Back Geodesics The metric and geodesic equations are in the form:
d s 2 = − ( 1 − 2 r ) d t 2 + ( 1 − 2 r ) − 1 d r 2 + r 2 d ϕ 2 , d 2 t d λ 2 + 2 r ( r − 2 ) d t d λ d r d λ = 0 , d 2 r d λ 2 + r − 2 r 3 ( d t d λ ) 2 − 1 r ( r − 2 ) ( d r d λ ) 2 − ( r − 2 ) ( d ϕ d λ ) 2 = 0 , d 2 ϕ d λ 2 + 2 r d r d λ d ϕ d λ = 0. \begin{align*} ds^2 = -\left(1 - \frac{2}{r}\right) dt^2 + \left(1 - \frac{2}{r}\right)^{-1} dr^2 + r^2 d\phi^2, \\ \frac{d^2 t}{d\lambda^2} + \frac{2}{r(r - 2)} \frac{dt}{d\lambda} \frac{dr}{d\lambda} &= 0, \\ \frac{d^2 r}{d\lambda^2} + \frac{r - 2}{r^3} \left(\frac{dt}{d\lambda}\right)^2 - \frac{1}{r(r - 2)} \left(\frac{dr}{d\lambda}\right)^2 - (r - 2) \left(\frac{d\phi}{d\lambda}\right)^2 &= 0, \\ \frac{d^2 \phi}{d\lambda^2} + \frac{2}{r} \frac{dr}{d\lambda} \frac{d\phi}{d\lambda} &= 0. \end{align*} d s 2 = − ( 1 − r 2 ) d t 2 + ( 1 − r 2 ) − 1 d r 2 + r 2 d ϕ 2 , d λ 2 d 2 t + r ( r − 2 ) 2 d λ d t d λ d r d λ 2 d 2 r + r 3 r − 2 ( d λ d t ) 2 − r ( r − 2 ) 1 ( d λ d r ) 2 − ( r − 2 ) ( d λ d ϕ ) 2 d λ 2 d 2 ϕ + r 2 d λ d r d λ d ϕ = 0 , = 0 , = 0. For incoming lightlike geodesics, we will take θ , ϕ \theta, \phi θ , ϕ
d s 2 = − ( 1 − 2 r ) d t 2 + ( 1 − 2 r ) − 1 d r 2 , d 2 t d λ 2 + 2 r ( r − 2 ) d t d λ d r d λ = 0 , d 2 r d λ 2 + r − 2 r 3 ( d t d λ ) 2 − 1 r ( r − 2 ) ( d r d λ ) 2 = 0. \begin{align*} ds^2 = -\left(1 - \frac{2}{r}\right) dt^2 + \left(1 - \frac{2}{r}\right)^{-1} dr^2, \\ \frac{d^2 t}{d\lambda^2} + \frac{2}{r(r - 2)} \frac{dt}{d\lambda} \frac{dr}{d\lambda} &= 0, \\ \frac{d^2 r}{d\lambda^2} + \frac{r - 2}{r^3} \left(\frac{dt}{d\lambda}\right)^2 - \frac{1}{r(r - 2)} \left(\frac{dr}{d\lambda}\right)^2 &= 0. \\ \end{align*} d s 2 = − ( 1 − r 2 ) d t 2 + ( 1 − r 2 ) − 1 d r 2 , d λ 2 d 2 t + r ( r − 2 ) 2 d λ d t d λ d r d λ 2 d 2 r + r 3 r − 2 ( d λ d t ) 2 − r ( r − 2 ) 1 ( d λ d r ) 2 = 0 , = 0. I will not be using the geodesic equations. I will instead be using the metric which is equal to zero for lightlike intervals:
0 = − ( 1 − 2 r ) d t 2 + ( 1 − 2 r ) − 1 d r 2 , ( r − 2 r ) d t 2 = ( r − 2 r ) − 1 d r 2 , d t 2 = ( r r − 2 ) 2 d r 2 , d t = ± r r − 2 d r , ∫ 0 t d t = ± ∫ r 0 r r r − 2 d r , t = ± ∫ r 0 r r r − 2 d r = ± ∫ r 0 r r − 2 + 2 r − 2 d r = ± ∫ r 0 r ( 1 + 2 r − 2 ) d r = ± [ r + 2 ln ( r − 2 ) ] r 0 r , \begin{align*} 0 = -\left(1 - \frac{2}{r}\right) dt^2 &+ \left(1 - \frac{2}{r}\right)^{-1} dr^2, \\ \left(\frac{r - 2}{r}\right) dt^2 &= \left(\frac{r - 2}{r}\right)^{-1} dr^2, \\ dt^2 &= \left(\frac{r}{r - 2}\right)^2 dr^2, \\ dt &= \pm \frac{r}{r - 2} dr, \\ \int_0^t dt &= \pm \int_{r_0}^r \frac{r}{r - 2} dr, \\ t &= \pm \int_{r_0}^r \frac{r}{r - 2} dr \\ &= \pm \int_{r_0}^r \frac{r - 2 + 2}{r - 2} dr \\ &= \pm \int_{r_0}^r \left(1 + \frac{2}{r - 2}\right) dr \\ &= \pm \left[r + 2 \ln (r - 2)\right]_{r_0}^r, \\ \end{align*} 0 = − ( 1 − r 2 ) d t 2 ( r r − 2 ) d t 2 d t 2 d t ∫ 0 t d t t + ( 1 − r 2 ) − 1 d r 2 , = ( r r − 2 ) − 1 d r 2 , = ( r − 2 r ) 2 d r 2 , = ± r − 2 r d r , = ± ∫ r 0 r r − 2 r d r , = ± ∫ r 0 r r − 2 r d r = ± ∫ r 0 r r − 2 r − 2 + 2 d r = ± ∫ r 0 r ( 1 + r − 2 2 ) d r = ± [ r + 2 ln ( r − 2 ) ] r 0 r , this applies outside the event horizon. For the inside, the expression is equal to:
