Conservation of energy in GR

  • #1
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I would like to understand better about the conservation of energy in GR.

Let us think of infinitesimal vacuum volume [tex]dr\ sin\theta d\theta d\phi[/tex] around the star in center.
Light emitted from the star hit the bottom surface, r, of the volume. Say violet light photons hit the area 1 photon/1 second of the bottom local time.
Ligth escapes the volume at the top surface, r+dr, with red-shifted color with the rate less than 1 photon/1 second of the top local time, I guess.

In coming energy and out going energy should be equal in this stationary case. How can I cofirm it for the case?
 
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  • #2
Cryo
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In coming energy and out going energy should be equal in this stationary case. How can I cofirm it for the case?
I am not an expert in GR, so I appologize in advance if I say something wrong, but why should the energy be conserved here? Energy of a particle is just the zero-th component of its four-momentum, here you clearly have a situation where the manifold is not trivial, so components of a four-vector will change if you parallel-transport it around the space (I am guessing), so why would you expect the energy conservation to apply?

Having said that. Looking at the Schwartzschild metric, it has no explicit time dependence, so, I guess, the relevant Lagrangian density will also not have explicit time dependence, and therefore there should be some quantity that is conserved as a result of this. But will it correspond to what we usually call energy?
 
  • #3
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Thanks Cryo.
I have understood conservation of energy as denial of perpetual motion machine that can produce as much energy as we wish.
If GR denies conservation of energy, I am interested in how it denies, e.g. do perpetual motion machines revive? (though I do not expect so.)
 
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  • #4
Cryo
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I have understood conservation of energy as denial of perpetual motion
That sounds rather abstract. How would you actually calculate the energy?
 
  • #5
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Rather abstract, I admit. I am not sure how I calculate the energy.

I am not sure also about definition of conservation.

[tex]a.\ T^{\mu\nu}_{,\nu}=0[/tex] , where "," means ordinary coordinate derivative, holds in classical and SR physices but doses not hold in GR
[tex]b.\ T^{\mu\nu}_{:\nu}=0[/tex], where ":" means covariant derivative, holds in general.
Which is the formula of "conservation" in definition ?

If it is a., energy is not coserved in GR. Based on our wish that something else of energy should be "conserved" non tensor t is invented , i.e.
[tex] (T^{\mu\nu}+t^{\mu\nu}){,\nu}=0[/tex]
 
  • #7
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Thanks Dale.
 
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  • #8
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I have studied GR by old text Dirac 1975 section 24 of which says,
"In flat spacetime equation (24.5) ( my b.) would become (my a.) and would then give conservation of energy and momentum. In curved space the conservation of energy and momentum is in only approxomation. The error is to be ascribed to the gravitational field working on the matter and having itself some energy and momentum."

I find the text available on the web at https://www.google.com/url?sa=t&rct...lativity.pdf&usg=AOvVaw108LkvdoRz-qsnip_6hwIP

This old interpretaiton was the reason why I thought that definition of conservation might be a.

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  • #9
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In elementary physics of motion in gravitational field we learn the consercvation of energy
[tex]\frac{1}{2}mv^2+mgh=Constant[/tex]
introducing gravitational potential energy.

After learning GR I think GR could give reinterpretaion of the conservation relation, not using potential energy that we cannot ascribe where it is because it does not have tensor property, but using curved spacetime or covariant derivative forluma of conservation relation. I would like to know it if there already is.

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  • #10
Dale
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I don’t know why it would be described as an approximation. Your b is exact in GR.
 
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  • #11
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Thanks again for reconfirming that b. stands in GR and it is called conservation relation, and a. is not called conservtion relation in GR and does not hold.
 
