Fluid Tensors and the Cosmological Constant

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

 

[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.

 

[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.

 

[Mentz114]

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 U_αU_β = g_αβ and p=-μ then the perfect fluid EMT

T_αβ = (μ+p)U_αU_β + pg_αβ becomes T_αβ = pg_αβ.

Presumably that's the 'uniform' perfect fluid.

 

[Messenger]

If U_αU_β = g_αβ and p=-μ then the perfect fluid EMT

T_αβ = (μ+p)U_αU_β + pg_αβ becomes T_αβ = pg_αβ.

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?

 

[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?

 

[Mentz114]

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?

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.

 

[Messenger]

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?

 

[Mentz114]

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)

 

[Messenger]

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?

 

[Mentz114]

"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.

 

[Messenger]

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!

 

[Messenger]

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?

 

[Mentz114]

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 p^kdV will give energy in the volume.

 

[PeterDonis]

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.

 

[Messenger]

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.

 

[PeterDonis]

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.

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.

 

[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!

 

[PeterDonis]

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, $\frac{dp}{dV}=\frac{d\rho}{dV}=\rho^{meas}=\frac{E}{V}$?

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.

 

[Messenger]

$p$ 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.

Above the quote I gave for the equations, it also states:

One type of matter studied extensively later in this book is a "perfect fluid". A perfect fluid is a fluid or gas that (1) moves through spacetime with a 4-velocity u which may vary from event to event and (2) exhibits a density of mass-energy $\rho$ and an isotropic pressure p in the rest frame of each fluid element. Shear stresses, anisotropic pressures, and viscosity must be absent, or the fluid is not perfect. The stress-energy tensor for a perfect fluid at a given event can be constructed from the metric tensor, g, the 4-velocity u, and the rest frame density and pressure $\rho$ and p:

The pressures don't make up the 4-momentum?

 

[Messenger]

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.

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.

 

[Messenger]

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.

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 $\frac{d\rho}{dV}$.

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

 

[PeterDonis]

The pressures don't make up the 4-momentum?

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.

What do you make of where Mentz shows that density and pressure have the same units, but that density and pressure/volume would not?

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?

 

[PeterDonis]

Thus integrating ρdV or p^kdV will give energy in the volume.

Went back and re-read this in the course of answering Messenger's post. If by p^k 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.

 

[Messenger]

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

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 $\frac{kg m}{sm^3}=\frac{kg }{sm^2}$?

 

[PeterDonis]

If we know the units for $\rho$, how can I figure out the units for $-\frac{dp^0}{dV}$?

If you're trying to work things out in conventional units, the units of $p^0$ are energy, not momentum, and the units of $V$ are volume, so the units of $dp^0 / dV$ are energy per unit volume, or energy density, just like $\rho$. Strictly speaking, the units of all components of 4-momentum have to be the same, so the spatial components are $c$ times the components of ordinary momentum, so that they also have units of energy. ("Momentum density" in the SET, in ordinary units, is really momentum density times $c$, so it has units of energy density, like the other SET components.)

One of the advantages of using "natural" units, where $c = G = 1$, is that you don't have to worry about all this except at the final step where you compare your theoretical results with experiments; and then it's just a matter of inserting appropriate factors of $c$ and $G$ to make the units come out right. Most relativity textbooks use "natural" units for this reason, which is why factors of $c$ and $G$ don't appear in equations like the one you wrote down. In "natural" units, all of the SET components (energy density, momentum density, pressure, stress, etc.) have units of curvature, or inverse length squared, since length is the only "fundamental" unit. Energy density, for example, has units of length (energy) divided by length cubed (volume), or inverse length squared.

 

[Messenger]

Hi Peter,

Looking through the relationships between one-forms and gradients. Forgetting about mass, momentum and energy for now I have a question on strictly fluid dynamics. If we take two volumes of a perfect fluid with each element at rest, with $\rho^1>\rho^2$ and each with a constant $\rho$ and p for each element in their respective volumes, what would I be measuring with one-forms and gradients? What is different between the two fluids that would give different stress-energy tensor values at anyone element?

 

[PeterDonis]

Forgetting about mass, momentum and energy for now

If you're going to ask things like "what would I be measuring" or "what is different between the two fluids", then trying to forget about mass, momentum, and energy is kind of counterproductive.

what would I be measuring with one-forms and gradients?

Not sure what you mean by this. One-forms and gradients are mathematical objects, just like tensors; you don't "measure" things with them. Measurements are numbers, and you get numbers by contracting tensors, vectors, one-forms, gradients, etc. to obtain scalars (things that have no indexes).

What is different between the two fluids that would give different stress-energy tensor values at anyone element?

Um, their energy density? (Meaning, in ordinary terms, their mass density. Remember that "energy density" in GR includes "rest energy density", i.e., what we ordinarily view as just mass density.) Also, the energy density would include the temperatures of the two fluid elements, which might differ as well.

 

[Messenger]

If you're going to ask things like "what would I be measuring" or "what is different between the two fluids", then trying to forget about mass, momentum, and energy is kind of counterproductive.



Not sure what you mean by this. One-forms and gradients are mathematical objects, just like tensors; you don't "measure" things with them. Measurements are numbers, and you get numbers by contracting tensors, vectors, one-forms, gradients, etc. to obtain scalars (things that have no indexes).

But with a gradient of say pressure, by definition it represents the magnitude and direction of the greatest change in pressure from a point. Even a plain derivative "measures" the best linear approximation of a function. The zero component of a stress energy tensor "measures" energy density. I think your definition of "measure" is narrower than mine.

Um, their energy density? (Meaning, in ordinary terms, their mass density. Remember that "energy density" in GR includes "rest energy density", i.e., what we ordinarily view as just mass density.) Also, the energy density would include the temperatures of the two fluid elements, which might differ as well.

But then if the cosmological constant represents a homogeneous perfect fluid, then the uniform density of this fluid presents a stress-energy tensor at all points in space-time where the zero component is the energy density of this fluid and so should gravitate, no?

 

[PeterDonis]

But with a gradient of say pressure, by definition it represents the magnitude and direction of the greatest change in pressure from a point. Even a plain derivative "measures" the best linear approximation of a function. The zero component of a stress energy tensor "measures" energy density. I think your definition of "measure" is narrower than mine.

I don't object to yours, I just didn't fully understand what it was. I would probably prefer the word "represents" for yours. It seems like you have a good handle on what the various mathematical objects measure/represent in the above quote.

But then if the cosmological constant represents a homogeneous perfect fluid, then the uniform density of this fluid presents a stress-energy tensor at all points in space-time where the zero component is the energy density of this fluid and so should gravitate, no?

Yes, and it does. A spacetime with nothing but a cosmological constant in it either expands (if the constant is positive) or contracts (if the constant is negative). In the expanding case "gravitates" may be the wrong word; the constant is acting like "dark energy" and causing "gravitational repulsion" instead of attraction. But in either case the spacetime is curved, not flat, so "gravity" is present.

The positive constant case is called de Sitter space, and the negative constant case is called Anti-de Sitter space. See these Wiki pages for a start:

http://en.wikipedia.org/wiki/De_Sitter_universe

http://en.wikipedia.org/wiki/Anti_de_Sitter_space