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Einstein's Field Equation StressEnergy 
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#1
Mar2107, 10:27 AM

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Hi,
as I try to understand Einstein's Field Equations, I realize that I don't understand the meaning of its components. I gather that Rtt=Ttt+Txx+Tyy+Tzz is the dominating parameter in cosmological models. (Sorry, I also have problems with tex.) Ttt means "mass energy density". Is this "relativistic mass" per unit volume? Txx etc. means pressure. Is this momentum flow per unit area? The net flow should be 0 (in a comoving frame), how do I calculate Txx. Example: A constant matter density gives Rtt=Ttt=rho in the comoving frame. What if all that matter annihilates? What remains constant, Rtt or Ttt? Rtt=rho+3*rho/3 or Rtt=rho/2+3*rho/6? Is Rtt equal to some total energy density? Thanks. 


#2
Mar2107, 12:38 PM

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It's really better, though, to say that T_tt is the energy density. This includes all sources of energy, including the energy in rest mass (mc^2). It can also be regarded more simply in the comoving frame as the pressure. I find it simpler to regard T_xx in the comoving frame as the pressure in the comoving frame. (This simple approach doesn't work well in noncomoving frames, which is why Baez formulates it in the manner above). While Baez's defintion may be a bit confusing, consider a pressure P in some cube. It's transporting negative x momentum in the x direction, and positive x momentum in the +x direction. The minus signs cancel each other out, and the net effect of the pressure is transport of the xmomentum in the x direction. If Baez's approach is too confusing, always work in the comvoing frame, and think of T_xx as the pressure. If you must work in a noncomvoving frame, tensors transform in a welldefined manner. You can thus use the tensor transformation rules to assign a meaning to T_xx (and the other components) in a noncomoving frame. Note that, using the "pressure in comoving frame" approach, pressure really is a rank 2 tensor in the general case. In ideal fluids, though, this pressure is always isotropic, which makes things a lot simpler (there is only one P, which appears on all the diagonal terms of the stressenergy tensor). In the general case, the pressure tensor can be diagonalized by aligning the axes correctly. 


#3
Mar2107, 07:15 PM

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Pete 


#4
Mar2107, 08:10 PM

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Einstein's Field Equation StressEnergy
http://math.ucr.edu/home/baez/physic...y/SR/mass.html is one source that supports my position. For more sources that I've found that support my view see http://www.physicsforums.com/showpos...0&postcount=28 Neither of us has succeeded in convincing the other, and no third party has volunteered to arbitrate the dispute over the definition of "relativistic mass". As far as I'm concerned, issues regarding the definition of the term are basically just another reason to avoid the term "relativistic mass" which is IMO depreciated (see the FAQ entry, the first URL I posted), and stick with "energy density". Hopefully this relatively minor disagreement won't get in the way of the main point, which is that T_00, being the total energy, includes terms of the form mc^2, where m is the rest mass or invariant mass. 


#5
Mar2107, 11:13 PM

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Grammar !
...this is an old argument between Pete and ME ... Luckily your physics is better than your grammar. Yes, I am a pedant. 


#6
Mar2107, 11:28 PM

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#7
Mar2207, 04:40 AM

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Thank you, pervect!
So, in my example, if all matter would annihilate, Ttt would remain constant (=rho) and there would Pressure Txx=rho/3 etc. ? Is therefore the statement "Rtt equals some total energy density" wrong, as we expect energy to be conserved? Is it correct then, considering the entire universe and ignoring lambda, that its expansion would decelerate twice as fast (I hope you know wht I mean) after annihilating all matter? Obviously, if we could annihilate a star, it would not double its gravitation. I guess there are some contributions of the boundary, if we confine the photon gas? IIRC, you did mention such things. Is it ok to call a decelerating lamba a positive energy density? Lots of questions... 


