# I Vacuum energy and Relativity

1. May 21, 2017

### davidge

Matter is responsible for the stress-energy tensor. What is included in "matter"? Does vacuum energy contributes somewhat to the stress-energy tensor? Would it be correct to include all forms of energy in the stress-energy tensor, through the mass-energy equivalence?

2. May 21, 2017

### Staff: Mentor

No, "matter" is a name we give to one particular kind of stress-energy tensor (roughly speaking, its pressure is much less than its energy density). It's not the only thing that contributes to it.

In certain models, yes.

Yes. But not just energy--or more precisely, energy density. The stress-energy tensor includes momentum density/energy flux, pressure, and other stresses such as shear stress. That's why it's called the stress-energy tensor instead of just the energy tensor.

3. May 22, 2017

### martinbn

Not all forms of energy, anything that is gravitational will not be included in the stress-energy tensor.

4. May 22, 2017

### davidge

What would be gravitational? Curvature itself?
So how many stress-energy tensors are there?
When deriving the EFE from variational principle, the action due matter is the only thing included in the stress-energy tensor. Why this is so?

5. May 22, 2017

### davidge

Or maybe I'm confusing stress-energy tensor with momentum-energy tensor?

6. May 22, 2017

### Staff: Mentor

Yes.

You mean, how many different "kinds" of stress-energy tensors are there? Remember, the SET is a general object; it has ten independent components, just like the Einstein tensor. A particular "kind" of SET is one that has particular relationships between the components; we use words like "matter" to describe particular kinds of SETs because particular relationships between the components are characteristic of particular kinds of "stuff" that have stress-energy. (Note also that the relationships between the components can be frame-dependent, since the components themselves are; the relationships used to define different "kinds" of SETs are usually given in the rest frame of the "stuff".)

For example, a very general kind of SET that is often used in models is that of a perfect fluid. In the rest frame of the fluid, this SET looks like this:

$$T^a{}_b = \begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \end{bmatrix}$$

where $\rho$ is the energy density and $p$ is the pressure, as measured by an observer at rest in the fluid. This is the kind of SET that is used, for example, in the FRW solutions in cosmology. We then give particular names to particular kinds of relationships between $\rho$ and $p$:

"Matter" means $p \ll \rho$ (usually approximated as $p = 0$).

"Radiation" means $p = \frac{1}{3} \rho$.

"Dark energy" means $p = - \rho$.

7. May 22, 2017

### davidge

Thanks Peter. I see what you mean. But what if we consider the most general stress-energy tensor, namely $$T_{\mu \nu} = - \frac{S_{\text{matter}}}{\sqrt{-g}}K$$ where $K$ is the other terms that go into the expression.

8. May 22, 2017

### Staff: Mentor

Where are you getting this expression from? It doesn't look right.

The most general stress-energy tensor is derived from a Lagrangian for whatever kind of "stuff" you are considering.

9. May 22, 2017

### davidge

That expression you obtain by varying the Einstein-Hilbert action and requiring the variation of the total action to be zero.

10. May 22, 2017

### Staff: Mentor

Please either give a reference or show your work. I know how the SET is supposed to be derived from a Lagrangian. If what you wrote looked like what you are supposed to get when you vary the Lagrangian for whatever non-gravitational "stuff" is present, I wouldn't be asking about it.

11. May 22, 2017

### davidge

$$S = S_{\text{gravity}} + S_{\text{matter}} \\ = \frac{1}{16 \pi G} \int_M R + \int_M \mathcal{L}_{\text{matter}} \\ \delta S = 0 = \delta \frac{1}{16 \pi G} \int_M R + \delta \int_M \mathcal{L}_{\text{matter}} \\ = \ ... \\ = \frac{1}{16 \pi G} \int d^4x (R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R) \sqrt{-g} \ \delta g^{\mu \nu} + \delta \int_M \mathcal{L}_{\text{matter}} \\ \frac{1}{\sqrt{-g}} \frac{1}{\delta g^{\mu \nu}} \delta S = \frac{1}{16 \pi G}(R_{\mu \nu} - \frac{1}{2}g_{\mu \nu}R) + \frac{1}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{\delta g^{\mu \nu}} \\ = 0 \\ \Rightarrow T_{\mu \nu} = -2 \frac{1}{\sqrt{-g}} \frac{\delta S_{\text{matter}}}{\delta g^{\mu \nu}}$$

12. May 22, 2017

### Staff: Mentor

Your derivation is wrong. I don't know where you're getting it from, but you need to go back and review. Also, the expression you end up with here for the SET is not the same as the one you wrote down before. I suspect you don't properly understand what the $\delta$ notation means in this context.

This Wikipedia article gives a correct derivation, and also references Carroll's textbook, which, like most GR textbooks, goes into this in more detail. Note that the expression for the SET is not the one you wrote down (either time).

13. May 22, 2017

### davidge

14. May 22, 2017

### Staff: Mentor

Well, that explains it--you're using somebody's thesis as a source instead of a textbook or peer-reviewed paper. There's a reason why PF's guidelines for acceptable sources do not include theses. I strongly suggest forgetting about this source and learning this material from a textbook. I've already given a good reference to one.

Ok, but what does that mean? More to the point, how do you compute it given a quantity? You don't appear to know how to do that. Learning this from a textbook will help teach you.

15. May 22, 2017

### Staff: Mentor

Aside from the issues with deriving the correct general form for this, what question were you trying to get answered here?

16. May 22, 2017

### Staff: Mentor

I also note that, having skimmed through the relevant section (starting on p. 24), the derivation does appear to be done correctly there (it's basically the same as the one in the Wikipedia article I linked to)--but you have either misunderstood it or mis-transcribed it, since your post does not match it.

17. May 22, 2017

### davidge

I thought the most general form of the SET was that that appears when varying the action. Thus I was asking what is included in $S_{\text{matter}}$ in that expression, that is, what is included in "matter".

18. May 22, 2017

### Staff: Mentor

Yes, but that "most general form" makes no assumptions at all about what is included in $S_{matter}$, or more precisely in $\mathcal{L}_{matter}$, since $S_{matter}$ is just $\mathcal{L}_{matter}$ integrated over spacetime. So asking what is included in that expression is pointless in the most general case, since there is no answer other than "everything that isn't gravity".

19. May 22, 2017

### Staff: Mentor

Because when you split things up that way, and only when you split things up that way, both sides of the EFE have vanishing covariant divergence. In other words, the Einstein tensor $G^{ab}$ and the stress-energy tensor $T^{ab}$ satisfy $G^{ab}{}_{; b} = T^{ab}{}_{; b} = 0$. This is because the Einstein tensor satisfies geometric identities called the Bianchi identities, and the two tensors are equal (except for constant factors) by the EFE, so the SET must satisfy the same identities as the Einstein tensor. And the Einstein tensor is the only second rank tensor derivable from the metric and its derivatives that satisfies these identities. So you can't move any "gravity" terms from the LHS to the RHS of the EFE and include them in "stress-energy" without violating those identities.

The reason this is important is that it enforces local conservation of stress-energy: $T^{ab}{}_{; b} = 0$ says that stress-energy cannot be created or destroyed anywhere in spacetime. It can only flow from one region of spacetime to another. All of the observations we ordinarily associate with conservation of energy and momentum follow from this. So the reason things are split up the way they are is so that stress-energy is locally conserved.

20. May 23, 2017

### davidge

Ah, ok. Thanks for the explanation.