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
$R_{\mu\nu}+\frac{1}{2}Rg_{\mu\nu}=\kappa T_{\mu\nu}^{matter}=\Omega g_{\mu\nu}-\kappa T_{\mu\nu}^{generic}$?

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Blog Entries: 3 Recognitions: Gold Member 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.
 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.

Blog Entries: 3
Recognitions:
Gold Member

 Quote by 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.
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.

 Quote by 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.
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?

 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/s...s/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?

Blog Entries: 3
Recognitions:
Gold Member
 Quote by 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/s...s/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?
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.

 Quote by 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.
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?

Blog Entries: 3
Recognitions:
Gold Member
 Quote by Messenger Ok, glad I haven't gone crazy. 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?
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)

 Quote by 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)
"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?

Blog Entries: 3
Recognitions:
Gold Member
 Quote by 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?
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.

 Quote by 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.
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!

Sorry, another question...

Found the following definition:
 In the fluid's rest frame, the components of this stress-energy tensor have the expected form (insert into a slot of T, as the 4-velocity of observer, just the fluid's 4-velocity): $T^\alpha_\beta \mu^\beta=[(\rho+p)\mu^\alpha \mu^\beta +p\delta^\alpha_\beta]\mu^\beta = -(\rho+p)\mu^\alpha +p\mu^\alpha=-\rho\mu^\alpha;$ i.e; $T^0_\beta \mu^\beta=-\rho=$-(mass energy density)$=-dp^0/dV,$ $T^j_\beta \mu^\beta=0=$-(momentum density)$=-dp^j/dV,$
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?

Blog Entries: 3
Recognitions:
Gold Member
 Quote by Messenger This part $T^0_\beta \mu^\beta=-\rho=$-(mass-energy density)$=-dp^0/dV$ bothers me.
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.

Blog Entries: 9
Recognitions:
Gold Member
 Quote by Messenger 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?
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.

 Quote by PeterDonis 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.
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?

 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.
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.

Blog Entries: 9
Recognitions:
Gold Member
 Quote by Messenger 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.
$p$ 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.