Chapter 6

Gravity

6.1 Newtonian Gravity

Let’s review the original theory of gravity, and see where and how it might be updated. Newton’s theory of gravity begins with a specified source of mass, ⇢(r), the mass density, as a function of position. Then the gravitational field g is related to the source ⇢ by

g = 4 ⇡G⇢, (6.1) r· where G is the gravitational constant (G 6.67 10 11 Nm2/kg2). Particles ⇠ ⇥ respond to the gravitational field through the force F = m g used in Newton’s second law (or its relativistic form):

m x¨(t)=m g, (6.2) for a particle of mass m at location x(t). The field g is a force-per-unit-mass (similar to E, a force-per-unit-charge), and it has g =0, so it is natural to introduce a potential with g = , r⇥ r just as we do in E&M. Then the field equation and equation of motion read (again, given ⇢) 2 =4⇡G⇢ r (6.3) m x¨(t)= m . r 6.1.1 Solutions From the field equation, we can immediately obtain (by analogy if all else fails) the field g for a variety of symmetric source configurations. Start with the point

187 188 CHAPTER 6. GRAVITY source solution – if ⇢ = m3(r) for a point source located at the origin, we have Gm = (6.4) r similar to the Coulomb potential V = q ,butwith 1 G (and 4 ⇡✏0 r 4 ⇡✏0 ! q m, of course). ! We can generate, from (6.1), the “Gauss’s law” for gravity – pick a domain ⌦, and integrate both sides of (6.1) over that domain, using the divergence theorem, to get

g da = 4 ⇡G ⇢(r0) d⌧ 0 . (6.5) · I@⌦ Z⌦ Menc ⌘ As usual, this form is always true, but not always| {z useful} for finding g. Suppose we have an infinite line of mass with constant mass-per-unit-length .Thensymmetrydemandsthatg = g(s) ˆs, and using a Gaussian cylinder of height h and radius s centered on the wire, we have

g da = g(s)2⇡hs, (6.6) · I@⌦ with Menc = h,sothat 2 G g(s)= (6.7) s which mimics the electrostatic result E(s)= , and the two are again 2 ⇡✏0 s related under 1 G. 4 ⇡✏0 ! Problem 6.1 Find the gravitational field above and below an infinite sheet of mass with constant mass-per-unit-area .

Problem 6.2 What is the potential for the infinite line of mass and for the infinite sheet?

Problem 6.3 What is the potential inside and outside a sphere of radius R with constant mass density ⇢0? 6.2. OPTIONS 189

6.2 Options

6.2.1 Three types of mass There are a few natural questions we might ask at this point. The first – is it clear that the mass appearing on the left and right of (6.2) are the same? In E&M, for example, Newton’s second law reads: m x¨(t)=q E and we don’t associate m and q (units aside), why do we have the same m appearing on both sides of (6.2)? We could imagine that the inertial mass m on the left is di↵erent from the “passive” gravitational mass, mp, on the right:

m x¨(t)=mp g. (6.8)

While experiment suggests m m , it is impossible (within the confines of ⇡ p Newtonian gravity and dynamics) to rule out the possibility that m = m . 6 p Einstein’s theory of general relativity requires that m = mp,soitscorrectness suggests that m = mp exactly, i.e. that the inertial and passive gravitational masses are identical. There is a third type of mass built into Newtonian gravity – the “active” gravitational mass that generates the gravitational field – in theory, then, there are three di↵erent masses that could be present in the equations of Newtonian gravity: g = 4 ⇡G⇢a r· (6.9) m x¨(t)=mp g, where ⇢a is the active mass density. Again, up to experimental error, the active and passive masses are the same, so we don’t worry about these distinctions (people worry about them all the time – if you could find, say, a di↵erence between m and mp, then general relativity would automatically be unavailable as the theory of gravity).

6.2.2 Negative mass In E&M, there are two types of charge, positive and negative – yet we only (typically) allow one sign for mass. What would happen if you had a negative mass? Take a pair of masses m` and mr (left and right) separated a distance x as shown in Figure 6.1. The force on each mass is:

Gm m Gm m F = ` r xFˆ = ` r xˆ, (6.10) ` x2 r x2 but if we assume all forms of mass are the same (inertial, active and passive), 190 CHAPTER 6. GRAVITY

m mr

xˆ x

Figure 6.1: Two masses separated by a distance x. then the acceleration of each mass is Gm Gm a = r xaˆ = ` xˆ. (6.11) ` x2 r x2 So far, we have made no assumptions about the signs of the masses – take mr > 0 and m` < 0, what happens? The details of the dynamics are, as always, complicated, but of interest is the direction of the acceleration – both accelerations point to the right (in the xˆ direction) – the masses “chase” each other. That implies that both masses are speeding up and traveling in the same direction. What happens to energy and momentum conservation? The total energy of the system is 1 1 Gm m E = m v2 + m v2 ` r , (6.12) 2 ` ` 2 r r x for v` and vr the velocity vectors. The energy can remain constant for a constant separation x while both v` and vr increase because m` < 0.Similarly, the momentum of the initial configuration is zero (assuming the masses start from rest) – if the two masses move in the same direction with increasing speed, it would appear that the linear momentum is no longer zero. Again, since m` < 0, we have total momentum p = m` v` + mr vr that is zero even though both masses are moving in the same direction.

