x α x β g α β = xα x β g αβ. (1.1)

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1 Physics 445 Solution for homework 4 Fall Cornell University NOTE We use the notion where four-vectors v are denoted by an arrow, and three-vectors v will be in bold. Hartle uses the opposite notation. NOTE We use units where c =. I. EXISTENCE OF LOCAL LOENTZ FAMES The transformation law for the metric components has been derived in lecture g α β Near x α =, the point P, our two coordinate systems are related by = xα x β g αβ..) x α x β x α = Λ α β xβ + Mα β γ xβ x γ + O[x ) 3 ]..) Equation.), which defines the tensor M, has the property that any piece of M which is antisymmetric in the indices β and γ does not contribute. So, without loss of generality, we can assume that M is symmetric on that pair of indices. The derivatives in Eq..) are then given by x α x α = Λ α α + xγ M α α γ + O[x ) ].3) where we have used x α,β = δα β, and Mα β γ = Mα γ β. We now substitute this and the metric expansion near P g αβ x µ ) = g αβ + g αβ,γx γ + O[x) ] = g αβ + g αβ,γλ γ γ xγ + O[x ) ],.4) into Eq..) to find the expected result written in a slightly different form then on the problem set) ) g α β = Λα α g αβλ β β + Λ α α g αβm β γ β + gαβm α α γ Λβ β + Λ α α Λβ β Λ γ γ g αβ,γ x γ + O[x ) ]..5) Now define matrices L, G, and N such that for each matrix, the element in row a and column b is given by L) ab = Λ a b, G ) ab = g ab, N) ab = η ab,.6) where η αβ is the Minkowski metric. The the requirement that the first term in Eq..5) is the Minkowski metric can now be written as a matrix equation L T G L = N,.7) where T is the transpose operator: A T ) ab = A ba, for any matrix A. One can show as follows that a matrix L satisfying Eq..7) can be found. First, since G is a symmetric real matrix, it has a complete set of orthonormal eigenvectors v i and real eigenvalues g i for i 4. We define the matrix L = v, v, v 3, v 4 ), i.e. the columns of L are the eigenvectors. Then it is easy to see that the similarity transform of G by L is a diagonal matrix: L T G L = diagg,g,g 3,g 4 )..8) Next, we define L to be diagd,d,d 3,d 4 ) where d i = / d i if g i and d i = if g i =, and we define L = L L. Under the combined similarity transformation we find L T G L = L T L T G L L = diagsgng ),sgng ),sgng 3 ),sgng 4 )),.9) where sgnx) = if x >, if x =, and if x =.

2 The number each of s, s and s that occur in the final diagonal form is a property of the original matrix that is invariant under similarity transformations. In general relativity the metric is assumed to be non-degenerate the determinant of the metric is non-vanishing), so there are no zeros. It is also a postulate of general relativity that the metric is in the class of matrices which has exactly one and three + s; this is necessary in order that general relativity reduces to special relativity locally. With this assumption, it is easy to find a coordinate transformation that makes the occur in the first position, so the final result is the Minkowski metric diag,,,). Now consider the order x γ term in Eq..5). We want to choose the components of M so that this term vanishes. First define matrices M γ, Z γ, and S γ M γ ) ab = M a bγ, Z γ ) ab = G M γ ) ab = g aξm ξ bγ, S γ ) ab = Λ α aλ β b Λγ γ g αβ,γ..) so that the vanishing requirement can be written as the following matrix equation L T Z γ + Zγ T L = S γ..) Inserting the suggested solution Z γ = L T ) S γ /, where the superscript denotes an inverse matrix, gives S γ + S γ [ L T ) ] T L = Sγ,.) which is an equality since [L T ) ] T = L. So we have shown that the suggested solution is indeed a solution. Then since Z γ = G M γ, we have M γ = which determines M α β γ as a function of g αβ and g αβ,γ. Now Eq..5) takes the form G ) L T ) Sγ,.3) g α β = η α β + O[xµ ) ]..4) This is simply an expansion of the metric g α β near P, similar to Eq..4). Since there is no order xγ term, the first derivatives g α β,γ vanish at P, and the primed frame is a local Lorentz frame. Note that equations.7) and.3) could now be used in Eq..) to determine the local Lorentz coordinates x α strictly in terms of the original coordinates x α, the original components of the local metric gαβ, and the derivatives g αβ,γ. II. ALTENATIVE ACTION PINCIPLE FO TIMELIKE GEODESICS Consider the functional S[x µ τ)] = dτ g αβ x µ )ẋ α ẋ β = dτlx µ,ẋ µ ),.) where an overdot represents d/dτ. We want to determine the function x µ τ) which extremizes S. This function will satisfy the Euler-Lagrange equation From Eq..) we have L,λ = g αβ,λ ẋ α ẋ β, L,λ = d L dτ ẋ λ..) L ẋ λ = g λµẋ µ, where we have used g αβ = g βα, and x α,β = δα β. Substituting into Eq..) then gives Contracting with g λσ gives g λµ ẍ µ = d L dτ ẋ λ = g λµẍ µ + g λµ,γ ẋ γ ẋ µ,.3) gαβ,λ ẋ α ẋ β g λµ,γ ẋ γ ẋ µ)..4) ẍ σ = gλσ g αβ,λ ẋ α ẋ β g λµ,γ ẋ γ ẋ µ)..5)

