# Energy-momentum tensor of a perfect fluid flowing at the speed of light?

The general energy-momentum tensor for a fluid is given by (working with $$c=1$$ convention)

$$$$\label{T}\tag{1}T^{\mu\nu}=\left(\rho+p\right)U^\mu U^\nu+p g^{\mu\nu}$$$$

with $$g^{\mu\nu}$$ the inverse of the Minkowski metric in mostly-plus signature. One can e.g. go to the rest frame of the fluid $$U^\mu=(1,0,0,0)$$ and consider the simplified rest frame result

$$T^{\mu\nu}_r= \left(\begin{matrix} \rho & 0 & 0 & 0 \\ 0 & p & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \\ \end{matrix}\right).$$

Instead of the rest frame I would like to consider the case where the fluid is flowing at the speed of light e.g. in the $$x$$-direction. This scenario makes sense when the fluid itself consists of massless particles, such as radiation ($$p=\frac{1}{3}\rho$$).

I can think of two different ways to write down an energy-momentum tensor in that case:

1. Simply take the general equation $$\eqref{T}$$ and instead of a time-like $$U^\mu$$ use a light-like $$U^\mu=(1,1,0,0)$$. This leads to:

$$T^{\mu\nu}_{c,1}=\left( \begin{array}{cccc} \rho & p+\rho & 0 & 0 \\ p+\rho & 2 p+\rho & 0 & 0 \\ 0 & 0 & p & 0 \\ 0 & 0 & 0 & p \\ \end{array} \right).$$

1. Start with the rest frame result $$T^{\mu\nu}_r$$ and perform a boost in the $$x$$-direction $$\Lambda^\mu{}_\alpha \Lambda^\nu{}_\beta T^{\alpha\beta}_r$$, with the boost Lorentz transformation matrix:

$$\Lambda^\sigma{}_\tau=\left( \begin{array}{cccc} \frac{1}{\sqrt{1-v^2}} & \frac{v}{\sqrt{1-v^2}} & 0 & 0 \\ \frac{v}{\sqrt{1-v^2}} & \frac{1}{\sqrt{1-v^2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{array} \right)$$

then consider the light-like limit $$v\to1$$. To do that properly, we apply a redefinition of $$\rho = \rho'\sqrt{1-v^2}$$ and $$p=p'\sqrt{1-v^2}$$ to preserve finite scale in the problem (note that the rescaling is actually dimensionless $$\sqrt{1-{v^2}/{c^2}}$$ when $$c$$ is written explicitly). This second approach leads to:

$$T^{\mu\nu}_{c,2}=\left( \begin{array}{cccc} p'+\rho' & p'+\rho' & 0 & 0 \\ p'+\rho' & p'+\rho' & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right).$$

My question is:

Which of the two energy-momentum tensors is correct? Or maybe they are both wrong and you can tell me the correct expression?

I tend to think that $$T^{\mu\nu}_{c,2}$$ is correct, since if all the constituent particles of the fluid are moving at the speed of light in the $$x$$-direction, any additional motion in the $$y$$- and $$z$$-directions would cause the particles to exceed the speed of light overall and is therefore not allowed. Which forces the pressure in $$y$$- and $$z$$-direction to be zero. What do you think?

Once the radiation fluid is traveling at the speed of light in a particular spatial direction, the situation effectively becomes equivalent to a mixture of plane waves of light traveling in the same direction.

To get an idea how the energy momentum tensor should look like in this situation, take an electromagnetic potential of a linearly polarized plane wave traveling in the $$x$$-direction with momentum $$p^\mu=(\omega,\omega,0,0)$$:

$$A_\mu = (0,0,0,\cos(t\omega-x\omega))$$

From this we obtain the electromagnetic field $$F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$$, and from that the usual electromagnetic energy momentum tensor

$$T_{\mu\nu}=-\frac{1}{\mu_0}\left(F_{\mu\alpha}\eta^{\alpha\beta}F_{\beta\nu}-\frac{1}{4}\eta_{\mu\nu}F_{\sigma\alpha}\eta^{\alpha\beta}F_{\beta\rho}\eta^{\rho\sigma}\right)$$ $$=\left( \begin{array}{cccc} \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & -\frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & 0 & 0 \\ -\frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

This shows that there are indeed only four non-zero components, all of the same magnitude. Raising the spacetime indices on this expression gives

$$T^{\mu\nu}=\eta^{\mu\alpha}\eta^{\nu\beta}T_{\alpha\beta}=\left( \begin{array}{cccc} \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & 0 & 0 \\ \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & \frac{\omega ^2 \sin ^2(\omega (t-x))}{\mu _0} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{array} \right)$$

such that the overall signs of all components are consistent.

Due to the superposition principle, summing over various energies and polarizations to compose a radiation "fluid" will not change the qualitative form of the energy momentum tensor. For radiation, traveling in one spatial direction only, it will always have four non-zero components of equal magnitude.