6.3 Predictions: Non-relativistic

First, we can go back to the original data that motivated Newton to introduce the 1/r2 force of gravity, associated with two massive bodies. The Kepler laws, obtained by Kepler through observation, are encoded in Newtonian gravity. Our first goal will be to identify the motion of a test mass induced by the gravitational field of a spherically symmetric central body of mass M. Then we’ll move on to cases in which the “test body” (that feels the e↵ect of the field) is not a point particle. 6.3. PREDICTIONS: NON-RELATIVISTIC 191

6.3.1 Elliptical Orbits For a particle of mass m that travels in the presence of the gravitational field of a spherical central body of mass M and radius R, the total energy is 1 H m r˙2 + r2 ˙2 + U(r)=E, (6.13) ⌘ 2 ⇣ ⌘ where we are working in the x y plane in spherical coordinates and U(r)= GMm. Angular momentum is conserved here, from p˙ = @H =0,sothat r @ J mr2 ˙ is a conserved quantity. Using this, we can write the total energy z ⌘ as 1 J 2 H = m r˙2 + z + U(r)=E. (6.14) 2 m2 r2 ✓ ◆ Because the potential energy here goes like 1/r,itisnaturaltodefinea new coordinate ⇢ 1/r to make the potential energy linear in ⇢.Withthis ⌘ substitution in place, our energy expression becomes

1 ⇢˙2 J 2 H = m + z ⇢2 + U(⇢)=E. (6.15) 2 ⇢4 m2 ✓ ◆ Finally, we want a geometric description of the trajectory – we don’t particularly care how long it takes the particle to move from one place to another, we’re more interested in the shape of its motion. To obtain a geometric description, re-parametrize in terms of instead of t –wehave:

d⇢ Jz 2 ⇢˙ = ˙ = ⇢0 ⇢ (6.16) d m using ˙ = Jz ⇢2 and defining ⇢ d⇢ .Now m 0 ⌘ d 2 1 Jz 2 2 H = ⇢0 + ⇢ + U(⇢)=E, (6.17) 2 m and we can take the derivative of both sides to get rid of the constant E, then solve for ⇢00:

m GMm2 ⇢00 = ⇢ 2 U 0(⇢)= ⇢ + 2 , (6.18) Jz Jz and GMm2 ⇢()=A cos()+B sin()+ 2 . (6.19) Jz 192 CHAPTER 6. GRAVITY

We don’t need to keep both the sine and cosine terms, either one suces, so take GMm2 ⇢ = 2 + A cos . (6.20) Jz Then the radius, as a function of angle, is

1 J 2 1 p r = = z (6.21) ⇢ GMm2 1+e cos ⌘ 1+e cos

2 2 where e A Jz is a new constant and p Jz .Fore =0,this ⌘ GMm2 ⌘ GMm2 6 expression for r() describes an ellipse with “eccentricity” e and “semi-latus rectum” p. The closest point to the central body (situated at the origin) occurs when =0, and is called the “perihelion” (technically, that’s the closest point in an orbit about the sun, as the name suggests), and the furthest point from the central body is at = ⇡, called the “aphelion”. The details of the orbit provide those values, and from them we can extract information about the central body.

6.3.2 Kepler’s Laws Kepler’s first law is that the motion of bodies around the sun is described by elliptical orbits, and we have demonstrated this directly with the solution (6.21). Kepler’s second law says that equal areas are swept out in equal times, and we dA can establish this by showing that dt = constant for area dA swept out in time dt. Since we’re working infinitesimally, we can approximate the motion along the ellipse, in time dt, as a straight line with length rdas shown in Figure 6.2. 1 The area of the region is approximately dA = 2 (rd) r, appropriate for a

yˆ

rd ⇠

r d xˆ

Figure 6.2: A particle goes through a small angle d in a time dt traveling along an elliptical orbit. 6.3. PREDICTIONS: NON-RELATIVISTIC 193 triangle – then the time-derivative is

dA 1 = r2 ˙. (6.22) dt 2

2 We know that the angular momentum is constant, with Jz = mr ˙,sothat dA 1 J = z (6.23) dt 2 m a constant. Conclusion: equal area is swept out by the orbiting body in equal time everywhere along its trajectory. As for Kepler’s third law, that is a statement about the relation between the period of the motion and the elliptical geometry. We want to find the time it takes an orbiting body to go all the way around once – we could get that by integrating dt, and to relate this integral in time to an integral in the geometrical parameter go back to the angular momentum relation:

d J J (1 + e cos )2 = z = z . (6.24) dt mr2 p2 m

Then we can write dt in terms of d

p2 md dt = 2 . (6.25) Jz (1 + e cos )

Integrating all the way around, from =0to 2 ⇡ gives the total period

2 ⇡ p2 md p2 m 2 ⇡ J 3 2 ⇡ T = = = z 2 3/2 2 3 2 3/2 J (1 + e cos ) Jz 2 G m M 2 Z0 z (1 e ) (1 e ) (6.26) Problem 6.4 For the ellipse written in the form (6.21), what is the value of the perihelion and aphelion in terms of p and e? The “major” axis of the ellipse is its total horizontal extent, and the “minor” axis is the total vertical extent. Write these in terms of p and e.

