ e are+ equivalent e ⇧ e (I= have g simply+ renamed g the(2.55) 3 free lower indices cyclically: @x↵ @x↵ @x↵ ⌫ µ @x↵ µ↵ ⌫ ⌫↵ µ µ↵ ⌫ ⌫↵ µ @g @g ↵µ = g + g and ⌫↵ = g + g (2.56) The following equations are equivalent (I have simply renamed@x⌫ ↵⌫ the 3µ freeµ lower⌫ ↵ indices cyclically:@xµ ⌫µ ↵ ↵µ ⌫

@g↵µ Add two of these equations@g⌫↵ and subtract the third and take advantage of the symmetry of the metric = gµ + g↵ and = g↵ + g⌫ (2.56) @x⌫ ↵⌫and the symmetryµ⌫ of the@x Christoµ ⌫↵µel symbol’s↵µ lower indices to simplify the result. Then multiply both 1 Add two of these equations andsides subtract by 2 theand third the and inverse take metric, advantage and of contract the symmetry over one of the upper metric index of that inverse metric to get and the symmetry of the Christoequation↵el symbol’s 2.26. lower indices to simplify the result. Then multiply both 1 sides by 2 and the inverse metric,2.3 Calculate and contract the Christo over↵ oneel symbols upper index for ✓, of coordinates that inverse on metric the surface to get of a two-dimensional sphere of equation 2.26. 2 2 radius R,wherethemetricisg✓✓ =1,g = R sin ✓,g✓ = g✓ = 0. 2.3 Calculate the Christo↵el symbols for ✓, coordinates on the surface of a two-dimensional sphere of 2 2 radius R,wherethemetricisg✓✓ =1,g = R sin ✓,g✓ = g✓ = 0. 14

14 We can now calculate the components of the particle’s four-velocity in the observer’s frame by calculating the necessary dot products in the global Schwarzschild coordinate system:

uT = ⌘Tµo ⇧ u = ⌘TTo ⇧ u =( 1)(o )↵g u = (o )tg ut µ T T ↵ T tt 1 2GM 1 1 =+ 1 = (2.53a) 1 2GM/r r 1 2GM/r 1 2GM/r ✓ ◆ z zµ zz ↵ r r u = ⌘ poµ ⇧ u = ⌘ oz ⇧ u = (+1)(oz) g↵u =+(poz) grru 2GM 1 2GM/r = 1 2GM/r = (2.53b) r 1 2GM/r 1 2GM/r r x xµ xx p ↵ u = ⌘ oµ ⇧ u = ⌘ ox ⇧ u = (+1)(ox) g↵up=+(ox) gu = 0 (2.53c) y yµ yy ↵ ✓ ✓ u = ⌘ oµ ⇧ u = ⌘ oy ⇧ u = (+1)(oy) g↵u =+(oy) g✓✓u = 0 (2.53d) Finally, we can evaluate the components of the particle’s ordinary speed in the observer’s frame as follows: ux uy uz 2GM/r 2GM vx = =0,vy = =0,vz = = 1 2GM/r = (2.54) uT uT uT 1 2GM/r r Summary: general method p for The particle’s measured speed approaches that of light asp the observer’s radial coordinate r 2GM. (An observer will not measure a speed larger than that of light at a position r<2GM because! an observer cannotcalculating be at rest when r<2GM.) quantities in a LOF/LIF So the general approach to calculating quantities in an observer’s LOF or LIF is as follows:

1. Use the observer’s four-velocity uobs as the the observer’s time-directed basis vector oT .

2. Construct a set of spatial basis vectors ox, oy, oz such that oµ ⇧ o⌫ = ⌘µ⌫ .

3. Find the components of oµ in whatever global coordinate system describes spacetime on the large scale. 4. Also determine the components of the four-vector A of interest in that global coordinate system.

µ µ⌫ 5. The components of the four-vector in the observer’s system are A0 = ⌘ o⌫ ⇧ A, where one evaluates ↵ the dot product oµ ⇧ A = g↵(oµ) A in the global coordinate system. One can generalize this scheme to tensor quantities as well (see the homework). This is a very powerful scheme for translating abstract tensor quantities expressed in arcane coordinates into quantities that one can more easily interpret physically.

Homework Problems 2.1 What is the metric for an r, ✓ polar-coordinate system whose basis vectors point in the same directions as those in the usual orthonormal polar coordinates, but whose magnitudes are appropriate for a coordinate basis? What are the magnitudes of the basis vectors in this case? 2.2 We can derive equation 2.26 as follows. Calculate the partial derivative of the metric and use the definition of the metric gµ⌫ = eµ ⇧ e⌫ and equation 2.17 that defines the Christo↵el symbol: @g @(eµ ⇧ e ) @e @e µ⌫ = ⌫ = µ ⇧ e + e ⇧ ⌫ = e ⇧ e + e ⇧ e = g + g (2.55) @x↵ @x↵ @x↵ ⌫ µ @x↵ µ↵ ⌫ ⌫↵ µ µ↵ ⌫ ⌫↵ µ The following equations are equivalent (I have simply renamed the 3 free lower indices cyclically: @g @g ↵µ = g + g and ⌫↵ = g + g (2.56) @x⌫ ↵⌫ µ µ⌫ ↵ @xµ ⌫µ ↵ ↵µ ⌫ Add two of these equations and subtract the third and take advantage of the symmetry of the metric and the symmetry of the Christo↵el symbol’s lower indices to simplify the result. Then multiply both 1 sides by 2 and the inverse metric, and contract over one upper index of that inverse metric to get equation 2.26. 2.3 Calculate the Christo↵el symbols for ✓, coordinates on the surface of a two-dimensional sphere of 2 2 radius R,wherethemetricisg✓✓ =1,g = R sin ✓,g✓ = g✓ = 0.

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