Solved Problems on Sound Waves - Final Exam | PHYS 691, Exams of Quantum Physics

Download Solved Problems on Sound Waves - Final Exam | PHYS 691 and more Exams Quantum Physics in PDF only on Docsity! PHYS691 Final Exam Attempt each of the following problems. Attach the resulting …le to an email to rhgowdy@vcu.edu. Due date: Thursday, May 12, 2005. 1 Problem 1: Sound Waves Use the stress-energy tensor conservation laws to …nd the speed of sound waves (as a fraction of the speed of light) in a medium that obeys an equation of state of the form p = f () Do the calculation for an arbitrary curved spacetime. 1.1 Answer: (‡at spacetime version 1) First get the basic idea by doing the problem out in components in ‡at Minkowski spacetime. The conservation law is then T; = 0 or, split into time and space parts T 00;0 + T 0m ;m = 0 Tm0;0 + T mn ;n = 0 Assume a ‡uid with isotropic stress and p = f () T 00 =  Tmn = f () gmn = f () nm Since we cannot have the coordinates follow the ‡uid (they are …xed) we have to allow the ‡uid to move in order to have sound waves. Thus, there must be small non-zero components T 0m = Tm0 = jm so that the conservation law becomes ;0 + j m ;m = 0 jm;0 + ( mnf ());n = 0 But (mnf ());n = @ @xm f () = df d @ @xm = f 0;m 1 and the second equation becomes jm;0 + f 0;m = 0 The essential trick is to eliminate the mass-energy ‡ow variables jm by taking the spatial divergence of this last result jm;0m + f 0;mm + f 0 ;m;m = 0 jm;0m + f 0;mm + f ";m;m = 0 and comparing that to the time derivative of the …rst conservation equation ;00 + j m ;m0 = 0 Subtract the equations and obtain jm;0m + f 0;mm + f ";m;m ;00 jm;m0 = 0 or f 0;mm ;00 + f ";m;m = 0 or @ 2 @t2 + f 0r2+ f "  ~r 2 = 0 Compare this equation to the wave equation with propagation velocity v @ 2 @t2 + v2r2 = 0 The signal propagation characteristics of the equation are determined by its second derivative terms, so the sound-speed is v = s df d = s dp d 1.2 Answer: (curved spacetime version) The straightforward approach is to replace commas by semicolons in the version 1 calculation above, thus introducing a mess of connection coe¢ cients. The coe¢ cients, but not their derivatives, can be made to go away by assuming a local Lorentz Frame. The remaining extra terms do not a¤ect the second derivatives of , so we get the same sound speed result as before. 2 Problem 2: Bosons in Curved Spacetime In Special Relativity, the wave function for a spin-zero massive particle obeys the Klein Gordon Equation @ 2 @t2 + @2 @x2 + @2 @y2 + @2 @z2 = m2 a) Suppose that such a particle is moving through a curved spacetime and use minimal coupling to …nd a candidate for its wave equation. 2 The di¤erential equations satis…ed by the functions ~X are obtained from the requirement A = 0 which is evaluated just as for the string action but with greek indexes being summed just from 1 to 2. A = 1 2 p jgHH jgHH  @ ~X @x
@ ~X @x ! = p jgHH jgHH @ ~X @x
@ ~X @x ! = @ @x p jgHH jgHH @ ~X @x !
 ~X ! @ @x p jgHH jgHH @ ~X @x !
 ~X so that A = Z Z dxdy ( @ @x p jgHH jgHH @ ~X @x !
 ~X ! @ @x p jgHH jgHH @ ~X @x !
 ~X ) Use Green’s Theorem on the total divergence term:Z Z dxdy ( @ @x p jgHH jgHH @ ~X @x !
 ~X !) = I d`n p jgHH jgHH @ ~X @x !
 ~X where d` is the line element along the wire boundary and n is the outward directed normal. So long as the variation is held …xed at the boundary,  ~X = 0 and this term vanishes. The condition is thenZ Z dxdy ( @ @x p jgHH jgHH @ ~X @x !
