1. Jul 23, 2012

### Messenger

Studying and looking through fluid tensors used in GR and have a question to make sure I understand correctly:

If I had an isotropic and homogeneous perfect fluid $\Omega g_{\mu\nu}$ and within this fluid I had a generic stress energy tensor $\kappa T_{\mu\nu}^{generic}$ but defined it so that $\Omega g_{\mu\nu}=\kappa T_{\mu\nu}^{generic}+\kappa T_{\mu\nu}^{matter}$ where the generic stress energy tensor was the inverse of the stress energy tensor of matter, would I still be able to equate this to successive contractions of the Riemann? Meaning is it still mathematically permissible to write
$R_{\mu\nu}+\frac{1}{2}Rg_{\mu\nu}=\kappa T_{\mu\nu}^{matter}=\Omega g_{\mu\nu}-\kappa T_{\mu\nu}^{generic}$?

2. Jul 24, 2012

### Mentz114

The RHS of the EFE can certainly be decomposed into components representing matter and an electric field, say,
$$R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=\kappa T_{\mu\nu}^{matter}+\kappa T_{\mu\nu}^{EM}$$
so I don't see why your decomposition would be a problem. Whether it is physically possible to have dust and a PF with pressure in the same emt is another question.

3. Jul 24, 2012

### Messenger

Hi Mentz,
Thanks for the reply. That is what I was thinking also. I had been reading about the cosmological constant $\Lambda g_{\mu\nu}$ and had seen it described as a uniform perfect fluid. This confused me so was trying to figure out what exactly is the relation between a constant with no subscripts and a stress-energy tensor when written as fluid tensors. I just wanted to know if anyone knew of any a priori reason this couldn't be done that I hadn't run across.

4. Jul 24, 2012

### Mentz114

If UαUβ = gαβ and p=-μ then the perfect fluid EMT

Tαβ = (μ+p)UαUβ + pgαβ becomes Tαβ = pgαβ.

Presumably that's the 'uniform' perfect fluid.

Last edited: Jul 24, 2012
5. Jul 24, 2012

### Messenger

I don't understand...If p isn't independent with respect to $U_{\alpha}U_{\beta}$, I can see getting an answer besides zero for any integration, but why is this called "uniform" (or perhaps that was your point )? Maybe I am confused on the original cosmological constant...didn't the magnitude of it not matter since it was constant with respect to $U_{\alpha}U_{\beta}$ and so had no bearing on curvature?

Last edited: Jul 24, 2012
6. Jul 25, 2012

### Messenger

Now I am really confused...on Wikipedia's page for Lambdavacuum solutions http://en.wikipedia.org/wiki/Lambdavacuum_solution it has
$G^{\alpha\beta}=-\Lambda diag(-1,1,1,1)$ but the problem I see with this is that it should always be zero if my $\Omega g_{\mu\nu}=\kappa T_{\mu\nu}^{generic}+\kappa T_{\mu\nu}^{matter}$ decomposition is to hold true. I can see a stress energy tensor with $\rho$ and p since those are actually functions of their component positions, but $\Lambda$ isn't, it is an independent constant so I don't understand how $G^{\alpha\beta}=-\Lambda diag(-1,1,1,1)$ can ever be anything other than zero.

On http://theoretical-physics.net/dev/src/fluid-dynamics/general.html it is explained that a stress energy tensor is $\frac{dp^\alpha}{dV}=-T_\alpha^\beta \mu^\beta$ but for constant $p^\alpha$ then the tensor should be zero, no?

7. Jul 25, 2012

### Mentz114

It is confusing. But I don't think the EMT of a Lambdavac is treatable as a physical, Eulerian, fluid. The energy and pressure is the same in all frames which is impossible for a material fluid. Also, the energy density and pressure (weirdly) have opposite signs. The fact that the Lambda stuff looks the same to everyone makes it a good candidate for the 'vacuum' of quantum physics.

8. Jul 25, 2012

### Messenger

Ok, glad I haven't gone crazy. (Woops on the plus sign in the first post.) Maybe it will become clearer after I start studying the Friedmann equations. I have read that the cosmological constant was used to keep the universe static to match the observations at the time, so it should be defined or explained somewhere in the derivation.

