# Conservation of energy in GR

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

Let us think of infinitesimal vacuum volume $$dr\ sin\theta d\theta d\phi$$ 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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Cryo
Gold Member
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?

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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Cryo
Gold Member
I have understood conservation of energy as denial of perpetual motion

That sounds rather abstract. How would you actually calculate the energy?

Rather abstract, I admit. I am not sure how I calculate the energy.

I am not sure also about definition of conservation.

$$a.\ T^{\mu\nu}_{,\nu}=0$$ , where "," means ordinary coordinate derivative, holds in classical and SR physices but doses not hold in GR
$$b.\ T^{\mu\nu}_{:\nu}=0$$, 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.
$$(T^{\mu\nu}+t^{\mu\nu}){,\nu}=0$$

Dale
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Which is the formula of "conservation" in definition ?
It is b.

• Cryo and sweet springs
Thanks Dale.

• Dale
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.

Best

• Cryo
In elementary physics of motion in gravitational field we learn the consercvation of energy
$$\frac{1}{2}mv^2+mgh=Constant$$
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

Dale
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I don’t know why it would be described as an approximation. Your b is exact in GR.

• sweet springs
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.

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

$$T^{\mu\nu}_{:\nu}=T^{\mu\nu}_{,\nu}+\Gamma^{\mu}_{\nu\alpha}T^{\nu\alpha}+\Gamma^{\nu}_{\nu\alpha}T^{\mu\alpha}=0$$,
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.

Matterwave
Gold Member
In elementary physics of motion in gravitational field we learn the consercvation of energy
$$\frac{1}{2}mv^2+mgh=Constant$$
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.

• sweet springs
PeterDonis
Mentor
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.

PeterDonis
Mentor
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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Matterwave
Gold Member
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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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.

PeterDonis
Mentor
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.

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.

Matterwave
Gold Member
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?

PeterDonis
Mentor
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.

Matterwave
Gold Member
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. :)

Re: post#5,#6, #8 and #10
It is b.
Landau-Lifshitz old text says in chapter of energy momentum pseudotensor, that $$\int T^k_i\sqrt{-g}dS_k$$ is conserved in condition that $$\frac{\partial \sqrt{-g}T^k_i}{\partial x^k}=0$$ 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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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.

Thnaks　Dale.

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

The formula of 3d volume integration
$$\int(t^0_\mu+Y^0_\mu)\sqrt{-g}dx^1dx^2dx^3$$ 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 ?

PAllen
The (integral conservation) equations Einstein, Landau-Lifschitz, and presumably Dirac, specify for the energy momentum pseudo-tensor are exact rigorous equations. They hold in any coordinates. The issue is that the pseudo-tensor is not covariant, nor tensor density or any other standard geometric object. Further, interpreting it as 'energy including gravity' in some given coordinates, typically leads to implausible results. Thus, the majority view is that the pseudotensor does not meaningfully localize energy, and the conservation laws are not meaningful. This majority view is further bolstered by Noether's theorems, which imply there cannot be well formed conservation laws in GR.

However, there are experts who disagree with this view. Nakanishi (google his papers) argues that the pseudotensor produces physically plausible results in harmonic coordinates, and this plus the fact that conservation equations hold in any coordinates, rescues the relevance of the pseudotensor.

Thnaks PAllen
Thus, the majority view is that the pseudotensor does not meaningfully localize energy, and the conservation laws are not meaningful. This majority view is further bolstered by Noether's theorems, which imply there cannot be well formed conservation laws in GR.
Noether's theorem tells us conservation laws cannot stand in GR. It's great.

Back to my post#9, I will be satisfied better if someone show me how conservation of energy in gravitational field
$$\frac{1}{2}mv^2+mgh=const.$$
in high school physics, should be reinterpreted in GR rigorously.

PeterDonis
Mentor
I will be satisfied better if someone show me how conservation of energy in gravitational field

$$\frac{1}{2}mv^2+mgh=const.$$

in high school physics, should be reinterpreted in GR rigorously.

I already answered that in post #15: the energy at infinity is a constant of geodesic motion in GR, for spacetimes in which it is well-defined. The Newtonian formula you give is an approximation to the GR energy at infinity for the case where the range of height ##h## is small compared to the range over which ##g## changes, and for which ##v## is much less than ##c##. (The approximation also requires subtracting off the rest energy ##mc^2##, which is part of the energy at infinity in GR, and adjusting the "zero point" of potential energy to a finite height.)

Post #15
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.
Let me take an example to understand your explanation. On the ground I pop up the baseball of mass m. The ball reaches height h and falls back into my glove. In this case of high school physics class, what, how much and where is "energy at infinity" and "potential energy" in rigorous GR?

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PeterDonis
Mentor
On the ground I pop up the baseball of mass m. The ball reaches height h and falls back into my glove. In this case of high school physics class, what, how much and where is "energy at infinity" and "potential energy" in rigorous GR?

Remember that the formula for energy at infinity (per unit mass) is ##\xi^a u_a = g_{ab} \xi^a u^b##, where ##\xi^a## is the timelike Killing vector field and ##u^b## is the 4-velocity. We can just multiply by the rest mass ##m## to get the energy total.

We will assume that the Earth is not rotating, so we can use the simple values from the Schwarzschild metric. Then, in standard Schwarzschild coordinates, we have

$$\xi^a = \left( 1, 0, 0, 0 \right)$$

The simplest place to calculate the energy at infinity is at the top of the ball's trajectory, where it is momentarily at rest. Then its 4-velocity in Schwarzschild coordinates is

$$u^b = \left( \frac{1}{\sqrt{1 - 2M / r}}, 0, 0, 0 \right)$$

The energy at infinity is then simply

$$E = m g_{ab} \xi^a u^b = m \left( 1 - \frac{2M}{r} \right) \frac{1}{\sqrt{1 - 2M / r}} = m \sqrt{1 - \frac{2M}{r}}$$

where ##r## here is the radial coordinate of the ball at the top of its trajectory. For this case we can approximate ##r = R + H##, where ##R## is the radius of the Earth and ##H## is the maximum height the ball reaches. So we have

$$E = m \sqrt{1 - \frac{2M}{R + H}}$$

Since ##H << R## and ##M << R## (note that we are using units in which ##G = c = 1##), we can further approximate this as

$$E = m \sqrt{1 - \frac{2M}{R} \left( 1 - \frac{H}{R} \right)} = m \left( 1 - \frac{M}{R} \left( 1 - \frac{H}{R} \right) \right) = m \left( 1 - \frac{M}{R} + \frac{M}{R^2} H \right)$$

Now we simply observe that ##g = M / R^2## near the Earth's surface, and subtract off the constant ##m \left( 1 - M/R \right)##, to obtain ##E = m g H##. Note that, since the ball is at rest at this point, the total energy is equal to the potential energy.

From here, since we are in a regime where Newtonian gravity is a good approximation, finding the general formula ##E = \frac{1}{2} m v^2 + m g h## is straightforward, since ##E## is a constant of the motion and we already know how ##v## and ##h## are related, given that the ball is at rest with ##v =0## at ##h = H##.

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