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,
<br /> R_{\mu\nu}-\frac{1}{2}Rg_{\mu\nu}=\kappa T_{\mu\nu}^{matter}+\kappa T_{\mu\nu}^{EM}<br />
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&gt;\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

 

[Messenger]

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.

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

Interesting...so my equation in the original post, \Omega g_{\mu\nu}=\kappa T_{\mu\nu}^{generic}+\kappa T_{\mu\nu}^{matter}, makes different assumptions about the zero components of tensors than deSitter space. I needed a differential \frac{d\rho}{dV} instead of just \rho for it to work. How can deSitter space still be flat with a \Lambda present?

 

[PeterDonis]

How can deSitter space still be flat with a \Lambda present?

De Sitter *spacetime* is not flat; it's curved. It can be foliated by flat spatial slices, but that's not the same thing as spacetime as a whole being flat.

 

[Messenger]

De Sitter *spacetime* is not flat; it's curved. It can be foliated by flat spatial slices, but that's not the same thing as spacetime as a whole being flat.

So the Wikipedia article is actually talking about the flat slices?

It models the universe as spatially flat and neglects ordinary matter, so the dynamics of the universe are dominated by the cosmological constant, thought to correspond to dark energy in our universe or the inflaton field in the early universe.

Does that mean the cosmological constant which they talk about here:
http://map.gsfc.nasa.gov/news/ counteracts the gravitational pull of baryonic and dark matter?

 

[PeterDonis]

So the Wikipedia article is actually talking about the flat slices?

Not sure exactly what in the Wikipedia article you are referring to, but yes, if the word "flat" is being used it can only refer to the flat spatial slices. Note that that is not the only possible slicing; the "mathematical" article about de Sitter space talks about other slicings:

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

Does that mean the cosmological constant which they talk about here:
http://map.gsfc.nasa.gov/news/ counteracts the gravitational pull of baryonic and dark matter?

Yes. For more detail on what that article calls "the standard model of cosmology (Cold Dark Matter and a Cosmological Constant in a flat universe)", Ned Wright's cosmology FAQ is a good place to start:

http://www.astro.ucla.edu/~wright/cosmology_faq.html#bestfit

Edit: In view of the comment I made above about different possible slicings, I should clarify that the flat spatial slicing in the standard model of cosmology is the natural slicing for "comoving" observers, i.e., observers that see the universe as homogeneous and isotropic. In other words, "flat universe" means these observers see the universe as spatially flat.

 

[Messenger]

Can you explain your view of the following. I have seen the statement in several books and probably need it spelled out to me.

thanks to Bianchi identities and energy-momentum conservation, Einstein and stress tensors have zero divergence.

How does conservation of energy require that there is no divergence of stress tensors (i.e. \nabla_\mu T^{\mu\nu}=0? I am assuming that they are talking about the divergence where \nabla \cdot F(p) where F must be a function of a point p, but that energy conservation does not require that F is a function of a point p and thus has zero divergence.

 

[PeterDonis]

I am assuming that they are talking about the divergence where \nabla \cdot F(p) where F must be a function of a point p, but that energy conservation does not require that F is a function of a point p and thus has zero divergence.

I'm not sure what you mean by this. Tensors are geometric objects, one of which exists at each point of the spacetime. (More precisely, each point in the spacetime has its own tangent space in which tensors defined at that point "live".)

How does conservation of energy require that there is no divergence of stress tensors (i.e. \nabla_\mu T^{\mu\nu}=0?

There are two answers to this. First, what "conservation of energy" means, in terms of the physical meaning of the divergence of the SET; second, how GR "enforces" it.

Take the second point first. The Bianchi identities guarantee that the covariant divergence of the *Einstein* tensor is identically zero. That is, \nabla_\mu G^{\mu \nu} = 0 is an identity. This forces \nabla_\mu T^{\mu \nu} = 0 to also be true because of the Einstein Field Equation.

Now, the first point; why do we say that \nabla_\mu T^{\mu \nu} = 0 means "conservation of energy"? First, we need to be precise about what kind of "energy" we are talking about; actually, what is being conserved is energy, momentum, stress, etc. due to *non-gravitational* sources, or non-gravitational "stress-energy". In other words, this law does *not* include "energy in the gravitational field"; there is no single well-defined way to capture that in GR (the reasons why are a whole other can of worms that I would rather not open for this thread).

Now, think about what the covariant divergence means. Consider some small 4-volume of spacetime surrounding a point. Stress-energy is flowing in and out of this 4-volume; you can think of it as a sort of "fluid" with flow lines going into the volume and flow lines coming out. If the fluid is "conserved"--that is, if there are no "sources" or "sinks" for the fluid inside the small 4-volume--then the number of flow lines going in must equal the number of flow lines coming out. In other words, the quantity (flow lines coming out) minus (flow lines going in) must be zero. But that's exactly what the covariant divergence of the SET captures; so "conservation" of the stress-energy "fluid" means that the covariant divergence of the SET must be zero.