Does EMT stand for electromagnetic tensor?

9. Jul 25, 2012

### Mentz114

EMT is energy-momentum tensor. Sometimes called SET, stress-energy tensor.

This link is the best succinct treatment I've seen of the Friedmann equations and cosmological constant,
http://ned.ipac.caltech.edu/level5/Carroll2/frames.html

(There are a couple of small typos in the text where he uses lower case λ instead of the upper case)

10. Jul 25, 2012

### Messenger

"EMT is energy-momentum tensor."

Am actually a fan of Carroll's videos, but I don't see anything more in depth than the introduction of equations (8) and (9). Any idea where I can find an English translation of Friedmann's papers?

11. Jul 25, 2012

### Mentz114

For me the merit of that article is that it shows clearly how the cosmological constant is introduced so that equations (8) and (9) have a solution with $\dot{a}=0$.

I can't help with Friedmann's publications, but the topic is covered in every decent GR textbook.

12. Jul 25, 2012

### Messenger

I did have a copy of Gravitation by MTW, but it is one of those texts for me that have to be read over and over (not that I actually have read the entire thing heh). I will get it again. Thanks Mentz!

13. Jul 25, 2012

### Messenger

Sorry, another question...

Found the following definition:
This part $T^0_\beta \mu^\beta=-\rho=$-(mass-energy density)$=-dp^0/dV$ bothers me. Sticking with the definition of a pf tensor, shouldn't this technically read
$T^0_\beta \mu^\beta=-d\rho/dV=-dp^0/dV,$ and then integrate this function $-d\rho/dV= \rho(V)$ with $\int (\rho(V))dV$ to find the total mass-energy density within a volume V? Meaning a perfect fluid with constant $\rho$ has no stress-energy and thus no mass-energy density?

14. Jul 25, 2012

### Mentz114

Me too. I don't know what it means. ρ has the dimensions of energy/volume, [ML2T-2 L-3] = [ML-1T-2]. So does pressure=force/area. Thus integrating ρdV or pkdV will give energy in the volume.

15. Jul 25, 2012

### Staff: Mentor

The units of $\rho$, as Mentz114 says, are energy/volume. The units of $d \rho / dV$ would therefore be energy/volume^2 (which doesn't really make physical sense).

The units of $p^{0}$ are energy (it's the zero component of the energy-momentum 4-vector), so the units of $dp^{0} / dV$ are also energy/volume.

All the equation $T^\alpha_\beta \mu^\beta = - \rho \mu^\alpha = -dp^\alpha/dV$ is really saying is that contracting the SET with the fluid's 4-velocity at an event gives you the 4-momentum of the fluid element at that event.

16. Jul 25, 2012

### Messenger

Something still isn't clicking in my head yet...
With a change in pressure with respect to a change in volume for the zero component $dp^0 / dV$, then pressure p was the original component as shown in the perfect fluid equation and this is equivalent to mass-energy. For $\rho$ it is saying the units coming out are Energy/volume=$\rho$. What was original component zero, pure energy? Shouldn't it be reading units of $d \rho / dV$ which is the change of density with respect to change in volume? Isn't that required in order to have stress-energy and to line up with $-\rho=p$ from the definition of the stress-energy tensor of a fluid?

The way it is written, one states that the energy within a volume is from the entire density of the fluid. It also states that the energy is equivalent to only the change in pressure within the volume.

17. Jul 25, 2012

### Staff: Mentor

$p$ in the equations I wrote (and you wrote) isn't pressure; it's 4-momentum. More specifically, the 4-vector $p^\alpha$ is the 4-momentum of a fluid element, and $p^0$ and $p^i$ are its time and space components in a given frame. Sorry for the confusing notation, but it's standard.