t = ± ∫ r 0 r r r − 2 d r = ± ∫ r 0 r − r 2 − r d r = ± ∫ r 0 r 2 − r − 2 2 − r d r = ± ∫ r 0 r ( 1 − 2 2 − r ) d r = ± ∫ r 0 r − ( − 1 + 2 2 − r ) d r , u = 2 − r , d u = − d r , t = ± ∫ 2 − r 0 2 − r ( − 1 + 2 u ) d u = ± [ − u + 2 ln u ] 2 − r 0 2 − r = ± [ − 2 + r + 2 ln ( 2 − r ) ] r 0 r constants cancel in definite integrals = ± [ r + 2 ln ( 2 − r ) ] r 0 r . \begin{align*} t &= \pm \int_{r_0}^r \frac{r}{r - 2} dr \\ &= \pm \int_{r_0}^r \frac{- r}{2 - r} dr \\ &= \pm \int_{r_0}^r \frac{2 - r - 2}{2 - r} dr \\ &= \pm \int_{r_0}^r \left(1 - \frac{2}{2 - r}\right) dr \\ &= \pm \int_{r_0}^r -\left(- 1 + \frac{2}{2 - r}\right) dr, \\ u &= 2 - r, \\ du &= -dr, \\ t &= \pm \int_{2 - r_0}^{2 - r} \left(- 1 + \frac{2}{u}\right) du \\ &= \pm \left[-u + 2 \ln u\right]_{2 - r_0}^{2 - r} \\ &= \pm \left[-2 + r + 2 \ln (2 -r)\right]_{r_0}^r \qquad \textrm{constants cancel in definite integrals} \\ &= \pm \left[r + 2 \ln (2 -r)\right]_{r_0}^r. \\ \end{align*} t u d u t = ± ∫ r 0 r r − 2 r d r = ± ∫ r 0 r 2 − r − r d r = ± ∫ r 0 r 2 − r 2 − r − 2 d r = ± ∫ r 0 r ( 1 − 2 − r 2 ) d r = ± ∫ r 0 r − ( − 1 + 2 − r 2 ) d r , = 2 − r , = − d r , = ± ∫ 2 − r 0 2 − r ( − 1 + u 2 ) d u = ± [ − u + 2 ln u ] 2 − r 0 2 − r = ± [ − 2 + r + 2 ln ( 2 − r ) ] r 0 r constants cancel in definite integrals = ± [ r + 2 ln ( 2 − r ) ] r 0 r . For incoming geodesics the sign is − - − + + +
When r > 2 r > 2 r > 2
g 00 = − ( 1 − 2 r ) < 0 , g 11 = ( 1 − 2 r ) − 1 > 0 , \begin{align*} g_{00} &= - \left(1 - \frac{2}{r}\right) < 0, \\ g_{11} &= \left(1 - \frac{2}{r}\right)^{-1} > 0, \end{align*} g 00 g 11 = − ( 1 − r 2 ) < 0 , = ( 1 − r 2 ) − 1 > 0 , meaning the t t t r r r r → 2 r \to 2 r → 2
g 00 = − ( 1 − 2 r ) = 0 , g 11 = ( 1 − 2 r ) − 1 → ∞ , \begin{align*} g_{00} &= - \left(1 - \frac{2}{r}\right) = 0, \\ g_{11} &= \left(1 - \frac{2}{r}\right)^{-1} \to \infty, \end{align*} g 00 g 11 = − ( 1 − r 2 ) = 0 , = ( 1 − r 2 ) − 1 → ∞ , meaning the t t t r r r only in this coordinate system ). When r < 2 r < 2 r < 2
g 00 = − ( 1 − 2 r ) > 0 , g 11 = ( 1 − 2 r ) − 1 < 0 , \begin{align*} g_{00} &= - \left(1 - \frac{2}{r}\right) > 0, \\ g_{11} &= \left(1 - \frac{2}{r}\right)^{-1} < 0, \end{align*} g 00 g 11 = − ( 1 − r 2 ) > 0 , = ( 1 − r 2 ) − 1 < 0 , meaning the t t t r r r
There are two constants of motion. The first can be derived from the t t t
d 2 t d λ 2 + 2 r ( r − 2 ) d t d λ d r d λ = 0 , d t d λ = U 0 , d U 0 d λ + 2 r ( r − 2 ) d r d λ U 0 = 0 , 1 U 0 d U 0 + 2 r ( r − 2 ) d r = 0 , ∫ 1 U 0 d U 0 + ∫ 2 r ( r − 2 ) d r = C , ln U 0 + ∫ ( 1 r − 2 − 1 r ) d r = C , ln U 0 + ln ( r − 2 ) − ln ( r ) = C , ln ( U 0 r − 2 r ) = C , U 0 r − 2 r = e C , \begin{align*} \frac{d^2 t}{d\lambda^2} + \frac{2}{r(r - 2)} \frac{dt}{d\lambda} \frac{dr}{d\lambda} &= 0, \qquad \frac{dt}{d\lambda} = U^0, \\ \frac{dU^0}{d\lambda} + \frac{2}{r(r - 2)} \frac{dr}{d\lambda} U^0 &= 0, \\ \frac{1}{U^0} dU^0 + \frac{2}{r(r - 2)} dr &= 0, \\ \int \frac{1}{U^0} dU^0 + \int \frac{2}{r(r - 2)} dr &= C, \\ \ln U^0 + \int \left(\frac{1}{r - 2} - \frac{1}{r}\right) dr &= C, \\ \ln U^0 + \ln (r - 2) - \ln(r) &= C, \\ \ln \left(U^0 \frac{r - 2}{r}\right) &= C, \\ U^0 \frac{r - 2}{r} &= e^C, \end{align*} d λ 2 d 2 t + r ( r − 2 ) 2 d λ d t d λ d r d λ d U 0 + r ( r − 2 ) 2 d λ d r U 0 U 0 1 d U 0 + r ( r − 2 ) 2 d r ∫ U 0 1 d U 0 + ∫ r ( r − 2 ) 2 d r ln U 0 + ∫ ( r − 2 1 − r 1 ) d r ln U 0 + ln ( r − 2 ) − ln ( r ) ln ( U 0 r r − 2 ) U 0 r r − 2 = 0 , d λ d t = U 0 , = 0 , = 0 , = C , = C , = C , = C , = e C , the constant on the right side is the energy per unit mass E = E m \mathcal{E} = \frac{E}{m} E = m E
