AY10/11 PC4248 RELATIVITY EXAM SOLUTIONS The answer for certain questions in this document is incomplete. Would you like to help us complete it? If yes, Please send your suggested answers to [email protected]. Thanks! ☺ Question 1 ∆ = 2 ∞ 1 2 1 − 1 − = ⇒ = − ∆ = 2 = 2 2 1 − 1 − 1 − 1 − 2 = 2 1 − 2 1 2 − 1− ≈ 2 1 + 2 − 1 + = 2 1 + − − 1+ ≈ 2 1 − − 1 + ≈ 2 1 2 = 2 + 1 − − 1 − − 4 1+ − 2 = 2 + 1 − − 1 − − 1 1 − − − 2 1 − − ! 1 AY10/11 PC4248 RELATIVITY EXAM SOLUTIONS 2 & 1 − = 2 "sin − + ' ≈ ( − 2 sin& − 2 2 + 4 =(+ ≈(+ 2 + 4 4. = / In the case of grazing the edge of the sun, 4.⊙ *+,- = / 1⊙ ∴ *+,- = ∆ − ( = Question 2 2 6 51 + 7 = ±4 − 1 − 2 Proper time after the particle crossed the horizon, 2 = − 4 − 1 − 2 6 + 1 For maximum proper time, 4 = 0, 6 = 0: 2= 2 − 1 = 9 , = 299 2= √ = √ 299 2 −1 9 29 √2 − 9 9 = − <29=2 − 9 > = √ √ 2=2 − 9 9 + √ 2=2 − 9 9 2 AY10/11 PC4248 RELATIVITY EXAM SOLUTIONS 9 = √2 sin ? , 9 = √2 cos ? ? B 2 = 2 =2 − 2 sin ? √2 cos ? ? B = 2 2 cos ? ? = 4 B 1 2 + cos 2? ? 2 B ? sin 2? = 4 " + ' 2 4 ( = 4 = ( ∴ 2CDE = ( Question 3 (a) F = −29G + H I9JK + I9JL 2 F 9 G K M= = − = 2 − H N N N N N − L N OM OM = O9 N O 9 N K L 1 N G 1 N − 2 ′ = P−2HH′ Q 2 N N 2 2 2 N 2 2 G K ′ = −HH 2 2 OM OM = OG N O G N 9 1 N 0= 2 2 N 2 2 9 I2J = 0, 2 L − 2 ′ , I1J 3 AY10/11 PC4248 RELATIVITY EXAM SOLUTIONS OM OM = OK N O K N K 1 N 0= "−2H ' N N 2 2 K = −H 2 2 9 K K = −2HH′ − H 2 2 2 ′ 2H 9 K K I3J =− , H 2 2 2 By symmetry, L 2 ′ 9 L =− , 2 2 2 I4J So the non-zero Christoffel symbols are ΓSEE = HH′ ΓSTT = ′ H′ H ′ = E E ΓUE = ΓEU = T T ΓUT = ΓTU Question 3 (b) X X X X 1VW = OX ΓVW − OW ΓVX + ΓVW ΓXZ − ΓVZ ΓWX Z Z E E E 1UU = −OU ΓUE − OU ΓUT − ΓUE ΓUE − ΓUT ΓUT T T T H[ H[[ [ [[ H′ ′ =5 − 7+5 − 7−5 7 −5 7 H H H = −5 1SS = 0, H[[ [[ + 7 H 1EE = 0, 1US = 1SU = 0, 1UE = 1EU = 0, 1SE = 1ES = 0, 1TT = 0 1ET = 1TE = 0 1UT = 1TU = 0 1ST = 1TS = 0 ∴ The only non-zero Ricci curvature tensor, 1UU H[[ [[ = −5 + 7 H 4 AY10/11 PC4248 RELATIVITY EXAM SOLUTIONS Question 3 (c) 1VW = 0 1UU = − 5 H[[ [[ 1 H 1 + 7=− − =0 H H 9 9 1 H 1 − =0 H 9 9 1 H 1 + =0 H 9 9 H H H + =0 H 9 9 H +H =0 9 9 IH J = 0 9 IH J = \ 9 − ∴ HI9J I9J = \ 9 + \ where \ , \ are constants. Question 3 (d) Solutions provided by: Dr Wang Qinghai (Q1-2) John Soo (Q3a-c) ☺ 2012, NUS Physics Society. 5
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