Physics 403: Relativity Takehome Final Examination Due 07 May 2007


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1 Physics 403: Relativity Takehome Final Examination Due 07 May Suppose that a beam of protons (rest mass m = 938 MeV/c 2 of total energy E = 300 GeV strikes a proton target at rest. Determine the largest mass of a particle X that could be produced in the following reaction: p+ p p+ p+x Solution: The proton projectile (with mass m and total energy E) is incident upon a proton at rest. Thus, the total initial energy and momentum in the laboratory frame are given as follows: E lab i = E + mc 2 == GeV c 2 p lab i 2 = E 2 m 2 c 4 The final state consists of two protons (with energies E 1 and E 2, respectively) and the Xparticle (with mass M X and energy E X ). Thus, the final energy and momentum are E f lab p f lab = E 1 + E 2 + E X = p 1 + p 2 + p X Let us view this collision in the center of momentum frame, in which the total momentum is zero before as well as after the collision. We make use of Lorentz invariance of the quantity E 2 c 2 p 2 to determine the energy in the center of momentum frame: (E cm i ) 2 = (E lab i ) 2 c 2 ( p lab i ) 2 = (E + mc 2 ) 2 E 2 + m 2 c 4 = 2mc 2 (E + mc 2 ) At the threshold for production of the Xparticle, each of the three final state particles will be at rest in that frame. As a consequence E cm f = 2mc 2 + m X c 2
2 Setting these two energies equal, we obtain E cm i = E cm f 2mc 2 (E + mc 2 ) = 2mc 2 + m X c GeV = GeV + mx c = m X c 2 m X c 2 = GeV 2. A rocket ship accelerates directly away from the earth (radius R) with a constant acceleration g, as seen in the rest frame of the rocket. Calculate the angular size of the earth, as viewed from the rocket, expressed in terms of the proper time of flight on the rocket. Show that, as the proper time gets large, the angular size approaches the limiting value 2gR/c 2. Solution: Let us restrict considerations to linear motion of the rocket along the xdirection, with the other spatial coordinates set to zero. The space time coordinates of the rocket ship in the frame of the earth at earth time t are x µ = (ct,x). The fourvelocity is given by u µ = dxµ dτ = This fourvelocity always satisfies the relation Let us define the fouracceleration as ( c dt dτ, dx ) ( = γ c, dx ) = γ (c,v) dτ dt u µ u µ = c 2 a µ = duµ dτ = d2 x µ dτ 2 = It follows directly from the definition that a µ u µ = 0 ( ) c d2 t x dτ 2,vd2 dτ 2
3 in every inertial frame. In the instananeous rest frame of the rocket (comoving frame) the speed of the rocket is zero, so that v µ = (c,0) and a µ = (0,g). we may calculate the frameinvariant quantity a µ a µ = g 2. Let us next calculate these two invariant quantities in the earth s frame: Thus a µ u µ = 0 a µ a µ = g 2 = ( = = γ c 2 d2 t ( c d2 t dτ 2 ) dτ 2 x vd2 dτ 2 ) 2 ( d 2 ) 2 x dτ 2 (1 v2 c 2 c 2 d2 t dτ 2 = v d2 x dτ ) 2 2 = g 2 ) ( d 2 x dτ 2 d 2 x dτ 2 = d (γ v) = γ g dτ γ dv ) (1+γ 2 v2 dτ c 2 = γ g γ 2 dv dτ = g dv 1 v 2 /c 2 = g dτ We begin with v = 0 at proper time τ = 0, so that v γ dx dτ = ctanh gτ c = cosh gτ c = γv = csinh gτ c Since the rocket starts at x 0 = R e at proper time τ = 0, its position at proper time τ is
