On Construction of Energy-Momentum Tensors

We may think of continuum version of energy and momentum as we learned particle geodesic. We may define the flow of energy and momentum which is called energy-momentum tensor, and talks about its conservation. Also, just like defining the energy and momentum of electromagnetic field in classical electrodynamics, we may define the energy-momentum tensor of force fields, in way of conserving the whole energy-momentum tensor. Then, we may obtain the tensor for some specific cases: one particle, perfect fluid, spin-1 field, and thermodynamical gas of classical particles and photons; and confirm the consistency between statistical mechanics and relativistic fluid theory, and between gravitational theory and electromagnetic field theory.

Introduction

In classical mechanics, we found some conservative quantities which is called momentum. You may have defined the momentum with some conservative quantities by evolution of time in Newtonian mechanics, or you may have directly derived from the Lagrangian itself in canonical way.

     We will talk about the conservative quantities which corresponds to energy and momentum in continuum and its conservation law. Then, interaction may break the conservation and revise the conservation to the equation of motion. We may define the energy-momentum tensor of the interaction so that still the energy-momenum is conserved, as we defined Maxwell stress tensor in classical electrodynamics.

Conservative Quantities

You may understand well the conservative law of charged particles (fluid): $$\begin{array}{}\text{(1)}& \frac{\partial \rho }{\partial t}+\mathrm{\nabla }\cdot (\rho \overrightarrow{v})=0.\end{array}$$ Here, we call $\int \rho d^{3}x$ the (conservative) charge and $\rho \overrightarrow{v}$ the current. Under the Lorentz boost, we can find that the density $\rho$ is not a proper scalar in the 4-D, but is magnified $\gamma =(1-\beta {)}^{-1/2}$ times by the length contraction. So, using the proper density ${\rho }_0=\rho /\gamma$, the density in the local frame of fluid, the equation $(1)$ may be expressed as $$\frac{\partial \gamma {\rho }_0}{\partial t}+\mathrm{\nabla }\cdot (\gamma {\rho }_0\overrightarrow{v})=0,$$ which also can be expressed with the 4-velocity $u^{\alpha }$  $$\begin{array}{}\text{(2)}& {\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha })=0.\end{array}$$Thus, we can understand that $j^{\alpha }={\rho }_0{u}^{\alpha }$ should be called the proper 4-current in the same manner above, and the conservative charge is $q=\int \rho d^{3}x=\int j^0{d}^{3}x$ for a flat time slice, in the special relativity. If we expand this naturally to the general relativity, the charge within an arbitrary time slice segment $R$ may be defined as ${\int }_R\rho (R)d^{3}x={\int }_R{j}^{\alpha }d{S}_{\alpha }$.

     This proposal works well for non-mass charges such as electric charge as above, but mass-related quantities defined using above make some catastrophe. Analogy to above, we might define mass current ${\rho }_0^{(\mathrm{mass})}{u}^{\alpha }$, but this quantity is not an actual physical quantity. In a physical frame, only energy and momentum is an observable. Thus, we may define a current of energy and momentum density, the zeroth component of which may be the density itself.

     First, the zeroth component, the energy current may be considered. Consequent discussion may progress in the flat spacetime. The energy conservation may be written as: $$\begin{array}{}\text{(3)}& \frac{\partial \rho }{\partial t}+{\partial }_k(\rho v^k)=0\end{array}$$where $\rho$ is the energy density. Distinctly from the electric example, the energy density may be transformed into ${\gamma }^2$ magnified, one $\gamma$ of which is from the length constraction and another one of which is from the 4-momentum transformation (mass increase), under the Lorentz transformation ($\rho ={\rho }_0{\gamma }^2$). Thus, using the rest mass density ${\rho }_0$, the equation may be expressed to $${\partial }_{\alpha }({\rho }_0\gamma u^{\alpha })=0.$$As $\gamma =u^0$ and generally the partial derivative is extended to the covariant derivative in the general relativity, the covariant form may be naturally $${\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^0{u}^{\alpha })=0.$$Then the zeroth component ${\rho }_0{u}^0{u}^0$ is the energy density as we proposed.

