Stress Energy Tensor: How Does It Implicate Flow of Energy?

[Silviu]

Hello! I am reading that in a perfect fluid we have no heat conduction, which implies that energy can flow out of a fluid element only if particles flow, so $T^{0i} = 0$. I am not sure I understand why. We have $\Delta E = \Delta Q - p \Delta V$. In our case as Q is constant, $\Delta E = -p \Delta V$. How does this implies that energy is changed only if particles flow. If the volume is changing, than we can have a change in energy and there is nothing preventing the volume of the fluid element to change, right? Can someone explain this to me please? Thank you!

 

[PeterDonis]

I am reading

Please give a specific reference.

 

[Silviu]

Please give a specific reference.

Sorry, "A first course in General Relativity", Bernard Schutz, section 4.6.

 

[Ibix]

There are two volumes here. One is a volume element that may or may not contain fluid particles. The other is the region of space currently occupied by some fluid.The volume $V$ that you are talking about is the volume occupied by the fluid, but the volume the stress-energy tensor cares about is the elementary volume. If $V$ changes ($\Delta V\neq 0$) then the particle density changes, which is to say that the number of particles in a specified volume (elementary or otherwise) changes. This is just particles flowing in or out of a volume in the sense that the stress-energy tensor cares about.

 

[Silviu]

There are two volumes here. One is a volume element that may or may not contain fluid particles. The other is the region of space currently occupied by some fluid.The volume $V$ that you are talking about is the volume occupied by the fluid, but the volume the stress-energy tensor cares about is the elementary volume. If $V$ changes ($\Delta V\neq 0$) then the particle density changes, which is to say that the number of particles in a specified volume (elementary or otherwise) changes. This is just particles flowing in or out of a volume in the sense that the stress-energy tensor cares about.

I am sorry, but I am still a bit confused. At the beginning of section 4.5 in second edition, he states that: "In the MCRF, we imagine that the fluid element is able to exchange energy with its surroundings in only two ways: by heat conduction (absorbing an amount of heat $\Delta Q$) and by work (doing an amount of work $p \Delta V$, where V is the three-volume of the element). " I am not sure I understand your distinction as based on Schuts, the volume element, can do work $p \Delta V$ and from what I see here he seems to identify the 2 volumes you are talking about. Could you please elaborate a bit? Thank you!

 

[PeterDonis]

I am reading that in a perfect fluid we have no heat conduction, which implies that energy can flow out of a fluid element only if particles flow, so $T^{0i} = 0$.

The way Schutz puts this seems a bit confusing to me. The reasoning is actually that, if we are in the fluid's MCRF, then "no heat conduction" means $T^{0i} = 0$. If we are in the fluid's MCRF, then there is no "particle flow" at all--that's the definition of the MCRF, the frame in which the fluid is (locally) at rest. So $T^{0i} = 0$ in the MCRF can't mean "no particle flow", because that's always true in the MCRF; what it means (as Schutz explains a bit earlier, on p. 96) is "no stress-energy flow", i.e., "no heat conduction".

It's possible that by "particle flow" in this particular case Schutz means to include the case of particles "flowing" from the past to the future, i.e., having nonzero energy density. In other words, he is saying that the only "energy flow" in a perfect fluid is due to the energy density $\rho$ being transported along the fluid's worldlines. But if that's what he means, I think it's a confusing way to say it.

 

[PeterDonis]

At the beginning of section 4.5 in second edition, he states that: "In the MCRF, we imagine that the fluid element is able to exchange energy with its surroundings in only two ways: by heat conduction (absorbing an amount of heat $\Delta Q$) and by work (doing an amount of work $p \Delta V$, where V is the three-volume of the element). "

I don't think Schutz means to exclude the possibility of a fluid element in a perfect fluid doing work. But a fluid element doing work doesn't involve any particles or any stress-energy crossing the boundary of the fluid element.