# Relative speed of orbiting rockets in Schwarzschild metric

## Homework Statement

Two rockets are orbiting a Schwarzschild black hole of mass M, in a circular path at some location R in the equatorial plane θ=π/2. The first (rocket A) is orbiting with an angular velocity Ω=dΦ/dt and the second (rocket B) with -Ω (they orbit in opposite directions).

Find the speed of B as measured by A, whenever they meet. Equivalently, find the relative Lorentz factor.

## Homework Equations

Schwarzschild metric with dr=dθ=0.

## The Attempt at a Solution

I have calculated the four-velocity components of each rocket in terms of M, R and Ω.
$$\frac{\mathrm{d}\phi}{\mathrm{d}\tau}=\frac{\pm\Omega}{\sqrt{1-2M/R-R^2\Omega^2}}$$
$$\frac{\mathrm{d}t}{\mathrm{d}\tau}=\frac{1}{\sqrt{1-2M/R-R^2\Omega^2}}$$

Obviously the other two components are zero.

According to my textbook (Hartle), observed quantities correspond to projections on the observers' basis vectors. I know I should compute the relative velocity with:

$$V=\frac{\vec{u_B}\cdot\vec{e_{\phi}}}{\vec{u_B}\cdot\vec{e_t}}$$

where e_t and e_Φ are the coordinate basis vectors of the observer (in this case of rocket A) and u_B the four-velocity of rocket B (with the -Ω). The problem is I don't understand why I should divide by the projection onto e_t. For example, when we calculate the measured energy, one only has to project on the relevant basis vector:

$$E=-\vec{p}\cdot\vec{e_t}$$

Or for the momentum

$$P=\vec{p}\cdot\vec{e_r}$$

Thanks for your help!

## Answers and Replies

Orodruin
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I suggest starting by computing the relative gamma factor. How would you do that in SR?

I suggest starting by computing the relative gamma factor. How would you do that in SR?
Actually I am not sure what that means ... Is it just:
$$\frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A}$$

Orodruin
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No. You can express it in a manifestly coordinate invariant fashion. What is the gamma factor?

No. You can express it in a manifestly coordinate invariant fashion. What is the gamma factor?

As Orodruin said, try expressing ##\frac{d\tau}{dt}## using the schwarzschild metric. It might be helpful to first write down the metric.

No. You can express it in a manifestly coordinate invariant fashion. What is the gamma factor?
The γ factor is dt/dτ, which is the time-component of the 4-velocity. I have already expressed it above in the derived equations.

As Orodruin said, try expressing ##\frac{d\tau}{dt}## using the schwarzschild metric. It might be helpful to first write down the metric.
I already did that, see my first post in the attempt at a solution part.

I already did that, see my first post in the attempt at a solution part.
Ok good.Now given ##\gamma_A## and ##\gamma_B## how do you relate them with ##\gamma_w## where ##\vec w## is the relative velocity? You can do that the standard way or using the rapidity approach.

Ok good.Now given ##\gamma_A## and ##\gamma_B## how do you relate them with ##\gamma_w## where ##\vec w## is the relative velocity? You can do that the standard way or using the rapidity approach.
I would think it is the ratio between the two as I previously posted. That would measure how the proper time of A ellapses relative to the proper time of B.

I would think it is the ratio between the two as I previously posted. That would measure how the proper time of A ellapses relative to the proper time of B.

(proper time of A)/(Proper time of B) is NOT proper time of B in A's frame.

You want ##\gamma_w##, proper time of B in A's frame. In short, ##\gamma_w \neq
\frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A} ##.

(proper time of A)/(Proper time of B) is NOT proper time of B in A's frame.

You want ##\gamma_w##, proper time of B in A's frame. In short, ##\gamma_w \neq
\frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A} ##.
Proper time of B in A's frame ? Isn't the notion of proper time tied to one reference frame ? i.e. there is just one proper time of B: that measured by B.
Or do you mean "how would A compute the proper time of B?" In which case I would guess by creating two new 4 velocity vectors, one in which A is stationary and one describing the relative motion of B to A. I have tried doing that but unsuccessfully...

Proper time of B in A's frame ? Isn't the notion of proper time tied to one reference frame ? i.e. there is just one proper time of B: that measured by B.
Or do you mean "how would A compute the proper time of B?" In which case I would guess by creating two new 4 velocity vectors, one in which A is stationary and one describing the relative motion of B to A. I have tried doing that but unsuccessfully...

Particle A has velocity ##\vec u##. Particle B has velocity ##\vec v##. Relative velocity (also B's velocity in A's rest frame) between them is ##\vec w##.

How do you find ##\gamma_w## in terms of ##\gamma_u, \gamma_v##?

A trick is to use invariance in 4-vectors. Given that the 4-momentum is ##P = \gamma(c,\vec u)## and using invariance ##U \cdot V = U^{'} \cdot V^{'}## (Hint: Which frames would you use?), how would you find ##\gamma_w## in terms of ##\gamma_u, \gamma_v##?

Orodruin
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As Orodruin said, try expressing ##\frac{d\tau}{dt}## using the schwarzschild metric. It might be helpful to first write down the metric.

I never said to do this. This is not a necessary step, although you will need the metric tensor.

The γ factor is dt/dτ, which is the time-component of the 4-velocity. I have already expressed it above in the derived equations.
No, expressing the relative gamma factor this way using the coordinate is too SR and not the way to go.

Can you think of an expression which is manifestly Lorentz invariant and evaluates to the relative gamma factor between the 4-velocities U and V regardless of the frame it is evaluated in?

No, expressing the relative gamma factor this way using the coordinate is too SR and not the way to go.

Can you think of an expression which is manifestly Lorentz invariant and evaluates to the relative gamma factor between the 4-velocities U and V regardless of the frame it is evaluated in?
Using the mass-energy equivalence principle?
$$E^2-\vec{p}\cdot\vec{p}=m^2$$

Orodruin
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Using the mass-energy equivalence principle?
$$E^2-\vec{p}\cdot\vec{p}=m^2$$
No, there are no masses involved. You only have the 4-velocities.

No, there are no masses involved. You only have the 4-velocities.
I know the norm of any 4-vector is Lorentz invariant, but the only expression which comes to mind that contains the gamma factor would be the energy.
$$E=-\vec{p}\cdot\vec{e_t}=m\gamma$$
I guess the mass from the momentum ##\vec{p}## and from the right hand side would cancel.

Orodruin
Staff Emeritus