Contribution of Pressure to the Energy–Momentum Density in a Moving Perfect Fluid: A Physical Perspective

Abstract

In the energy–momentum density expressions for a relativistic perfect fluid with a bulk motion, one comes across a couple of pressure-dependent terms, which though well known, are to an extent, lacking in their conceptual basis and the ensuing physical interpretation. In the expression for the energy density, the rest mass density along with the kinetic energy density of the fluid constituents due to their random motion, which contributes to the pressure as well, are already included. However, in a fluid with a bulk motion, there are, in addition, a couple of explicit, pressure-dependent terms in the energy–momentum density, whose presence to an extent, is shrouded in mystery, especially from a physical perspective. We show here that one such pressure-dependent term appearing in the energy density, represents the work done by the fluid pressure against the Lorentz contraction during transition from the rest frame of the fluid to another frame in which the fluid has a bulk motion. This applies equally to the electromagnetic energy density of electrically charged systems in motion and explains in a natural manner an apparently paradoxical result that the field energy of a charged capacitor system decreases with an increase in the system velocity. The momentum density includes another pressure-dependent term, that represents an energy flow across the system, due to the opposite signs of work being done by pressure on two opposite sides of the moving fluid. From Maxwell’s stress tensor we demonstrate that in the expression for electromagnetic momentum of an electric charged particle, it is the presence of a similar pressure term, arising from electrical self-repulsion forces in the charged sphere, that yields a natural solution for the notorious, more than a century old but thought by many as still unresolved, 4/3 problem in the electromagnetic mass.

Appendix

1.1 Energy Momentum in the Electromagnetic Fields of a Charged Particle in Motion

We assume the charged particle to be a thin spherical shell of radius $\epsilon$ of a uniform surface charge density, with total charge e, stationary in a frame $\mathcal{S}_0$, called the rest frame (Fig. 5). The electric field is zero inside the shell, and is radial outside the shell, with magnitude $E_0=e/r^2$.

Fig. 5

A uniformly charged, thin spherical shell, has only a radial electric field outside the shell, but no fields inside the shell, as seen in the rest frame $\mathcal{S}_0$. The charged shell moves with velocity $\mathbf{v}$ along the x-axis with respect to the lab frame $\mathcal{S}$, where both electric and magnetic fields exist outside the spheroid-shaped shell, but nil fields on the inside

The volume integral of the energy density in rest frame $\mathcal{S}_0$ is easily calculated to be

$$\begin{aligned} \mathcal{E}_{0}=\int _\epsilon ^\infty \frac{E_0^{2}}{8\pi }\,\mathrm{d}V_0 = \frac{e^2}{2\epsilon }. \end{aligned}$$

(66)

The momentum is zero as the magnetic field is zero throughout

The volume integrals of the energy and momentum densities of the electromagnetic fields of the charge in frame $\mathcal{S}$, where it is moving with a velocity $\mathbf{v}$,  say, along x-axis, are then computed, making use of Eqs. (50) and (51), as

$$\begin{aligned}&\mathcal{E}=\int \frac{E^{2}+B^{2}}{8\pi }\,\mathrm{d}V, \end{aligned}$$

(67)

$$\begin{aligned}&\mathbf{P}=\int \frac{\mathbf{E} \times \mathbf{B}}{4\pi c}\,\mathrm{d}V, \end{aligned}$$

(68)

where $\mathbf{E}$ and $\mathbf{B}$ are the electric and magnetic fields of the uniformly moving charge.

These integrals can be evaluated in a simple way by noting that $\mathbf{B}=\mathbf{v} \times \mathbf{E}/c$, and using the transformation relations between $\mathcal{S}$ and $\mathcal{S}_0$, for the electromagnetic fields as well as for the volume element as

$$\begin{aligned} E_{\Vert }= \,\,& E_{0\Vert }\,, \end{aligned}$$

(69)

$$\begin{aligned} E_{\bot }=\, & {} \gamma E_{0\bot }\,,\end{aligned}$$

(70)

$$\begin{aligned} \mathrm{d}V= \,& {} \mathrm{d}V_{0}/\gamma . \end{aligned}$$

(71)

From that we get

$$\begin{aligned} \mathcal{E}=\int \frac{E^{2}_{0\Vert }+(\gamma ^{2}+\gamma ^{2}v^2/c^{2}) E^{2}_{0\bot }}{8\pi \gamma }\,\mathrm{d}V_0. \end{aligned}$$

(72)

Now because of the circular-cylindrical symmetry of the system about the x-axis, the direction of motion, we have

$$\begin{aligned} \int E^{2}_{0\bot }dV_0=\frac{2}{3}\int E_0^{2}\,\mathrm{d}V_0. \end{aligned}$$

(73)

Then, after integration, we get for the field energy in $\mathcal{S}$

$$\begin{aligned} \mathcal{E}=\gamma \mathcal{E}_{0}\left( 1+\frac{v^{2}}{3c^2}\right) . \end{aligned}$$

(74)

In the same way, we have an expression for the field momentum

$$\begin{aligned} \mathbf{P}=\int \frac{\gamma ^{2}{} \mathbf{v} E^{2}_{0\bot }}{4\pi c^2 \gamma }\,\mathrm{d}V_0, \end{aligned}$$

(75)

which, after integration, yields

$$\begin{aligned} \mathbf{P}=\frac{4}{3}\,\mathcal{E}_{0} \frac{\gamma \mathbf{v}}{c^{2}}. \end{aligned}$$

(76)