3 3 To clean up a bit, rename dummy indices µ α and γ β in the second term on the right hand side ẍ σ = gλσ g αβ,λ g λα,β ) ẋ α ẋ β..6) Now contract the known expression for the connection coefficients [see Eq. 8.9) in Hartle] with ẋ α ẋ β, Γ σ αβẋ α ẋ β = gλσ g αβ,λ + g λα,β + g βλ,α ) ẋ α ẋ β = gλσ g αβ,λ + g λα,β ) ẋ α ẋ β,.7) where in the last term we have switched dummy indices α β and used g αβ = g βα. We then recognize Eq..6) as the geodesic equation ẍ σ = Γ σ αβẋ α ẋ β..8) III. HATLE CHAPTE 6, POBLEM 4 We will calculate the proper time along worldlines near the Earth in the static weak field Newtonian) limit. In this limit, the line element is [see Eq. 6.) in Hartle, using c = ] ds = [ + Φx)]dt + [ Φx)]dx + dy + dz ), 3.) where Φx) = GM /x + y + z ) is usual Newtonian potential for the the the Earth s gravitational field. The proper time between events A and B is then given by τ AB = B A dτ = B A ds = B A dt [ + Φ) Φ)v ] P / = dt + Φ v + Φv ) /, 3.) where v = dx/dt) + dy/dt) + dz/dt), and where the events A and B will always be such that tb) ta) = P. In the limit where v and Φ, we have P τ AB = P + dt Φ ) v + OΦ ) + Ov 4 ) + OΦv ). 3.3) A. For a circular worldline at constant radius with period P, both Φ = GM /, and v = π/p), are constant. Using Eq. 3.3) the proper time from A to B along this worldline is given by τab c GM = P + π ) P P dt = P π P GM ). 3.4) Kepler s law is valid in the Newtonian limit. For circular Earth orbits this law takes the form P = 4π 3 /GM ), see Example 3. on page 4 in Hartle. Substituting this into Eq. 3.4) gives τab c = P 3 ) GM 3.5) In the Newtonian limit, this worldline is a geodesic and must therefore be a path of extremal proper time. B. For a stationary observer at constant position x + y + z =, once again both Φ = GM /, and v =, are constant. The proper time from A to B along this worldline is given by Eq. 3.3) τab s = P GM P dt = P GM ). 3.6) This worldline is not a geodesic. It might be a bit of a surprise then that τ c AB < τs AB. However the circular orbit, which is a geodesic, need not be a global maximum of the proper time integral. It need only be an extrema.

4 4 C. By definition, the proper time vanishes along any photon worldline, or null curve. The proper time from A to B along this worldline is then τ γ AB =. 3.7) Note that Eq. 3.3) cannot be used in this case since the assumption v is invalid. It is likely that the word photon was a typo in the text, and should have been replaced with particle. That is, Hartle might have meant for us to consider a particle that goes radially outward with an initial velocity that is a little bit less than the escape velocity, chosen so that the particle goes out and returns back to the staring point in a time P. For a radially moving particle that travels radially outwards to a maximum distance r = h and falls back to the starting point r =, we start from the conservation of energy ) dr GM = GM dt r h. 3.8) By the symmetry of the problem, it is sufficient to consider one leg of the particle s path and multiply by two. We will consider the return trip when the particle falls from rest from r = h back to the starting point at r =. Then, we have from Eq. 3.8) and solving for dt gives for the time t for the entire trip h3/ t = GM λ dr dt = GM r ) /, 3.9) h [ ] x λ dx x = h3/ λ) + GM cos λ ), 3.) where x = r/h and λ = /h. We now solve for λ by equating the time taken to the period P: and numerically solving this equation gives π = λ λ + λ 3/ cos λ ) 3.) λ = h ) The elapsed proper time can be found by using Eq. 3.8) in Eq. 3.3): P [ τab r = P + GM dt r + ) ] h r P GM ).55GM P.65 GM = P + GM h P + GM h λ dx x x ) / ). 3.3) D. The worldline of a particle on any eccentric periodic orbit which includes the events A and B) with period P, and other worldlines of particles which move only radially are also extrema of the proper time integral. Although there is a theorem of uniqueness for geodesics between any two nearby events, in this case the events A and B are sufficiently separated that the extrema of the proper time integral are not unique. That is, the extrema are local rather than global. If you enjoyed this problem, you might also want to try problem in chapter 6 of Hartle.

