I Conservation of stress-energy tensor

[KleinMoretti]

TL;DR

does the equation ∇𝜇𝑇𝜇𝜈=0. imply conservation of energy

I came across this statement
"The covariant energy-momentum conservation lawis
∇𝜇𝑇𝜇𝜈=0.
Be careful though: "convariant conservation" equations do not imply that any component of the energy or momentum is actually conserved.
To get actual conserved quantities you need a symmetry. In particular an isometry associated with a Killing vector field."

now when by "convariant conservation" equations do not imply that any component of the energy or momentum is actually conserved." they are talking about global conservation right? because I thought that ∇𝜇𝑇𝜇𝜈=0. did imply energy is conserved locally

 

[Orodruin]

No, they are talking about local conservation. Local conservation means that, locally, a continuity equation is satisfied. It does not mean that - for example - energy density is conserved. In a small enough region, the time derivative of the energy (or momentum) in that region (in some local Minkowski frame) is equal to the flux of energy (or momentum) into that region.

 

[Ibix]

Note that $\nabla_\mu T_{\mu\nu}$ doesn't actually makes sense - you can't sum across a pair of lower or upper indices. You want $\nabla_\mu T^{\mu\nu}$.

MTW have a good discussion of this. Roughly what it means is that stress-energy that enters a small region of spacetime leaves it again and vice versa. They give the example of a cubic box full of powder, and consider a short time of it (so for example a 1 foot cube with a 1ns duration). If you do not disturb any powder in that time, all the stress-energy enters from the past boundary and leaves through the future boundary. If you open one side and scoop out some powder, all of it enters from the past, but what you scooped out leaves through a spatial boundary and only what remains leaves through the future boundary. All such possibilities are enforced by the continuity equation. It won't let powder just vanish or just appear in the box.

But it does not mean that $T^{00}$ (for example) is conserved. This is in contrast to something like the components of the four momentum in SR, where the individual components are conserved separately. That only happens when you have a Killing field (the symmetries it's talking about) and you use it as one of your basis vectors.

 

[KleinMoretti]

No, they are talking about local conservation. Local conservation means that, locally, a continuity equation is satisfied. It does not mean that - for example - energy density is conserved. In a small enough region, the time derivative of the energy (or momentum) in that region (in some local Minkowski frame) is equal to the flux of energy (or momentum) into that region.

but aren't they saying then that energy is not conserved locally

 

[KleinMoretti]

Note that $\nabla_\mu T_{\mu\nu}$ doesn't actually makes sense - you can't sum across a air of lower or upper indices. You want $\nabla_\mu T^{\mu\nu}$.

MTW have a good discussion of this. Roughly what it means is that stress-energy that enters a small region of spacetime leaves it again and vice versa. They give the example of a cubic box full of powder, and consider a short time of it (so for example a 1 foot cube with a 1ns duration). If you do not disturb any powder in that time, all the stress-energy enters from the past boundary and leaves through the future boundary. If you open one side and scoop out some powder, all of it enters from the past, but what you scooped out leaves through a spatial boundary and only what remains leaves through the future boundary. All such possibilities are enforced by the continuity equation. It won't let powder just vanish or just appear in the box.

But it does not mean that $T^{00}$ (for example) is conserved. This is in contrast to something like the components of the four momentum in SR, where the individual components are conserved separately. That only happens when you have a Killing field (the symmetries it's talking about) and you use it as one of your basis vectors.

but when we talk about conservation of energy in GR, it's always said that energy is conserved only locally and I though that's what this eq. ∇𝜇𝑇𝜇𝜈=0. meant.

 

[Ibix]

It's a continuity condition that means that you can't have stress-energy just pop into or out of existence (in particular, it forbids the "what if the sun vanished" question). And it is the closest thing GR has to a law of conservation of energy. But iy is not the same as "every component of the stress-energy tensor is individuallly and separately conserved.

Note that you do need to get the indices correct. What you are writing, $\nabla_\mu T_{\mu\nu}=0$, is nonsense.

 

[PeterDonis]

I came across this statement

Where? Please give a reference.

 

[KleinMoretti]

It's a continuity condition that means that you can't have stress-energy just pop into or out of existence (in particular, it forbids the "what if the sun vanished" question). And it is the closest thing GR has to a law of conservation of energy. But iy is not the same as "every component of the stress-energy tensor is individuallly and separately conserved.

Note that you do need to get the indices correct. What you are writing, $\nabla_\mu T_{\mu\nu}=0$, is nonsense.

ok just to clarify when we talk about conservation of energy in GR we are talking about the stress-energy tensor right? also like I said before I have always heard that in general relativity energy is only conserved locally and global conservation of energy can only be obtained in specific cases(spacetimes with a killing field being one of them) so what do you mean by is not the same as "every component of the stress-energy tensor is individuallly and separately conserved.

 

[PeterDonis]

but aren't they saying then that energy is not conserved locally

No. See below.

when we talk about conservation of energy in GR, it's always said that energy is conserved only locally and I though that's what this eq. ∇𝜇𝑇𝜇𝜈=0. meant.

That is what that equation means (when you get the indexes correct). But "energy is conserved only locally" doesn't mean what you think it means.

First, it is a general statement in GR. There are particular solutions in GR where there are other conservation laws you can state. But those are only particular solutions. In general in GR, energy is conserved only locally; there are no other conservation laws that apply generally, to all solutions, in GR.