     Second, we may remind the Navier-Stokes equation to write the equation of momentum conservation i.e. equation of motion in fluid^[1,2]: $$\begin{array}{}\text{(4)}& \rho \frac{\partial v^k}{\partial t}+\rho v^j{\partial }_j{v}^k=f^k={\partial }_j{\sigma }^{jk}\end{array}$$where $\overrightarrow{f}$ is the force density and the density is expressed by stress tensor $\sigma$. The stress tensor expression can generally give non-isotropic pressure and sheer forces. The first term may be changed to$$\rho \frac{\partial v^k}{\partial t}=\frac{\partial }{\partial t}(\rho v^k)-v^k\frac{\partial \rho }{\partial t}=\frac{\partial }{\partial t}(\rho v^k)+v^k{\partial }_j(\rho v^j)$$using the equation $(3)$ above. Then the Navier-Stokes equation  $(4)$  may be$$f^k=\frac{\partial }{\partial t}(\rho v^k)+\rho v^j{\partial }_j{v}^k+v^k{\partial }_j(\rho v^j)=\frac{\partial }{\partial t}(\rho v^k)+{\partial }_j(\rho v^j{v}^k).$$Using $\rho ={\rho }_0(u^0{)}^2$,$$f^k=\frac{\partial }{\partial t}({\rho }_0{u}^0{u}^k)+{\partial }_j({\rho }_0{u}^j{u}^k)={\partial }_{\alpha }({\rho }_0{u}^{\alpha }{u}^k).$$

     Thus, if we write the energy and momentum equations in covariant form, with the natural extension of stress tensor into 4-D ${\mathrm{\nabla }}_{\alpha }{\sigma }^{\alpha \beta }=f^{\beta }$, $$\begin{array}{}\text{(5)}& {\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha }{u}^{\beta })=f^{\beta }=-{\mathrm{\nabla }}_{\alpha }{\sigma }^{\alpha \beta }.\end{array}$$The minus sign is come from $(+,-,-,-)$ sign of the metric in the space-time while we used $(+++)$ in the 3-D. In the classical limit, only the spatial ($3\times 3$) part of ${\sigma }^{\alpha \beta }$ may be the same to the classical stress tensor and the other components may be zero as the energy and momentum are conserved.

     Hence, the current of energy and momentum, called energy-momentum tensor, may be defined $$\begin{array}{}\text{(6)}& T^{\alpha \beta }={\rho }_0{u}^{\alpha }{u}^{\beta }\end{array}$$if there is no interaction i.e. $f^{\beta }=0$. The tensor may satisfy the conservation law ${\mathrm{\nabla }}_{\alpha }{T}^{\alpha \beta }=0$ as shown above.

     Using the obtained tensor above, we can re-confirm that the fluid follows the geodesic equation in covariant way.^[3] Written the conservation law explicitly, $$\begin{array}{}\text{(7)}& {\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha }{u}^{\beta })=u^{\beta }{\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha })+{\rho }_0{u}^{\alpha }{\mathrm{\nabla }}_{\alpha }{u}^{\beta }=0.\end{array}$$Contracted with $u_{\beta }$, $${\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha })+{\rho }_0{u}^{\alpha }{u}_{\beta }{\mathrm{\nabla }}_{\alpha }{u}^{\beta }=0.$$The normalisation $u_{\beta }{u}^{\beta }=1$ gives $u_{\beta }{\mathrm{\nabla }}_{\alpha }{u}^{\beta }=0$ so the equation is reduced to $${\mathrm{\nabla }}_{\alpha }({\rho }_0{u}^{\alpha })=0$$Applying this result to the original equation $(7)$ again, we get $$\begin{array}{}\text{(8)}& {\rho }_0{u}^{\alpha }{\mathrm{\nabla }}_{\alpha }{u}^{\beta }=0\end{array}$$which is the geodesic equation. In conclusion, we can show that the world lines of the free fluid follows the geodesic.

