stevebd1 said:
Hello Peter
While I understand that the conventional trace for the stress-energy tensor for an object of mass is tr(T^{\alpha \beta})=\rho+3p
I realized on reading over this thread again that I should also comment on the above statement. As you've stated it, it's incorrect: the trace of the perfect fluid stress-energy tensor is *not* \rho + 3p. However, that does not mean that that expression doesn't have a key role to play in how a fluid acts as a gravitational source; it does. Let me run through how that comes about.
(Btw, what I'm going to say is based on
John Baez' excellent tutorial on http://math.ucr.edu/home/baez/einstein/" . He gives a lot more details than I'm giving here.)
First of all, I was a bit cavalier in an earlier post when I said that the trace of the stress-energy tensor is the key quantity in determining how an object acts as a gravitational source. It is, for the particular example I was considering (the isolated moving free particle), but that's a special case (I'll show below how that special case arises from the more general analysis we're about to do). In the general case, the quantity you need to look at is the 0-0 (or t-t) component of the Ricci tensor, in the rest frame of the object (or fluid, or whatever) whose behavior as a "source" of gravity you want to investigate. (Baez' tutorial explains why this is true; I won't go into the details here.)
The Ricci tensor is related to the stress-energy tensor via the Einstein field equation, in the following form:
R^{ab} = \frac{8 \pi G}{c^4} \left( T^{ab} - \frac{1}{2} g^{ab} T \right)
where G is Newton's gravitational constant and c is the speed of light. (Those constants have to appear as they do on the RHS of the above in order for us to use ordinary units for both the Ricci and stress-energy tensors; the Ricci tensor is a "geometric" object, meaning its normal units are units of length, but the stress-energy tensor's normal units are energy density, or pressure--the two are really the same. The \frac{G}{c^4} coefficient in front of the RHS is the "conversion factor" between energy density and length. In many textbooks on general relativity that part of the RHS is left out, and "geometric units" are used for all quantities--meaning everything is written in terms of length, even things like energy and pressure.)
So if we have an expression for the stress-energy tensor in the rest frame of the "source" object, we can write down the Ricci tensor in that same frame. The 0-0, or t-t, component is the one we want; it is:
R^{tt} = \frac{8 \pi G}{c^4} \left( T^{tt} - \frac{1}{2} g^{tt} T \right)
With the sign conventions I've been using, g^{tt} = 1, and the trace of the stress-energy tensor for a perfect fluid is
tr \left( T^{ab} \right) = T = T^{tt} - T^{xx} - T^{yy} - T^{zz} = \rho c^2 - 3 p
where I've used \rho for the *mass* density of the fluid, so its energy density (in its rest frame) is \rho c^2. Now we have all we need to find the expression for the t-t component of the Ricci tensor:
R^{tt} = \frac{8 \pi G}{c^4} \left[ \rho c^2 - \frac{1}{2} \left( \rho c^2 - 3 p \right) \right] = \frac{4 \pi G}{c^4} \left( \rho c^2 + 3 p \right)
That's where the \rho c^2 + 3 p comes from, and as you can see, it *is* the appropriate expression to describe the "strength of the source of gravity" for a perfect fluid.
Now, let's suppose we had a fluid that had *zero* pressure. Then we would have simply:
R^{tt} = \frac{4 \pi G}{c^4} \left( \rho c^2 \right) = \frac{1}{2} T^{tt} = \frac{1}{2} T
In other words, in this special case (where the only non-zero component of the stress-energy tensor in the rest frame of the object is the t-t component), the quantity we need for "strength of source", the t-t component of the Ricci tensor, is equal to half the trace of the stress-energy tensor. Of course, it's also equal to half the t-t component of the stress-energy tensor. But it's nicer to express it as the trace because the trace is an invariant quantity; it's the same in any frame of reference.
So, for example, if we happen to have an expression for the stress-energy tensor in a frame in which the object is in motion (aha!), instead of having to go to all the trouble of transforming the whole tensor back into the object's rest frame, we just take its trace and we're done--which is, of course, what I did with the isolated free particle example above. (I'll leave it as an exercise to show that, in the rest frame of the isolated free particle, the only non-zero stress-energy tensor component *is* the t-t one, so that the special case applies.)
(I should also note that the expression given on the Wikipedia page for the isolated free particle stress-energy tensor, which I used above, sweeps some technicalities under the rug. As I noted above, the "normal" units for the stress-energy tensor are energy density units--but as the isolated free particle tensor is given above, its units are just energy-- m c^2 --not energy density. To be rigorously correct, the m should be a \rho, for mass density, but you can't really define the "mass density" of an isolated point particle. These technicalities don't affect the main point I was making, so I didn't mention them in my earlier post. The perfect fluid stress-energy tensor expression avoids the problem by just assuming that the fluid is a continuous substance, not a bunch of isolated points; as long as the distance scales you're working on are much larger than the actual size of the particles, that assumption works fine.)
