Momentum conservation for a free-falling body in GR

In summary: So the result is zero.In summary, the conversation discusses a conservation law for the covector and the equation that represents it. When contracting the terms in the equation, some terms within the bracket cancel due to symmetry between the indices being contracted with the momentum. This leads to a simplified equation with fewer terms.
  • #1
complexconjugate
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Homework Statement
The conservation law for the energy-momentum vector of a free-falling massive particle is: $$m\frac{\mathrm{d}p^\alpha}{\mathrm{d}t} = -\Gamma^\alpha_{\beta\mu}p^\beta p^\mu$$. Show the conservation law for the covector reads: $$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}g_{\mu\nu,\beta}p^\beta p^\mu$$
Relevant Equations
I: $$m\frac{\mathrm{d}p^\alpha}{\mathrm{d}t} = -\Gamma^\alpha_{\beta\mu}p^\beta p^\mu$$
II: $$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}g_{\mu\nu,\beta}p^\beta p^\mu$$
Hello everyone!
It seems I can't solve this exercise and I don't know where I fail.
By inserting the metric on the lefthand side of I. and employing the chain rule, the equation eventually reads (confirmed by my notes from the tutorial):
$$m\frac{\mathrm{d}p_\delta}{\mathrm{d}t} = \Gamma^\gamma_{\beta\delta}g_{\mu\gamma}p^\beta p^\mu$$
Now contracting the metric with the Christoffel symbol and renaming indices gives $$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}\left[g_{\mu\beta,\nu}+g_{\nu\mu,\beta} - g_{\nu\beta,\mu} \right]p^\nu p^\mu$$
Now I don't understand why two terms in the brackets vanish. Is there some symmetry in the indices I'm missing?
Thanks for any hints.
 
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  • #2
Welcome to PF!

complexconjugate said:
Homework Statement:: Show the conservation law for the covector reads: $$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}g_{\mu\nu,\beta}p^\beta p^\mu$$
Something's wrong here. On the left side, ##\beta## is a lone index. But on the right side, ##\beta## is a summation index.

$$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}\left[g_{\mu\beta,\nu}+g_{\nu\mu,\beta} - g_{\nu\beta,\mu} \right]p^\nu p^\mu$$
Now I don't understand why two terms in the brackets vanish. Is there some symmetry in the indices I'm missing?
Thanks for any hints.
In general, none of the terms within the bracket vanish or cancel. However, after contracting the terms in the bracket with ##p^\nu p^\mu##, you will see that there will be some cancellation.
 
  • #3
TSny said:
Something's wrong here.
I noticed, I copied the Latex code and forgot to swap out the beta... the upper beta should be a nu.
TSny said:
you will see that there will be some cancellation
I think I see what you mean, can you tell me how to justify it properly? Something like symmetry between nu and mu because it's both times contracted with the momentum?

Thank you a lot!
 
  • #4
complexconjugate said:
I think I see what you mean, can you tell me how to justify it properly? Something like symmetry between nu and mu because it's both times contracted with the momentum?
Yes

$$m\frac{\mathrm{d}p_\beta}{\mathrm{d}t} = \frac{1}{2}\left[g_{\mu\beta,\nu}+g_{\nu\mu,\beta} - g_{\nu\beta,\mu} \right]p^\nu p^\mu$$

To see what happens, consider the first and last terms in the bracket of the right-hand side and contract them with ##p^\nu p^\mu##. Write out these terms explicitly; i.e., carry out the sum over ##\mu## and ##\nu##. You'll probably never need to do that again :oldsmile:

The idea is that ##g_{\mu\beta,\nu} - g_{\nu\beta,\mu}## is antisymmetric with respect to ##\mu## and ##\nu## and it's being contracted with ##p^\nu p^\mu## which is symmetric with respect to ##\mu## and ##\nu##
 
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Likes DEvens and complexconjugate

1. What is momentum conservation in general relativity (GR)?

Momentum conservation in GR is a fundamental principle that states that the total momentum of a system remains constant, unless acted upon by an external force. This means that in a closed system, the total momentum before an event is equal to the total momentum after the event.

2. How does momentum conservation apply to a free-falling body in GR?

In GR, momentum conservation still holds true for free-falling bodies. However, the concept of momentum is modified in GR to include both linear and angular momentum, as well as the energy and momentum of gravitational fields. This means that the total momentum of a free-falling body, including its gravitational interaction with other bodies, remains constant.

3. What is the significance of momentum conservation in GR?

Momentum conservation in GR is significant because it allows us to make predictions about the behavior of objects in gravitational fields. It also helps us understand the fundamental principles that govern the universe, such as the conservation of energy and the equivalence of mass and energy.

4. Are there any exceptions to momentum conservation in GR?

There are no known exceptions to momentum conservation in GR. However, the principle may break down in extreme situations, such as near black holes or during the early stages of the universe. In these cases, we need to use more complex theories, such as quantum mechanics or general relativity, to fully understand the behavior of the system.

5. How does momentum conservation in GR relate to the conservation of other physical quantities?

Momentum conservation in GR is closely related to the conservation of other physical quantities, such as energy and angular momentum. In fact, momentum conservation is often referred to as the "consequence" of these other conservation laws. This means that if momentum is conserved, then energy and angular momentum must also be conserved.

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