 

[Messenger]

Now, think about what the covariant divergence means. Consider some small 4-volume of spacetime surrounding a point. Stress-energy is flowing in and out of this 4-volume; you can think of it as a sort of "fluid" with flow lines going into the volume and flow lines coming out. If the fluid is "conserved"--that is, if there are no "sources" or "sinks" for the fluid inside the small 4-volume--then the number of flow lines going in must equal the number of flow lines coming out. In other words, the quantity (flow lines coming out) minus (flow lines going in) must be zero. But that's exactly what the covariant divergence of the SET captures; so "conservation" of the stress-energy "fluid" means that the covariant divergence of the SET must be zero.

Right, this is what I meant in the first part. This is saying that you can't differentiate the stress-energy flow in any 4-volume from any other 4-volume, even if the other is vacuum energy. Even if two particle pairs come into existence and then recombine, it seems they wouldn't have an effect on the stress-energy tensor while they were in existence since there were never any sources or sinks there at all, only constant energy density.

 

[Messenger]

Around pg. 55 of MTW, they speak of the "bongs" of a bell where each one represents a 1-form and the constant phase of deBroglie wavelength. If each 1-phase represents a point in space-time, then it doesn't seem with the stress-energy tensor you would ever be able to determine anything about the phase at the point you are at. You would only be able to determine an absolute magnitude for your energy level since stress-energy from a deBroglie phase does not flow from 1-form to 1-form (would it be a discontinuous jump?). It seems like it would require divergence in order to be able to do this.
Looking at Fig. 2.4 closer, it states that the phase difference from one event to another would be (4.4) wavelengths. But how can there be any portion of a wavelength (phase) unless there are different energy levels for each portion of a phase? Wouldn't this imply that if each part of a wavelength (phase) occupies a different point in space-time and each phase point has a different energy level (by definition), then the stress-energy tensor is not constant over the space-time volume that a particle occupies and thus has divergence?

 

[PeterDonis]

Right, this is what I meant in the first part. This is saying that you can't differentiate the stress-energy flow in any 4-volume from any other 4-volume, even if the other is vacuum energy.

I'm still not sure I understand what you mean by this. Differentiation is a local operation--when you differentiate a tensor, you are performing a mathematical operation in the tangent space at a particular point. So "differentiate the flow in any 4-volume from any other 4-volume" doesn't make sense to me the way you are stating it.

Even if two particle pairs come into existence and then recombine, it seems they wouldn't have an effect on the stress-energy tensor while they were in existence since there were never any sources or sinks there at all, only constant energy density.

The SET is a classical object; virtual particles coming into existence and then recombining is a quantum process. The SET corresponding to vacuum energy is not intended to capture this process; as you correctly note, it can't. The SET corresponding to vacuum energy is a sort of average over many virtual particle processes. I say "sort of" because nobody has derived such an SET directly from the first principles of quantum field theory, by averaging or any other method, that actually matches observation. You can derive an effective SET for the quantum vacuum state based on virtual particle processes from quantum field theory, but the answer you get is that the effective energy density should be something like 10^120 times greater than the observed "dark energy" density.

For more on this, John Baez' page on "what's the energy density of the vacuum" is worth reading:

http://math.ucr.edu/home/baez/vacuum.html

Real particles, as opposed to virtual ones, are not created from nothing and don't disappear into nothing when they recombine; if they are created, it must be from something else, and if they recombine, they change into something else. For example, an electron-positron pair might be created from a pair of photons, and the electron and positron might annihilate each other and turn into a pair of photons. Those processes can be described by a conserved SET.

 

[PeterDonis]

If each 1-phase represents a point in space-time

It doesn't; it represents a *surface* in spacetime. (More precisely, it represents a surface in the tangent space at a particular point in spacetime. But for lots of purposes you can think of it as representing a surface in the actual spacetime--the surface on which the phase takes a particular value.)

Looking at Fig. 2.4 closer, it states that the phase difference from one event to another would be (4.4) wavelengths. But how can there be any portion of a wavelength (phase) unless there are different energy levels for each portion of a phase?

I don't understand what you mean by this. The "phase" is just a way of picking out where on the waveform you are, so to speak. The energy has to do with the frequency of the wave--how many waveforms pass a given point per unit time.

I'm also not sure how you are trying to link this to the stress-energy tensor, which is a second rank tensor, not a 1-form (first rank tensor).

 

[Messenger]

I don't understand what you mean by this. The "phase" is just a way of picking out where on the waveform you are, so to speak. The energy has to do with the frequency of the wave--how many waveforms pass a given point per unit time.

