Physics 480/581. Homework No. 10 Solutions: due Friday, 19 October, 2018

Transcription

1 Physics 480/58 Homework No. 0 Solutions: due Friday, 9 October, 208. Using the coordinate bases for -forms, and their reciprocal bases for tangent vectors, and the usual form of the Schwarzschild metric, namely g = J(r)dr 2 + r 2 (dθ 2 + sin 2 θdφ 2 ) H(r)dt 2, calculate the connections in this case, which are also called the Christöffel symbols, { α, β γ which are symmetric in their two lower indices, and are generated by the formula involving the first derivatives of the components of the metric tensor. [5 pts] We first recall the formula for the connections in a coordinate basis: { µ ν β 2 gµα ( g νβ,α + g αν,β + g αβ,ν ), where we also recall, or note from the formula, that they are symmetric in the two lower indices, so that we only have to compute at most 40 of them. We begin with the ones with upper index µ = r: { r ν β 2 grr ( g νβ,r + g rν,β + g rβ,ν ). Remembering that all the non-zero elements of the metric are along the diagonal, we can quickly see that only these connection coefficients where ν and β are equal can be non-zero, we proceed: { r ( ) = H2 r r 2 H 2 = m/r2,r H 2 = m/r2 2m/r ; { r = H2 θ θ 2 ( g θθ),r = r( 2m/r) = 2m r ; { r = H2 φ φ 2 ( g φφ),r = 2r 2 sin 2 θ( 2m/r) = (2m r) sin 2 θ ; { r = H2 t t 2 ( g tt),r = (m/r 2 )( 2m/r).

2 Now we consider the next set, where we choose µ = θ: { θ ν β 2 gθθ ( g νβ,θ + g θν,β + g θβ,ν ). This is clearly only non-zero when nu and β are r or φ, giving us { θ r θ 2r 2 (g θθ),r r, { θ = φ φ 2r 2 (g φφ),θ = sin θ cos θ. Proceeding onwards now to the cases where µ = φ, we note that no metric components depend on φ, so that we may write { φ ν β 2 gφφ (g νφ,β + g βφ,ν ). As g φφ depends on both r and θ, we acquire non-zero terms for { φ = r φ { φ = θ φ 2r 2 sin 2 θ (g φφ),r r ; 2r 2 sin 2 θ (g φφ),θ = cot θ. Lastly we consider the cases where µ = t, again remembering that nothing depends on t: which only allows one non-zero case: { t ν β 2H 2 (g tν,β + g tβ,ν ), { t t r 2H 2 (g tt),r = m/r r 2m, which constitutes some 3 which are non-zero remembering the symmetry on the two lower indices. 2. Killing s equations determine the generators, called Killing vectors, for transformations of the metric that leave it invariant. There are more than one way to determine Killing vectors; however, 2

3 for the moment we will use the following requirement, which are Prof. W. Killing s original equations: K (µ;ν) = 0, which says that the symmetric sum of the covariant derivative, in an arbitrary direction, of the components, K µ, of a Killing vector must vanish. [Do notice that this requirement constitutes some 0 independent equations, all of which must be verified. And also note that K µ = g µν K ν, i.e., the equation considers the components of the -form version of the Killing vector.] It is a mathematical fact that it is much easier to show that an object is a Killing vector if we study its metric and its components in a (holonomic) coordinate-based basis for tangent vectors. Therefore, please use the Schwarzschild metric and its standard coordinate-based basis for tangent vectors, { r, θ, φ, t, and show that the following 2 tangent vectors are indeed independent Killing vectors: K 4 = t, K = φ, I note that this metric actually has a total of 4 Killing vectors. Can you guess what are the other two? You will need to have the Christöffel symbols for the Schwarzschild metric in order to work in this mode; [0 pts] The equations to be verified are 0 in number for each alleged Killing vector; therefore, let s give these equations a name, so that we can keep track of them in the general case, and then spell them out, inserting the Christöffel symbols as above, where we use the known symmetry of the Christöffel symbols in the two lower indices to justify the 2 below, and I note that the sum over λ, in the term with the Christöffel symbols never has more than 2 entries, and that only once: { λ W µν K µ;ν + K ν;µ = K µ,ν + K ν,µ 2 µ ν { r 2 W rr = K r,r r r K λ ; K r = K r,r + m/r2 2m/r K r, { θ W rθ = K r,θ + K θ,r 2 K θ = K r,θ + K θ,r 2 r θ r K θ, 3