Single Particle

As an example, we may obtain the energy-momentum tensor of a particle to confirm the discrete-continuum correspondence; we will put the delta function to $\rho$ and confirm if $(0,\alpha )$-component corresponds to the momentum. In the local rest frame of a particle, the particle should be at rest, so the tensor may be $$T_{(\mathrm{rest})}^{\alpha \beta }=m{\delta }^3(\overrightarrow{x})u^{\alpha }{u}^{\beta }=m\delta (x)\delta (y)\delta (z)u^{\alpha }{u}^{\beta }$$where $u^{\alpha }={\delta }_0^{\alpha }$ here, of course. Then, we may apply Lorentz transformation to express moving particle. $$T^{\alpha \beta }=\delta (\gamma x-\gamma \beta t)\delta (y)\delta (z)u^{\alpha }{u}^{\beta }$$where $u^{\alpha }=(\gamma ,\gamma \beta ,0,0)$ here. Now, we will obtain the momentum by integrating the tensor by the local time slice.$$\begin{array}{rl}{p}^{\beta }& ={\int }_{t=0}{T}^{0\beta }dxdydz\\ & =\int m\delta (\gamma x)\delta (y)\delta (z)\gamma u^{\beta }dxdydz=m{u}^{\beta }\end{array}$$So, here we confirmed the correspondence.

Maxwell-Boltzmann Dust

If we integrate such single particles which follow the Maxwell-Boltzmann distribution, we may define the energy-momentum tensor of non-interacting particles which follow the Maxwell-Boltzmann distribution. Among the particles, we can make a pair of two comoving groups of particles the velocities of which are opposite each other. Let the velocity be $u^{\alpha }=(\gamma ,\gamma \beta ,0,0)$ without loss of generality. Then, the energy-momentum tensor may be $$\begin{array}{rl}{T}^{\alpha \beta }& =\frac{\rho }{2}\left[\begin{array}{cccc}1& \beta & & \\ \beta & {\beta }^2& & \\ & & 0& \\ & & & 0\end{array}\right]+\frac{\rho }{2}\left[\begin{array}{cccc}1& -\beta & & \\ -\beta & {\beta }^2& & \\ & & 0& \\ & & & 0\end{array}\right]\\ & =\rho \left[\begin{array}{cccc}1& & & \\ & {\beta }^2& & \\ & & 0& \\ & & & 0\end{array}\right]\end{array}$$where $\rho ={\gamma }^2{\rho }_0$ is the total energy density of the two groups. Here, the trace of spatial part is $\rho {\beta }^2$. This will never be changed if the direction of the velocity changes.

     Also, we can find easily that the off-diagonal term will not survive; if you want to eliminate the $(i,j)$-th term, which is proportional to $v^i{v}^j$, you can always find the group reflected in $i$-direction from the original group to cancel the $(i,j)$-th term when the two are summed. Thus, if we sum them all in every direction equally, the energy-momentum tensor of an isotropic gas may be: $$\begin{array}{}\text{(9)}& \begin{array}{rl}{T}^{\alpha \beta }& =\left[\begin{array}{cccc}\rho & & & \\ & ⟨\rho v_x^2⟩& & \\ & & ⟨\rho v_y^2⟩& \\ & & & ⟨\rho v_z^2⟩\end{array}\right]\\ & \approx \rho \left[\begin{array}{cccc}1& & & \\ & ⟨v_x^2⟩& & \\ & & ⟨v_y^2⟩& \\ & & & ⟨v_z^2⟩\end{array}\right]\end{array}\end{array}$$where $\rho$ is the total energy density of the whole group and, of course, $⟨v_x^2⟩=⟨v_y^2⟩=⟨v_z^2⟩=\frac{1}{3}⟨v^2⟩$.

