4-acceleration in a circular orbit around a black hole

[Thales Castro]

Homework Statement

A rocked describes a circular orbit around a black hole with angular velocity $\Omega$ (measured by a static observer at infinity) and Schwarzschild radius $r=R$. Calculate the 4-acceleration felt by the rocket.

Relevant Equations

Schwarzschild metric:
$$
ds^{2} = -\left( 1 - \frac{2M}{r} \right) + \frac{1}{1-\frac{2M}{r}} dr^{2} + r^{2}d\Omega^{2}
$$

Kepler law for GR:
$$
\Omega^{2} = \frac{M}{r^{3}}
$$

Christoffel symbols:
$$
\Gamma ^{\alpha}_{\mu \nu} = \frac{1}{2}g^{\alpha \beta}\left( \partial_{\mu}g_{\nu \beta} + \partial_{\nu}g_{\mu \beta} - \partial_{\beta}g_{\mu \nu}\right )
$$

4-acceleration:
$$
a^{\mu} = u^{\alpha} \nabla_{\alpha} u^{\mu} = \frac{d u^{\mu}}{d\tau} + \Gamma^{\mu}_{\alpha \beta}u^{\alpha}u^{\beta}
$$

In a circular orbit, the 4-velocity is given by (I have already normalized it)
$$
u^{\mu} = \left(1-\frac{3M}{r}\right)^{-\frac{1}{2}} (1,0,0,\Omega)
$$Now, taking the covariant derivative, the only non vanishing term will be

$$
a^{1} = \Gamma^{1}_{00}u^{0}u^{0} + \Gamma^{1}_{33}u^{3}u^{3}
$$

Evaluating the Christoffel symbols, we have:

$$
\Gamma^{1}_{00} = \frac{M}{r^2}\left( 1 -\frac{2M}{r} \right )
$$

$$
\Gamma^{1}_{33} = -r\left( 1 -\frac{2M}{r} \right )
$$

By putting these values in the equation for a^1, I get

$$
a^{1} = \left( 1 - \frac{2M}{R} \right )\left(1 - \frac{3M}{R} \right )^{-1} \left[\frac{M}{R^{2}} - R\Omega^{2} \right ] = 0
$$

Now, I don't see why the acceleration should be zero in this problem, but still I can't find what I have done wrong in my calculations. Can anyone help me? Thanks in advance.

 

[TSny]

Now, I don't see why the acceleration should be zero in this problem, ...

The ship is in free-fall. Therefore, it follows a time-like geodesic in spacetime. The equation for geodesic motion tells you something about the four acceleration.

Alternately, imagine going to a "local inertial frame" of the ship. What is the four-acceleration of the ship in this frame? From this, what can you conclude about the four-acceleration of the ship in any other frame of reference (such as the frame using the Schwarzschild coordinates)?

 

[Thales Castro]

The ship is in free-fall. Therefore, it follows a time-like geodesic in spacetime. The equation for geodesic motion tells you something about the four acceleration.

Alternately, imagine going to a "local inertial frame" of the ship. What is the four-acceleration of the ship in this frame? From this, what can you conclude about the four-acceleration of the ship in any other frame of reference (such as the frame using the Schwarzschild coordinates)?

Thanks for the reply. That makes sense. One thing that I discussed with my colleagues later is that the relation between Omega and R is not necessarily the one I gave in the text. The rocket could be using his propulsion to keep himself at the "wrong" speed in his orbit, and in this his worldline would not be a geodesic.

 

[phyzguy]

Are you assuming that it is orbiting at the Schwarzschild radius R? Or is the orbital radius r a parameter? In this case, the acceleration is not zero.

 

[TSny]

... the relation between Omega and R is not necessarily the one I gave in the text. The rocket could be using his propulsion to keep himself at the "wrong" speed in his orbit, and in this his worldline would not be a geodesic.

Oh. In that case, you can't assume M = Ω²r³ so the 4-acceleration would be nonzero.

In the last line of your first post, you seem to have let r = R. Did you mean to do that? I think @phyzguy was wondering about that also.

 

[Thales Castro]

Are you assuming that it is orbiting at the Schwarzschild radius R? Or is the orbital radius r a parameter? In this case, the acceleration is not zero.

Oh. In that case, you can't assume M = Ω²r³ so the 4-acceleration would be nonzero.

In the last line of your first post, you seem to have let r = R. Did you meant to do that? I think @phyzguy was wondering about that also.

Yeah, R is an arbitrary fixed radial coordinate (not necessarily the Schwarzschild radius 2M) in which the circular orbit takes place. r is the radial coordinate (in the same sense I would have an x coordinate and a fixed point x_{0})

 

[TSny]

Yeah, R is an arbitrary fixed radial coordinate (not necessarily the Schwarzschild radius 2M) in which the circular orbit takes place. r is the radial coordinate (in the same sense I would have an x coordinate and a fixed point x_{0})

OK. It was just a little confusing since the problem statement uses R for the Schwarzschild radius.

 

[George Jones]

$$
a^{1} = \left( 1 - \frac{2M}{R} \right )\left(1 - \frac{3M}{R} \right )^{-1} \left[\frac{M}{R^{2}} - R\Omega^{2} \right ] = 0
$$

According to what I calculate, the
$$\left(1 - \frac{3M}{R} \right )^{-1}$$
factor in the acceleration is not correct for non-geodesic uniform circular motion.

The 4-velocity is
$$u =\left(\frac{dt}{d\tau},0,0,\frac{d \phi}{d\tau} \right) = \frac{dt}{d\tau} \left(1,0,0,\Omega \right) ,$$
where $\Omega = d \phi /dt$ is constant.

Use the normalization of the 4-velocity to eliminate $dt / d \tau$.