# 4-acceleration in a circular orbit around a black hole

• Thales Castro
In summary: Omega \right) $$According to what I calculate, the$$\left(1 - \frac{3M}{R} \right )^{-1}$$factor in the acceleration is not correct for non-geodesic uniform circular motion. Thales Castro Homework Statement A rocked describes a circular orbit around a black hole with angular velocity \Omega (measured by a static observer at infinity) and Schwarzschild radius r=R. Calculate the 4-acceleration felt by the rocket. Relevant Equations Schwarzschild metric:$$
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. Thales Castro said: 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 and Cryo TSny said: 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. 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. Thales Castro said: ... 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. Last edited: phyzguy said: 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 said: 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}) Thales Castro said: 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. Thales Castro said:$$
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##.

Thales Castro and TSny

## What is 4-acceleration?

4-acceleration is a mathematical concept used in the theory of relativity to describe the change in velocity (both magnitude and direction) of an object as it moves through spacetime.

## How is 4-acceleration calculated in a circular orbit around a black hole?

In a circular orbit around a black hole, 4-acceleration is calculated using the Schwarzschild metric, which takes into account the strong gravitational pull of the black hole. It is also affected by the velocity and position of the object in the orbit.

## What does 4-acceleration tell us about the motion of an object in a circular orbit around a black hole?

4-acceleration can tell us about the rate at which an object's velocity is changing in both magnitude and direction, as well as the forces affecting its motion. In a circular orbit around a black hole, it can also give insight into the effects of the strong gravitational pull of the black hole on the object's motion.

## How does 4-acceleration change as the object gets closer to the black hole?

As the object gets closer to the black hole, the 4-acceleration increases due to the stronger gravitational pull. This increase can be seen in the curvature of the orbit and the change in the object's velocity direction as it moves around the black hole.

## Can 4-acceleration be used to predict the behavior of objects in a circular orbit around a black hole?

Yes, 4-acceleration can be used to predict the behavior of objects in a circular orbit around a black hole. By calculating the 4-acceleration at different points in the orbit, we can understand how the object's velocity and direction will change and how it will respond to the gravitational pull of the black hole.

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