Relativistic Euler equations

In the case of flat space, that is ${\displaystyle \nabla _{\mu }=\partial _{\mu }}$ and using a metric signature of ${\displaystyle (-,+,+,+)}$, the equations of motion are,^[6] $${\displaystyle \left(e+p\right)u^{\mu }\partial _{\mu }u^{\nu }=-\partial ^{\nu }p-u^{\nu }u^{\mu }\partial _{\mu }p}$$

Where ${\displaystyle e=\rho (c^{2}+\varepsilon )}$ is the energy density of the system, with ${\displaystyle p}$ being the pressure, and ${\displaystyle u^{\mu }=\gamma (1,\mathbf {v} /{c})}$ being the four-velocity of the system.

Expanding out the sums and equations, we have, (using ${\displaystyle {\frac {d}{dt}}}$ as the material derivative) $${\displaystyle \left(e+p\right){\frac {\gamma }{c}}{\frac {du^{\mu }}{dt}}=-\partial ^{\mu }p-{\frac {\gamma }{c}}{\frac {dp}{dt}}u^{\mu }}$$

Then, picking ${\displaystyle u^{\nu }=u^{i}={\frac {\gamma }{c}}v_{i}}$ to observe the behavior of the velocity itself, we see that the equations of motion become $${\displaystyle \left(e+p\right){\frac {\gamma }{c^{2}}}{\frac {d}{dt}}{\left(\gamma v_{i}\right)}=-\partial _{i}p-{\frac {\gamma ^{2}}{c^{2}}}{\frac {dp}{dt}}v_{i}}$$

Note that taking the non-relativistic limit, we have ${\textstyle {\frac {1}{c^{2}}}\left(e+p\right)=\rho +{\frac {1}{c^{2}}}\rho \varepsilon +{\frac {1}{c^{2}}}p\approx \rho }$. This says that the energy of the fluid is dominated by its rest energy.

In this limit, we have ${\displaystyle \gamma \to 1}$ and ${\displaystyle c\to \infty }$, and can see that we return the Euler Equation of ${\displaystyle \rho {\frac {dv_{i}}{dt}}=-\partial _{i}p}$.

In order to determine the equations of motion, we take advantage of the following spatial projection tensor condition: $${\displaystyle \partial _{\mu }T^{\mu \nu }+u_{\alpha }u^{\nu }\partial _{\mu }T^{\mu \alpha }=0}$$

We prove this by looking at ${\displaystyle \partial _{\mu }T^{\mu \nu }+u_{\alpha }u^{\nu }\partial _{\mu }T^{\mu \alpha }}$ and then multiplying each side by ${\displaystyle u_{\nu }}$. Upon doing this, and noting that ${\displaystyle u^{\mu }u_{\mu }=-1}$, we have ${\displaystyle u_{\nu }\partial _{\mu }T^{\mu \nu }-u_{\alpha }\partial _{\mu }T^{\mu \alpha }}$. Relabeling the indices ${\displaystyle \alpha }$ as ${\displaystyle \nu }$ shows that the two completely cancel. This cancellation is the expected result of contracting a temporal tensor with a spatial tensor.

Now, when we note that $${\displaystyle T^{\mu \nu }=wu^{\mu }u^{\nu }+pg^{\mu \nu }}$$

where we have implicitly defined that ${\displaystyle w\equiv e+p}$, we can calculate that $${\displaystyle {\begin{aligned}\partial _{\mu }T^{\mu \nu }&=\left(\partial _{\mu }w\right)u^{\mu }u^{\nu }+w\left(\partial _{\mu }u^{\mu }\right)u^{\nu }+wu^{\mu }\partial _{\mu }u^{\nu }+\partial ^{\nu }p\\[1ex]\partial _{\mu }T^{\mu \alpha }&=\left(\partial _{\mu }w\right)u^{\mu }u^{\alpha }+w\left(\partial _{\mu }u^{\mu }\right)u^{\alpha }+wu^{\mu }\partial _{\mu }u^{\alpha }+\partial ^{\alpha }p\end{aligned}}}$$

and thus $${\displaystyle u^{\nu }u_{\alpha }\partial _{\mu }T^{\mu \alpha }=(\partial _{\mu }w)u^{\mu }u^{\nu }u^{\alpha }u_{\alpha }+w(\partial _{\mu }u^{\mu })u^{\nu }u^{\alpha }u_{\alpha }+wu^{\mu }u^{\nu }u_{\alpha }\partial _{\mu }u^{\alpha }+u^{\nu }u_{\alpha }\partial ^{\alpha }p}$$

Then, let's note the fact that ${\displaystyle u^{\alpha }u_{\alpha }=-1}$ and ${\displaystyle u^{\alpha }\partial _{\nu }u_{\alpha }=0}$. Note that the second identity follows from the first. Under these simplifications, we find that $${\displaystyle u^{\nu }u_{\alpha }\partial _{\mu }T^{\mu \alpha }=-(\partial _{\mu }w)u^{\mu }u^{\nu }-w(\partial _{\mu }u^{\mu })u^{\nu }+u^{\nu }u^{\alpha }\partial _{\alpha }p}$$

and thus by ${\displaystyle \partial _{\mu }T^{\mu \nu }+u_{\alpha }u^{\nu }\partial _{\mu }T^{\mu \alpha }=0}$, we have $${\displaystyle (\partial _{\mu }w)u^{\mu }u^{\nu }+w(\partial _{\mu }u^{\mu })u^{\nu }+wu^{\mu }\partial _{\mu }u^{\nu }+\partial ^{\nu }p-(\partial _{\mu }w)u^{\mu }u^{\nu }-w(\partial _{\mu }u^{\mu })u^{\nu }+u^{\nu }u^{\alpha }\partial _{\alpha }p=0}$$

We have two cancellations, and are thus left with $${\displaystyle (e+p)u^{\mu }\partial _{\mu }u^{\nu }=-\partial ^{\nu }p-u^{\nu }u^{\alpha }\partial _{\alpha }p}$$

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