General parameters of the stress energy tensor in local inertial frame

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?

-

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.
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. –  Ondřej Čertík Jan 2 '12 at 19:16