5 5 Consider the spacetime with line element IV. HATLE CHAPTE 7, POBLEM 5 ds = xdv + dvdx. 4.) A. We can find the null geodesics by demanding that ds =. This condition is satisfied by lines of constant v for which dv =, and also by solutions of the differential equation dv/dx = /x, or rather dx/dv = x/, which have the form where the initial position x is an integration constant. xv) = x e v/ 4.) B. Some representative null curves of the form 4.) are shown in Fig v x FIG. : Null curves for the spacetime with line element given by Eq. 4.). The shape of the light cone at any point is given by intersecting null curves at that point. C. A timelike geodesic must have ds < or dx dv < x. 4.3)

6 In other words, starting at some event P, for a given small future directed positive) change in v, the small change in x along any timelike geodesic originating at P must always be less than the associated small change in x along a null geodesic originating at P. We can draw such curves in Fig. ) as follows: Start at some point on one of the curved null geodesics. Move a little bit along that curve in the positive v direction, making sure to stay to the left of the original null curve. In this way, we can draw curves which cross from positive x to negative x, however we can t draw any which cross from negative x to positive x. 6 V. HATLE CHAPTE 8, POBLEM We want to find the timelike geodesics for the indler spacetime with line element ds = x dt +dx. We follow the method described in problem II. That is, we will find the functions x α τ) = [tτ),xτ)], which extremizes the action dτ g αβ ẋ α ẋ β = dτ x ṫ + ẋ ) = dτ L, 5.) using the Euler-Lagrange equation.). Note that we are again using an overdot to denote a derivative with respect to proper time τ. To use the Euler-Lagrange equation, we will need L,λ =, xṫ ), Since L, =, we know L/ ṫ is constant. The equation of motion for ṫ is then L ẋ λ = x ṫ,ẋ). 5.) ṫ = E x, 5.3) where E is constant. Note that we could have also found this result by noting that t is a Killing vector, which means t ẋ = E, is constant. The Euler-Lagrange equation also gives ẍ = xṫ, or rather ẍ = E x 3, if we use Eq. 5.3). However, there is an easy way to get a first order equation. ecall that d x/dτ) = g αβ ẋ α ẋ β =. This requirement, called 4-velocity normalization, gives x ṫ + ẋ =, which can be simplified using Eq. 5.3) ẋ = E x. 5.4) We are only looking for the shape xt) of the geodesics. We can use ẋ = ṫdx/dt to combine Eqs. 5.4) and 5.3) into a single differential equation for xt) dx dt = ± x x 4 E. 5.5) If we define x = ±z/e we have dz/dt = z z 4. The solutions are zt) = secht + C), or xt) = ±E secht + C), 5.6) where C is a constant related to the initial position x) = ±E sechc). You may be surprised that our solutions are determined by only two integration constants: E and C. After all, the geodesic equations give a second order equation for each dimension, and there are two dimensions. However 4-velocity normalization, which is not guaranteed by the geodesic equation, always eliminates one constant. We eliminated another constant by asking only for the shape xt) of the geodesics, rather than the full parametrized solution x α τ). In Newtonian mechanics, we sometimes also eliminate a constant by seeking only the shape of an orbit, however there is no analog to 4-velocity normalization. Lastly note that, as was suggested in problem II, it would be a simple matter to find the connection coefficients from the equations of motion. Comparing ẍ = xṫ to Eq..8) gives Γ x tt = x, Γ x tx = Γ x xt = Γ x xx =. 5.7) Similarly, differentiating E yields ẗ = /x)ṫẋ, which when compared to Eq..8) gives Γ t tx = Γ t xt = x, Γt tt = Γ t xx =. 5.8) See problem set 5.