     According to the Maxwell-Boltmann distribution, which is the random group by thermodynamics in classical limit, $⟨v^2⟩=\frac{3kT}{m}$ where $m$ is the mass of a particle.^[4] Applied this, $$\begin{array}{}\text{(10)}& T^{\alpha \beta }=\left[\begin{array}{cccc}\rho & & & \\ & nkT& & \\ & & nkT& \\ & & & nkT\end{array}\right]\end{array}$$where $n$ is the number density of the gas and $k$ is the Boltzmann constant,
as $\rho =mn+\mathrm{\Theta }(v^2)$. Then, actually the extra term may be on the spatial diagonal as $nkT+\mathrm{\Theta }(v^2)nkT$, but as $nkT$ is also in dimension of classical kinetic energy $\mathrm{\Theta }(v^4)$, the additional term may be $\mathrm{\Theta }(v^4)$ which is ignored in classical limit. (which is reason for the approximation in $(9)$)

     In addition, the $v^2$ term of $(0,0)$-th component cannot be ignored as we consider until $v^2$. As we know the kinetic energy density is $\frac{3}{2}nkT$, the total energy density may be $\rho =mn+\frac{3}{2}nkT$.

     We conclude that the dynamical equivalence makes an different result from the absolute rest case, and this random motion gives the diagonal terms which later corresponds to pressure; surprisingly, you may notice that the diagonal term $nkT$ is the pressure predicted by thermodynamics in ideal gas.

With Interaction

With interactions, we may define the energy-momentum tensor as $$\begin{array}{}\text{(11)}& T^{\alpha \beta }={\rho }_0{u}^{\alpha }{u}^{\beta }+{\sigma }^{\alpha \beta }\end{array}$$so still the conservation ${\mathrm{\nabla }}_{\alpha }{T}^{\alpha \beta }=0$ is still satisfied. We expanded the stress tensor to 4-D maintaining the 3-D part the same in the classical limit, but, Nevertheless, the other components, $(0,\alpha )$ and $(\alpha ,0)$ component, is not known.

     We will obtain the energy-momentum tensor of perfect fluid as an ideal example, where no sheer force can exist and only isotropic pressure exists. We know that in classical ${\mathbb{E}}^3$, the stress tensor is ${\sigma }^{\alpha \beta }=p{\delta }^{\alpha \beta }$ where $p$ is the pressure. Thus, if the fluid is nearly at rest, the energy-momentum tensor may be$$\begin{array}{}\text{(12)}& T^{\alpha \beta }=\left[\begin{array}{cccc}{\rho }_0& & & \\ & p& & \\ & & p& \\ & & & p\end{array}\right].\end{array}$$By definition of tensor, covariant expression of a tensor value of which in one frame is only given is unique. $$\begin{array}{}\text{(13)}& T^{\alpha \beta }=({\rho }_0+p)u^{\alpha }{u}^{\beta }-p{g}^{\alpha \beta }\end{array}$$where the metric is in $(+---)$ convention. You can easily confirm that this will give $(12)$ at rest: $u^{\alpha }=(1,0,0,0)$.

Electromagnetic Wave

In the classical electromagnetism, the energy density, the momentum density, and the Maxwell stress tensor of electromagnetic field is suggested.^[5] $$\begin{gathered} u=\frac{1}{2}({ϵ}_0{E}^2+\frac{1}{{\mu }_0}{B}^2) \\ \overrightarrow{S}=\frac{1}{{\mu }_0}(\overrightarrow{E}\times \overrightarrow{B}) \\ {T}_{ij}={ϵ}_0(E_i{E}_j-\frac{1}{2}{E}^2{\delta }_{ij})+\frac{1}{{\mu }_0}(B_i{B}_j-\frac{1}{2}{B}^2{\delta }_{ij}) \end{gathered}$$We know that the stress tensor is $(i,j)$-th component and the energy and momentum is $(0,\alpha ),(\alpha ,0)$-th component of the energy-momentum tensor, we can combine to make the energy-momentum tensor. Suggested below may give those if you apply each indices: $$\begin{array}{}\text{(14)}& T^{\mu \sigma }=-\frac{1}{{\mu }_0}(F^{\mu \nu }{F^{\sigma }}_{\nu }-\frac{1}{4}{F}^2{g}_{\mu \sigma })\end{array}$$where $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$ and $F^2=F_{\mu \nu }{F}^{\mu \nu }$.