4 x = c2 ( cosh gτ ) g c 1 + R e According to an observer inside the rocket, the distance to earth is Lorentzcontracted: x = x γ = c2 (coshgτ/c 1)+g R e gcoshgτ The angular size of the earth, as seen on the rocket, is θ, where tan θ 2 = y x = R x = g R e coshgτ/c c 2 (coshgτ/c 1)+g R e In the limit of large proper time τ we obtain the limiting angular size: tan θ 2 g R e c 2 The rocketeer thus always has a (small) glimpse of the receding earth at least in principle! 3. A black hole at the center of our galaxy has a radio source moving in a circular orbit about it with a visual radius θ of about 0.2 arc secon, and with a period T of about 30 years. The galactic center is a distance D of about 10 kiloparsecs away from us. Using Newtonian gravity, determine the mass of the black hole. Solution: We shall use nonrelativistic kinematics and Newtonian gravity, under the assumption that the orbiting radio source (mass m and orbit radius r)is much lighter than the black hole (mass M), as well as far away from it. Consequently F = GMm [ 2π T r 2 = mω 2 r ] 2 = ω 2 = GM r 2 T 2 = 4π2 r 3 GM This formula would also apply for the orbit of the earth about the sun (mass M S ), with its (average) radius r E of 1 AU m and period T E = 1 year. That is
5 T 2 E = 4π2 r 3 E GM S Taking the ratio of these two expressions, we obtain [ ] 2 TE [ re ] 3 M = T r M S [ ] 2 [ ] TE r 3 M = M S T r E We may thus determine M (in solar masses) in terms of T in years and r in AU by using the fomula M = T 2 r 3 The time T is 30 years. What is the distance r? First we express the distance D to the galactic center in AU: D = 10 4 parsec Next, we obtain the visual radius θ in radians: θ = 0.2 arcsec 1 AU parsec = AU 1 deg 3600 arcsec π rad 1 deg = radians Thus the orbit radius is r = Dθ = 1900 AU and M = /30 2 = solar masses. Since the solar mass is kg, we may express the mass as M = kg. The Schwarzschild radius of the black hole may then be computed 2R 0 = 2GM c 2 = m Since 1 AU = m, we obtain r = 1900 AU = m
6 The orbiting radio source thus lies at a radius of more than ten thousand times the Schwarzschild radius, so that nonrelativistic kinematics and Newtonian gravitation are justified. 4. A comet starts at infinity, goes around a relativistic star, and goes out to infinity. The impact parameter of the comet at infinity is b. The Schwarzschild radius of closest approach is r 0. What is the speed of the comet at closest approach as seen by an observer at that point? Solution: The comet trajectory in the equatorial plane θ = π/2 is a timelike geodesic, with the effective action [ Z S = c 2 ( dt ) 2 ( 1 2R r The three constants of the motion are ( ) dt 2 ( c 2 1 2R r ) ) ( dr ) 2 1 2R/r r2 dt ( dr ) 2 1 2R/r r2 ( ) dφ r 2 ( 1 2R ) r ( ) ] dφ 2 = B = A ( ) dφ 2 = 1 We substitute the first two equations into the third one to obtain c 2 A 2 = Then we use the relation ( ) dr 2 +( 1 2R r dr = dr dφ dφ and the first constant of the motion to obtain c 2 A 2 B 2 = 1 ( ) dr 2 +( r 4 1 2R dφ r We make the replacement u = 1/r to get ) ] [1+ B2 r 2 ) ( 1 B ) r 2
