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I am in a General relativity course and I am pretty new to tensors. My professor says that he will leave learning about how to manipulate tensors up to our reading (as there is a section in the book describing it, namely the book by Carroll) but I am still confused about a few things. Here is a problem I just solved but I want like to have more detail as to what is going on.

The problem says,"For a group of particles all moving with the same velocity, $\tilde{\beta} = \beta \tilde{e}_x$ as seen in an inertial reference frame S with a rest-mass density of $\rho_0$, calculate the stress-energy tensor."

So again, I have the solution and I was able to obtain it, but I don't understand it. So for a perfect fluid, the stress energy tensor only has one component, namely, \begin{equation} T^{\mu \nu} = \left( \begin{array}{cccc} \rho_0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation}

The 4-velocity of an intertial frame moving with velocity $\tilde{\beta}$ is given by:

\begin{equation} U^{\mu} = \left( \begin{array}{cccc} \gamma \\ \gamma \beta \\ 0 \\ 0 \end{array} \right) \end{equation}

So in the solution it says, "By transforming to a frame moving in the x-direction with velocity $-\beta$, we find the nonzero components."

So to obtain the solution they provide, I have determined, purely by mathematical manipulation that I need to do the following (I do not think my notation is correct):

First write: \begin{equation} U^{\mu}U^{\nu}= \left( \begin{array}{cccc} \gamma^2 & \gamma^2\beta & 0 & 0 \\ \gamma^2\beta & \gamma^2\beta^2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation}

Then I did this this:

\begin{equation} T^{\mu\nu}\otimes U^{\mu}U^{\mu}= \left( \begin{array}{cccc} \gamma^2\rho_0 & \gamma^2\rho_0\beta & 0 & 0 \\ \gamma^2\rho_0\beta & \gamma^2\rho_0\beta^2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation} which is the answer. But why does this represent a transformation to the moving frame? Also, I think there may also be something with the sign of $\beta$ as it is moving $-\beta$ with respect to the particles rest frame. Also, I am not sure my use of $\otimes$ is correct. All I did was matrix multiply $U^{\mu}U^{\nu}$ with $T^{\mu\nu}$. Any help would be appreciated!

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  • $\begingroup$ If the frame is moving with $-β$, the matter is moving with $β$, so the sign is correct. $\endgroup$ – hebetudinous Sep 3 '16 at 3:09
  • $\begingroup$ Hi. For a perfect fluid the stress-energy tensor is $T_{αβ}=ρu^{α}u^{β} $. Lorentz transform your objects from the inertial frame you already have to the moving. I think you will then find out why the result is the same. $\endgroup$ – Constantine Black Sep 3 '16 at 15:46
  • $\begingroup$ It may (or may not) help to consider the metric transpose $g_{\mu \kappa} T^{\kappa \nu}$ rather than $T$, so you have an object whose coordinate representation truly is a $4 \times 4$ matrix. If you're careful about rows vs. columns, the coordinate representation of $T^{\mu \nu}$ should be a $16 \times 1$ matrix partitioned into four $4 \times 1$ blocks. $\endgroup$ – Hurkyl Sep 6 '16 at 4:12
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Your method here is incorrect and your calculations are a bit off (although you got the right answer). So let's go through the correct procedure.

For any arbitrary tensor in flat spacetime (so we still live in special relativity), the transformation law can be described by the following equation: $$ T^{\mu'\nu'} = T^{\mu\nu} \Lambda^{\mu'}_{\mu} \Lambda^{\nu'}_{\nu}$$ Where the $\Lambda$s represent Lorentz transformations from the original frame (unprimed) to the frame we are transforming to (the primed frame). So $T^{\mu\nu}$ are the components of the stress-energy tensor in the unprimed frame, while $T^{\mu'\nu'}$ are the corresponding components in the primed frame.

Now since you expressed the stress-energy tensor in its matrix form, I will do the same. The matrix form of the transformation equation is: $$ T' = \Lambda T \Lambda^T$$ Where $\Lambda$ is the matrix representations of the Lorentz transformation and $\Lambda^T$ is the transpose (exchanging rows and columns) of $\Lambda$.

For a boost in the $x$ direction with velocity $- \beta$ (your case), the $\Lambda$ matrix looks like this: \begin{equation} \Lambda = \left( \begin{array}{cccc} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation} The transpose is actually the same: \begin{equation} \Lambda^T = \left( \begin{array}{cccc} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation} So we can then calculate $T'$: \begin{equation} T' = \left( \begin{array}{cccc} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} \rho_0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \left( \begin{array}{cccc} \gamma & \gamma \beta & 0 & 0 \\ \gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) = \left( \begin{array}{cccc} \gamma^2 \rho_0 & \gamma^2 \rho_0 \beta & 0 & 0 \\ \gamma^2 \rho_0 \beta & \gamma^2 \rho_0 \beta^2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation}

Surprise! I used a different method but obtained the same answer. So what is the coincidence here? Well, the coincidence is simply due to the special form of a perfect fluid stress-energy tensor: $$T^{\mu\nu} = \rho_0 U^{\mu} U^{\nu}$$ So when we apply the Lorentz transformation to the tensor, we get: $$ T^{\mu'\nu'} = T^{\mu\nu} \Lambda^{\mu'}_{\mu} \Lambda^{\nu'}_{\nu} = \rho_0 U^{\mu} \Lambda^{\mu'}_{\mu} U^{\nu} \Lambda^{\nu'}_{\nu}$$ Which is equivalent to applying Lorentz transformations to the four velocities $U^{\mu}$ and $U^{\nu}$ separately and then taking a tensor product: $$ T^{\mu'\nu'} = \rho_0 U^{\mu’} U^{\nu’}$$ In tensor form, this is equivalent to: $$ T' = \rho_0 U' \otimes U'$$ So your use of the tensor product was incorrect. If we carry out the calculation this way, we get:

\begin{equation} U' = \left( \begin{array}{cccc} \gamma \\ \gamma \beta \\ 0 \\ 0 \end{array} \right) \end{equation} So we can then calculate $T'$: \begin{equation} T^{\mu'\nu'} = \rho_0 U^{\mu'} U^{\nu'} = \rho_0 \left( \begin{array}{cccc} \gamma^2 & \gamma^2 \beta & 0 & 0 \\ \gamma^2 \beta & \gamma^2 \beta^2 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} \right) \end{equation}

I hope this clears up your confusion!

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  • $\begingroup$ Wow thanks so much! This clears up my confusion perfectly! $\endgroup$ – user41178 Sep 10 '16 at 1:27

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