Relative speed of orbiting rockets in Schwarzschild metric

  • Thread starter zimo123
  • Start date
  • #1
18
0

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.
[tex]\frac{\mathrm{d}\phi}{\mathrm{d}\tau}=\frac{\pm\Omega}{\sqrt{1-2M/R-R^2\Omega^2}}[/tex]
[tex]\frac{\mathrm{d}t}{\mathrm{d}\tau}=\frac{1}{\sqrt{1-2M/R-R^2\Omega^2}}[/tex]

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:

[tex]V=\frac{\vec{u_B}\cdot\vec{e_{\phi}}}{\vec{u_B}\cdot\vec{e_t}}[/tex]

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:

[tex]E=-\vec{p}\cdot\vec{e_t}[/tex]

Or for the momentum

[tex]P=\vec{p}\cdot\vec{e_r}[/tex]

Thanks for your help!
 

Answers and Replies

  • #2
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,308
7,149
I suggest starting by computing the relative gamma factor. How would you do that in SR?
 
  • #3
18
0
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:
[tex]\frac{\gamma_B}{\gamma_A}=\frac{(\mathrm{d}t/\mathrm{d}\tau)_B}{(\mathrm{d}t/\mathrm{d}\tau)_A}[/tex]
 
  • #4
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,308
7,149
No. You can express it in a manifestly coordinate invariant fashion. What is the gamma factor?
 
  • #5
1,734
13
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.
 
  • #6
18
0
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.
 
  • #7
18
0
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.
 
  • #8
1,734
13
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.
 
  • #9
18
0
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.
 
  • #10
1,734
13
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} ##.
 
  • #11
18
0
(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...
 
  • #12
1,734
13
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##?
 
  • #13
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,308
7,149
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?
 
  • #14
18
0
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?
[tex]E^2-\vec{p}\cdot\vec{p}=m^2[/tex]
 
  • #15
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,308
7,149
Using the mass-energy equivalence principle?
[tex]E^2-\vec{p}\cdot\vec{p}=m^2[/tex]
No, there are no masses involved. You only have the 4-velocities.
 
  • #16
18
0
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.
[tex]E=-\vec{p}\cdot\vec{e_t}=m\gamma[/tex]
I guess the mass from the momentum ##\vec{p}## and from the right hand side would cancel.
 
  • #17
Orodruin
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
Gold Member
17,308
7,149
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?
 
  • Like
Likes unscientific

Related Threads on Relative speed of orbiting rockets in Schwarzschild metric

Replies
1
Views
979
Replies
0
Views
2K
  • Last Post
Replies
4
Views
1K
Replies
6
Views
1K
  • Last Post
Replies
10
Views
397
  • Last Post
Replies
4
Views
3K
Replies
4
Views
2K
Replies
1
Views
1K
T
Replies
2
Views
920
Replies
0
Views
1K
Top