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  General parameters of the stress energy tensor in local inertial frame

+ 2 like - 0 dislike

A general 4x4 symmetric tensor has 10 independent components. How many components are we free to prescribe in the local inertial frame?

For example, relativistic dust is $\mbox{diag}(\rho c^2, 0, 0, 0)$ in local inertial frame (thus 1 parameter) which gives $$T_{\mu\nu} = \rho v_\mu v_\nu$$ that has 4 parameters (one is $\rho$ and then 3 parameters in $v_\mu$ due to the condition $v_\mu v^\mu = c^2$).

Another example is perfect fluid with $\mbox{diag}(\rho c^2, p, p, p)$ (thus 2 parameters) which gives $$ T_{\mu\nu} = \left(\rho+{p\over c^2}\right) v_\mu v_\nu + p g_{\mu\nu} $$ That has 5 parameters ($\rho$, $p$ and the spatial components $v_i$).

As such, it seems to me that there are only 7 independent parameters in the local inertial frame, as the other 3 degrees of freedom are given by the velocity (which is zero in the inertial frame). Is that correct?

This post has been migrated from (A51.SE)
asked Jan 1, 2012 in Theoretical Physics by Ondřej Čertík (55 points) [ no revision ]

1 Answer

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No, it is surely not possible to write every general tensor in terms of 7 parameters. Clearly, the space of possible stress-energy tensors is 10-dimensional in $d=4$ (10 parameters), so you can't make it 7-dimensional (7 parameters). The special choices of the tensor that you mentioned are isotropic in a frame – treating $x,y,z$ on equal footing (they are invariant under an $SO(3)$). But general stress-energy tensors are not isotropic.

You may always diagonalize a symmetric tensor in $d=4$, i.e. replace it by 4 eigenvalues which I may call $\rho, p_{xx}, p_{yy}, p_{zz}$. However, the data needed to specify in which coordinate systems the tensor gets diagonal are equivalent to an element of $SO(3,1)$ – the Lorentz transformation needed to switch from a given basis to the basis of eigenvectors of the matrix – which has the remaining 6 parameters (the dimension of the Lorentz group), so if you also want to remember the information about the directions – and the fact that you included the components of $v^\mu$ shows that you do want to count them – then you're back to 10 parameters.

This is of course generalized to $d$ dimensions. A symmetric tensor has $d(d+1)/2$ components which may be decomposed as $d$ eigenvalues and $d(d-1)/2$ elements of an antisymmetric matrix whose exponentiation gives the right rotation or Lorentz transformation for which the tensor diagonalizes.

This post has been migrated from (A51.SE)
answered Jan 2, 2012 by Luboš Motl (10,278 points) [ no revision ]
Ah, you are of course right. Thanks for the clarification. It's 4 eigenvalues + 6 parameters of the Lorentz group. The velocity is related to the parameters of the Lorentz group -- that was actually my question, what role the velocity is playing in this (I didn't know how to formulate it the best).

This post has been migrated from (A51.SE)
To be specific for perfect fluid: we want the tensor to get diagonal ($\mbox{diag}(\rho, p_{xx}, p_{yy}, p_{zz})$) in the local inertial frame. The (restricted) Lorentz group is generated by 3 spatial rotations and 3 boosts, and we don't want this to depend on the spatial rotations, so we get 3 boosts (that can be parametrized by the velocity $v_i$), $\rho$ and isotropic $p=p_{xx}=p_{yy}=p_{zz}$, total of 5 parameters. All makes sense. I marked your answer as accepted.

This post has been migrated from (A51.SE)
Thanks for your understanding, Ondřeji.

This post has been migrated from (A51.SE)

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