Is energy conserved in cosmology according to the laws of thermodynamics?

[WannabeNewton]

I'd say L is a 3-form.

A 3 - form is of the type (0 , 3). Your L defined by contracting the energy - momentum tensor with the killing vector is a one - form.

 

[Ben Niehoff]

OK, so I think what you meant to write is

\int_\Sigma \ast L

in which case I would agree that this integral gives the same value on any spacelike hypersurface \Sigma by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above). Hence this is a globally conserved charge (modulo some topological considerations, if the spacetime manifold has spacelike hypersurfaces that cannot be continuously deformed into one another).

So now I think I agree, that if you have a timelike Killing vector, you can define a globally conserved energy.

In the FLRW metric you do not generally have a timelike Killing vector, except in the special case of the inflation period, which is de Sitter space and hence maximally symmetric.

Edited to add: This globally-conserved energy is the energy of stuff in the spacetime. So for vacuum solutions such as Schwarzschild and de Sitter, this energy is zero anyway!

Are there any solutions with a finite density of matter that admit a timelike Killing vector? I'm not very familiar with fluid and dust solutions.

 

[WannabeNewton]

OK, so I think what you meant to write is

\int_\Sigma \ast L

in which case I would agree that this integral gives the same value on any spacelike hypersurface \Sigma by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above).

Just to clarify for myself, you are integrating over a space - like hyper surface because the hodge dual of L will give a 3 - form right?

 

[TrickyDicky]

OK, so I think what you meant to write is

\int_\Sigma \ast Lin which case I would agree that this integral gives the same value on any spacelike hypersurface \Sigma by the divergence theorem (given L is constructed from T and a timelike Killing vector as shown above). Hence this is a globally conserved charge (modulo some topological considerations, if the spacetime manifold has spacelike hypersurfaces that cannot be continuously deformed into one another).

So now I think I agree, that if you have a timelike Killing vector, you can define a globally conserved energy.

That's right Ben, I just didn't know how to Latex that asterisk. (Thought the Hodge op. was implicit in the way I wrote it anyway)

In the FLRW metric you do not generally have a timelike Killing vector

As I have noted several times too.

Are there any solutions with a finite density of matter that admit a timelike Killing vector? I'm not very familiar with fluid and dust solutions.

There are, but are not considered mainstream.

 

[PAllen]

There are, but are not considered mainstream.

There are some simple boring ones. A perfect fluid ball (nothing else in the universe) , with its vacuum part coinciding with exterior Schwarzschild is one. It's got everything, static, asymptotic flatness, you name it.

 

[TrickyDicky]

Still waiting for the page number from Penrose.

I don't ave access to the book right now, but I'm quite sure I saw this derivation there.

 

[bcrowell]

I don't ave access to the book right now, but I'm quite sure I saw this derivation there.

You wrote down something and claimed you had a reference to support it. Now other people have pointed out that it was incorrect, and you say you don't have the reference after all.

What is the expression for the global, scalar mass-energy that you claim to have found? You haven't supplied it yet. Once you provide it, it will remain to be shown that it can be interpreted as a measure of mass-energy, e.g., by showing that for a localized distribution of matter, it corresponds properly to the proper acceleration of a distant, static observer. I think it's extremely unlikely that you can demonstrate this, since what you'd presumably be reinventing would be the Komar mass, and I don't think the Komar mass can be reinvented in three lines of math, as you seem to be claiming in #45.

Note that the flux L that you defined in #45 is a vector, not a scalar. That means that any conserved charge you get is going to be a vector, not a scalar. That means you can't define a global conservation law for it, for the reasons given in the first reference in the FAQ -- but you still haven't shown any signs of having looked at any of the references in the FAQ.

Of course none of this has much to do with your statement in #1 that 'I've just read the FAQ about this and IMO it is "not even wrong" to say that energy conservation doesn't apply to Cosmology.' Cosmological spacetimes don't have timelike Killing vectors. I'm still waiting to hear you clearly admit that your assertion in #1 was wrong.

 

[TrickyDicky]

Why should I admit is wrong when it's not just because you say so?
You've been incorrect so many times before without admitting it (like for instance right now even if others with more knowledge than you have clearly pointed out like in post #52 after your #44) that I or anyone can't really take you seriously when you say something is wrong.

Be water, my friend.

 

[TrickyDicky]

BTW, in the MTW cited page, the infamous misleading phrase "conservation of energy doesn't apply to cosmology" (which is basically what I've been saying that needs to be better expressed) is nowhere to be found, actually it only refers to closed universes.

 

[TrickyDicky]

To come back to the OP once some things have been clarified, I must say that it is very hard for me to understand how fundamental physics laws are so easily dismissed by some.

I guess when the claim is made that energy is not conserved in cosmology it is not realized that not only the first law of thermodynamics is rescinded but also the second law since both laws are interconnected.

The measurement of energy is always in relative terms since there is no absolute measure of energy, only the transition of a system from one state into another can be defined, and it takes place through the flow or time-asymmetry of the second law, and the same is true of every measurement process.
Every measurement process also demonstrates the first law as well since the reasoning and relations that hold in the math require something that remains invariant (energy) over those relations (or else one could not get invariant results). The first and second laws are thus automatically entailed in every measurement process.

So I guess according to this, the claim is actually made in the FAQ that Thermodynamics doesn't apply to cosmology which is quite risible since it is used all the time .