Energy is inversely proportional to the wavelength, which as far as I am aware occupies more than one point in space-time. Phase is more than just picking out where on the waveform you are at, at least for electromagnetic waves. It still takes a finite amount of time for a wavelength to pass a given point. Does 100% of a photon's energy (say a large radio wave in the meters range) pass a point when only half the time for a full wavelength has passed? That energy should still generate a stress-tensor, I just don't understand how all the energy can be described as one 4-dimensional point. My introduction with tensors started with 4th rank elasticity tensors so that may have something to do with my inability to grasp this.

 

[PeterDonis]

Energy is inversely proportional to the wavelength, which as far as I am aware occupies more than one point in space-time.

The wavelength is not the wave; it's a number we derive from the wave. The wavelength doesn't occupy spacetime; the wave does.

Phase is more than just picking out where on the waveform you are at, at least for electromagnetic waves.

Huh? That's how phase is defined. See, for example, here:

http://en.wikipedia.org/wiki/Phase_(waves))

It still takes a finite amount of time for a wavelength to pass a given point.

No, it takes a finite amount of time for a full cycle of the *wave* to pass a given point. See above. Also note that "point" here means a point in space, not spacetime--see below.

Does 100% of a photon's energy (say a large radio wave in the meters range) pass a point when only half the time for a full wavelength has passed?

The wave is different from the SET; if you are trying to match up the SET of an EM wave with the wave picture of that wave, I think you are just going to get yourself very confused. If you want to look at an expression for the SET of an electromagnetic field, see for example here:

http://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

Note that this expression is general; it applies to *any* EM field, not just one that happens to be an EM wave.

Also, when you think about the energy of the wave "passing a point", you are thinking about energy moving through space in time; the "point" is a point in space, *not* a point in spacetime. A point in spacetime is something like "here in the center of my lab at 12:00 Noon local time on July 30, 2012". An EM wave does not "move through" that point in spacetime--it simply has a certain set of properties at that point (values for the E and B field, or the EM field tensor if you want to write it in relativistic form, or amplitude and phase of an EM wave if you want to write it in that form).

I just don't understand how all the energy can be described as one 4-dimensional point.

The SET doesn't actually describe energy, it describes energy *density* (and momentum density, and pressure, and stress--but all of those have units of energy density, if you multiply momentum density by c as you have to in relativity if you're not using "natural" units). Energy *density* can be defined at a point (strictly speaking, by taking the limit of energy/volume as volume goes to zero).

Also, the fact that the SET has some definite value at a particular point in spacetime does not mean that "energy is being described as one 4-dimensional point". Stress-energy is a continuous substance (at least, it is in the classical approximation we use in GR), like a "fluid", as I've said before. The SET simply gives the properties of the fluid at each spacetime point; it does not say that the fluid is entirely "contained" in one spacetime point.

 

[Messenger]

Energy *density* can be defined at a point (strictly speaking, by taking the limit of energy/volume as volume goes to zero).

Also, the fact that the SET has some definite value at a particular point in spacetime does not mean that "energy is being described as one 4-dimensional point". Stress-energy is a continuous substance (at least, it is in the classical approximation we use in GR), like a "fluid", as I've said before. The SET simply gives the properties of the fluid at each spacetime point; it does not say that the fluid is entirely "contained" in one spacetime point.

Question 1: Is what tom.stoer states (and the links in the thread) in his original post in the following thread incorrect then? He and some of the responses seem to think that energy density is not well defined.

I know that the definition of energy E[V] as a volume integral of T°°(x) over a spacelike slice = 3-volume V is not (always) possible in general relativity. The basic reason is that the usual current conservation dj=0 is replaced by the covariant equation DT=0 (D is the covariant derivative; T is the energy momentum tensor). Therefore the usual trick making use of integration by parts does no longer work because additional terms (not vanishing on the boundary) are present. In addition the requirement that the integral E[V] must be the 0-component of a 4-vector would no longer hold.

As the energy E[V] is no longer well-defined, global energy conservation becomes meaningless in general relativity.

https://www.physicsforums.com/showthread.php?t=344288

Question 2: If \nabla_k G^{ij}= \frac {\partial G^{ij}}{\partial x^k} + \Gamma^i_{lk}G^{lj}+\Gamma^j_{lk}G^{il}=0, does the term \frac{\partial G^{ij}}{\partial x^k}=0?

 

[PeterDonis]

Question 1: Is what tom.stoer states (and the links in the thread) in his original post in the following thread incorrect then? He and some of the responses seem to think that energy density is not well defined.