#8
Mar2207, 02:28 PM

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R_tt is closely related, but not quite the same as, one particular measure of energy, the Komar energy. (It's often called the Komar mass, which is just the Komar energy / c^2). This concept of energy is not completely general, however, it applies only to stationary or static systems. I've written a few wikipedia articles about this http://en.wikipedia.org/wiki/Mass_in_general_relativity talks about some of the concepts of mass in GR, including the Komar mass. http://en.wikipedia.org/wiki/Komar_mass goes into more detail on the Komar mass. To get the the contribution of R_tt to the Komar mass of a body, you have to multiply R_tt by another factor, the square root of the metric coefficient g_tt, which can be regarded as the "redshift factor" at that location. But we still haven't gotten to the heart of the matter. So let's take an example closely related to yours. This is closely related to an example I worked out in the first wiki article. http://en.wikipedia.org/wiki/Mass_in...ral_relativity Suppose we have a very strong steel sphere. Inside this we have matter and antimatter. The matter and antimatter annihilate. This causes the matter m in the sphere to become energy. The average energy density inside the sphere T_00 stays constant during this process. The average pressure, T_xx, T_yy, and T_zz each go from 0 to some high value (1/3 T_00, as I recall). What happens to the (Komar) mass of the system, and its gravitational field? We have the same energy, and we've added pressure. So it looks like it should go up (double when we work out the details). But when we actually consider the complete system, including the pressure vessel, the answer is that nothing happens to the mass. The reason that the answer is nothing is that while the average pressure inside the sphere is high, there is a counterbalancing average tension in the walls of the container that exactly counterbalances it. (A tension is just a negative pressure. It subtracts from the Komar mass, just as pressure adds to it). The walls of the container are needed to keep the system "stationary". So this answer is a bit long, but you can see that while R_00, even when corrected by the appropriate redshift factor to become the Komar mass, is not a general measure of energy in GR. It does work correctly for a stationary system, but it wouldn't work right if we didn't have the pressure vessel there. If the Komar mass doesn't work, is there some other formulation that does? To a limited extent, the answer is yes. One of the most general defintions for a conserved energy exists is the ADM energy, which applies to any asymptotically flat spacetime. The Komar mass is simpler to compute, but the ADM mass is more general. While it is more general, the ADM mass is not completely general. The ADM mass requires an asymptotically flat spacetime. (Note that any isolated system surrounded by an infinite vacuum region will have this property of asymptotic flatness). If you have a totally arbitrary spacetime, there is no truly general defintion of conserved energy in GR. (see the wikipedia article I quoted for more details, and some of its references, such as the sci.physics.faq entry on energy in GR http://math.ucr.edu/home/baez/physic...energy_gr.html) The problem with cosmology is that standard cosmologies aren't asymptotically flat. The observable universe is assumed to be some part of a very large (possibly infinite) system, for instance, so it isn't required to have (and in fact doesn't have) the property of asymptotic flatness needed to have a conserved ADM energy. 


#9
Mar2307, 07:35 AM

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Thanks, it becomes clearer.
Should I post my questions about expanison of the universe and definition of "Energy density" in the complete universe, including lambda, in the cosmology section? After all, it comes down to GR. 


#10
Mar2307, 08:55 AM

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Pete 


#11
Mar2307, 09:48 AM

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Does this have any bearing on my OP?



#12
Mar2307, 10:41 AM

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Pete 


#13
Mar2307, 10:52 AM

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Could you explain?
Does it mean that you count some components of T (other than Ttt) among "relativistic mass"? 


#14
Mar2307, 11:23 AM

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http://www.geocities.com/physics_wor...gy_vs_mass.htm Eq. (2) is the momentum density. To obtain the mass density divide that by the velocity. You'll see that you don't end up with something proportional to T^{00}. However that term will disappear when you calculate the entire mass of the whole/isolated system. Pete 


#15
Mar2307, 02:44 PM

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The energymomentum of an object with finite volume is not in general a covariant entity, see for instance http://arxiv.org/abs/physics/0505004. (The total energymomentum of an object with finite volume is covariant if the object is an isolated system, but in general we are interested in nonisolated systems).
The energy momentum density, i.e. the stressenergy tensor, is always a covariant entity, however. (Roughly speaking, this is because it is a density, so it takes the limit as the volume approaches zero, where there is no issue.) So the correct way to work out the momentum of a nonisolated body is to perform the appropriate integral of the stress energy tensor. One finds that because of the Lorentz transform, pressure in one frame can convert to energy or momentum in another, just as for the simpler rank 1 tensor, energy can convert to momentum, or space into time. If one does a "boost" of a stationary ball of fluid with a density of [itex]\rho[/itex] and an isotropic pressure of P, one finds the momentum of such a fluid element in the boosted frame as being [tex]u \gamma^2 (\left \rho + P \right) [/tex] where u is the 4velocity of the boost (and [itex]\gamma = 1/\sqrt{1u^2}[/itex]). See for instance Rindler, Introduction to special relativity, pg 132, where this is worked out. Rindler actually works out the more complex case of an anisotropic pressure, with all the T^ij in the stressenergy tensor differnet. Note that Rindler uses nongeometric units, so he replaces P with P/c^2 There are several important points to be made, in my opinion. 1) The pressure does contribute to the momentum in the direction of motion, via the Lorentz transform. 2) The contribution is not of the form [itex]\rho + 3P[/itex]. In fact, only the component of the pressure term in the direction of motion contributes to the momentum density, hence [itex]\rho + P[/itex] 3) In general, momentum is a vector that doesn't point in the same direction as velocity. Therefore in general the concept of momentum / velocity is a tensor, not a scalar. The previous quantites mentioned (invariant mass and E/c^2 which is often called relativistic mass) were scalars, while the ratio of momentum / velocity is not a scalar. 


#16
Mar2307, 02:50 PM

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#17
Mar2307, 02:55 PM

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It also relates as to how seriously one can treat [itex]\rho + 3P[/itex], i.e. R_00, as some sort of energy density. 


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