General Relativity Tutorial - The Stress-Energy Tensor

John Baez

G_ab = T_ab

(in nice units). The thing on the left is the Einstein tensor, which summarizes some information about spacetime curvature.

Ed Green asks: "What stress tensor?"

The stress-energy tensor, aka energy-momentum tensor, T_ab, where a,b go from 0 to 3. This tells you everything about what energy and momentum are doing at your given point of spacetime, as follows:

T_ab is the flow in the a direction of momentum in the b direction!

To understand this, remember that a,b=0,1,2,3 correspond to t,x,y, and z respectively. Also, remember that "energy" is the same as "momentum in the time direction", and that "density" is the same as "flow in the time direction". Thus the top row of the stress-energy tensor keeps track of the density of energy --- that's T₀₀ --- and the density of momentum in the x,y, and z directions --- those are T₀₁, T₀₂, and T₀₃ respectively. On the other hand, T₁₀, T₂₀, and T₃₀ represent the flow of energy in the x, y and z directions, respectively. The other entries keep track of the flow of spatial momentum in various spatial directions. For example, T₁₂ keeps track of the flow in the x direction of momentum in the y direction.

Ed Green asks: "What language are you talking?"

Physics, ca 20th century.

(I sure can be snide. Sorry.)

Ed frowns and says: "You wouldn't care to give me a crash descriptive course in GR, would you? (That is the language you are speaking, isn't it?)."

Well, the stress-energy tensor is a basic gadget throughout physics, since we all want to know where energy and momentum are going and how much there is sitting around, right? But it's only in general relativity where the stress-energy tensor is sitting proudly on the right side of an equation, telling spacetime how to curve.

Michael Weiss elaborates:

"I remember one thing about T_ij that puzzled me when I first encountered it, and the "aha" mental lightbulb that cleared things up.

Here's the puzzle: if T_ij measures the amount of i-momentum being transferred in the j-direction, then how come T_ij isn't identically zero in the rest frame of the fluid, when the momentum vanishes?

Let's polish off a couple of points quickly. As Bruce Scott points out [below], in general a fluid doesn't have a global rest frame. We can say we're just talking about an infinitesimal neighborhood of a point, or the special case of a motionless fluid.

Next, T₀₀ is a special case. In the rest frame of the fluid, the velocity 4-vector of a fluid element is (E,0,0,0) -- you can make the v_i components vanish for i=1,2,3 by picking the right frame of reference, but not the v₀ component, aka the energy. (I'm being sloppy about the difference between tensors and tensor-densities, but let's ignore that. What's a determinant among friends?) So it's not surprising that T₀₀ doesn't vanish, but why doesn't T_ii vanish for i=1,2,3? (Once we've plopped ourselves down in the rest frame of the fluid, that is.)

Answer: think microscopically, and remember that T depends *quadratically* on v. Consider the yz-plane, for yucks. Particles of fluid constantly zip through this plane in all directions. If a particle with 4-vector (E, v_x, v_y, v_z) passes across the plane, headed in the +x direction (i.e., v_x>0), then it transfers a bit of v_x momentum in that direction, and likewise v_y momentum and v_z momentum.

OK, now here's the kicker. If we look at the *average* transfer of v_y and v_z across the yz-plane, we get zero. That's because v_x is uncorrelated with v_y and v_z, so the average of (v_x.v_y) is zero, ditto (v_x.v_z). (If we weren't sitting in the rest-frame of the fluid--- if we felt a breeze--- then this wouldn't necessarily be true.) But the average of (v_x.v_x) is positive, of course; it doesn't take much imagination to see why this represents the pressure."

Edward Green replies:

"Thank you for your answers! I still feel there must be something special about the top-row, though where it falls between the mathematics and the physics I couldn't say."

Bruce Scott comments:

"The way you make the top row unique is to look at it in the rest frame of the thing whose stress tensor you're studying. For example, a dissipationless fluid has a stress tensor of

T^ab = n e u^a u^b + p (g^ab + u^a u^b)

(n is the particle density in the fluid's rest frame, e is the thermal energy per particle, and p is the pressure), which looks a bit complicated. In fact, there is _nothing_ special about the top row of this if g^ab has time-space components (example: Kruskal coordinates for the black hole problem) and u is relativistic (especially if it has significant variation in any of the four coordinates). Note that the only physical requirement on u is that it must be a timelike _unit_ vector: u_a u^a = -1 (in units with c = 1).

But if you contract this with the fluid's velocity you get a peek at the fluid's rest frame:

- T^ab u_b = n e u^a.

Now, that looks a lot more like an energy flux, doesn't it? It is in fact a four-vector. But what about the pressure? You can find the fluid's rest frame by transforming such that u^a = (1,0,0,0). Note you have to do a local Lorentz transformation: one that is defined for the small neighborhood about a single point. This is because u for a fluid is itself a field variable. But in the local rest frame you find

T^ab = diag ( ne, p, p, p ),

where "diag" denotes a diagonal matrix. The 00-component is the thermal energy density and the ii-components are the pressure (one assumes an isotropic pressure for a ideal fluid).

Note that "thermal energy" here includes rest energy.

Your question was actually aimed at distinguishing among the ii-components. This is, as you surmise, not possible without arbitrariness, unless there is a physical phenomenon (such as a local magnetic field; gee, isn't plasma physics nice :-] ) to show you the way.

There is a popular book I saw once where the question of explaining left and right to an extraterrestrial was discussed, to show the arbitrariness in our convention. A complicated set of mental gymnastics was offered by which one could use symmetry violation in kaon decays to do this. No, I don't remember the details :-)"