E = d t d λ ( 1 − 2 r ) . \mathcal{E} = \frac{dt}{d\lambda} \left(1 - \frac{2}{r}\right). E = d λ d t ( 1 − r 2 ) . The other constant of motion can be derived from the ϕ \phi ϕ
d 2 ϕ d λ 2 + 2 r d r d λ d ϕ d λ = 0 , d ϕ d λ = U 3 , d U 3 d λ + 2 r d r d λ U 3 = 0 , 1 U 3 d U 3 + 2 r d r = 0 , ∫ 1 U 3 d U 3 + ∫ 2 r d r = C , ln U 3 + 2 ln r = C , ln ( U 3 r 2 ) = C , U 3 r 2 = e C , \begin{align*} \frac{d^2 \phi}{d\lambda^2} + \frac{2}{r} \frac{dr}{d\lambda} \frac{d\phi}{d\lambda} &= 0, \qquad \frac{d\phi}{d\lambda} = U^3, \\ \frac{dU^3}{d\lambda} + \frac{2}{r} \frac{dr}{d\lambda} U^3 &= 0, \\ \frac{1}{U^3} dU^3 + \frac{2}{r} dr &= 0, \\ \int \frac{1}{U^3} dU^3 + \int \frac{2}{r} dr &= C, \\ \ln U^3 + 2\ln r &= C, \\ \ln (U^3 r^2) &= C, \\ U^3 r^2 &= e^C, \\ \end{align*} d λ 2 d 2 ϕ + r 2 d λ d r d λ d ϕ d λ d U 3 + r 2 d λ d r U 3 U 3 1 d U 3 + r 2 d r ∫ U 3 1 d U 3 + ∫ r 2 d r ln U 3 + 2 ln r ln ( U 3 r 2 ) U 3 r 2 = 0 , d λ d ϕ = U 3 , = 0 , = 0 , = C , = C , = C , = e C , the constant on the right side is the angular momentum per unit mass L = L m \mathcal{L} = \frac{L}{m} L = m L
L = d ϕ d λ r 2 . \mathcal{L} = \frac{d\phi}{d\lambda} r^2. L = d λ d ϕ r 2 . Taking the limit as r → ∞ r \to \infty r → ∞ E \mathcal{E} E λ = τ \lambda = \tau λ = τ
lim r → ∞ E m = lim r → ∞ d t d τ ( 1 − 2 r ) = d t d τ = γ , E → γ m . \begin{align*} \lim_{r \to \infty} \frac{E}{m} &= \lim_{r \to \infty} \frac{dt}{d\tau} \left(1 - \frac{2}{r}\right) \\ &= \frac{dt}{d\tau} \\ &= \gamma, \\ E &\to \gamma m. \end{align*} r → ∞ lim m E E = r → ∞ lim d τ d t ( 1 − r 2 ) = d τ d t = γ , → γm . Consider an orbit where θ = π 2 \theta = \frac{\pi}{2} θ = 2 π
where v r = d r d t v_r = \frac{dr}{dt} v r = d t d r v p e r p = r d ϕ d t v_{perp} = r \frac{d\phi}{dt} v p er p = r d t d ϕ
E = 1 2 m v 2 − G M m r = 1 2 m [ ( d r d t ) 2 + r 2 ( d ϕ d t ) 2 ] − G M m r = 1 2 m ( d r d t ) 2 + 1 2 m r 2 ( d ϕ d t ) 2 − G M m r = 1 2 m ( d r d t ) 2 + 1 2 m r 2 m 2 r 4 ( d ϕ d t ) 2 − G M m r = 1 2 m ( d r d t ) 2 + L 2 2 m r 2 − G M m r = 1 2 m ( d r d t ) 2 + U e f f ( r ) , \begin{align*} E &= \frac{1}{2} m v^2 - G \frac{Mm}{r} \\ &= \frac{1}{2} m \left[\left(\frac{dr}{dt}\right)^2 + r^2 \left(\frac{d\phi}{dt}\right)^2\right] - G\frac{Mm}{r} \\ &= \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \frac{1}{2} m r^2 \left(\frac{d\phi}{dt}\right)^2 - G\frac{Mm}{r} \\ &= \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \frac{1}{2 m r^2} m^2 r^4 \left(\frac{d\phi}{dt}\right)^2 - G\frac{Mm}{r} \\ &= \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \frac{L^2}{2 m r^2} - G\frac{Mm}{r} \\ &= \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + U_{eff}(r), \end{align*} E = 2 1 m v 2 − G r M m = 2 1 m [ ( d t d r ) 2 + r 2 ( d t d ϕ ) 2 ] − G r M m = 2 1 m ( d t d r ) 2 + 2 1 m r 2 ( d t d ϕ ) 2 − G r M m = 2 1 m ( d t d r ) 2 + 2 m r 2 1 m 2 r 4 ( d t d ϕ ) 2 − G r M m = 2 1 m ( d t d r ) 2 + 2 m r 2 L 2 − G r M m = 2 1 m ( d t d r ) 2 + U e ff ( r ) , where U e f f ( r ) = L 2 2 m r 2 − G M m r = L 2 2 m r 2 − m r U_{eff}(r) = \frac{L^2}{2 m r^2} - G\frac{Mm}{r} = \frac{L^2}{2 m r^2} - \frac{m}{r} U e ff ( r ) = 2 m r 2 L 2 − G r M m = 2 m r 2 L 2 − r m L = m r 2 d ϕ d t L = m r^2 \frac{d\phi}{dt} L = m r 2 d t d ϕ G = M = m = 1 G = M = m = 1 G = M = m = 1 L = 0 L = 0 L = 0 L = 0.5 L = 0.5 L = 0.5 L = 0.75 L = 0.75 L = 0.75 L = 1 L = 1 L = 1
