# Timelike geodesic equations for the Schwarzschild metric

#### Deadstar

I'm following a slightly confusing set of notes in which I can't tell what exactly the timelike geodesic equations for the Schwarzschild metric are (seems to have about 3 different equations for them).

How are these derived, or alternatively, does anyone have a link to a site in which they are derived? (Or even a site which just states what they are!)

Related Special and General Relativity News on Phys.org

#### Deadstar

Ok maybe I'll explain further...

We have, for timelike geodesics... L=1 where

$$L = (1 - \frac{2MG}{r})\.{t}^2 - (1 - \frac{2MG}{r})^{-1}\.{r}^2 - r^2( \.{\theta}^2 + \sin^2(\theta) \.{\phi}^2)$$

But what part does this 1 actually play?

When we go on and write out the Euler-Lagrange equations this 1 doesn't do anything. So is the timelike geodesic just the same as the null geodesic (in which L=0)? I can't think this is so otherwise there would be no distinction between them.

Edit: Maybe it's just me but the Latex isn't showing right. Im sure you'lll know the rest of the equation anyway but click the code if you need to check.

#### Altabeh

Ok maybe I'll explain further...

We have, for timelike geodesics... L=1 where

$$L = (1 - \frac{2MG}{r})\.{t}^2 - (1 - \frac{2MG}{r})^{-1}\.{r}^2 - r^2( \.{\theta}^2 + \sin^2(\theta) \.{\phi}^2)$$

But what part does this 1 actually play?

When we go on and write out the Euler-Lagrange equations this 1 doesn't do anything. So is the timelike geodesic just the same as the null geodesic (in which L=0)? I can't think this is so otherwise there would be no distinction between them.

Edit: Maybe it's just me but the Latex isn't showing right. Im sure you'lll know the rest of the equation anyway but click the code if you need to check.
Timelike curves in any spacetime are characterized by setting $$ds^2>0$$. This inequality lets the problem be discussed inside the future directed light-cones so it gives all the possible trajectories for particles falling into the black hole. If you take

$$ds^2=(1-2\mu /r)dt^2-(1-2\mu /r)^{-1}dr^2-r^2d\theta^2-r^2\sin^2(\theta)d\phi^2,$$

to be the Schwarzschild metric, then for all massive particles, assuming only those which are radially moving, the inequality

$$(1-2\mu /r)dt^2>(1-2\mu /r)^{-1}dr^2$$

is determinant of the trajectories they follow when falling into the BH. To solve this differential equation, one simply acts as follows:

$$dt>\pm\frac{1}{(1-2\mu /r)}dr,$$

$$dt>\pm\frac{r}{(r-2\mu)}dr,$$

$$dt>\pm(\frac{1}{(r-2\mu)}dr+dr),$$

which when integrated from $$(t_0=0; r_0)$$ to $$(t;r)$$ gives

$$t>-2\mu\ln|\frac{r-2\mu}{(r_0-2\mu)}|-r+r_0+C,$$

for incoming particles, and

$$t>2\mu\ln|\frac{r-2\mu}{(r_0-2\mu)}|+r-r_0+C,$$

for outgoing particles with C being constant of integration and obviously $$r>R$$ where R is the radius of gravitating body. Remembering that the black hole is obsorbing everything toward itself we must single out the second solution only and just think of incoming (or falling) particles. Hence, one can for example set $$r_0=4\mu$$ to get

$$t>-2\mu\ln|{r/2\mu-1}|-r+C',$$

where now $$C'=C+4\mu$$. This inequality represents all the possible trajectories of massive particles (thus timelike trajectories) falling radially into the BH from $$t_0=0,r_0=4\mu$$. In case the particles turn out to be massless, then the inequality becomes equality which has, of course, more interest among physicists in GR. There one must take into account the Hawking radiation as the probable source of outgoing photons so the second solution is valid when a photon is to be discussed.

AB

#### pervect

Staff Emeritus
Science Advisor
I'm following a slightly confusing set of notes in which I can't tell what exactly the timelike geodesic equations for the Schwarzschild metric are (seems to have about 3 different equations for them).

How are these derived, or alternatively, does anyone have a link to a site in which they are derived? (Or even a site which just states what they are!)
The end results are given at http://www.fourmilab.ch/gravitation/orbits/, though they don't do the derivation.

Sean Caroll provides a bit more detail at http://arxiv.org/abs/gr-qc/9712019 in section 7. But you'll need to skip around a bit to follow the solution. The computationally easy way to solve the geodesic equations, which Caroll takes advantage of, is to take advantage of the symmetries of the problems, which are called Killing vectors.

The bare results of this approach, also presented by Caroll , are as follows

Let t(tau), r(tau), and phi(tau) be the worldline, paramaterized by it's proper time, on the equatorial plane of a geodesic in the Schwarzschild metric (so theta = 0).

Then
g_tt (dt/dtau) = -(1-2M/r) (dt/dtau) is constant everywhere along the wordline, and can be thought of as a sort of "conserved energy" , -$\tilde E$ in the fourmilab webpage

g_\phi\phi (dphi/dtau) = r^2 (dphi/dtau) is also constant everywhere along the worldine, and can be thought of as a sort of "conserved angular momentum", $\tilde L$ in the fourmilab webpage.

Finally one knows that g_tt (dt/dtau)^2 + g_rr (dr/dtau)^2 + g_\phi\phi (d phi/dtau)^2 = constant = -1 (it could be +1 with a different sign convention) for a timelike worldline, because this is a general property of any four vector. Why is the magnitude of this expression one? The simple argument is that the expression is a tensor quantity, independent of the coordinates used. If we know the value in one coordinate system, we know it in all. So lets find the value of the expression in coordinates that represent a locally Lorentz frame where the observer is at rest.

In that case dt/dtau = 1 (coordinate time is the same as proper time), all the spatial components of the velocity vanish, and g_00 = -1. Thus we see that in this case the magnitude of the four-velocity is g_00 (dt/dtau)^2 = -1. Because the expression is a tensor, it's true in all coordinate systems, not just the locally Lorentz one where we derived it.

Caroll derives the geodesic equations for a general metric much earlier in section 3, at 3.47 for the "parallel transport" definition of geodesics, and shows that it minimizes the action in the following section. (Perhaps this is the confusing set of notes you have?).

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