In the equation contracting the SET with the fluid's 4-velocity, pressure does not appear at all; only the energy density $\rho$. As I said before, that's because what the equation is telling you is that contracting the SET with the fluid's 4-velocity gives you the 4-momentum of the fluid element. In the fluid's rest frame, the fluid element is at rest (of course), so the only nonzero component of its 4-momentum is the 0 (time) component, the energy density, i.e., $\rho$. Pressure doesn't appear in the 4-momentum at all.

To extract the pressure from the SET, you have to project the SET into a spatial hypersurface that's orthogonal to the fluid's 4-velocity. I believe MTW goes into this, but I can't find the reference right now.

18. Jul 25, 2012

### Messenger

To make it more clear, shouldn't it be stating that measurable mass-energy density $\rho^{meas}$ is equivalent to the stress energy of a perfect fluid, $\frac{dp}{dV}=\frac{d\rho}{dV}=\rho^{meas}=\frac{E}{V}$?

Ok, saw your reply and will dig into MTW so that I can interpret it, heh. Thanks!

19. Jul 25, 2012

### Staff: Mentor

I don't understand why you would write $\frac{d\rho}{dV}$ for "energy density". Energy density is just $\rho$. You can call it $\rho^{meas}$ if you want to emphasize that its value can be measured, I suppose, but I don't see the point; it's still just one thing.

As for the other equalities, $dp^0 / dV$ is just the local, differential version of $E / V$; i.e., it is the limit of $E / V$ as the volume goes to zero, centered on a particular event of interest (where "volume" is measured in the spacelike hypersurface orthogonal to the fluid's 4-velocity at that event). In other words, it's just another way of writing the energy density $\rho$. So I don't see any problem with what you wrote except for the issue with $\frac{d\rho}{dV}$ that I discussed above.

20. Jul 25, 2012

### Messenger

Above the quote I gave for the equations, it also states:
The pressures don't make up the 4-momentum?

21. Jul 25, 2012

### Messenger

Ok, it was hanging me up when it seemed that the $dp^0 / dV$ was from the perfect fluid equation, and I couldn't see how $\rho$ wouldn't also be of the exact same form.

22. Jul 25, 2012

### Messenger

What do you make of where Mentz shows that density and pressure have the same units, but that density and pressure/volume would not? That was another thing that was hanging me up about needing the same form of $\frac{d\rho}{dV}$.

$p=\frac{N}{A}=\frac{kg m}{sec^2 m^2}=\frac{kg}{sec^2m}$
$\rho=\frac{kg m^2}{sec^2 m^3}=\frac{kg}{sec^2 m}$

23. Jul 25, 2012

### Staff: Mentor

No. The spatial components of 4-momentum are momentum densities, not pressures; they are only nonzero for an object that is moving in the particular frame of reference you are using. Fluid elements are at rest in the coordinates being used in this example, so their momentum is zero.

Pressure has to do with momentum exchange across surfaces, yes, but it's not the same as momentum density.

Yes, pressure (force per unit area) and energy density (energy per unit volume) have the same units. That should be obvious just from the observation that energy (work) is force times distance.

Pressure / volume, pressure per unit volume, doesn't make physical sense; force per unit area per unit volume?

24. Jul 26, 2012

### Staff: Mentor

Went back and re-read this in the course of answering Messenger's post. If by pk you mean a spatial component of 4-momentum, integrating it over a volume gives momentum, not pressure. (Integrating the 0 component of 4-momentum over a volume gives energy.)

Multiplying pressure times a small change in volume, PdV (using capital P here to avoid confusion with momentum), gives the work done in the course of the volume change. So integrating PdV from starting volume to ending volume would give total work done in the course of some process of expansion or compression. But if there is no volume change, there is no work done; so just integrating pressure over some unchanging volume doesn't really give anything meaningful.

25. Jul 26, 2012

### Messenger

If we know the units for $\rho$, how can I figure out the units for $-\frac{dp^0}{dV}$ in
$T^0_\beta \mu^\beta=-\rho=$-(mass energy density)$=-dp^0/dV$? Seems the units for momentum are $\frac{kg m}{s}$ so the units for $-dp^0/dV$ would be $\frac{kg m}{sm^3}=\frac{kg }{sm^2}$?