How can you add pressure and internal energy in relativistic enthalpy?

Question

In relativistic fluid dynamics the relativistic enthalpy in natural units is defined as: \begin{equation} h = \frac{e+p}{\rho}, \end{equation} Where $e$ is the total energy density and $p$ is the thermodynamic pressure. In contrast, the Newtonian enthalpy is defined as: \begin{equation} h_N = \epsilon+\frac{p}{\rho}, \end{equation} Where $\epsilon$ is the specific internal energy and $\rho$ is the rest mass density.

The question: What are the units of $e$ and $p$ so that I can add them in the definition of $h$ above?

Answer 1

So, following @Ghoster's comments, $e$ and $p$ simply have the same dimensions in both systems of units: \begin{equation} e = \underbrace{(M\ L^2\ T^{-2})}_{energy} \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = \left( \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L}_{times\ length} \right) \cdot \underbrace{L^{-3}}_{per\ unit\ volume} = M\ L^{-1}\ T^{-2} \\ p = \underbrace{(M \ L \ T^{-2})}_{force} \cdot \underbrace{L^{-2}}_{per\ unit\ area} = M\ L^{-1}\ T^{-2} = e \end{equation}