7 c 2 A 2 B 2 = ( ) du 2 + u 2 (1 2R u)+ 1 (1 2R u) dφ B2 First we evaluate this expression at r =, where u = 0 and du/dφ = 1/b. We obtain c 2 A 2 B 2 = 1 b B 2 We also evaluate it at the distance of closest approach, u = u 0 = 1/r 0, for which du/dφ = 0: c 2 A 2 B 2 = u 2 0 (1 2R u 0)+ 1 B 2 (1 2R u 0) We set these two expressions equal, and solve for 1/B 2 : 1 B 2 = 1 (u 20 2Ru (1 2R u 0) 1b ) 2 0 We then obtain c 2 A 2 B 2 = 1 2Ru 0 (u 20 1b ) 2Ru 2 0 The velocity v 0 at the distance of closest approach is transverse, with magnitude v T = r 0 dφ dt = r 0 dφ/ dt/ = B/r 0 A/(1 2R/r 0 ) = B u 0 (1 2Ru 0 ) A Now we use the formula just obtained for (ca/b) 2 to get v T = c 2Ru 0 (1 2Ru 0 ) 1 1/(b 2 u 2 0 ) 5. For the two dimensional metric 2 = (1+gx) 2 (cdt) 2 dx 2
8 compute the Christoffel symbols and the Riemann tensor. In addition, find a full set of linearly independent Killing vectors. Solution: The components of the metric tensor are g 00 = (1+gx) 2 and g 11 = 1. The Christoffel symbols are computed as follows: Γ a bc = 1 2 gad [ b g dc + c g bd d g bc ] Γ 1 00 = 1 2 g11 [ 1 g 00 ] = g (1+gx) Γ 0 10 = Γ0 01 = 1 2 g00 [ 1 g 00 ] == g 1+gx The rest of the Christoffel symbols are zero. For the Riemann curvature tensor we have R a bcd = c Γ a bd dγ a bc + Γa ce Γe bd Γa de Γe bc R = 0 Γ 1 01 = 0 R = 1 Γ 0 10 Γ0 10 Γ0 10 = g 2 (1+gx) 2 g 2 (1+gx) 2 = 0 All components of the Riemann tensor vanish, so that the space is flat. The covariant Killing vectors (k 0 (x 0,x),k 1 (x 0,x)) satisfy the following equations: D a k b + D b k a = 0 a k b + b k a = 2Γ c ab k c We thus have the following three equations: 1 k 1 = Γ c 11 k c = 0 0 k 0 = 2Γ 1 00 k 1 = g(1+gx)k 1 0 k k 0 = 2Γ 0 10 k 0 = 2g 1+gx k 0 The first equation requires that k 1 is independent of x; k 1 = f(x 0 ). The second equation then requires that
9 0 k 0 = g(1+gx) f(x 0 ) k 0 = h(x)+g(1+gx)d(x 0 ) d (x 0 ) = f(x 0 ) We insert these results into the third equation to obtain d (x 0 )+h (x)+g 2 2g d(x 0 ) = 1+gx h(x)+2g2 d(x 0 ) h (x) 2g 1+gx h(x) = g2 d(x 0 ) d (x 0 ) In the last relation, the left side depen only on x 0, whereas the right side depen only on x. Therefore, they are each equal to a constant, λ: For the left side: For the right side: (1+gx) 2 2gh (1+gx) 3 = [ ] d h dx (1+gx) 2 = h (1+gx) 2 h g 2 d(x 0 ) d (x 0 ) = λ We may then determine the Killing vectors: λ (1+gx) 2 λ (1+gx) 2 = h 0 λ g 1 1+gx h(x) = h 0 (1+gx) 2 λ g (1+gx) d(x 0 ) = λ g 2 + Aegx 0 + Be gx 0 k 0 = h 0 (1+gx) 2 + g (1+gx) ( Ae gx 0 + Be gx 0 ) k 1 = gae gx 0 gbe gx 0
10 Note that the parameter λ cancels out the expressions, and that there are three linearly independent Killing vectors. Here is a full set of linearly independent Killing vectors: (k 0,k 1 ) = 1+gx) 2 (1,0) = (1+gx) e gx 0 (1,g) = (1+gx) e gx 0 (1, g) The transformtion from this metric to two dimensional Minkowski space is given by the transformation (x 0,x) (x 0,x ), where x 0 = (1/g+x) sinhgx 0 x = (1/g+x) coshgx 0 The infinitesimal changes in coordinates are given by Thus, dx 0 = dx sinhgx 0 + dx 0 (1+gx) coshgx 0 dx = dx coshgx 0 + dx 0 (1+gx) sinhgx 0 2 = (dx 0 )2 (dx ) 2 = (1+gx) 2 (dx 0 ) 2 dx 2 This result may be shown directly from the transformation formula for the Killing vectors: k ν = u µ u ν k µ
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