Problem 6.5 Write Kepler’s third law (6.26) in terms of the semi-major (half the major axis) axis of the ellipse (and the mass of the central body, plus any constants that remain). 194 CHAPTER 6. GRAVITY

6.3.3 Tidal Forces One of the implications of the equality of all of the inertial, passive and active masses is the carefully orchestrated stretching and squeezing of extended bodies as they approach a central mass. Suppose you were falling towards a spherically symmetric central object (the sun, for example) – di↵erent parts of your body would experience acceleration of di↵erent magnitudes. Your feet accelerate more than your head (if you’re going in feet first), so you would get stretched. You would also get squeezed – physically, all the pieces of your body are falling in directly towards the center of the sun, and so your left and right sides would have to get closer together. We can quantify the di↵erence in acceleration from top to bottom and left to right – as is typical, we will model our body as a rectangle of width w and height h (our depth doesn’t matter for the calculation, by symmetry). The setup, for a central body of mass M (assume spherical) is shown in Figure 6.3, with the accelerations of interest marked. The di↵erence in acceleration between the top and bottom is

1 1 2 GMh a a = GM yˆ yˆ (6.27) t b (r + h)2 r2 ⇡ r3  where we have assumed that h r in the approximation above. If the object is ⌧ to fall rigidly, there must be internal force to oppose this acceleration discrepancy from the top to the bottom – those internal forces are provided by your body, up to a point . . . For the “squeezing” portion, we’ll look at the di↵erence between the left and right accelerations:

2 GM GMw a a = sin ✓ yˆ = yˆ r ` 2 3/2 r2 + w 2 w 2 2 r + 2 (6.28) ⇣GMw ⌘ ⇣ ⌘ yˆ, ⇡ r3 and once again we’d have to oppose this di↵erence in accelerations in order to avoid a relative squeeze.

6.3.4 Cosmology Cosmology is the study of the large-scale structure of the universe from a grav- itational point of view. That gravity is the dominant force over large scales (spatial and temporal) is clear from its relative weakness. If, for example, there 6.3. PREDICTIONS: NON-RELATIVISTIC 195

w

at

h

ab a ar

r ✓

zˆ

yˆ

M

Figure 6.3: A massive body of width w and height h falls towards a spherical central body of mass M (at the origin) – we want to find the di↵erence in accelerations from top to bottom and left to right. 196 CHAPTER 6. GRAVITY was a pocket of charge lying around in the universe, it would “quickly” (relative to gravity) attract opposite charge and neutralize. Following a line of reasoning from Newton’s time, we assume that the uni- verse is infinite in extent (if the universe were made of a finite set of masses, it would collapse on itself and therefore have a finite lifetime). This immediately leads to problems – assuming the universe is uniformly filled with light-emitting stu↵(stars), and is of infinite extent means that light should be everywhere – the darkness of the night sky is a statement of Olbers’ paradox – if stars are all around emitting light, the night sky should be (infinitely) bright. There are ways around this – maybe the universe has large pockets of stars, and then mass that does not emit light, so that the night sky is really just the lack of light coming from the “empty” (of starlight) pieces. Another option is that the universe is finitely old, and the lack of light comes from the lack of stu↵prior to some point in the past. Newtonian gravity can be used to make predictions about the dynamics of the universe, provided we make some reasonable assumptions. We’ll assume that the universe is everywhere “the same”, so that there are no preferred points or directions – a preferred point is clearly out, we want mass distributed uniformly without clumping (the universe, then, has a spatially constant mass density ⇢), that’s what we see (at the large scale) in our vicinity, and there’s no reason to imagine other locations are significantly di↵erent. No preferred directions means that there’s not, for example, an electric field pointing in one direction throughout the universe. Finally, there’s the observation that the galaxies in the universe are moving radially away from each other – we need to encorporate that somehow. The typical model is points on a balloon – as the balloon inflates, the points get further away from one another – each galaxy sees all the others receding. We’ll start with a simple model, in which at time t,theuniversehasuniformdensity ⇢(t), and generate the function that describes the balloon’s “inflation” as a function of time. For a sphere of uniform density ⇢, we know that a particle of mass m located adistancer from the center of the sphere experiences a force due to all the mass enclosed by the sphere, while the mass at points >rdoes not contribute, that’s the “shell” theorem familiar from Newtonian gravity (and E&M, common to all 1/r2 forces). At time t, the total mass enclosed by a sphere of radius r is 4 3 M = 3 ⇡r ⇢(t). For a particle of mass m at r = r ˆr, the total energy is 1 GMm 1 4 E = m r˙ r˙ = m r˙ r˙ Gm⇡r2 ⇢. (6.29) 2 · r 2 · 3 The particle of mass m can either move in or out, in this scenario, so that its 6.3. PREDICTIONS: NON-RELATIVISTIC 197 position at time t can be thought of as a function r(t)=S(t) w where S(t0)=1 for a particle with initial vector location w (and note that this assumed form e↵ects all particles at radius r at time t in the same way). So in general, the energy (which is, of course, constant) is given by 1 4 E = mw2 S˙ 2 ⇡Gmw2 ⇢S2, (6.30) 2 3 and this is typically rearranged to read

2 S˙(t) 2 E 8 2 E 8 = 2 2 + ⇡G⇢(t)= 2 + ⇡G⇢(t), (6.31) S(t)! mw S(t) 3 mr(t) 3 which is known as Friedman’s equation. We write the equation in terms of S/S˙ because of the interpretation of this quantity. Suppose we have two points on the sphere, r1 = S(t) w1 and r2 = S(t) w2 (where w1 and w2 are the locations at time t0)–thevectorthat points from 2 to 1 is given by: r S(t)(w w ). The magnitude of this 12 ⌘ 1 2 vector is the distance between the two points at time t.Thenr˙ 12 gives the relative velocity of the two points on the sphere, and we can relate it to the distance vector at time t via:

