# Stress Energy Tensor

What is a good intuitive way to think of the stress-energy tensor outside of Einstein's Ric-(1/2)Sg = 8pi T? I'm trying to understand the concept, but coming entirely from a math background I'm not quite getting it.

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pervect
Staff Emeritus
If you have a math background, you might be familiar with Clifford algebra.

If you are, then you know that a volume can be represented as a 3-form, and has a clifford dual, which is a 1-form.

The point of this is that the total amount of energy and momentum contained in any particular volume in any frame is given by the volume element (a 1-form) multiplied by the stress-energy tensor. This is a coordinate independent formulation for the "density" of energy and momentum.

You might not think that you need a rank-2 tensor to describe the density of energy and momentum, but it turns out to be required.

If you don't have a background in Clifford algebra, think of a volume element as being defined by a vector which is orthogonal to the volume element. (This implies that the choice of the reference frame determines the direction of the vector). The length of the vector defines the size of the volume element.

When we look at this tensor do we consider a volume element and consider how it changes in time, like divergence?

Or, would we keep a static volume element and consider how the energy inside the volume is changing?

George Jones
Staff Emeritus
Gold Member
What is a good intuitive way to think of the stress-energy tensor outside of Einstein's Ric-(1/2)Sg = 8pi T? I'm trying to understand the concept, but coming entirely from a math background I'm not quite getting it.
Overview.

The stress-energy tensor T is a second rank tensor so it can be viewed, for example, as a bilinear map from pairs of tangent vectors to the real. The metric tensor allow T also to be viewed a linear map from tangent vectors to tangent vector, as a linear map from cotangent vectors to tangent vector, etc. Depending on the viewpoint taken, and on which arguments (like 4-velocities) are used for the linear maps, various physical quantities (like pressure) appear the image of a particular linear mapping.

Hopefully, I can specifics tomorrow.

On the other hand, the stress-energy tensor used in some cosmological models is that of a perfect fluid -

$$T_{\mu\nu} = pg_{\mu\nu} + \mu u_{\mu}u_{\nu} + \frac{p}{c^2}u_{\mu}u_{\nu}$$

pervect
Staff Emeritus
As George alludes, there are several different interpretations possible of the stress-energy tensor. These are discussed in, for example, MTW's "Gravitation". IMO, however, the most useful interpretation of the stress-energy tensor is that it is a bi-linear map from a 4-vector representation of a volume element to the vector representation of the 4-momentum contained within that volume element. The 4-momentum is of course a 4-vector.

Thus the stress-energy tensor is a map from a 4-vector to a 4-vector, which makes it a rank 2 tensor.

Clifford algebra comes in handy in demonstrating that a vector *is* the correct way to represent a volume.

On the other hand, the stress-energy tensor used in some cosmological models is that of a perfect fluid -

$$T_{\mu\nu} = pg_{\mu\nu} + \mu u_{\mu}u_{\nu} + \frac{p}{c^2}u_{\mu}u_{\nu}$$
Could someone, please, explain me, the conflict in the stress-energy tensor units. So for the perfect fluid we have:

$$T_{\mu\nu} = pg_{\mu\nu} + (\rho + \frac{p}{c^2})u_{\mu}u_{\nu}$$

But on the other hand

$$u_{\mu} = g_{\mu\nu}u^{\nu}$$.

Now, in the comoving frame

$$u^{\nu} = \frac{dx^{\nu}}{d\tau}$$

where

$$d\tau^2 = - ds^2$$ (-, +, +, + signature).

that follows:

$$d\tau = \sqrt{-g_{00}}dx^0$$

(remember, we are in comoving frame). Here the real problem starts. For Schwarzchild solution, $$g_{00} = - e^{2\nu}$$ and $$dx^0 = cdt$$. Then

$$d\tau = e^{\nu} c dt$$

This makes

$$u^{\nu} = \frac{dx^{\nu}}{d\tau} = e^{-\nu} (1, 0, 0, 0)$$

and

$$u_{\mu} = g_{\mu\nu}u^{\nu} = - e^{\nu} (1, 0, 0, 0)$$

Now let us see, the diagonal elements of energy momentum tensor:

$$T^0_0 = p \delta^0_0 + (\rho +p/c^2)u^0u_0 = p - \rho - p/c^2$$?

And this makes no sense. Could someone help me, where am I doing wrong? Thanks.

$$d\tau^2 = - ds^2$$ (-, +, +, + signature).

I think i found my mistake:

$$c^2 d\tau^2 = - ds^2$$