# Relative speed of orbiting rockets in Schwarzschild metric

1. Apr 27, 2015

### zimo123

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
Schwarzschild metric with dr=dθ=0.

3. 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

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

2. Apr 28, 2015

### Orodruin

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

3. Apr 28, 2015

### zimo123

Actually I am not sure what that means ... Is it just:
$$\frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A}$$

4. Apr 28, 2015

### Orodruin

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

5. Apr 28, 2015

### unscientific

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

6. Apr 28, 2015

### zimo123

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.

7. Apr 28, 2015

### zimo123

I already did that, see my first post in the attempt at a solution part.

8. Apr 28, 2015

### unscientific

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.

9. Apr 28, 2015

### zimo123

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.

10. Apr 28, 2015

### unscientific

(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, $\gamma_w \neq \frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A}$.

11. Apr 28, 2015

### zimo123

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...

12. Apr 28, 2015

### unscientific

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 $\gamma_u, \gamma_v$?

13. Apr 28, 2015

### Orodruin

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

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?

14. Apr 28, 2015

### zimo123

Using the mass-energy equivalence principle?
$$E^2-\vec{p}\cdot\vec{p}=m^2$$

15. Apr 28, 2015

### Orodruin

Staff Emeritus
No, there are no masses involved. You only have the 4-velocities.

16. Apr 28, 2015

### zimo123

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.

17. Apr 28, 2015

### Orodruin

Staff Emeritus
You really should drop the masses, the only thing they do here is to obfuscate what is going on. How would you express the gamma factor using the 4-velocity? (Hint: how is the 4-momentum and 4-velocity related.) What gamma factor is it that you get out of doing so? It needs to be relative to something. What vector does $\vec e_t$ represent for an observer at rest in the particular frame?