1 1 S˙(t) r˙ = S˙(t) r r = r , (6.32) 12 S(t) 1 S(t) 2 S(t) 12 ✓ ◆ so that S/S˙ is the relation between the relative speed of the two points and their distance apart at time t. This combination is known as Hubble’s constant, and as the t-dependence suggests, it is not necessarily a constant. One feature of Friedman’s equation is that there is no way to get S˙ =0 – the universe must be expanding or contracting, it cannot be static. This was problematic (before people realized that the universe can’t be static) – yet there is no natural fix. One way to “fudge” the original (6.30) is to introduce an additional potential energy term, but this new term must be quadratic in r – the energy E which is constant in time must go like r2 so that (6.31) has no r dependence (the left-hand-side is r-independent, so the right-hand side must be as well), and every term in E goes like r2, as can be seen in (6.30). That means the only thing you could add would be something like ⇤ r2 =⇤w2 S(t)2 for constant ⇤, and this is the so-called “cosmological constant” – it corresponds to a new force, linear in r, that acts in addition to gravity. Start with 1 4 1 E = mw2 S˙ 2 ⇡Gmw2 ⇢S2 + ⇤ w2 S2, (6.33) 2 3 6 198 CHAPTER 6. GRAVITY then you get an updated form of the Friedman equation: 2 S˙(t) 2 E 8 1 ⇤ = 2 + ⇡G⇢(t)+ . (6.34) S(t)! mr(t) 3 3 m Problem 6.6 Suppose the Hubble constant increases linearly with time, H = ↵t,whatis the density ⇢(t) in that case? What if ⇤ =0? 6

Problem 6.7 Show that S˙ =0is allowed in the Friedmann equation with cosmological constant.

6.4 Predictions: Relativistic

There are a variety of interesting predictions that we can make by introducing a “little” special relativity, and some of these are bourne out in the full theory of gravity. The following requires a slight suspension of disbelief, but it’s worth it for building physical intuition.

6.4.1 Event Horizon When we define the escape speed of an object in a potential energy landscape, we use the minimum speed necessary for the object to get to spatial infinity – that generally means we want the speed to be zero at spatial infinity. For example, if a mass m blasts o↵from the surface of a spherically symmetric central body of mass M and radius R at speed vesc, the total energy is 1 GMm E = mv2 . (6.35) 2 esc R At spatial infinity, the potential energy term is zero, and we assume the kinetic term is also zero (the mass is at rest out there as in Figure 6.4), so E =0and we can solve (6.35) for vesc: 2 GM vesc = . (6.36) r R We can turn the escape speed problem around and ask: For what spherical radius R is vesc the escape speed? Then 2 GM R = 2 . (6.37) vesc 6.4. PREDICTIONS: RELATIVISTIC 199

m at rest

1

vesc

m

M M R R

Figure 6.4: The initial configuration on the left: a mass m leaves the surface of a spherically symmetric body of mass M and radius R with speed vesc.On the right: the mass escapes to spatial infinity where it is at rest.

And finally: For what R is c the escape speed? For massive spheres of this radius or less, then, even light would not escape the gravitational attraction of the central body. The answer, although it is flawed (can you spot the hole?) is

2 GM R = . (6.38) c2

This radius, first predicted by Laplace, represents the “event horizon” of a massive body – if all of the mass of a spherical body is squeezed beneath this radius, light cannot escape from the surface of the central body. This is an example of a “black hole”, and is both qualitatively and quantitatively correct1, the result from general relativity is identical (there, this radius is known as the “Schwarzschild radius”, and it shows up in a very di↵erent way). Problem 6.8 What is the Schwarzschild radius for the Sun? For the Earth?

6.4.2 Bending of Light

Light will respond to a gravitational field – this makes sense, light has energy associated with it, and so has some e↵ective “mass” which could be used to

1This is unfortunate, ironically – I would prefer if the result was qualitatively correct, but was o↵by a factor of 2 or 4,asmostpseudo-classicalresultsingravityare. 200 CHAPTER 6. GRAVITY interact with a gravitational field. Start with the general solution to the grav- itational equations of motion, written in ⇢() form as in (6.19). There was nothing explicitly non-luminal about that motion’s setup, so we’ll start with it, assuming motion in the x y plane as before. The setup is shown in Figure 6.5. Take M =0for a moment, then we know the motion must be a straight line,

2 ✓ ⌘ v = ✓ v xˆ r ✓v M = ✓v R

Figure 6.5: The motion of a particle near a central body – the solid line is an approximate trajectory, the dashed line is the trajectory if there was no central mass, and the dotted lines define the straight lines along which we perceive the motion of light. and we’ll arrange that line to graze the central body right at R (corresponding to the dashed line in Figure 6.5), in this case, we can set A =0and B =1/R in (6.19) (so that r =1/⇢ = R when M =0there). If we wrote out the x and y coordinates of this motion, we’d have x = R cot y= R (6.39) which does indeed describe a line a distance R from the x-axis – the cot parameter goes from to as goes from ⇡ to 0 while y is fixed at R. 1 1 Using this M =0description of the line in (6.19), we have 1 GMm2 ⇢()= sin + 2 . (6.40) R Jz Suppose we are at spatial infinity in the x direction so that ⇢ =1/ =0, 1 and the massive object arrives at our location making an angle ✓v as shown in Figure 6.5, that angle would be defined by 2 2 1 GMm 1 GMm R 0= sin( ✓ )+ ✓ =sin . (6.41) R v J 2 ! v J 2 z ✓ z ◆ 6.4. PREDICTIONS: RELATIVISTIC 201

