Radial geodesics in Schwarzschild spacetime

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Lizbeth Phillips
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1 Rdil geodesics in Schwzschild spcetime Spheiclly symmetic solutions to the Einstein eqution tke the fom ds dt d dθ sin θdϕ whee is constnt. We lso hve the connection components, which now tke the fom using e ν e λ, nd theefoe ν ln nd ν, e ν Γ Γ 0 0 Γ 0 0 Γ 00 Γ Γ 0 Γ 0 0 Γ 0 0 Γ Γ Γ Γ 3 3 Γ 3 3 Γ 33 sin θ Γ 3 3 Γ 3 3 cos θ sin θ Γ 33 sin θ cos θ Now conside the geodesic eqution fo pticle which stts fom est t time τ t 0. 0 du Γ bcu b u c It might seem tht the initil velocity 4-vecto is u, 0, 0, 0, but this is not llowed. Using the line element, ds, we must hve dt c d c dθ c sin θ dϕ whee c is the speed of light. If thee is no initil motion in the sptil diections we must hve dt u whee 0 is the initil dil position. The geodesic eqution fo ech component is then t the initil time, 0 c du 0 Γ0 0u 0 u

2 0 du c Γ 00u 0 u 0 Γ u u Γ u u Γ 33u 3 u 3 0 du c Γ u u Γ 33u 3 u 3 0 du 3 c Γ3 3u u 3 Γ 3 3u u 3 Notice tht if u 0 o u 3 0 then the coesponding cceletions lso vnish, so they emin zeo. Howeve, u cnnot emin zeo. Fo dil motion, substituting fo the connection coefficients, we theefoe hve 0 c 0 c du 0 du u 0 u u0 u 0 We lso hve the eltion given by the line element, dt c d u u Fist eqution Integte the fist eqution, 0 du0 u 0 d ln u 0 d ln u 0 d ln u 0 ln ln ln b fo some constnt b, nd theefoe u 0 b Evluting t the initil condition, we see tht b 0. Rdil component of velocity fom the line element Now substitute into the line element eqution to find u, u 0 u b u u b u b 0

3 Now integte to find τ, cτ d 0 d 0 d Let y. Then dy, so Now let y 0 cosh ξ so tht cτ y dy y 0 cτ 0 cosh ξ 0 sinh ξdξ sinh ξ 0 3/ 0 3/ 0 3/ cosh ξdξ cosh ξ sinh ξ dξ cosh ξ dξ 0 3/ sinh ξ ξ 0 3/ sinh ξ cosh ξ cosh 0 0 3/ 0 0 cosh / cosh 0 Second eqution We cn lso get this esult fom the second eqution, Substituting fo u 0, 0 c du c du u0 u 0 0 c du b b u u u u u u Then, binging the u dependent fcto to the left nd multiplying both sides by u c u du b u u d 3 d,

4 we cn integte, u du b ln ln u u u du ln ln b u u dy b y ln b y ln d Theefoe, Comping to the pevious esult, u Solution to lowest ode b y y u d To lowest ode in 0, the components of the velocity e which gives t τ nd b d b we see tht the esults gee povided d. u 0 b 0 0 u 0 Notice tht the Newtonin cceletion in the diection gees with the lowest ode limit if we fix the vlue of the constnt, GM c d c dt du c b u u 0 GM c GM c 4 0

5 This finlly estblishes the Schwzschild line element, ds GM c dt d GM dθ sin θdϕ c Obits The Keple poblem Fo compison, we fist compute the obits in Newtonin gvity. We stt fom the consevtion lws. Since the velocity is v ṙ θθ sin θ ϕ ϕ the ngul momentum is L p m ṙ θθ sin θ ϕ ϕ m θ θ m sin θ ϕ ϕ Since this is conseved in both mgnitude nd diection, the obit emins in the plne pependicul to L. Without loss of genelity, we my tke the obit to lie in the θ π plne, so tht L m ϕk The enegy is lso conseved, v ṙ ϕ ϕ E m ṙ ϕ GMm Define the ngul momentum pe unit mss, l ϕ nd the enegy pe unit mss, E E m. Then we hve ϕ l so tht E ṙ l GM ṙ E l GM To find n eqution fo the obit, ϕ, divide by ϕ l nd integte: ld ϕ E l GM Now set u Let so tht ϕ ldu E l u GMu ldu GM l lu E G M l y GM l lu A E G M 5 l