 ~X ) = 0 for arbitrary  ~X or @ @x p jgHH jgHH @ ~X @x ! = 0 An equivalent form of this expression is just (2)r2 ~X = 0 where (2)r2 is the covariant Laplacian on the surface. The famous result of the Plateau problem follows by noticing that this equation is also the condition that X;Y; Z are each real or imaginary parts of analytic functions of x + iy. Thus, we can construct soap bubble …lms from triplets of analytic functions. As 5 a result, the entire problem is solved exactly. For example, choose the analytic functions, X = x; Y = y; Z = Re  (x+ iy) 2  = x2 y2 and obtain a "saddle" shaped …lm of extremal area. Similarly, X = x; Y = y; Z = Re  (x+ iy) 3  describes an extremal area surface that is sometimes called a "monkey saddle". This expression is already enough for this part, but we will need a bit more for the next part of the problem. Notice that a variation that satis…es the constraint H ~X = 0 will not change the surface. It will only change the coordinates x; y on the surface. Thus the area will not change under such a variation. Thus, the following equation is an identity:Z Z dxdy ( @ @x p jgHH jgHH @ ~X @x !
H ~X ) = 0 But that is the same asZ Z dxdy ( H @ @x p jgHH jgHH @ ~X @x !
 ~X ) = 0 Since  ~X is arbitrary, we have the identity H @ @x p jgHH jgHH @ ~X @x ! = 0 so that the equation that constrains the surface can also be written as V @ @x p jgHH jgHH @ ~X @x ! = 0 Use Leibniz’s product rule to obtain V ( @ @x p jgHH jgHH  @ ~X @x + p jgHH jgHH @2 ~X @x @x + ) = 0 But the vectors @ ~X@x lie in the surface, so V @ ~X @x = 0 and we get the equation for the soap …lm surface in the simple form V gHH @2 ~X @x @x = 0: c) Find the condition(s) satis…ed by the second fundamental form of such a soap …lm. 6 3.3 Answer c) The straightforward way to do this problem is to take the expression for the projection curvature tensor from the notes hH c da = Hjd @Xj @x gg  @2Xp @x@x @Xk @x VpaH c k This expression has way too many uncontracted indexes to compare with the di¤erential equations that we obtained in part b. Those equations had just one Cartesian index. The only obvious way to get rid of indexes is to contract them and the only way that does not give zero is to contract the …rst two (the third is projected in the complementary direction). hHa = hH c ca = Hjc @Xj @x gg  @2Xp @x@x @Xk @x VpaH c k = Hjk @Xj @x gg  @2Xp @x@x @Xk @x Vpa = @Xj @x Hjk @Xk @x gg  @2Xp @x@x Vpa = @ ~X @x
@ ~X @x gg  @2Xp @x@x Vpa = g g  @ 2Xp @x@x Vpa =  g  @ 2Xp @x@x Vpa = g @2Xp @x @x Vpa Now we recognize the result of the previous problem and see that the condition on the second fundamental form is just that it be trace-free: hHa = 0. 4 Problem 4: Gravitational Wave Sources a) Use the results found in class, but ignore polarization e¤ects and derive an approximate relationship between detector strain, source luminosity, source distance, and source frequency. For this part, just leave everything in Planck units. 4.1 Answer a) The detector strain is given by the expression hjk x0; xi
= 2 r P jrP k s Irs x0 r
7 where v; P are polynomials P = v2 + (1 2m=r) r4 v = Kr3=3H By the way, this is the metric of a black hole of mass m in peculiar coordi- nates. 