We can define the “deflection angle”, , for the motion – that is the di↵erence between the actual starting point of the trajectory and the straight-line defined by our viewing angle (the deflection angle is then =2✓v as shown in Fig- ure 6.5) – this is a strange quantity for actual particle motion since we have no reason to assume the particle travelled in a straight line. But for light, we always assume constant velocity vector with speed c,so2 ✓v is the di↵erence between the perceived location of the particle (of light) and the actual location. So far, there has been no reference to the type of particle we have – it is with the definition of the constant Jz that we can actually tailor our solution to refer to light – at the = ⇡/2 grazing point, all of the motion is in the ˆ 2 direction, and the light is traveling at speed c,soJz = mR ˙ = mRc,and this must be its value at all times. We can put this into our deflection angle expression 1 GM GM =2✓ =2sin 2 , (6.42) v Rc2 ⇡ Rc2 ✓ ◆ where the approximation comes from assuming that GM/(Rc2) 1. ⌧ The point of all this is that we would say that light moves along trajectories that are not straight lines in the presence of Newtonian gravity. The defining characteristic of light, here, is that it travels at c (the fact that it must be massless spoils the derivation). This result again holds in general relativity, although the current prediction for the deflection angle is o↵by a factor of 2 (thank goodness!). There are all sorts of other interesting motions that light can execute – circular motion is an option, although you’d never see it (why?). Problem 6.9 Is the approximation GM/(Rc2) 1 justified for the Sun? What is the ⌧ deflection angle in this case?

Problem 6.10 Find the radius at which light can orbit in a circle around a (spherical) central body of mass M.

6.4.3 Energy Coupling and Charge Moving along to some more drastic modifications that are motivated by special relativity, suppose we go back and re-imagine the original field equation, g = 4 ⇡G⇢, (6.43) r· for ⇢ a mass density. Special relativity provides a relation between mass and energy, so that if you had a source of energy density u, you could write the 202 CHAPTER 6. GRAVITY e↵ective mass density of it as ⇢ = u/c2, and that energy density would show up in the field equation u g = 4 ⇡G . (6.44) r· c2 Consider a point charge with mass M and charge Q –theelectricfieldhas 1 2 energy density u = 2 ✏0 E , so for a point source’s field, the energy density is Q2 u = 2 4 . (6.45) 32 ⇡ ✏0 r We assume that the gravitational field is spherically symmetric, g(r)=g(r) ˆr, and then we have to solve the ODE, from (6.44)

1 GQ2 2 0 2 r g(r) = 2 4 , (6.46) r 8 ⇡✏0 c r which gives GQ2 g = 2 3 ˆr. (6.47) 8 ⇡✏0 c r Note that this is just the contribution due to the extended energy source, we must also include the massive point-source (with g =0away from the origin) r· piece, so the full field is really

GM GQ2 g = + ˆr. (6.48) r2 8 ⇡✏ c2 r3 ✓ 0 ◆ It is interesting that in theory you could detect the charge (up to sign) of a body by measuring its gravitational force – take a neutral mass m moving with speed v in a circular orbit around a point source at radius R – the charge of the central body is given by

Rv2 Q = 8 ⇡✏ c2 r M + . (6.49) ± 0 G s ✓ ◆ Of course, this presumes we know the mass M of the source, otherwise we have no way to separate the e↵ective mass contribution Rv2/G from M. Problem 6.11 What is the gravitational potential associated with (6.48)? What would the classical event horizon look like here?

Problem 6.12 6.4. PREDICTIONS: RELATIVISTIC 203

Just as the electric field stores energy, the gravitational field does as well – find the expression for u g g. ⇠ ·

Problem 6.13 How long would it take a test particle of mass m to fall from rest at R to 1 2 R moving under the influence of (6.48)?

6.4.4 Perihelion Precession

We know, from Bertrand’s theorem, that closed stable orbits can only come from 1/r or r2 potentials – the potential associated with (6.48) has an extra 1/r2 piece. But it is easy to get potential contributions that are of the form 1/rn from less exotic considerations – the multipole moments of a central body start with 1/r (monopole), then have cos ✓/r2 (dipole) etc. so by including contributions from the various multipole moments, we can get di↵erent radial (and angular) dependence.

m1

r1 r12

m2 r M 2

Figure 6.6: A central body of mass M and an orbiting body of mass m2 act gravitationally on m .Wecantaker r and expand to describe the e↵ect 1 1 2 of m2 perturbatively.

Another way to get perturbing 1/rn contributions is by considering the interaction of masses orbiting a central body. As an example, suppose we have a spherically symmetric central body of mass M at the origin, and then a mass m1 at r1 and m2 at r2. The force on the m1 mass is

GMm1 Gm1 m2 F1 = 2 ˆr 2 ˆr12 (6.50) r1 r12 204 CHAPTER 6. GRAVITY where r r r . The potential energy that this force comes from is 12 ⌘ 1 2

GMm1 Gm1 m2 V1 = , (6.51) r1 r12 and the second term can be expanded for r r (as depicted in Figure 6.6) 1 2 to read Gm m 1 r cos ✓ 1 2 Gm m + 2 + ... (6.52) r ⇡ 1 2 r r2 12 ✓ 1 1 ◆ where ✓ is the angle between r1 and r2. This approach treats the masses M and m2 as the “source”, and the expansion above represents the monopole and dipole contributions of m2. We can consider a small perturbing potential (from any of these mecha- nisms) on top of the dominant central body potential. We have a test particle of mass m moving with potential energy described by

