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GR191511

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GR191511

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malawi_glenn

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Have you tried to verify this, by say, doing a Lorentz boost in the x direction?

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$$T{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu} - P \eta^{\mu \nu}$$

in a local inertial frame, where the four-velocity of the fluid element is ##u^{\mu}##. So finally in an arbitrary frame you have

$$T^{\mu \nu}=(\epsilon+P) u^{\mu} u^{\nu}-P g^{\mu \nu}.$$

Note that ##\epsilon## and ##P## are scalars since they are defined in a specific "preferred" frame of reference, i.e., the local rest frame of the fluid cell.

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malawi_glenn

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Momentarily comoving rest frame

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martinbn

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The diagonal elements of a diagonal matrix are the eigenvalues. If it is diagonal in all frames it means that any basis of the vector spaces will consist of eigen vectors. If you have two different eigenvalues, say ##\alpha\not=\beta## and ##v## and ##w## are corresponding eingenvectors, then choose a basis with first basis vector ##v+w##. When you apply the matrix to it you get ##\alpha v+\beta w##, which is not a multiple of ##v+w##.The only matrix diagonal in all frames is a multiple of the identity: all its diagonal terms are equal.

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$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

This is diagonal, but it'll be non-diagonal if you boost it - i.e. in any other frame than the MCRF frame.

Now let's consider a case with pressure.

If the whole stress energy tensor is T[0:3, 0:3], the submatrix representing pressure, also called the stress tensor, will be T[1:3, 1:3]. If the pressure is isotropic, the pressure will be diagonal under rotational transformations. But if the pressure is not isotropic, I.e. if we have

$$\begin{bmatrix} p1 & 0 & 0 \\ 0 & p2 & 0 \\ 0 & 0 & p3 \end{bmatrix}$$

where p1, p2, and p3 are not all equal, a rotational transformation will make the stress tensor non-diagonal.

A perfect fluid, which was discussed in a previous post, will always have an isotropic pressure by definition.

Note that we have not considered all possible casses of the stress energy tensor, such as those containing radiation and not matter.

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$$T^{\mu \nu}=\rho u^{\mu} u^{\nu},$$

where ##u^{\mu}## is the four-velocity field of the dust distribution, which is of course not diagonal. Again the quantity ##\rho## is a scalar field, i.e., the energy density as measured in the rest frame of the medium.

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martinbn

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I am not sure why you are trying to consider examples! The whole point is that if it is diagonal in all frames it has to be a scalar times the identity.

$$\begin{bmatrix} \rho & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{bmatrix}$$

This is diagonal, but it'll be non-diagonal if you boost it - i.e. in any other frame than the MCRF frame.

Now let's consider a case with pressure.

If the whole stress energy tensor is T[0:3, 0:3], the submatrix representing pressure, also called the stress tensor, will be T[1:3, 1:3]. If the pressure is isotropic, the pressure will be diagonal under rotational transformations. But if the pressure is not isotropic, I.e. if we have

$$\begin{bmatrix} p1 & 0 & 0 \\ 0 & p2 & 0 \\ 0 & 0 & p3 \end{bmatrix}$$

where p1, p2, and p3 are not all equal, a rotational transformation will make the stress tensor non-diagonal.

A perfect fluid, which was discussed in a previous post, will always have an isotropic pressure by definition.

Note that we have not considered all possible casses of the stress energy tensor, such as those containing radiation and not matter.

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malawi_glenn

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Beucase it is illustrative to see how examples play out?I am not sure why you are trying to consider examples!

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martinbn

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In general yes, but why here? His reply started with "I'm not sure what the author is trying to say." The point is that the matter content is irrelevant.Beucase it is illustrative to see how examples play out?

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martinbn

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The reference is section 4.6 Perfect Fluids, pages 100-101 in the second edition.

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$$T^{\mu \nu}=\mathrm{diag}(\epsilon,P,P,P).$$

In an arbitrary frame, all you have to express this in a covariant way is the metric tensor and the fluid four-velocity field, i.e.,

$$T^{\mu \nu} = (\epsilon+P) u^{\mu} u^{\nu} - P g^{\mu \nu},$$

using the mainly-minus convention of the metric, i.e., ##(\eta_{\mu \nu})=(1,-1,-1,-1)## in the local inertial frame.

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malawi_glenn

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Author and publisher could be helpfulThe reference is section 4.6 Perfect Fluids, pages 100-101 in the second edition.

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malawi_glenn

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Right, I was thinking it was in a book called "perfect fluids". Too hot today.

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One can visualize a symmetric matrix using a quadric (for example, an ellipsoid or hyperboloid).“The only matrix diagonal in all frames is a multiple of the identity：all its diagonal terms are equal.”

When rotating the axes to make the matrix diagonal, these axes are along the axes of symmetry of the quadric.

(Think about the moment of inertia tensor.)

If the matrix is diagonal in all frames, then the quadric is a sphere.

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One can visualize a symmetric matrix using a quadric (for example, an ellipsoid or hyperboloid).

When rotating the axes to make the matrix diagonal, these axes are along the axes of symmetry of the quadric.

(Think about the moment of inertia tensor.)

If the matrix is diagonal in all frames, then the quadric is a sphere.

Yes, the stress-energy tensor is a symmetric bilinear form, which means that it can be represented by a quadratic form. And the principal axis theorem applies to quadratic forms, thus one can always diagonalize them. (This is from memory, I might have missed a few subtle points).

In the diagonal form, the diagonal elemetns are the eigenvalues, and if all three eignevalues are the same, one has the spherical symmetry you note.

Since the digaonal form of the stress-energy tensor always exists, it's simplest to imagine choosing a preferred coordinate system where the stress-energy tensor is diagonal. This makes it a lot easier to grasp, at least for me. Various techniques exist to visualize diagonal matrices, such as representing them via ellispoids.

One then can eventually realize that if the three eigenvalues are identical, the form has spherical symmetry, and rotation won't change the diagonalization property. If the three eigenvalues are not identical, spherical symmetry is lacking, and rotational transformations will not preserve the diagonalization.

Because the context of the original text is apparently restricted to perfect fluids, spherical symmetry exists. But this won't necessarily be true for stress-energy tensors that are not perfect fluids.

As for examples, I always find it useful to consider specific examples of abstract theorems. Your mileage may vary, I suppose.

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