Second, what is conserved locally is not "energy" but stress-energy. That doesn't just include energy density. It includes momentum density, pressure, and shear stresses. It also is affected by the local geometry of spacetime, since that affects how you count the "flow" of stress-energy in and out of a small spacetime 4-volume.

The statement that seems to be confusing you is this one:

"convariant conservation" equations do not imply that any component of the energy or momentum is actually conserved.
To get actual conserved quantities you need a symmetry. In particular an isometry associated with a Killing vector field."

The first part of this is just saying what I said above, that stress-energy is not just energy density. Energy density is just one component. The local conservation law includes multiple components, and you have to look at all of them.

The second part is talking about particular solutions in GR where you can find additional conservation laws that are not just local. But it only applies to those particular solutions. And it does not mean that the local conservation law stops being true in those solutions. The local conservation law is always true, in every solution.

I will say that what is quoted above is not very well stated by whatever reference it came from, IMO.

 

[KleinMoretti]

No. See below.


That is what that equation means (when you get the indexes correct). But "energy is conserved only locally" doesn't mean what you think it means.

First, it is a general statement in GR. There are particular solutions in GR where there are other conservation laws you can state. But those are only particular solutions. In general in GR, energy is conserved only locally; there are no other conservation laws that apply generally, to all solutions, in GR.

Second, what is conserved locally is not "energy" but stress-energy. That doesn't just include energy density. It includes momentum density, pressure, and shear stresses. It also is affected by the local geometry of spacetime, since that affects how you count the "flow" of stress-energy in and out of a small spacetime 4-volume.

The statement that seems to be confusing you is this one:


The first part of this is just saying what I said above, that stress-energy is not just energy density. Energy density is just one component. The local conservation law includes multiple components, and you have to look at all of them.

The second part is talking about particular solutions in GR where you can find additional conservation laws that are not just local. But it only applies to those particular solutions. And it does not mean that the local conservation law stops being true in those solutions. The local conservation law is always true, in every solution.

I will say that what is quoted above is not very well stated by whatever reference it came from, IMO.

okay so if I understand correctly you are saying that energy density, momentum density, pressure, and shear stresses are conserved together in what is stress-energy no?, but that does include energy and say energy is conserved right?. so what confuses me is the comment I referenced seems to be saying that conservation of stress-energy doesn't mean energy and momentum are conserved.

also even those specific cases with a symmetry where its possible to get other conservation laws what would be conserved would still be the stress-energy right?

 

[KleinMoretti]

also are there conservation laws in general relativity that are non-covariant

 

[PeterDonis]

you are saying that energy density, momentum density, pressure, and shear stresses are conserved together in what is stress-energy

Basically, yes. As I said, you also have to include the effects of spacetime geometry, since that affects how you count those various things.

that does include energy and say energy is conserved right?

Not by itself, no. Only stress-energy overall.

the comment I referenced seems to be saying that conservation of stress-energy doesn't mean energy and momentum are conserved.

No. As I said before, that comment is talking about something different when it talks about conservation of energy and momentum. It is talking about a global conservation based on a symmetry, which only holds in certain particular solutions--so it's something different from the local conservation of stress-energy that holds in all solutions.

even those specific cases with a symmetry where its possible to get other conservation laws what would be conserved would still be the stress-energy right?

No.

are there conservation laws in general relativity that are non-covariant

No.

 

[KleinMoretti]

Basically, yes. As I said, you also have to include the effects of spacetime geometry, since that affects how you count those various things.


Not by itself, no. Only stress-energy overall.


No. As I said before, that comment is talking about something different when it talks about conservation of energy and momentum. It is talking about a global conservation based on a symmetry, which only holds in certain particular solutions--so it's something different from the local conservation of stress-energy that holds in all solutions.


No.


No.

but does the stress-energy tensor have to be covariant in order to be conserved?

 

[Orodruin]

but does the stress-energy tensor have to be covariant in order to be conserved?

I think you need to take two steps back in terms of what you are trying to understand. As it is right now you seem to be throwing terminology around at random with a hit-and-miss philosophy. Why don’t you start by breaking down and explaining what you mean by the stress-energy tensor being covariant. (Because there is a way to make sense of that, although likely not what you think you are asking)

 

[KleinMoretti]

I might be wrong but from my mediocre understanding of physics covariance means that physical laws are true for all observers no matter where they are or how they’re moving.

 

[Orodruin]

What you are writing, $\nabla_\mu T_{\mu\nu}=0$, is nonsense.

They are writing $\nabla\mu T\mu\nu =0$. That’s just sloppy.

 

[Orodruin]

I might be wrong but from my mediocre understanding of physics covariance means that physical laws are true for all observers no matter where they are or how they’re moving.

That’s not the tensor being covariant. That is the relation being covariant.

 

[dextercioby]

They are writing $\nabla\mu T\mu\nu =0$. That’s just sloppy.

I reported the OP for the LaTex part asking a mod to fix it. Got rejected, the OP should fix his math formatting by himself. But he can't do that because he doesn't understand the difference between indices "upstairs" or "downstairs". You say sloppy, I say a word beginning with <<ig>>> which I cannot write completely, because its usual perception is offensive.

 

[Orodruin]

You say sloppy, I say a word beginning with <<ig>>> which I cannot write completely, because its usual perception is offensive.

There can be several underlying reasons for sloppiness. Either one just doesn't care about accuracy in delivering a message, or one does not understand the subject matter. There are probably more.

 

[PeterDonis]

@dextercioby the correct formatting appears in post #3. That's good enough for readers of the thread.