What is the relationship between energy and speed in a Schwarzschild geometry?

In summary, the conversation discusses the relationship between the energy and speed of a rock thrown by a stationary observer in a Schwarzschild geometry. The formula E = \frac{m}{\sqrt{1-V^2}} is mentioned, but there is confusion about whether it applies to non-inertial frames. It is clarified that the formula applies when the rock's speed is measured with respect to an orthonormal frame on the observer's worldline, and the positions of the observer and rock coincide. The formula also applies in a static frame, such as the Schwarzschild metric, where there is a conserved quantity known as total energy.
  • #1
stephenmitten
3
0
If a stationary observer throws a rock out in the radial direction in a Schwarzschild geometry, what is the relationship between the energy of the rock and its speed in the observer's frame?

I'm a bit confused because the book seems to say it's [tex]E = \frac{m}{\sqrt{1-V^2}}[/tex], but I thought that only applied to inertial frames in special relativity.

Thanks.
 
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  • #2
That formula only makes sense if it is assumed that [itex]c=1[/itex], otherwise it could be written

[tex]E = \frac{mc^2}{\sqrt{1-V^2/c^2}}[/tex]

which I don't recognise. Over to someone else.
 
  • #3
If, with respect to an orthonormal frame on the observer's worldline, the rock has speed [itex]V[/itex] on release, then, at release and with respect to this frame, the total energy of the rock is given by the same expression as in special relativity.
 
  • #4
Thanks for the replies.

Is that because at release the positions of the observer and the rock coincide and since the spacetime is tangentially flat, the formula applies as in special relativity?

So is it true that if the rock's position was different from the observer's, the formula wouldn't apply?
 
  • #5
Can someone tell me why the formula applies in this situation?
 
  • #6
If, with respect to an orthonormal frame on the observer's worldline, the rock has speed LaTeX Code: V on release, then, at release and with respect to this frame, the total energy of the rock is given by the same expression as in special relativity.
Is that because at release the positions of the observer and the rock coincide and since the spacetime is tangentially flat, the formula applies as in special relativity?
Yes.
"Energy", especially "potential Energy" is a bit tricky, because for the latter there is not always a straightforward definition in GR.
So if you use local standards to measure the velocity of the stone, said formula holds. Energy is not conserved because you use a different frame for every point of the stone's worldline.
If you use a single (static) frame, like the Schwarzschild metric, there is a conserved quantity which one can identif with energy. If [tex]E_{kin}[/tex] is the kinetic energy as defined above, [tex]E_{kin}\sqrt{1-\frac{2M}{r}}[/tex] is the total energy, a constant.
 

Related to What is the relationship between energy and speed in a Schwarzschild geometry?

1. What is the formula for energy in General Relativity (GR)?

The formula for energy in GR is E = mc2, where E is energy, m is mass, and c is the speed of light in a vacuum.

2. How does the formula for energy in GR differ from the classical formula for energy?

The formula for energy in GR takes into account the effects of gravity and the curvature of space-time, while the classical formula does not. In GR, energy is not conserved in the same way it is in classical physics.

3. Can the formula for energy in GR be used to calculate the energy of an object in any situation?

No, the formula for energy in GR is only applicable in situations where gravity is a significant factor, such as near massive objects or in regions of strong gravitational fields.

4. How was the formula for energy in GR discovered?

The formula for energy in GR was derived by Albert Einstein in his theory of general relativity, which he published in 1915. It was a result of his efforts to unify the theories of gravity and special relativity.

5. Why is the formula for energy in GR important?

The formula for energy in GR is important because it allows us to understand and predict the behavior of energy in the presence of strong gravitational fields. It has been confirmed by numerous experiments and is a fundamental equation in modern physics.

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