Transforming Stress-Energy Tensors in Different Frames

[Silviu]

Homework Statement

In an inertial frame O calculate the components of the stress–energy tensors of the following systems:

1. (a) A group of particles all moving with the same velocity $v = \beta e_x$, as seen in O.
   Let the rest-mass density of these particles be $\rho_0$, as measured in their comoving frame. Assume a sufficiently high density of particles to enable treating them as a continuum.

Homework Equations

$T^{\alpha \beta} =\rho_0 U^{\alpha} U^\beta$

The Attempt at a Solution

I used the above equation, and I got the same results as in the book (as the particles can be assumed to be "dust"). However, in the MCRF, the tensor has $T^{00} = \rho_0$ and all the other components equal to 0. If I try to calculate the tensor in another frame moving with speed $\beta$ along the x-axis of this MCRF using $T^{\alpha ' \beta '} = \Lambda^{\alpha '}_\alpha \Lambda^{\beta '}_\beta T^{\alpha \beta}$. I don't get the same result. Why is this approach wrong? Thank you!

 

[TSny]

Both methods should give the correct answer. Can you show more detail of your calculations?

 

[Silviu]

Both methods should give the correct answer. Can you show more detail of your calculations?

We have $\Lambda = \begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$ so $T' = \begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rho_0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ =
$\begin{pmatrix} \gamma & -\beta \gamma & 0 & 0 \\ -\beta \gamma & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \rho_0\gamma & 0 & 0 & 0 \\ -\beta\rho_0\gamma & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$ =
$\begin{pmatrix} \rho_0 \gamma^2 + \beta^2 \gamma^2 \rho_0 & 0 & 0 & 0 \\ -2\beta \rho_0 \gamma^2 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$

 

[TSny]

Is this the matrix that transforms from the comoving frame to frame O, or does it transfrom from O to the comoving frame?

 

[Silviu]

Is this the matrix that transforms from the comoving frame to frame O, or does it transfrom from O to the comoving frame?

Sorry, I replied to your previous post. That is what you obtain going from MCRF to the one that is observing it (I hope I didn't do mistakes). However, when you do the other method you have the 01 and 11 entries non-zero, so I am doing something wrong here.

 

[TSny]

It helps to write the components of $\Lambda$ as $\Lambda^{\alpha '} \, _\mu$ so that the first index $\alpha '$ is the row index and the second index $\mu$ is the column index. Then you have $T^{\alpha ' \beta '} = \Lambda^{\alpha '} \, _\mu \Lambda^{\beta '} \, _\nu T^{\mu \nu}$. You can then see that this is not the same as matrix multiplication $\Lambda \Lambda T$. But note that

$T^{\alpha ' \beta '} = \Lambda^{\alpha '} \, _\mu \Lambda^{\beta '} \, _\nu T^{\mu \nu} = \Lambda^{\alpha '} \, _\mu T^{\mu \nu} \Lambda^{\beta '} \, _\nu = \Lambda^{\alpha '} \, _\mu T^{\mu \nu} \left( \Lambda^T \right) _\nu \, ^{\beta '}$.

Here, $\Lambda ^ T$ is the transpose of $\Lambda$. So the matrix multiplication is $\Lambda T \Lambda^T$.

The other issue is whether or not you have the correct entries for $\Lambda$ for transforming from the comoving frame to frame O.