GMm U(r)= + V (r) (6.53) r where V (r) GMm for all r. Going back to the -parametrized equation of ⌧ r motion for ⇢ 1/r that we developed in (6.18), we have ⌘ m GMm2 m ⇢00 = ⇢ 2 U 0(⇢)= ⇢ + 2 2 V 0(⇢). (6.54) Jz Jz Jz

For V =0, we get back (6.21), the original elliptical orbits. With the perturbing potential in place, the ellipse description changes. Take ⇢ = ⇢0 +✏⇢1() for constant ⇢0 (we are perturbing about a circular orbit), then let V (⇢ + ✏⇢ ) V (⇢ )+✏V (⇢ ) ⇢ – we can insert this into (6.54) and 0 1 ⇡ 0 0 0 1 collect in powers of ✏:

0 1 m ✏ :0= ⇢ + V 0(⇢ ) 0 p J 2 0 z (6.55) 1 m ✏ : ⇢00 = ⇢ 1+ V 00(⇢ ) . 1 1 J 2 0 ✓ z ◆

The first equation can, in principle, be solved for the constant ⇢0 –andthen the second equation is solved by

m ⇢ = A cos 1+ V (⇢ ) (6.56) 1 J 2 00 0 ✓r z ◆ 6.4. PREDICTIONS: RELATIVISTIC 205 again keeping just one of the two (cosine and sine) solutions. Then our full solution looks like

m ⇢ = ⇢ + A✏ cos 1+ V (⇢ ) (6.57) 0 J 2 00 0 ✓r z ◆ where ⇢0 is just a constant. This gives a radius that is of the form

1 r = = (6.58) ⇢ m 1+˜✏ cos 1+ 2 V 00(⇢0) Jz ✓q ◆ for constants and ✏˜.

Our solution would be an ellipse if V 00(⇢0)=0, and this form of r() does define an elliptical trajectory of sorts – look at the point of closest approach, though – at =0, we start o↵at 1+˜✏ , but we don’t return to this location until m 1+ V (⇢ ) =2⇡, (6.59) J 2 00 0 r z and that gives

2 ⇡ m = 2 ⇡ 1 V 00(⇢0) 2 ⇡ (6.60) m ⇡ 2 J 2 ⌘ 1+ 2 V 00(⇢0) z Jz ✓ ◆ q where the approximation comes from assuming that V 00(⇢0) is small. We return, then, to the same closest approach distance, but not at =2⇡,insteadit’sa little less than that (assuming V (⇢ ) > 0), at 2 ⇡ with defined in (6.60). 00 0 This motion of the perihelion is known as “precession”, and it occurs for any potential with V (⇢ ) =0.Ateachpass,wemaintainanellipticalshape,but 00 0 6 pick up an additional deficit of .Thepicture,forthefirsttwoorbits,isshown in Figure 6.7. The story can be complicated further by considering the angular dependence of perturbing potentials, i.e. additional terms of the form V (⇢,✓),buteven this simplest case makes the point. In particular, you could use the measured precession to pin down the charge of a central body as in the previous section. Problem 6.14 Use the e↵ective potential from (6.48) to calculate the deficit in (6.60). 206 CHAPTER 6. GRAVITY

yˆ

xˆ

Figure 6.7: A particle starts on the xˆ axis and moves around in an ellipse, returning to the point of closest approach (drawn as a circle of radius p/(1 e)) at an angle 2 ⇡ ,asshown.

6.5 Issues

We have looked at some of the physical predictions of Newtonian gravity – they are not so di↵erent qualitatively from the predictions of general relativity, espe- cially once we introduce a little special relativity. That introduction, however, brings up some immediate issues with the form of Newtonian gravity.

6.5.1 Lorentz Covariant Form

Our immediate objection is with the form of the field equation – working with the potential, 2 =4⇡G⇢has 2 which is not a Lorentz scalar – the fix is easy, r r though, we just move to 2 @ @µ, and the static limit is recovered trivially. ⌘ µ Similarly, we know how to update Newton’s second law so as to work with special relativity. Our field equation and equation of motion for the potential of Newtonian gravity now read

2 =4⇡G⇢

d m x˙ (6.61) = m . dt 2 x˙ x˙ 3 r 1 ·2 c 4q 5 6.5. ISSUES 207

The field equation now comes, correctly, from a scalar Lagrange density that looks like 1 1 = @ @µ + ⇢. (6.62) L 4 ⇡G 2 µ There is still a relatively serious relativistic problem, though – think back to the infinite line of charge from E&M described in Section 1.10, and let’s tell the story in the context of gravity. In a lab frame, we have an infinite line of mass moving with constant speed v, and with constant mass-per-unit-length = where is the rest mass density of the line and (1 v2/c2) 1/2 v 0 0 v ⌘ – the setup is shown in Figure 6.8. The force on the test particle is

yˆ v

d

xˆ u m

Figure 6.8: An infinite line of mass moves with constant speed v –itexertsa force on a test mass moving with speed u

2 G m F = 0 yˆ. (6.63) v d

If we analyze the problem in the test particle’s rest frame, as in Figure 6.9, where its charge density is ¯ = w¯ 0, we’d get a force

2 G m F¯ = 0 yˆ. (6.64) w¯ d The problem is that according to the transformation law for forces in special

¯ = w¯ 0 v u yˆ w¯ = 1 uv c2 d

xˆ m

Figure 6.9: Infinite line of mass viewed in the test particle’s rest frame. 208 CHAPTER 6. GRAVITY relativity from (1.77), we must have

1 F = F¯ (6.65) u as the relation between the force in the lab frame and the force in the rest frame. Instead, we have: 1 F = v F¯ = F¯. (6.66) w¯ 6 u What’s missing from the gravitational picture is, of course, a magnetic coun- terpart. This argument implies the existence of a gravitational “magnetic” field and force that mimics the usual magnetic one. Working from the rest frame force, and using the correct transformation (6.65), we must have

w¯ 2 G0 m 2 Gm uv F = yˆ = 1 2 yˆ (6.67) u d d c ⇣ ⌘ (using (1.87)) in the lab (not (6.63)). We could get this type of force if there was a gravitomagnetic field h with h = 4 ⇡G/c2 J for mass-current r⇥ density J = ⇢ v and if this gravitomagnetic force acted on moving masses via F = m v h. h ⇥ Problem 6.15 Find h for the infinite line of mass.