4 { φ W rφ = K r,φ + K φ,r 2 r φ K φ = K r,φ + K φ,r 2 r K φ, { t W rt = K t,r + K r,t 2 K t = K t,r 2m/r2 r t 2m/r K t, { r 2 W θθ = K θ,θ K r = K θ,θ + (r 2m)K r, { θ θ φ W θφ = K θ,φ + K φ,θ 2 K φ = K θ,φ + K φ,θ 2 cot θk φ, θ φ { r 2 W φφ = K φ,φ K r φ φ W θt = K θ,t + K t,θ, K θ = K φ,φ + (r 2m) sin 2 θk r + sin θ cos θk θ, { θ φ φ W φt = K φ,t + K t,φ, { r 2 W tt = K t,t K r = K t,t m/r2 t t 2m/r K r. a. We then begin with, hopefully, the simplest one: K4 = t, which we will refer to as just K to simplify the notation. It is clear that K µ = δ µ t, which tells us that K µ = g µν K ν = g µt = g tt δ t µ = ( 2m/r)δ t µ. We may then calculate the 0 equations above. Since only K t is different from zero, we immediately see that only those W µν can possibly be non-zero when one of µ or ν equals t; we show those 4 below: K t = ( 2m/r), W rt = K t,r 2m/r2 2m/r K t = 2m/r 2 + 2m/r 2 = 0, { r W θt = K t,θ = 0, W φt = K t,φ = 0 W tt = K t,t 0 = 0. t t Proceeding on to the next alleged Killing vector, we consider K = φ, which we will refer to as just L, as before to simplify the notation. It is clear that K µ = δ µ φ, which tells us that L µ = g µν L ν = g µν δ ν φ = g µφ = r 2 sin 2 θδ φ µ. Since only L φ is different from zero, and it is independent of both t and φ, we see that, again, many of the requirement zero s are just identities, leaving behind for us to check the following ones: L φ = (r sin θ) 2, { φ W rφ = L φ,r 2 L φ = 2r sin 2 θ 2 φ r r (r sin θ)2 = 0, { φ W θφ = L φ,θ 2 K φ = 2r 2 sin θ cos θ 2 cot θ(r sin θ) 2 = 0, θ φ 4

5 c. Since the solution in question has spherical symmetry, one should expect the entire angular momentum vector to be a generator for symmetries. Furthermore, / φ is the sphericalcoordinate presentation of the ẑ-component of angular momentum; therefore, it is reasonable that the other two Killing vectors are the other two components of the angular momentum, i.e., L x and L y, which, preferably, should be written out in spherical coordinates, although I do not do so here. 3. Consider the conformal transformations that were recently discussed in the problem session. Define new coordinates (R, T, θ, φ) via the following equations: tan p v t + r, tan q w t r. Since we know that r is never negative, this allows us to see that 0 2r = v w = v w. a. First show that there is a single scalar factor, which we will call Φ, such that the metric, in polar coordinates in 4-dimensional, flat Minkowski space may be written as g = Φ 2 g = Φ 2 ( 4 dp dq + sin 2 (p q) dω 2), dω 2 dθ 2 + sin 2 θ dφ 2. b. Next, show that both p and q are allowed to vary only in the region [ 2 π, + 2π], but that, also, we must always have p q, and then define yet another pair of variables, (R, T ): R p q [0, π], T p + q [ π, +π], and re-write the metric in terms of these variables, {R, θ, φ, T. Since this is simply flat, 4-dimensional space of special relativity, we know that radial lightrays have paths that are at 45 degree angles in {r, t-space. What do they look like in {R, T -space? [0 pts]