  • #12
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Light emitted from the star hit the bottom surface, r, of the volume. Say violet light photons hit the area 1 photon/1 second of the bottom local time.
Ligth escapes the volume at the top surface, r+dr, with red-shifted color with the rate less than 1 photon/1 second of the top local time, I guess.
In elementary physics of motion in gravitational field we learn the consercvation of energy
12mv2+mgh=Constant​
\frac{1}{2}mv^2+mgh=Constant
introducing gravitational potential energy.
In these cases of motion in gravitational field, the formula of covariant derivative divergence of stress energy tensor

[tex]T^{\mu\nu}_{:\nu}=T^{\mu\nu}_{,\nu}+\Gamma^{\mu}_{\nu\alpha}T^{\nu\alpha}+\Gamma^{\nu}_{\nu\alpha}T^{\mu\alpha}=0[/tex],
is applicable.

The second and the third term in RHS are proportional to T in rough saying.
Christoffel symbol Gammma should correspond to traditional g, gravitationla accerelation constant.

Gravitation force in hypothetical flat space in classical mechanics and SR is to be reinterpreted as the part of covariant divergence term in curved space.
 
  • #13
Matterwave
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In elementary physics of motion in gravitational field we learn the consercvation of energy
[tex]\frac{1}{2}mv^2+mgh=Constant[/tex]
introducing gravitational potential energy.

After learning GR I think GR could give reinterpretaion of the conservation relation, not using potential energy that we cannot ascribe where it is because it does not have tensor property, but using curved spacetime or covariant derivative forluma of conservation relation. I would like to know it if there already is.

Best
Unfortunately, moving to GR, you lose such a neat and convenient definition of "gravitational potential energy". In GR, such a concept is simply not well defined. A potential energy has to do with (conservative) forces - in GR, gravity is no longer a force so it becomes much harder to define its potential energy.
 
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  • #14
PeterDonis
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moving to GR, you lose such a neat and convenient definition of "gravitational potential energy". In GR, such a concept is simply not well defined
This is true in general, but there is a class of spacetimes, stationary spacetimes, in which there is a well-defined gravitational potential energy. The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.
 
  • #15
PeterDonis
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After learning GR I think GR could give reinterpretaion of the conservation relation
There is; the general concept you are looking for here is called "energy at infinity" and is a constant of the motion for free-falling test objects in a stationary spacetime.

Note, however, that this has nothing to do with your question about conservation laws for the stress-energy tensor; that is a whole different subject.
 
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  • #16
Matterwave
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This is true in general, but there is a class of spacetimes, stationary spacetimes, in which there is a well-defined gravitational potential energy. The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.
Could you remind me how the gravitational potential energy is defined in stationary space times? I do not recall a general definition. All I could say is given a time-like Killing field ##\xi^a##, there is a conserved quantity ##\xi^a u_a## for geodesics ##u^a##...but how that conserved quantity relates to a gravitational potential energy is escaping me at the moment.

EDIT: Specifically, I know one could associate a "total energy" (including gravitational) to the quantity ##E=-g_{ab}\xi^a u^b## and this quantity is conserved along geodesics. But how to deconstruct that into part Gravitational Potential energy and part "other" energy (kinetic energy/rest energy) is not apparent to me.
 
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  • #17
PeterDonis
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given a time-like Killing field ##\xi^a##, there is a conserved quantity ##\xi^a u_a## for geodesics ##u^a##
Yes. This is the energy at infinity, which can be thought of as the total of kinetic energy + potential energy. (More precisely, it's the energy per unit mass at infinity, assuming ##u^a## is a 4-velocity and not a 4-momentum.) The potential energy (more precisely, the potential energy per unit mass) is just ##V = \vert \xi^a \vert = \sqrt{ g_{ab} \xi^a \xi^b}##, i.e., the norm of the Killing field.
 
  • #18
PeterDonis
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The potential energy (more precisely, the potential energy per unit mass) is just ##V = \vert \xi^a \vert = \sqrt{ g_{ab} \xi^a \xi^b}##, i.e., the norm of the Killing field.
Actually, this definition makes the potential energy at infinity equal to ##1## for asymptotically flat spacetimes; the usual convention is for it to be ##0##, so that potential energy at any finite radius is negative. So ##V = \vert \xi^a \vert - 1## is the more usual definition.
 