5.1 Answer The metric components have the form [g] = 2664 1 (1 2m=r) vP1=2 0 0 vP1=2 r4P1 0 0 0 0 r2 0 0 0 0 r2 sin2  3775 while the inverse metric components are  g1  = 2664 1 (1 2m=r) vP1=2 0 0 vP1=2 r4P1 0 0 0 0 r2 0 0 0 0 r2 sin2  3775 1 or :  g1  = 26664 r4r4+v2+2mr3 P v p Pr4+ p Pv2+2 p Pmr3 0 0 P v p Pr4+ p Pv2+2 p Pmr3 2PmPr r5+2mr4+rv2 0 0 0 0 1r2 0 0 0 0 1 r2 sin2  37775 Compare these expressions to the ones given in the notes: [g] = 0@ ~N 2 N2 h ~NT h  ~N  [h] 1A  g1  =  1=N2 ~NT =N2 ~N=N2 [h] 1 ~N ~NT =N2  The spacelike metric is evidently [h] = 24 r4P1 0 00 r2 0 0 0 r2 sin2  35 and we can read o¤ the shift vector components with their indexes lowered by this metric: Ni = g0i 10 or N1 = vP 1=2 N2 = N3 = 0 Raise the index using h1 N1 = r4PvP1=2 = v r4 p P N2 = N3 = 0 The lapse function is obtained by comparing the inverse metric expressions 1=N2 = r4 r4 + v2 + 2mr3 = 1 v2r4 (1 2m=r) N = p v2r4 (1 2m=r) = r2 p v2 (1 2m=r) r4 Collect the …nal non-zero results in the form N1 = v r4 p v2 + (1 2m=r) r4 N = r2 p v2 (1 2m=r) r4 The important thing to notice about these results is that they are well-behaved at r = 2m. Thus, these t =constant surfaces are regular across the black hole event horizon and t is a well-behaved time coordinate there. One might not have guessed that from the original form of the metric tensor. The example is a static, regular slicing of a black hole by a set of hyperbolic constant-time surfaces that are asymptotically lightlike. 6 Problem 6: Initial Data Suppose that you wish to set up time-symmetric initial data for two black holes of identical mass separated by about ten Schwarzschild radii. The data is to be set up on a Cartesian coordinate grid (x; y; z) with the holes on the z-axis. For a single black hole the horizon corresponds to the minimal area r = constant surface at the instant of time symmetry. Assume that this relationship is approximately true for these interacting black holes so that their minimal area surfaces (now somewhat distorted) correspond to their horizons and give the spacetime Cartesian metric tensor components as functions of the coordinates (x; y; z). 6.1 Answer The spatial metric tensor is taken to be gij =  4ij 11 where a single black hole with mass m would be represented by  = 1 + 2m ~r with ~r the radius in terms of the ‡at space metric ~r = p x2 + y2 + z2 The area at constant ~r is just A (~r) = 4  1 + 2m ~r 4 ~r2 which goes through a minimum at ~r = 2m so that is the location of the "throat" in these coordinates. For two identical black holes, we would have  = 1 + 2m ~r1 + 2m ~r2 where ~r1 and ~r2 are distances from di¤erent points calculated using the ‡at metric. The simplest starting assumption to make is that these distance are calculated from points a distance 20m apart in the ‡at metric. Put one at z = 10m and the other at z = 10m along the z-axis so that ~r1 = q x2 + y2 + (z + 10m) 2 ~r2 = q x2 + y2 + (z 10m)2 and thus the proposed initial metric is gij = 0@1 + 2mq x2 + y2 + (z + 10m) 2 + 2mq x2 + y2 + (z 10m)2 1A4 ij Of course, the ‡at metric does not measure physical distances, so we still need to check what the actual distance between these black holes is. Along the z-axis, ds2 = 0@1 + 2mq (z + 10m) 2 + 2mq (z 10m)2 1A4 dz2 so that the distance element is ds =  1 + 2m jz + 10mj + 2m jz 10mj 2 dz 12 8.1 Answer a) The picture is actually in the notes. The initial singularity is the bottom branch of the hyperbola. It is replaced by the spacetime geometry inside the star. Thus, the initial singularity is not actually present in this spacetime. b) Suppose that a clock is on the surface of the star and is sending out light signals are regular intervals. Use the Kruskal Diagram to explain what the r = 3m observer will see in terms of the time, t for which the external 15 geometry is static. The Schwarzschild time coordinate goes to in…nity near the horizon, so the signals from regular events reach the r = 3m hyperbola at increasing time intervals as the surface nears the horizon. Light from the star would be red-shifted until it becomes undetectable. 16