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- Thread starter Philosophaie
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- #2

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Start by writing down the Schwarzschid metric for a spherically symmetric, nonrotating body:

[tex]

ds^2 = -\left( 1 - \frac{2GM}{r} \right) \textrm{d} t^2 + \left( 1 - \frac{2GM}{r} \right)^{-1} \textrm{d} r^2 + r^2 ( \textrm{d} \theta^2 + \sin^2(\theta) \textrm{d} \varphi^2 ) \textrm{.}

[/tex]

To get the geodesic equations from this, you can either calculate all the Christoffel symbols (long and tedious, but straightforward) or use the calculus of variations (my favorite method). In either case, I'd suggest setting [tex] e^{f(r)} = \left( 1 - \frac{2GM}{r} \right) [/tex] for ease of notation. You'll end up with four very ugly-looking equations (one for each of [tex] t, r, \theta [/tex], and [tex] \varphi [/tex]). By symmetry, you can set [tex] \theta = \frac{\pi}{2} [/tex] (i.e., [tex] \theta' = 0 [/tex]). The [tex] t [/tex] equation will then give you [tex] e^{f(r)} t' = E [/tex], where [tex] E [/tex] is a constant of motion. The [tex] \varphi [/tex] equation will give [tex] r^2 \varphi' = L [/tex], where [tex] L [/tex] is another constant of motion ("angular momentum"). The [tex] r [/tex] equation gives

[tex]

e^{-f(r)} (r')^2 + \frac{L^2}{r^2} - e^{-f(r)} E^2 = \epsilon \textrm{,}

[/tex]

where [tex] \epsilon [/tex] is another consant of motion. For massive particles, you can set [tex] \epsilon = -1 [/tex] without loss of generality (why?). Then you can rearrange the above equation to get

[tex]

\frac{1}{2} (r')^2 + V(r) = \mathcal{E} \textrm{,}

[/tex]

where [tex] V(r) = \frac{1}{2} \left(1 + \frac{L^2}{r^2} \right) e^{f(r)} [/tex] is a "potential energy" and [tex] \mathcal{E} = \frac{1}{2} E^2 [/tex] is a "total energy." This looks exactly like the form of a classical equation of motion, and you can analyze it using similar techniques. In particular, determining values for [tex] E [/tex], [tex] L [/tex], and [tex] r_{0} [/tex] that lead to bounded or unbounded paths is straightforward.

[tex]

ds^2 = -\left( 1 - \frac{2GM}{r} \right) \textrm{d} t^2 + \left( 1 - \frac{2GM}{r} \right)^{-1} \textrm{d} r^2 + r^2 ( \textrm{d} \theta^2 + \sin^2(\theta) \textrm{d} \varphi^2 ) \textrm{.}

[/tex]

To get the geodesic equations from this, you can either calculate all the Christoffel symbols (long and tedious, but straightforward) or use the calculus of variations (my favorite method). In either case, I'd suggest setting [tex] e^{f(r)} = \left( 1 - \frac{2GM}{r} \right) [/tex] for ease of notation. You'll end up with four very ugly-looking equations (one for each of [tex] t, r, \theta [/tex], and [tex] \varphi [/tex]). By symmetry, you can set [tex] \theta = \frac{\pi}{2} [/tex] (i.e., [tex] \theta' = 0 [/tex]). The [tex] t [/tex] equation will then give you [tex] e^{f(r)} t' = E [/tex], where [tex] E [/tex] is a constant of motion. The [tex] \varphi [/tex] equation will give [tex] r^2 \varphi' = L [/tex], where [tex] L [/tex] is another constant of motion ("angular momentum"). The [tex] r [/tex] equation gives

[tex]

e^{-f(r)} (r')^2 + \frac{L^2}{r^2} - e^{-f(r)} E^2 = \epsilon \textrm{,}

[/tex]

where [tex] \epsilon [/tex] is another consant of motion. For massive particles, you can set [tex] \epsilon = -1 [/tex] without loss of generality (why?). Then you can rearrange the above equation to get

[tex]

\frac{1}{2} (r')^2 + V(r) = \mathcal{E} \textrm{,}

[/tex]

where [tex] V(r) = \frac{1}{2} \left(1 + \frac{L^2}{r^2} \right) e^{f(r)} [/tex] is a "potential energy" and [tex] \mathcal{E} = \frac{1}{2} E^2 [/tex] is a "total energy." This looks exactly like the form of a classical equation of motion, and you can analyze it using similar techniques. In particular, determining values for [tex] E [/tex], [tex] L [/tex], and [tex] r_{0} [/tex] that lead to bounded or unbounded paths is straightforward.

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- #3

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[tex]ds^{2}=-(1-2GM/r)dt^{2}+(1-2GM/r)^{-1}dr^{2}[/tex]...

I already calculated Affinity (Torsion=0), Riemann Tensor and the Ricci Tensor from the metric. I got the four ugly looking equations from the diagonal of the Geodesic Equation. This is where I get stuck. I would like to have more than just [tex]\theta=\pi/2[/tex]. Is there a solution or a set of solution to these four Geodesic Equations no matter how difficult for all [tex]\theta[/tex]?

- #4

- 139

- 12

The Schwarzchild metric is:

LaTeX Code: ds^{2}=-(1-2GM/r)dt^{2}+(1-2GM/r)^{-1}dr^{2} ...

Yeah...sorry about that. \facepalm

Due to spherical symmetry, you can always change coordinates so that [tex] \theta = \frac{\pi}{2} [/tex], [tex] \theta' = 0 [/tex]. Briefly, this is because, since we have invariance under rotations, the "direction of angular momentum" is conserved, hence the motion of a particle must be planar. (More rigorously, to each Killing vector on a manifold we can associate a constant of geodesic motion. The timelike Killing vector of Schwarzschild yields conservation of "energy," while the three spherical Killing vectors yield conservation of the three components of the "angular momentum.")

I don't know of any analytical solutions to these equations, except for very special cases. Like I said, however, they're (fairly) straightforward to analyze nonetheless; for example, you can derive, without too much work, the rate of precession of the perihelion of a planet using perturbation theory (to a first approximation). You should get [tex] \Delta \varphi = \frac{6 \pi G^2 M^2}{L^2} [/tex], where [tex] \Delta \varphi [/tex] indicates the amount the perihelion advances per orbital period.

- #5

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Thanks for the info!

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