  • #19
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The OP's scenario can be modeled with a stationary spacetime, so there is a well-defined notion of gravitational potential energy for that scenario.
Thanks Peter. I wonder where is or in other words distribution of gravitation energy as for this well-defined potential energy for my scenario.
 
  • #20
Matterwave
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Actually, this definition makes the potential energy at infinity equal to ##1## for asymptotically flat spacetimes; the usual convention is for it to be ##0##, so that potential energy at any finite radius is negative. So ##V = \vert \xi^a \vert - 1## is the more usual definition.
That's an interesting definition. I don't think I can recall seeing it anywhere, but admittedly it has been a while since I've done too much in GR. Could you give some motivating background for why it's a good definition, or perhaps a source that talks about it?
 
  • #21
PeterDonis
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Could you give some motivating background for why it's a good definition, or perhaps a source that talks about it?
I think Wald discusses it, but it's been a while since I looked at the sources for this.

As a quick heuristic motivation for the definition, consider that ##\vert \xi^a \vert## is the "redshift factor". So, for example, it gives the factor by which the energy of a photon emitted by an observer at rest at a finite radius is redshifted when it is received by an observer at infinity. Then consider Einstein's thought experiment for why there must be gravitational redshift, with the two observers being the ones just mentioned: the observer at infinity takes an object at rest relative to him and drops it; it free-falls down to the observer at finite radius, who takes it, converts it entirely into photons, and emits them back up to the observer at infinity. You will see that the "redshift factor" must also give the potential energy change between the observer at infinity and the observer at the finite radius.
 
  • #22
Matterwave
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I think Wald discusses it, but it's been a while since I looked at the sources for this.

As a quick heuristic motivation for the definition, consider that ##\vert \xi^a \vert## is the "redshift factor". So, for example, it gives the factor by which the energy of a photon emitted by an observer at rest at a finite radius is redshifted when it is received by an observer at infinity. Then consider Einstein's thought experiment for why there must be gravitational redshift, with the two observers being the ones just mentioned: the observer at infinity takes an object at rest relative to him and drops it; it free-falls down to the observer at finite radius, who takes it, converts it entirely into photons, and emits them back up to the observer at infinity. You will see that the "redshift factor" must also give the potential energy change between the observer at infinity and the observer at the finite radius.
I don't recall seeing this in Wald, but I will have a look. Thanks for these posts, they've been insightful. :)
 
  • #23
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Re: post#5,#6, #8 and #10
It is b.
Landau-Lifshitz old text says in chapter of energy momentum pseudotensor, that [tex]\int T^k_i\sqrt{-g}dS_k[/tex] is conserved in condition that [tex]\frac{\partial \sqrt{-g}T^k_i}{\partial x^k}=0[/tex] so the integral is not conserved in case covariant derivative of T equals zero. How should I conciliate one ane the other?
 
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  • #24
Dale
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I tend to avoid the pseudo tensor entirely. It isn’t a real tensor and I would rather stick with true tensors.

The conservation law I identified above applies for the actual stress energy tensor. If you use the pseudotensor then you should not expect it to apply.
 
  • #25
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Thnaks Dale.

The formula of differention,
[tex]T^{\mu\nu}_{:\nu}=0[/tex]
,is a equation of continuity (of energy-momentum).

The formula of 3d volume integration
[tex] \int(t^0_\mu+Y^0_\mu)\sqrt{-g}dx^1dx^2dx^3[/tex] Dirac(31.4)
, is a equation of conservation of energy momentum of the system where no flow in 3d surface boundary.

Differential thus local continuation of T is not enough and it needs t to expalin integral thus global conservation of energy momentum. Am I right? The formula of continuation and the formula of conservtion do not have to be the same one ?
 

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