6 so tht so with y A sin θ we hve ϕ dy y A A cos θdθ ϕ A cos θ θ y csin A Solving fo we hve, A sin ϕ y GM l lu GM l l so GM l A l sin ϕ l GM l E G M sin ϕ To see tht this descibes n ellipse, let e sin ϕ Then, chnging to Ctesin coodintes, x e e sin ϕ ey ey e y x y ey e y x y ey ey y e e e e Finlly, setting y 0 e e, b e e nd c b e we hve the stndd fom fo n ellipse centeed t x, y 0, y 0 : x b y y 0 c An exmintion of the mgnitudes of the constnts shows tht this solution is vlid fo bound sttes, with E < 0. Fo positive enegy, the finl integl gives hypebolic function nd the eqution descibes hypebol. Now conside the obits descibed by genel eltivity. 6

7 Geodesic obits The geodesic equtions e: 0 c 0 c du 0 du u 0 u u0 u 0 0 du c sin θ cos θu3 u 3 u u 0 du 3 c cos θ sin θ u u 3 u3 u We lso hve the eltion given by the line element, ds dt c u u u u sin θu 3 u 3 d c dθ c sin θ dϕ Becuse of the spheicl symmety, obits will emin in plne, s seen by choosing initil conditions with θ π nd u 0 0. Then the geodesic equtions become 0 c 0 c du 0 du 0 du c 0 du 3 c u3 u u 0 u u0 u 0 so tht u emins zeo. We my integte the ϕ eqution immeditely, du 3 u 3 d u 3 l 0 dϕ l 0 which simply sttes consevtion of ngul momentum. We lso hve the sme esult fo u 0 s we did fo dil geodesics, u 0 b though b diffes. How we poceed depends on the type of obit we desie. Timelike obits nd peihelion dvnce We cn compute u fom the line element fo timelike obit, ds, b c u u u 3 u 3 d l 0 c 7

8 c To find b let t τ 0 when d b d c l 0 c d b l 0 c 0. Then so tht with this vlue fo b nd GM c b l c b l 0 0 GM c 0 c b l 0 GM 0 c 0 c GMl 0 0 3c4 we hve d GM GM l GM 0 c l 0 GM c We find the eqution fo the obit by dividing by u 3 l 0 4, d 4 GM dϕ 0 l0 GM l0 0 GM 0 c d dϕ GM 0l 0 GM l 0 0 GM c GM 0c GM c In tems of u this becomes dϕ du dϕ u 0 GMu0 l 0 GM l 0 GM c u 0 u u du u u 0 GM c GMu 0 l 0 u 0 u GMu c GM l0 u u GMu3 c which is the usul eqution fo n ellipse except fo the lst tem. Mke the definitions Then we hve u du dϕ u 0 GM c u 0 b GM l 0 c GM c u bu u cu 3 GMu 0 l 0 8

9 o dϕ du bu u cu 3 This gees with the Keple esult, except fo the lst, cubic tem in the denominto, which is vey smll fo odiny sts o plnets. If we integte this bout hlf of one obit without the cubic tem, we just get π. Multiplying by we get complete cicuit, tht is, π u u du bu u whee u ± e the extemes of the obit. The peihelion dvnce is the diffeence between this integl nd π, u du ϕ π bu u cu 3 u Unfotuntely, estimting this stightfowdly involves n elliptic integl. I hven t found ny wy to do it simple thn Weinbeg s clcultion in Gvittion nd Cosmology. I won t epet his clcultion hee. The esult fo the peihelion dvnce of Mecuy is bout 43 seconds of c pe centuy, in excellent geement with the expeimentl esult. Null geodesics Now etun to the solution fo the geodesic, but conside the null cse, ds 0. Then we still hve u 0 b m 0 dϕ l 0 0 c whee 0 is the dius of closest ppoch. But now we hve Theefoe, with m 0 dt c 0 b d c c dθ c sin θ d l 0 c d b c c 0 m At the dius of closest ppoch, this becomes 0 b c l 0 0 m 0 0 b m 0 dϕ 9

10 The pocedue is the sme. We convet to n eqution elting nd ϕ. Setting A bc l 0 0 m 0, this leds to d dϕ dϕ 0 c b c l 0 m d 0c b c l 0 m d b l 0 0 m 0 c d m 0 d 0 m 0 m 0 0 The totl chnge in ngle twice the integl fom infinity to 0. No devition coesponds to chnge in ϕ of π, so d ϕ π m Now expnd the second sque oot in powes of m 0 nd substitute 0 sin θ, d ϕ π m m O d π m π π π m 0 0 π 0 dθ sin θ m 0 m 0 4m 0 m sin 3 θ sin θ 0 dθ m0 sin θ sin θ sin θ The integl ws esily hndled by the Wolfm online integto. Fo the sun, 4m 4GM 0 R cos θ sin θ cos θ cos θ cos θ sin θ π

11 d deg/d 3600 sec/deg.748 sec so light is deflected by.75 seconds of c. This is confimed by expeiment.

This immediately suggests an inverse-square law for a "piece" of current along the line.

This immediately suggests an inverse-square law for a piece of current along the line. Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iot-svt Lw T nvese-sque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line