Problem 6.16 Introduce a potential for h, and combine with the potential for Newtonian gravity to form a four-vector with as the zero component, call it Aµ and ⇠ µ write its field equation. Given how this must couple to particles (as Aµ x˙ ), find the equation of motion.

Once h is on the scene, we have an augmented Newtonian theory of gravity that looks just like Maxwell’s equations with a few sign changes. Again, this result, that gravity looks like E&M is a prediction of general relativity. It is precisely the “weak-field” regime of linearized gravity, structurally identical to Maxwell’s equations, and all of the physics you get from those carries over. That means that wrong-signed E&M provides great intuition for the weak-field (far away from sources) regime of gravity – close enough to be di↵erent from Newtonian gravity, but far enough away to avoid the full general relativistic description. 6.5. ISSUES 209

There is a fundamental di↵erence between the sources in E&M and gravity, one that spoils the above story slightly (but also provides a provocative hint about the general relativistic fix). When we have a moving line of charge, we have to account for the length contraction of special relativity, so that if a line of charge has constant 0 when at rest, it has constant = 0 when moving. That’s not quite right gravitationally, though – for an infinite line of mass with constant 0, when the line of mass is moving, there are two e↵ects: length contraction, and also the energy (which is really the source here) picks up a factor of . We’ll come back to this, but file it away for now, it’s an indication that this “wrong-signed” E&M approximation can’t be the end of the story.

6.5.2 Self Sourcing There is a major component of gravity missing both from the Newtonian theory and its naive relativistic (vector) extension. We saw how the energy stored in the electric field could act as a source for gravity, but what about the energy stored in the gravitational field itself? That should be counted, too. The first pass, which involves taking the energy required to assemble a configuration of mass, gives an energy density of u = 1 g g or, in terms of the gravitational 8 ⇡G · potential, u = 1 . These negative energy densities are easy to 8 ⇡G r ·r understand in terms of the sign of the work we do in building, say, a spherical shell of charge (we have to push away from the partially built configuration as it attracts mass), but it’s harder to think about as a gravitational source where it is akin to negative mass sources. Still, if we simply plow ahead – given mass density ⇢, we would write the (static) field equation as

1 2 =4⇡G⇢ . (6.68) r 2 c2 r ·r

Problem 6.17 What is the point source solution to (6.68)? How about a spherical shell of radius R and mass M spread uniformly over its surface? Does the “shell” theorem hold in the latter case (i.e. is the field inside the shell zero)?

The bigger problem, if you go back to the derivation of the work required to build a configuration of mass, is that we assume that the field equation itself is just 2 =4⇡G⇢in developing the expression for u, and this is no longer the r field equation. You can generate an infinite chain and sum it to find a consistent expression for the energy required to build a configuration, but even that falls 210 CHAPTER 6. GRAVITY short since it still does not really count the field energy. We know how to get that in a field theory – it’s the 00 component of the stress tensor for the theory. Suppose we start with the free relativistic Lagrangian, for arbitrary function f(): 1 f() = @ ⌘µ⌫ @ . (6.69) L 2 4 ⇡G µ ⌫ The stress tensor is given by f() 1 T µ⌫ = µ ⌫ ⌘µ⌫ ↵ , (6.70) 4 ⇡G , , 2 , ,↵  and the 00 component, for static fields (with ˙ =0), is

1 f() T 00 = = (6.71) 2 4 ⇡Gr ·r L so that the Lagrange density is the energy density here. In order to require self-consistent self-coupling, our target will be a vacuum 2 T 00 field equation of the form: =4⇡G 2 =4⇡G L2 . The field equation, r c c from the Lagrangian (6.69) (and again staying with static fields) is

2 1 f() + f 0() =0 (6.72) r 2 r ·r and we could write this as 1 f () 2 = 0 . (6.73) r 2 f() r ·r

We want the right-hand side to equal 4 ⇡GcL2 , giving us an ODE for f(): 1 f () 1 0 = f() (6.74) 2 f() r ·r 2 c2 r ·r

c2 with solution f()= . The field equation, in vacuum, is then 1 2 = , (6.75) r 2 r ·r or if we introduce material sources with mass density ⇢ (or energetic ones other than gravity), the field equation can be written 2 ⇡G 2 = ⇢ . (6.76) r c2 ⇣p ⌘ p 6.5. ISSUES 211

Problem 6.18 Use the fixed form of f() in the Lagrange density (6.69) with external sources, ⇢, to find the field equation that updates (6.75) to include sources. Show that this is equivalent to the compact (and linear!) form (6.76).

Problem 6.19 Find the point source solution for a mass M in the self-coupled theory (6.76).

Problem 6.20

A sphere of radius R has constant ⇢0 inside, find the interior and exterior solutions to (6.76).

Problem 6.21 What is the energy stored in the fields of a point particle in the self-coupled case?