Plotting the case where L = 0.5 L = 0.5 L = 0.5
The body will always have the potential energy less than or equal the total energy (as shown by the pink line for E 2 E_2 E 2 E 1 E_1 E 1 d r d t \frac{dr}{dt} d t d r E 2 E_2 E 2 r r r E 3 E_3 E 3
The energy and angular momentum stay constant. They are given by initial distance r 0 r_0 r 0 v 0 v_0 v 0 ω 0 \omega_0 ω 0
L = m r 0 2 ω 0 , E = 1 2 m v 0 2 + U e f f ( r 0 ) . \begin{align*} L &= m r_0^2 \omega_0, \\ E &= \frac{1}{2} m v_0^2 + U_{eff}(r_0). \end{align*} L E = m r 0 2 ω 0 , = 2 1 m v 0 2 + U e ff ( r 0 ) . Every orbit has two points where d r d t = 0 \frac{dr}{dt} = 0 d t d r = 0
E = L 2 2 m r 2 − m r , 0 = L 2 2 m r 2 − m r − E , 0 = − E r 2 − m r + L 2 2 m , D = ( − m ) 2 − 4 ( − E ) L 2 2 m = m 2 + 2 E L 2 m , r = − ( − m ) ± D 2 ( − E ) = − m ± D 2 E . \begin{align*} E &= \frac{L^2}{2 m r^2} - \frac{m}{r}, \\ 0 &= \frac{L^2}{2 m r^2} - \frac{m}{r} - E, \\ 0 &= -E r^2 - mr + \frac{L^2}{2 m}, \\ D &= (-m)^2 - 4(-E)\frac{L^2}{2 m} \\ &= m^2 + 2\frac{E L^2}{m}, \\ r &= \frac{-(-m) \pm \sqrt{D}}{2(-E)} \\ &= -\frac{m \pm \sqrt{D}}{2E}. \end{align*} E 0 0 D r = 2 m r 2 L 2 − r m , = 2 m r 2 L 2 − r m − E , = − E r 2 − m r + 2 m L 2 , = ( − m ) 2 − 4 ( − E ) 2 m L 2 = m 2 + 2 m E L 2 , = 2 ( − E ) − ( − m ) ± D = − 2 E m ± D . If the radial velocity d r d t \frac{dr}{dt} d t d r
E = L 2 2 m r 2 − m r , E = \frac{L^2}{2 m r^2} - \frac{m}{r}, E = 2 m r 2 L 2 − r m , this has to stay constant, taking the derivative:
d d t ( L 2 2 m r 2 − m r ) = − L 2 m r 3 + m r 2 = 0 , \frac{d}{dt} \left(\frac{L^2}{2 m r^2} - \frac{m}{r}\right) = - \frac{L^2}{m r^3} + \frac{m}{r^2} = 0, d t d ( 2 m r 2 L 2 − r m ) = − m r 3 L 2 + r 2 m = 0 , and solving for r r r
− L 2 m r 3 + m r 2 = 0 , r m 2 − L 2 = 0 , r = L 2 m 2 , \begin{align*} - \frac{L^2}{m r^3} + \frac{m}{r^2} &= 0, \\ r m^2 - L^2 &= 0, r &= \frac{L^2}{m^2}, \end{align*} − m r 3 L 2 + r 2 m r m 2 − L 2 = 0 , = 0 , r = m 2 L 2 , or if given radius, we can solve for angular velocity:
r = m 2 r 4 m 2 ( d ϕ d t ) 2 , ( d ϕ d t ) 2 = 1 r 3 , d ϕ d t = ± 1 r r , v p e r p = ± 1 r . \begin{align*} r &= \frac{m^2 r^4}{m^2} \left(\frac{d\phi}{dt}\right)^2, \\ \left(\frac{d\phi}{dt}\right)^2 &= \frac{1}{r^3}, \\ \frac{d\phi}{dt} &= \pm \frac{1}{r\sqrt{r}}, \\ v_{perp} &= \pm \frac{1}{\sqrt{r}}. \end{align*} r ( d t d ϕ ) 2 d t d ϕ v p er p = m 2 m 2 r 4 ( d t d ϕ ) 2 , = r 3 1 , = ± r r 1 , = ± r 1 . Consider a path parametrized by λ \lambda λ
ϵ = − ( 1 − 2 r ) ( d t d λ ) 2 + ( 1 − 2 r ) − 1 ( d r d λ ) 2 + r 2 ( d ϕ d λ ) 2 , \epsilon = -\left(1 - \frac{2}{r}\right) \left(\frac{dt}{d\lambda}\right)^2 + \left(1 - \frac{2}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2 + r^2 \left(\frac{d\phi}{d\lambda}\right)^2, ϵ = − ( 1 − r 2 ) ( d λ d t ) 2 + ( 1 − r 2 ) − 1 ( d λ d r ) 2 + r 2 ( d λ d ϕ ) 2 , where ϵ \epsilon ϵ