6 a. We need dr 2 and dt 2 to compute the metric in these new coordinates: 2r = tan p tan q = 2dr = sec 2 p dp sec 2 q dq, 2t = tan p + tan q = 2dt = sec 2 dp + sec 2 q dq. Therefore the relevant part of the flat Minkowski metric gives us dr 2 dt 2 { [sec 2 p dp sec 2 q dq] 2 [sec 2 p dp + sec 2 q dq] 2 dp dq = 4 cos 2 p cos 2 q 4Φ2 dp dq. However, for the angular portion we need just r 2 4 (tan p tan q)2, but we would like to see it with the overall conformal factor, Ω, factored out, so we re-write this r 2 multiplied by the inverse of that factor: 4 cos2 p cos 2 q(tan p + tan q) 2 ( 4 sin p cos q sin q cos p) 2 = sin 2 (p q) ). The result is then ( ) 2 g = dr 2 + r 2 dω 2 dt 2 { = 4 dp dq + sin 2 (p q) dω 2, 2 cos p cos q which agrees with the requested form, with Φ 2 cos p cos q. b. To determine the allowed ranges of p and q we ignore or hold fixed the two spherical angles. Then, if we were to allow both r and t to vary over all possible real numbers, then clearly t ± r also vary over that same range. Therefore, in principle, both p and q, being arctangents, vary from π/2 to +π/2, so that our entire manifold now fits into a rectangle of that size, where we have changed the metric by ignoring the conformal factor, Φ 2 in front of the metric; i.e., we are dealing with a conformally-transformed metric. NOTE to grader: the discussion below is meant to be educational, BUT is surely much more detailed than a proper response to the question needs to be Since we must always have p q, when q = π/2, we can have all the allowed values of p; 6

7 however, as the values of q increase, the allowed values of p become fewer. For instance, when q = 0, p may now only vary from 0 to +π/2, and, lastly, when q = +π/2, p has only one allowed value, namely +π/2. This sort of constraint gives us a triangle inside of our previously-allowed rectangle. If we draw that triangle with p horizontal and q vertical, then the triangle in question is the triangle below the line drawn from the lower-left corner where p = π/2 = q to the upper-right corner where p = +π/2 = q. However, as both p and q are obviously null coordinates, i.e., paths in Minkowski space with one of them constant and the other one varying are null rays, it is much more common to draw them on a diagram so that they make angles of ±45 with the horizontal and/or vertical. Therefore, we define the quantities R and T, as given in the problem, and draw all this on an axis where R is horizontal and T vertical. In this case, since T = p+q, then it must vary from the value in that lower-left corner, namely T = π there, to the value at the upper-right corner, namely T = +π. So, now let us rotate the above rectangle, and its included triangle, so that the axis labelled T is vertical. This causes lines of constant values of either p or q to be at ±45 to the vertical. As well, the variable R can never have a value that is negative, since p is never less than q, so that R varies from 0 to π along the horizontal axis. The now-rotated triangle, in this view, has one side as the complete T -axis, while the lower side is the line q = π/2, which runs from the corner T = π, R = 0 to the farthest right corner, with T = 0, R = π. The third side of our triangle is the line p = +π/2, which can be seen to begin at the very top of our figure, at the point T = +π, R = 0, and crossing over at a 45 direction down to meet the other side at that corner, where T = 0, R = +π. One can see this overall situation in the figure included just below, which has labels for some of the various parts that we have been discussing: 7

8 The left-hand side, vertical axis is the line r = 0 = R, with t = ± at the two ends. The horizontal axis through the middle of the triangle is the line t = 0 = T, with r = 0 at the left end and r = at the right end, at the corner. i.) The top end of the triangle (or the top of that T -axis), with p = +π/2 = q is usually referred to as future timelike infinity, and often denoted by the symbol i +. ii.) Analogously the bottom of the T -axis, or the bottom end of the triangle, with p = π/2 = q, is referred to as past timelike infinity, and denoted by the symbol i. iii.) Then the right-hand end of the horizontal axis in the middle, i.e., the R-axis, with p = +π/2 = q, is usually referred to as spatial infinity, and denoted by the symbol i 0. iv.) The line labeled q = 0 is a null ray, ending on the upper, right-hand boundary of the triangle, which has p = +π/2. Other parallel null rays begin at t 0, with a different value of q along them. They end on that same upper, right-hand boundary of the triangle, which we refer to as future null infinity, and use symbol I + to denote it. v.) The line labeled p = 0 is also a null ray, coming in toward r = 0 = t. It came from a point on the lower, right-hand boundary of the triangle, with q = π/2 and p = 0. For a different, fixed p the null ray would be parallel and arrive at r = 0 at some other value for t. Therefore all incoming light rays originate on that boundary of the triangle, and we refer to it as past, null infinity, denoted by I. 8