That thread is about a separate issue from what we are talking about here. The posters in that thread are saying that energy, viewed as an integral of an "energy density" over a finite volume (strictly speaking, a finite volume large enough to show the effects of spacetime curvature), is not well-defined. (This fact is more often stated as: "energy of the gravitational field" is not well-defined; which makes it clearer that we are not talking about the SET, but about something related to spacetime curvature.) None of that affects what I said before, that the energy density at a specific point (strictly speaking, the limit that I described before, of energy E in a small volume V as the volume goes to zero) is well-defined.

Question 2: If \nabla_k G^{ij}= \frac {\partial G^{ij}}{\partial x^k} + \Gamma^i_{lk}G^{lj}+\Gamma^j_{lk}G^{il}=0, does the term \frac{\partial G^{ij}}{\partial x^k}=0?

Not in general, no. That's why you have to take the covariant derivative--to make sure that you have correctly accounted for coordinate effects, which is what the Christoffel symbol terms account for.

 

[Messenger]

Not in general, no. That's why you have to take the covariant derivative--to make sure that you have correctly accounted for coordinate effects, which is what the Christoffel symbol terms account for.

So the covariant divergence for a constant, i.e. \nabla_s g^{ks} \Omega \neq 0 (since \frac{\partial \Omega}{\partial x^s}=0) or are the Christoffel symbol terms also zero?

 

[PeterDonis]

So the covariant divergence for a constant, i.e. \nabla_s g^{ks} \Omega \neq 0 (since \frac{\partial \Omega}{\partial x^s}=0) or are the Christoffel symbol terms also zero?

What you've shown isn't the covariant divergence of a constant; it's the covariant divergence of a constant *times the metric*. Big difference. (Hint: what are the partial derivatives of the metric coefficients?)

 

[Messenger]

What you've shown isn't the covariant divergence of a constant; it's the covariant divergence of a constant *times the metric*. Big difference. (Hint: what are the partial derivatives of the metric coefficients?)

Thought the answer to your hint was a simple zero for the coefficients of (-1,1,1,1) but then found this \Gamma^\lambda_{\mu\nu}=\frac{1}{2}g^{\lambda\rho}(\frac{\partial g_{\rho\mu}}{\partial x^\nu}+\frac{\partial g_{\rho\nu}}{\partial x^\mu}-\frac{\partial g_{\mu\nu}}{\partial x^\rho}) but this would seem to make the Christoffel symbols zero also. Thought the constant was just incorporated in with the metric when taking the partial derivative. Not getting what you meant by big difference

 

[PeterDonis]

Thought the answer to your hint was a simple zero for the coefficients of (-1,1,1,1)

Why are you assuming that those are the metric coefficients? Have you actually solved the Einstein Field Equation? I'm assuming the answer is "no", because if you had, you would have realized that, if there is a cosmological constant present, the metric *cannot* be (-1, 1, 1, 1). (Technically, you *can* make the coefficients (-1, 1, 1, 1) at a single point by an appropriate choice of coordinates, but only at a single point.)

Thought the constant was just incorporated in with the metric when taking the partial derivative. Not getting what you meant by big difference

You appeared to be assuming that, in order to take the partial derivative of the expression g^{ks} \Omega, you just had to take the derivative of \Omega, which is a constant. But g^{ks} is *not* a constant, so the partial derivative of g^{ks} \Omega will be \Omega times the partial derivative of g^{ks}, which is not zero. See above.

 

[Mentz114]

So the covariant divergence for a constant, i.e. \nabla_s g^{ks} \Omega \neq 0 (since \frac{\partial \Omega}{\partial x^s}=0) or are the Christoffel symbol terms also zero?

\nabla_s \left( g^{ks} \Omega \right) = \Omega \nabla_s g^{ks} = 0

because the covariant derivative of the metric vanishes*. So the divergence of the Lambda vacuum SET is zero.

*See https://www.physicsforums.com/showthread.php?t=479553

 

[Messenger]

Why are you assuming that those are the metric coefficients? Have you actually solved the Einstein Field Equation? I'm assuming the answer is "no", because if you had, you would have realized that, if there is a cosmological constant present, the metric *cannot* be (-1, 1, 1, 1). (Technically, you *can* make the coefficients (-1, 1, 1, 1) at a single point by an appropriate choice of coordinates, but only at a single point.)



You appeared to be assuming that, in order to take the partial derivative of the expression g^{ks} \Omega, you just had to take the derivative of \Omega, which is a constant. But g^{ks} is *not* a constant, so the partial derivative of g^{ks} \Omega will be \Omega times the partial derivative of g^{ks}, which is not zero. See above.

So the metric for the Lambdavacuum solution is not diag(-\Lambda,\Lambda,\Lambda,\Lambda)? Woops. And also the partial derivatives are not zero, but according to Mentz post the covariant derivative is zero thus making the divergence of the Lambdavacuum SET zero.