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- Thread starter Fubini
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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.

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

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George Jones

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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.

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[tex] T_{\mu\nu} = pg_{\mu\nu} + \mu u_{\mu}u_{\nu} + \frac{p}{c^2}u_{\mu}u_{\nu} [/tex]

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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.

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Could someone, please, explain me, the conflict in the stress-energy tensor units. So for the perfect fluid we have:

[tex] T_{\mu\nu} = pg_{\mu\nu} + \mu u_{\mu}u_{\nu} + \frac{p}{c^2}u_{\mu}u_{\nu} [/tex]

[tex] T_{\mu\nu} = pg_{\mu\nu} + (\rho + \frac{p}{c^2})u_{\mu}u_{\nu} [/tex]

But on the other hand

[tex]u_{\mu} = g_{\mu\nu}u^{\nu}[/tex].

Now, in the comoving frame

[tex] u^{\nu} = \frac{dx^{\nu}}{d\tau}[/tex]

where

[tex] d\tau^2 = - ds^2 [/tex] (-, +, +, + signature).

that follows:

[tex] d\tau = \sqrt{-g_{00}}dx^0 [/tex]

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

[tex] d\tau = e^{\nu} c dt [/tex]

This makes

[tex] u^{\nu} = \frac{dx^{\nu}}{d\tau} = e^{-\nu} (1, 0, 0, 0)[/tex]

and

[tex]u_{\mu} = g_{\mu\nu}u^{\nu} = - e^{\nu} (1, 0, 0, 0) [/tex]

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

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

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

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[tex] d\tau^2 = - ds^2 [/tex] (-, +, +, + signature).

I think i found my mistake:

[tex]c^2 d\tau^2 = - ds^2 [/tex]

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