ϵ < 0 , for timelike geodesics , ϵ = − 1 , for timelike geodesics where λ = τ , ϵ = 0 , for lightlike geodesics , ϵ > 0 , for spacelike geodesics , \begin{align*} \epsilon &< 0, & &\textrm{for timelike geodesics}, \\ \epsilon &= -1, & &\textrm{for timelike geodesics where \(\lambda = \tau\)}, \\ \epsilon &= 0, & &\textrm{for lightlike geodesics}, \\ \epsilon &> 0, & &\textrm{for spacelike geodesics}, \\ \end{align*} ϵ ϵ ϵ ϵ < 0 , = − 1 , = 0 , > 0 , for timelike geodesics , for timelike geodesics where λ = τ , for lightlike geodesics , for spacelike geodesics , but I will not be considering space-like geodesics.
Substiting the constants of motion into the equation for ϵ \epsilon ϵ
ϵ = − ( 1 − 2 r ) ( E 1 − 2 r ) 2 + ( 1 − 2 r ) − 1 ( d r d λ ) 2 + r 2 ( L r 2 ) 2 , ϵ = − r r − 2 E 2 + r r − 2 ( d r d λ ) 2 + L 2 r 2 , 0 = − r r − 2 E 2 + r r − 2 ( d r d λ ) 2 + L 2 r 2 − ϵ , 0 = − E 2 + ( d r d λ ) 2 + r − 2 r L 2 r 2 − r − 2 r ϵ , 0 = − E 2 + ( d r d λ ) 2 + ( 1 − 2 r ) ( L 2 r 2 − ϵ ) , E 2 = ( d r d λ ) 2 + ( 1 − 2 r ) ( L 2 r 2 − ϵ ) , = ( d r d λ ) 2 + ( L 2 r 2 − 2 L 2 r 3 + 2 r ϵ − ϵ ) , = ( d r d λ ) 2 + L 2 r 2 − 2 L 2 r 3 + 2 r ϵ − ϵ , \begin{align*} \epsilon &= -\left(1 - \frac{2}{r}\right) \left(\frac{\mathcal{E}}{1 - \frac{2}{r}}\right)^2 + \left(1 - \frac{2}{r}\right)^{-1} \left(\frac{dr}{d\lambda}\right)^2 + r^2 \left(\frac{\mathcal{L}}{r^2}\right)^2, \\ \epsilon &= -\frac{r}{r - 2} \mathcal{E}^2 + \frac{r}{r - 2} \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2}, \\ 0 &= -\frac{r}{r - 2} \mathcal{E}^2 + \frac{r}{r - 2} \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \epsilon, \\ 0 &= - \mathcal{E}^2 + \left(\frac{dr}{d\lambda}\right)^2 + \frac{r - 2}{r} \frac{\mathcal{L}^2}{r^2} - \frac{r - 2}{r} \epsilon, \\ 0 &= - \mathcal{E}^2 + \left(\frac{dr}{d\lambda}\right)^2 + \left(1 - \frac{2}{r}\right) \left(\frac{\mathcal{L}^2}{r^2} - \epsilon\right), \\ \mathcal{E}^2 &= \left(\frac{dr}{d\lambda}\right)^2 + \left(1 - \frac{2}{r}\right) \left(\frac{\mathcal{L}^2}{r^2} - \epsilon\right), \\ &= \left(\frac{dr}{d\lambda}\right)^2 + \left(\frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} + \frac{2}{r} \epsilon - \epsilon\right), \\ &= \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} + \frac{2}{r} \epsilon - \epsilon, \end{align*} ϵ ϵ 0 0 0 E 2 = − ( 1 − r 2 ) ( 1 − r 2 E ) 2 + ( 1 − r 2 ) − 1 ( d λ d r ) 2 + r 2 ( r 2 L ) 2 , = − r − 2 r E 2 + r − 2 r ( d λ d r ) 2 + r 2 L 2 , = − r − 2 r E 2 + r − 2 r ( d λ d r ) 2 + r 2 L 2 − ϵ , = − E 2 + ( d λ d r ) 2 + r r − 2 r 2 L 2 − r r − 2 ϵ , = − E 2 + ( d λ d r ) 2 + ( 1 − r 2 ) ( r 2 L 2 − ϵ ) , = ( d λ d r ) 2 + ( 1 − r 2 ) ( r 2 L 2 − ϵ ) , = ( d λ d r ) 2 + ( r 2 L 2 − r 3 2 L 2 + r 2 ϵ − ϵ ) , = ( d λ d r ) 2 + r 2 L 2 − r 3 2 L 2 + r 2 ϵ − ϵ , For timelike geodesics, λ = τ \lambda = \tau λ = τ ϵ = − 1 \epsilon = -1 ϵ = − 1