6.5.3 Universal Coupling We’ll conclude by thinking about what the full theory of gravity (as it is currently known) must look like given the deficiencies described in the previous subsection. Our first question is what type of field theory (for the gravitational “potential”) we should be investigating. Newtonian gravity starts as a scalar field, then we expanded the gravitational force to include a gravitomagnetic component, based on special relativity. The resulting theory of gravity is a vector field theory, like E&M. Finally, we considered self-coupling of gravity back in its scalar form. Each has flaws, but which, if any of these, is the correct starting point? Let’s think about sources first – in E&M, the source is charge and moving µ µ charge, combined in a current J which is conserved @µ J =0(a statement of charge conservation). For gravity, the source is energy and moving energy at the very least – what we want, then, is a structure that combines energy and moving energy, and is conserved. The natural object is the stress-energy tensor, T µ⌫,sinceT 00 is the energy density and T 0i the “moving energy” density. As an added bonus, we know that every physical theory has a conserved stress- energy tensor owing to the equivalence of physics in all coordinate systems (the definition and conservation are developed in Section 5.4), making it an ideal candidate for the universal coupling of gravity (every form of mass/energy µ⌫ µ⌫ sources and responds to gravity). So T is always available, and @µ T =0 212 CHAPTER 6. GRAVITY is its conservation statement. Back to the form of gravity – can a scalar be made out of T µ⌫?Sure,the µ⌫ µ µ⌫ trace of T is a scalar: T T µ = T ⌘ . In a sense, that was the motivation ⌘ µ⌫ for making a self-consistent scalar theory of gravity. In the previous section, we coupled the field to its energy density, T 00. But for static fields, the trace of the stress tensor is directly proportional to T 00. The problem with a scalar theory is that, at least in four-dimensional space-time, E&M has T =0,meaningthata scalar theory of gravity would have no way of coupling to E&M. We know that electromagnetic fields must source gravity – you might be tempted to use just u from the electric field as we did in the static case, but that would not lead to field equations that are generally covariant (meaning that they have the same form in all coordinate systems), and there would be preferred frames – if you started with T 00 2 =4⇡G , (6.77) c2 where T 00 is the energy density from E&M, then a Lorentz boost would change the value of T 00 while leaving the rest of the equation unchanged. So you’d have a preferred reference frame in which to evaluate T 00, and all others would pick up factors of 2. That is not in keeping with the first postulate of special relativity – the component T 00 is not a scalar. So a pure scalar field is out, how about vector? It’s not hard to see that there is no vector component of T µ⌫, no way to massage the coordinates so as to generate an object that transforms with one factor of the Lorentz boost (for example). The single-indexed T 0⌫, which appears to be available, does not µ transform correctly. Consider a boost with matrix entries L ⌫,thestresstensor transforms as ¯µ⌫ ↵ µ ⌫ T = T L ↵ L , (6.78) and then T¯0⌫ = T ↵ L0 L⌫ = T 0 L⌫ (6.79) ↵ 6 where the inequality at the end is what you would need (as equality) to make a vector field theory like: T 0⌫ 2h⌫ =4⇡G (6.80) c2 generally covariant. Evidently, the most natural description of a theory of gravity is as a symmet- ric second rank tensor, one that couples to the complete stress-tensor, something like: 4 ⇡G 2hµ⌫ = T µ⌫, (6.81) c2 6.5. ISSUES 213 or if we wanted to consider a dimensionless field (hµ⌫ has dimension of speed2),

hµ⌫ 4 ⇡G 2 = T µ⌫. (6.82) c2 c4 ✓ ◆ The field hµ⌫ is a symmetric (since T µ⌫ is), second rank tensor. This will do it, in the end, once we take the field hµ⌫/c2 and couple it to its own stress-tensor, just as we did with the scalar field above. With that necessary augmentation, we will recover general relativity, including its geometric interpretation. Problem 6.22 µ⌫ Using the field equation (6.82), argue that you must have @µ h =0.Take ↵ ik↵ x hµ⌫ = Pµ⌫ e and find the constraints placed on kµ and Pµ⌫ (both are made up of constants) such that the vacuum form of (6.82). How many elements do you get to specify in Pµ⌫ (this gives the number of independent degrees of µ⌫ freedom)? Be careful, even after @µ h =0is set, there is additional gauge freedom as in Section 7.5.1.

The interpretation of hµ⌫ comes from its use in particle actions – for E&M, we argued that a particle Lagrangian, written in proper time parametrization, µ should couple to the electromagnetic field Aµ via a scalar of the form x˙ Aµ. Given the symmetric tensor field hµ⌫, the natural analogue is a scalar term of µ ⌫ the form x˙ hµ⌫ x˙ . But that term will combine with the “usual” kinetic term – for a relativistic Lagrangian, we can start with: 1 L = m x˙ µ ⌘ x˙ ⌫ + ↵ x˙ µ h x˙ ⌫, (6.83) 2 µ⌫ µ⌫ where the first term involves the Minkowski metric, and ↵ is a constant that sets the coupling strength. In this form, the story is the normal one – we have afieldhµ⌫ that acts on the particle. We could rewrite the Lagrangian as 1 L = x˙ µ (m⌘ + ↵h )˙x⌫, (6.84) 2 µ⌫ µ⌫ and in this form, we could consider the particle to be moving freely (no potential) in a background metric with g m⌘ + ↵h . In a sense, any second rank µ⌫ ⌘ µ⌫ µ⌫ field theory has the potential for this interpretation. 214 CHAPTER 6. GRAVITY