3.4 notes on notation!

a c a a partial derivative ∂A /∂x = ∂cA = A ,c (I won’t use the latter as its too easy to lose a comma!) a a covariant derivative ∇cA = A ;c (I will use the semicolon notation because physicists tend to freeze when they see del!!) a a absolute derivative ∇sA = DA /ds a a Christoffel symbols (or connection coefficients) Γ bc or n bc o or {a, bc}

3.5 Example: 2D flat space

The metric for flat space in cartesian coordinates gAB = diag(1, 1) DOES NOT DEPEND ON POSITION. So the partial derivatives of the metric are A ZERO. so Γ BC = 0. BUT this is not true if we’d done it in terms of polars, ds2 = dx2 + dy2 = 2 2 2 2 dr + r dθ where gAB = diag(1,r ). Here the derivaties are NOT zero. So Christoffel symbols are like the metric - they do tell us about curvature (what we are interested in) but they also tell us about what coordinate system we have chosen (which isn’t at all fundamental). so if we are going to bother calculating them, we may as well choose a real curved surface to play with.

3.6 Surface of a sphere before we worked out the metric components for flat space ds2 = dr2 + r2dθ2 + r2 sin2 θdφ2 we can describe a sphere just by saying we are at fixed radius r = a so dr = 0. so ds2 = a2dθ2 + a2 sin2 θdφ2 2 2 2 i.e. gAB = diag(a , a sin θ) Christoffel symbols are defined as 1 Γf = gfb(∂ g + ∂ g − ∂ g ) ca 2 c ab a bc b ca AB AB so we need to also find the covariant metric components g from g gBC = A δC . In general this leads to a set of simultaneous equations to solve AB A1 A2 A g gBC = g g1C + g g2C = δC

1 BUT for diagonal metrics we don’t have to do a matrix inverse to solve these! AB AA in a diagonal metric all cross terms are zero so we get g gBA = g gAA =1 AA or g =1/gAA ONLY FOR DIAGONAL METRICS!

gAB = diag[1/a2, 1/(a2 sin2 θ)]

So in our 2D space we have

θ θB Γθθ =1/2g (∂θgθB + ∂θgBθ − ∂Bgθθ) as gθB = 0 except for B = θ then this collapses to

θ θθ 2 Γθθ =1/2g (∂θgθθ + ∂θgθθ − ∂θgθθ)=1/2 . 1/a . 0=0

θ θ θB Γθφ =Γφθ =1/2g (∂φgθB + ∂θgBφ − ∂Bgθφ) again gθB is only nonzero for B = θ since the metric is diagonal so

θ θ θθ 2 Γθφ =Γθφ =1/2g (∂φgθθ + ∂θgθθ − ∂θgθφ)=1/2 . 1/a . 0=0

θ θB 2 Γφφ =1/2g (∂φgφB + ∂φgBφ − ∂Bgφφ)=1/2 . 1/a (0+0 − ∂θgφφ)

=1/2 . 1/a2(−2a2 sin θ cos θ)= − sin θ cos θ φ φB φφ Γθθ =1/2g (∂θgθB + ∂θgBθ − ∂Bgθθ)=1/2g (∂θgθφ + ∂θgθφ − ∂φgθθ)=0 φ φ φB φφ Γθφ =Γφθ =1/2g (∂φgθB + ∂θgBφ − ∂B gθφ)=1/2g (∂φgθφ + ∂θgφφ − ∂φgθφ) 2 2 2 2 2 =1/2 1/(a sin θ)∂θgφφ =1/2 1/(a sin θ)2a sin θ cos θ = cot θ φ φB Γφφ =1/2g (∂φgφφ + ∂φgφφ − ∂φgφφ)=0

3.7 Geodesic equations And now we have everything we need to get the geodesic paths (inertial frames). If we were differential geometers, then we would define a geodesic as the shortest distance between two points. But we are phycisics so we can use the alternative definition that these are inertial frames. and in an inertial frame the velocity DOES NOT CHANGE - there are no forces to prodsce α α an acceleration. So velocity v = v eα = dx /dτeα If there is no change in

2 this then its derivative is zero but we also saw that there can be swings in a vector which arise from curved space. In other words we have to cast this in tensor language and say its the ABSOLUTE derivative which is zero in order to simply parallel transport the velocity vector. So the components can change like

dvα dxβ dxγ +Γα =0 dτ βγ dτ dτ d dxα dxβ dxγ +Γα =0 dτ dτ βγ dτ dτ d2xα dxβ dxγ +Γα =0 dτ 2 βγ dτ dτ or using notation where dot means derivative w.r.t. τ

α α β γ x¨ +Γ βγ x˙ x˙ =0

This works for any parameter u linearly related to path length s i.e. for u = A + Bs. Then the affinely parameterized geodesic equation in an N dimensional manifold is d2xa dxb dxc +Γa =0 ds2 bc ds ds or in slighly cleaner notation, where dot denotes derivative w.r.t. s

a a b c x¨ +Γ bcx˙ x˙ =0

So now we know how to do geodesic paths, as well as derivatives.

3.8 Example: Geodesics in flat space

i we saw that in flat space all the Γ jk = 0 so the geodesic equations becomes

x¨a =0 were dot denotes derivative wrt path length s. integrate to getx ˙ a = A and xa = As + B so geodsics in flat space have constant velocity and direction - Newtonian inertial frame (constant motion, straight line!)

3 3.9 Example: Geodesics on a sphere in general its too hard to do geodesics in full generality. so often we just choose a path and see if its a geodesic! and a natural one to choose is a path defined by only one of the parameters - its called a parameter curve. so on our sphere, we know that for a sphere the only non-zero christoffel θ φ φ symbols are Γφφ = − sin θ cos θ, and Γθφ =Γφθ = cot θ. Geodesic paths satisfy the equation

dxA dxB dxC +ΓA =0 ds2 BC ds ds suppose the path s is just a change in θ then s = aθ so there is no dependance on φ i.e. x2 = φconstant so dφ/ds = 0. While for θ we have dθ/ds = 1/a and d2θ/ds2 = 0. For θ then

d2θ dxB dxC d2θ dφ dφ +Γθ = +Γθ =0 ds2 BC ds ds ds2 φφ ds ds so this looks good. Check also that our condition for φ holds

d2φ dxB dxC d2φ dθ dφ dφ dθ +Γφ = +Γφ +Γφ =0 ds2 BC ds ds ds2 θφ ds ds φθ ds ds so all such paths are geodesics. - these are GREAT CIRCLES.

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