E 2 = ( d r d τ ) 2 + L 2 r 2 − 2 L 2 r 3 − 2 r + 1 , ( E m ) 2 = ( d r d τ ) 2 + L 2 m 2 r 2 − 2 L 2 m 2 r 3 − 2 r + 1 , 1 2 m ( E m ) 2 = 1 2 m ( d r d τ ) 2 + L 2 2 m r 2 − L 2 m r 3 − m r + m 2 , 1 2 ( E 2 m − m ) = 1 2 m ( d r d τ ) 2 + L 2 2 m r 2 − L 2 m r 3 − m r = 1 2 m ( d r d τ ) 2 + ( L 2 2 m r 2 − m r − L 2 m r 3 ) , E = 1 2 m ( d r d τ ) 2 + ( m L 2 2 r 2 − m r − m L 2 r 3 ) , = 1 2 m ( d r d τ ) 2 + U e f f ( r ) , \begin{align*} \mathcal{E}^2 &= \left(\frac{dr}{d\tau}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} - \frac{2}{r} + 1, \\ \left(\frac{E}{m}\right)^2 &= \left(\frac{dr}{d\tau}\right)^2 + \frac{L^2}{m^2 r^2} - \frac{2 L^2}{m^2 r^3} - \frac{2}{r} + 1, \\ \frac{1}{2} m \left(\frac{E}{m}\right)^2 &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + \frac{L^2}{2 m r^2} - \frac{L^2}{m r^3} - \frac{m}{r} + \frac{m}{2}, \\ \frac{1}{2} \left(\frac{E^2}{m} - m\right) &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + \frac{L^2}{2 m r^2} - \frac{L^2}{m r^3} - \frac{m}{r} \\ &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + \left(\frac{L^2}{2 m r^2} - \frac{m}{r} - \frac{L^2}{m r^3}\right), \\ \mathscr{E} &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + \left(\frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r} - \frac{m \mathcal{L}^2}{r^3}\right), \\ &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + U_{eff}(r), \end{align*} E 2 ( m E ) 2 2 1 m ( m E ) 2 2 1 ( m E 2 − m ) E = ( d τ d r ) 2 + r 2 L 2 − r 3 2 L 2 − r 2 + 1 , = ( d τ d r ) 2 + m 2 r 2 L 2 − m 2 r 3 2 L 2 − r 2 + 1 , = 2 1 m ( d τ d r ) 2 + 2 m r 2 L 2 − m r 3 L 2 − r m + 2 m , = 2 1 m ( d τ d r ) 2 + 2 m r 2 L 2 − m r 3 L 2 − r m = 2 1 m ( d τ d r ) 2 + ( 2 m r 2 L 2 − r m − m r 3 L 2 ) , = 2 1 m ( d τ d r ) 2 + ( 2 r 2 m L 2 − r m − r 3 m L 2 ) , = 2 1 m ( d τ d r ) 2 + U e ff ( r ) , where E = 1 2 ( E 2 m + m ) = 1 2 ( m E 2 + m ) \mathscr{E} = \frac{1}{2} \left(\frac{E^2}{m} + m\right) = \frac{1}{2} \left(m \mathcal{E}^2 + m\right) E = 2 1 ( m E 2 + m ) = 2 1 ( m E 2 + m ) E = E m \mathcal{E} = \frac{E}{m} E = m E L = L m = r 2 d ϕ d τ \mathcal{L} = \frac{L}{m} = r^2 \frac{d\phi}{d\tau} L = m L = r 2 d τ d ϕ U e f f ( r ) = m L 2 2 r 2 − m r − m L 2 r 3 U_{eff}(r) = \frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r} - \frac{m \mathcal{L}^2}{r^3} U e ff ( r ) = 2 r 2 m L 2 − r m − r 3 m L 2
Comparing with the newtonian potential equation, we can see the difference on the left side and the difference between potentials on the right side:
E = 1 2 m ( d r d τ ) 2 + ( L 2 2 m r 2 − m r − L 2 m r 3 ) , E = 1 2 m ( d r d t ) 2 + ( L 2 2 m r 2 − m r ) , \begin{align*} \mathscr{E} &= \frac{1}{2} m \left(\frac{dr}{d\tau}\right)^2 + \left(\frac{L^2}{2 m r^2} - \frac{m}{r} - \frac{L^2}{m r^3}\right), \\ E &= \frac{1}{2} m \left(\frac{dr}{dt}\right)^2 + \left(\frac{L^2}{2 m r^2} - \frac{m}{r}\right), \\ \end{align*} E E = 2 1 m ( d τ d r ) 2 + ( 2 m r 2 L 2 − r m − m r 3 L 2 ) , = 2 1 m ( d t d r ) 2 + ( 2 m r 2 L 2 − r m ) , we see that the right sides are almost identical apart from the extra term in the potential. The last extra term vanishes at further distances.
Below is a plot of the effective potential energy where m = 1 m = 1 m = 1 L = 2 \mathcal{L} = 2 L = 2 L = 10 \mathcal{L} = 10 L = 10 L = 15 \mathcal{L} = 15 L = 15 L = 20 \mathcal{L} = 20 L = 20
I will use the green line where L = 15 \mathcal{L} = 15 L = 15
for E 1 E_1 E 1 r 1 r_1 r 1 r 2 r_2 r 2 r 1 r_1 r 1 r 2 r_2 r 2 E 2 E_2 E 2 unstable circular orbit. If we change the total energy by even a small amount, then the body either falls into the other body or flies away towards infinity.
