-1
$\begingroup$

Good day to everyone. I have a problem with the stress energy tensor of a perfect fluid. In the frame of reference of the fluid the stress energy tensor is

$T^{\mu \nu} = \left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & P & 0 & 0 \\ 0 & 0 & P & 0 \\ 0 & 0 & 0 & P \end{array}\right)$

Instead, in a general inertial frame the stress energy tensor is

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

(I am working in the convention such that $\eta^{00}=-1$ and $\eta^{11}=\eta^{22}=\eta^{33}=1$, and $u^{\mu}=(\gamma, \gamma\vec{\beta})$). This general form of the stress energy tensor should be verified, for example, applying a Lorentz boost along the x direction: $\Lambda^{\mu}_{\nu}=\left( \begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \gamma \end{array}\right)$

So

$T'^{\mu\nu}=\Lambda^{\mu}_{\rho}\Lambda^{\nu}_{\sigma}T^{\rho\sigma} \to T'=\Lambda T \Lambda^T$

$T'^{\mu\nu}=\left( \begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \gamma \end{array}\right) \left( \begin{array}{cccc} \rho & 0 & 0 & 0 \\ 0 & P & 0 & 0 \\ 0 & 0 & P & 0 \\ 0 & 0 & 0 & P \end{array}\right) \left( \begin{array}{cccc} \gamma & -\beta\gamma & 0 & 0 \\ -\beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \gamma \end{array}\right)=$

$=\left( \begin{array}{cccc} \gamma^2\rho+\beta^2\gamma^2P & -\beta\gamma^2\rho-\beta\gamma^2P & 0 & 0 \\ -\beta\gamma^2\rho-\beta\gamma^2P & \beta^2\gamma^2\rho+P\gamma^2& 0 & 0 \\ 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \gamma \end{array}\right)$

For the components $T'^{00}$ and $T'^{11}$ I have effectively obtained

$T'^{00} = (\rho+P)\gamma^2 - P$

and

$T'^{11} = (\rho+P)\gamma^2\beta^2 + P$

But my result for the component $T'^{01}$ is

$T'^{01} = -(\rho+P)\gamma\cdot\gamma\beta $

So I have obtained the opposite sign respect to the form $T'^{01}=(\rho+P)u^0u^1.$

Where am I wrong ? If someone makes me understand where I have made any mistakes I will be infinitely grateful.

$\endgroup$

closed as off-topic by AccidentalFourierTransform, Kyle Kanos, glS, sammy gerbil, Emilio Pisanty Aug 10 '18 at 23:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – AccidentalFourierTransform, Kyle Kanos, glS, sammy gerbil, Emilio Pisanty
If this question can be reworded to fit the rules in the help center, please edit the question.

0
$\begingroup$

All of your calculations are correct, however, you simply used the incorrect sign for $\beta \gamma$ in your Lorentz transformation. In other words, your Lorentz transformation should be

$$ \Lambda^\mu_\nu= \left( \begin{array}{cccc} \gamma & \beta\gamma & 0 & 0 \\ \beta\gamma & \gamma & 0 & 0 \\ 0 & 0 & \gamma & 0 \\ 0 & 0 & 0 & \gamma \end{array}\right)$$

Now, the physical interpretation of this is that, originally, we were in the "moving frame" of our perfect fluid which gives us the familiar diagonal stress energy tensor. However, if we want to go into a frame where the four velocity of the perfect fluid is $u^{\mu}=(\gamma, \gamma\vec{\beta})$, then we have to boost to a frame moving at a three velocity of $-\vec{\beta} $ relative to our initial frame. With this change, the calculation works out as expected.

$\endgroup$
  • $\begingroup$ So the general form of the stress energy tensor is valid for a perfect fluid that has four velocity $u^{\mu}=(\gamma, \gamma\vec{\beta})$ respect to the observer, and so I had to go in the frame with three velocity $-\vec{\beta}$. Now I have understood, I thank you very much! $\endgroup$ – Pico Aug 3 '18 at 16:50

Not the answer you're looking for? Browse other questions tagged or ask your own question.