How can the following relation be justified, $$ P = \frac{p^2}{3E} $$ where $P$ is pressure, $p$ is momentum and $E$ is energy? In what convention is the above relation acceptable?
The source paper can be found here. The relation is not explicitly mentioned there but in the equations it is evident. Please refer to the 4$^\text{th}$ page for equations of state. Note that that speed of light $c$ and Planck's constant $h/2\pi$ are both taken as unity (one).
The equation $P=p^2/(3E)$ makes no sense (it connects a thermodynamic variables, $P$, to the momentum and energy of a single particle). It is also not dimensionally correct. With $c=1$ momentum $p$ and energy $E$ have the same units, but pressure has units of energy/volume (=energy*momentum$^3$).
The equations in the paper look o.k. Pressure is $$ P = \int d^3p\, p \frac{p}{E} \, f(p) $$ where $f$ is a dimensionless distribution function. The integral $d^3p$ gives 1/volume, $E=\sqrt{p^2+m^2}$ is an energy, $v=p/E$ is a velocity. This means we are integrating over $vp$, which is indeed the standard definition of pressure in kinetic theory.
For energy density $$ {\cal E} = \int d^3p\, E \, f(p) $$ which clearly makes sense.
Postscript: Pressure is defined as (in kinetic theory) $$ \Pi_{ij} = \int d^3p\, p_i \frac{p_j}{E}\, f(p) $$ with $\Pi_{ij}=\frac{1}{3}\delta_{ij}P$. Then $P= \frac{1}{3}\int d^3p\, p^2/E\, f(p)$ and now we have a rotationally invariant integrand $d^3p=4\pi p^2dp$.
Using dimensional analysis, and the relations $p=mv$ and $E=mgh$, we can write: $$[P]=\dfrac{kg^2\cdot m^2\cdot s^{-2}}{kg\cdot m^2\cdot s^{-2}}=kg$$
And kilogrammes are obviously not a unit of pressure, therefore your relation doesn't have the correct dimensions.