There are also two other energy levels that cannot be seen on the image above:
E 3 E_3 E 3 stable circular orbit and E 4 E_4 E 4
Similarly, the constants of motion can be given by the initial distance r 0 r_0 r 0 v 0 v_0 v 0 ω 0 \omega_0 ω 0
L = r 0 2 ω 0 , E = 1 2 m v 0 2 + U e f f ( r 0 ) , \begin{align*} \mathcal{L} &= r_0^2 \omega_0, \\ \mathscr{E} &= \frac{1}{2} m v_0^2 + U_{eff}(r_0), \end{align*} L E = r 0 2 ω 0 , = 2 1 m v 0 2 + U e ff ( r 0 ) , where U e f f ( r ) = m L 2 2 r 2 − m r − m L 2 r 3 U_{eff}(r) = \frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r} - \frac{m \mathcal{L}^2}{r^3} U e ff ( r ) = 2 r 2 m L 2 − r m − r 3 m L 2 d r d τ = 0 \frac{dr}{d\tau} = 0 d τ d r = 0
E = m L 2 2 r 2 − m r − m L 2 r 3 , 0 = − E − m r + m L 2 2 r 2 − m L 2 r 3 , 0 = − r 3 E − r 2 m + r m L 2 2 − m L 2 , \begin{align*} \mathscr{E} &= \frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r} - \frac{m \mathcal{L}^2}{r^3}, \\ 0 &= - \mathscr{E} - \frac{m}{r} + \frac{m \mathcal{L}^2}{2 r^2} - \frac{m \mathcal{L}^2}{r^3}, \\ 0 &= - r^3 \mathscr{E} - r^2 m + r \frac{m \mathcal{L}^2}{2} - m \mathcal{L}^2, \\ \end{align*} E 0 0 = 2 r 2 m L 2 − r m − r 3 m L 2 , = − E − r m + 2 r 2 m L 2 − r 3 m L 2 , = − r 3 E − r 2 m + r 2 m L 2 − m L 2 , this can be solved by the cubic equation.
For circular orbits d r d τ = 0 \frac{dr}{d\tau} = 0 d τ d r = 0
E = m L 2 2 r 2 − m r − m L 2 r 3 , 0 = d d r ( − m L 2 r 3 + m L 2 2 r 2 − m r ) = 3 m L 2 r 4 − m L 2 r 3 + m r 2 , 0 = r 2 m − r m L 2 + 3 m L 2 , D = ( − m L 2 ) 2 − 4 ( m ) ( 3 m L 2 ) = m 2 L 4 − 12 m 2 L 2 , r = − ( − m L 2 ) ± m 2 L 4 − 12 m 2 L 2 2 m = m L 2 ± m L 4 − 12 L 2 2 m = L 2 ± L 4 − 12 L 2 2 , \begin{align*} \mathscr{E} &= \frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r} - \frac{m \mathcal{L}^2}{r^3}, \\ 0 &= \frac{d}{dr}\left(- \frac{m \mathcal{L}^2}{r^3} + \frac{m \mathcal{L}^2}{2 r^2} - \frac{m}{r}\right) \\ &= 3 \frac{m \mathcal{L}^2}{r^4} - \frac{m \mathcal{L}^2}{r^3} + \frac{m}{r^2}, \\ 0 &= r^2 m - r m \mathcal{L}^2 + 3 m \mathcal{L}^2, \\ D &= \left(-m \mathcal{L}^2\right)^2 - 4 (m) (3 m \mathcal{L}^2) \\ &= m^2 \mathcal{L}^4 - 12 m^2 \mathcal{L}^2, \\ r &= \frac{-(-m \mathcal{L}^2) \pm \sqrt{m^2 \mathcal{L}^4 - 12 m^2 \mathcal{L}^2}}{2m} \\ &= \frac{m \mathcal{L}^2 \pm m\sqrt{\mathcal{L}^4 - 12 \mathcal{L}^2}}{2m} \\ &= \frac{\mathcal{L}^2 \pm \sqrt{\mathcal{L}^4 - 12 \mathcal{L}^2}}{2}, \end{align*} E 0 0 D r = 2 r 2 m L 2 − r m − r 3 m L 2 , = d r d ( − r 3 m L 2 + 2 r 2 m L 2 − r m ) = 3 r 4 m L 2 − r 3 m L 2 + r 2 m , = r 2 m − r m L 2 + 3 m L 2 , = ( − m L 2 ) 2 − 4 ( m ) ( 3 m L 2 ) = m 2 L 4 − 12 m 2 L 2 , = 2 m − ( − m L 2 ) ± m 2 L 4 − 12 m 2 L 2 = 2 m m L 2 ± m L 4 − 12 L 2 = 2 L 2 ± L 4 − 12 L 2 , where − - − + + +
The stable and unstable orbits can converge when discriminant is zero:
D = m 2 L 4 − 12 m 2 L 2 = 0 , m 2 L 4 = 12 m 2 L 2 , L 2 = 12 , \begin{align*} D &= m^2 \mathcal{L}^4 - 12 m^2 \mathcal{L}^2 = 0, \\ m^2 \mathcal{L}^4 &= 12 m^2 \mathcal{L}^2, \\ \mathcal{L}^2 &= 12, \end{align*} D m 2 L 4 L 2 = m 2 L 4 − 12 m 2 L 2 = 0 , = 12 m 2 L 2 , = 12 , substituting into the equation for r r r
r = L 2 ± L 4 − 12 L 2 2 = L 2 2 = 12 2 = 6 = 3 r s , \begin{align*} r &= \frac{\mathcal{L}^2 \pm \sqrt{\mathcal{L}^4 - 12 \mathcal{L}^2}}{2} \\ &= \frac{\mathcal{L}^2}{2} \\ &= \frac{12}{2} \\ &= 6 \\ &= 3 r_s, \end{align*} r = 2 L 2 ± L 4 − 12 L 2 = 2 L 2 = 2 12 = 6 = 3 r s , this is the inner most stable circular orbit.
