## Question about fluid tensors

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.

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 Quote by Messenger 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".)

 Quote by Messenger 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.

 Quote by PeterDonis 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.

 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?

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 Quote by Messenger 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.

 Quote by Messenger 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.

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 Quote by Messenger 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.)

 Quote by Messenger 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).

 Quote by PeterDonis 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.

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 Quote by Messenger 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.

 Quote by Messenger 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))

 Quote by Messenger 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.

 Quote by Messenger 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/Electro...3energy_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).

 Quote by Messenger 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.

 Quote by PeterDonis 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.

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 $\frac{\partial G^{ij}}{\partial x^k}=0$?

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 Quote by Messenger 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.

 Quote by Messenger 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.

 Quote by PeterDonis 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?

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 Quote by Messenger 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?)

 Quote by PeterDonis 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

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 Quote by Messenger 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.)

 Quote by Messenger 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.

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 Quote by Messenger 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.

 Quote by PeterDonis 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.

 Quote by Mentz114 $\nabla_s \left( g^{ks} \Omega \right) = \Omega \nabla_s g^{ks} = 0$ because the covariant derivative of the metric vanishes*. So the divergence of the Lambda vacuum SET is zero. *See http://www.physicsforums.com/showthread.php?t=479553
 A covariant derivative is a derivative from which we've removed any contribution that is purely a result of the choice of coordinates. The first derivatives of the metric are the way they are purely as a result of the choice of coordinates.
Hmmm. Perhaps I would be less confused just reading the forum instead of asking questions. Thanks for the link Mentz.