To predict the nature of orbit, we can use the derivative of the effective potential:
U e f f ′ ( r ) = 3 m L 2 r 4 − m L 2 r 3 + m r 2 U'_{eff}(r) = 3\frac{m \mathcal{L}^2}{r^4} - \frac{m \mathcal{L}^2}{r^3} + \frac{m}{r^2} U e ff ′ ( r ) = 3 r 4 m L 2 − r 3 m L 2 + r 2 m M ( k g ) M\ (kg) M ( k g ) m ( k g ) m\ (kg) m ( k g ) r 0 ( m ) r_0\ (m) r 0 ( m ) v 0 ( m s − 1 ) v_0\ (m\ s^{-1}) v 0 ( m s − 1 ) v p e r p 0 ( m s − 1 ) v_{perp0}\ (m\ s^{-1}) v p er p 0 ( m s − 1 ) L = 4.45 ⋅ 1 0 15 k g m 2 s − 1 , E = − 2.6 ⋅ 1 0 33 J , r 1 = 2.92 ⋅ 1 0 3 m , r 2 = 7.54 ⋅ 1 0 10 m , r 3 = N a N ⋅ 1 0 N a N m .
\begin{align*}
L &= 4.45 \cdot 10^{15}\ kg\ m^2\ s^{-1}, \\
E &= -2.6 \cdot 10^{33}\ J, \\
r_1 &= 2.92 \cdot 10^{3}\ m, \\
r_2 &= 7.54 \cdot 10^{10}\ m, \\
r_3 &= NaN \cdot 10^{NaN}\ m.
\end{align*}
L E r 1 r 2 r 3 = 4.45 ⋅ 1 0 15 k g m 2 s − 1 , = − 2.6 ⋅ 1 0 33 J , = 2.92 ⋅ 1 0 3 m , = 7.54 ⋅ 1 0 10 m , = N a N ⋅ 1 0 N a N m . Ellipse Data
Semi-major axis: 3.77 ⋅ 1 0 10 m 3.77 \cdot 10^{10} m 3.77 ⋅ 1 0 10 m
Distance of star from center: 3.77 ⋅ 1 0 10 m 3.77 \cdot 10^{10} m 3.77 ⋅ 1 0 10 m
For lightlike geodesics, ϵ = 0 \epsilon = 0 ϵ = 0
E 2 = ( d r d λ ) 2 + L 2 r 2 − 2 L 2 r 3 + 2 r ϵ − ϵ = ( d r d λ ) 2 + L 2 r 2 − 2 L 2 r 3 . \begin{align*} \mathcal{E}^2 &= \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3} + \frac{2}{r} \epsilon - \epsilon \\ &= \left(\frac{dr}{d\lambda}\right)^2 + \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3}. \end{align*} E 2 = ( d λ d r ) 2 + r 2 L 2 − r 3 2 L 2 + r 2 ϵ − ϵ = ( d λ d r ) 2 + r 2 L 2 − r 3 2 L 2 . Consider circular orbits (d r d λ \frac{dr}{d\lambda} d λ d r
E 2 = L 2 r 2 − 2 L 2 r 3 , \mathcal{E}^2 = \frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3}, E 2 = r 2 L 2 − r 3 2 L 2 , taking the derivative:
d d r ( L 2 r 2 − 2 L 2 r 3 ) = − 2 L 2 r 3 + 3 2 L 2 r 4 = 0 , \frac{d}{dr} \left(\frac{\mathcal{L}^2}{r^2} - \frac{2\mathcal{L}^2}{r^3}\right) = -2\frac{\mathcal{L}^2}{r^3} + 3 \frac{2\mathcal{L}^2}{r^4} = 0, d r d ( r 2 L 2 − r 3 2 L 2 ) = − 2 r 3 L 2 + 3 r 4 2 L 2 = 0 , and solving for r r r
− 2 L 2 r 3 + 3 2 L 2 r 4 = 0 , 2 L 2 r 3 − 3 2 L 2 r 4 = 0 , 2 r L 2 − 6 L 2 = 0 , r − 3 = 0 , r = 3 = 3 2 r s , \begin{align*} -2\frac{\mathcal{L}^2}{r^3} + 3 \frac{2\mathcal{L}^2}{r^4} &= 0, \\ 2\frac{\mathcal{L}^2}{r^3} - 3 \frac{2\mathcal{L}^2}{r^4} &= 0, \\ 2r \mathcal{L}^2 - 6\mathcal{L}^2 &= 0, \\ r - 3 &= 0, \\ r &= 3 \\ &= \frac{3}{2} r_s, \\ \end{align*} − 2 r 3 L 2 + 3 r 4 2 L 2 2 r 3 L 2 − 3 r 4 2 L 2 2 r L 2 − 6 L 2 r − 3 r = 0 , = 0 , = 0 , = 0 , = 3 = 2 3 r s , this is the radius of the photon sphere. Light rays at this distance orbit the body in circles.