     Furthermore, the trace of the energy-momentum tensor should be $$T=-\frac{1}{{\mu }_0}(F^2-F^2)=0$$for any field given. This result, in addition, may simplify the Einstein's equation of pure gravity-electromagnetism system: $$R_{\mu \nu }=T_{\mu \nu },$$ as the equation is alternatively expressed as $R_{\mu \nu }=T_{\mu \nu }-\frac{1}{2}T{g}_{\mu \nu }$.

Planck Gas

Electromagnetic wave also has no shear, so in the same sense in the Maxwell-Boltzmann section, only diagonal terms of the energy-momentum tensor may remain at the centre of mass frame if we sum up about the photon gas. Then, the $(0,0)$-th component may be the energy density and the others are pressure. Statistical mechanics says that the energy density is $\rho =\frac{{\pi }^2}{15}{T}^4$ and the pressure is $p=\frac{1}{3}\rho$ where $T$ is the temperature if we use the natural units: $c=\hslash =k=1$.^[6]

     The pressure also can be obtained by the relation $T=0$ if we know that the energy-momentum tensor only has diagonal terms; relativity and electromagnetism gives the same result with the statistical mechanics. $$\begin{array}{}\text{(15)}& T_{(\mathrm{rest})}^{\alpha \beta }=\frac{{\pi }^2}{15}{T}^4\left[\begin{array}{cccc}1& & & \\ & 1/3& & \\ & & 1/3& \\ & & & 1/3\end{array}\right]\end{array}$$

Field Theory Side

The Noether theorem in field theory gives that if an action$$S(\varphi ;x^{\mu })=\int \mathfrak{L}{d}^{4}x$$has the translational symmetry, the canonical energy-momentum tensor: $$t_{\sigma }^{\mu }=\frac{\partial \mathfrak{L}}{\partial {\varphi }_{,\mu }}{\varphi }_{,\sigma }-\mathfrak{L}{\delta }_{\sigma }^{\mu }$$should satisfies the conservation ${\mathrm{\nabla }}_{\mu }{t}_{\sigma }^{\mu }=0$ when the action is extremised.^[7] It is natural to be called energy-momentum tensor, as $\mu$ is the index of conservative current, and $\sigma$ is the index of corresponding translational symmetry. Also, this canonical tensor may give the real energy-momentum tensor which meets in the classical limit if we symmetrise it with right choice of gauge. Let us give an example of electromagnetism.

     If we give the action of the electromagnetism which gives the Maxwell equations and which is invariant by coordinate transform and gauge transform: $$S(A_{\mu })=\int F^{\mu \nu }{F}_{\mu \nu }{d}^{4}x,$$the canonical energy-momentum tensor is$$t_{\sigma }^{\mu }=4F^{\mu \nu }{A}_{\nu ,\sigma }-F^2{\delta }_{\sigma }^{\mu }$$which satisfies the conservation law ${\mathrm{\nabla }}_{\mu }{t}_{\sigma }^{\mu }=0$.

     However, this tensor is not symmetric. If we expand out and carefully exchange terms with identities and symmetries which still satisfy the conservation law, we may get the symmetric energy-momentum tensor.$$T^{\sigma \mu }=4F^{\mu \nu }{F_{\nu }}^{\sigma }-F^2{g}^{\sigma \mu }$$Remind the energy-momentum tensor of electromagnetic field. This is the exactly same if you multiply some coefficient.

G-EM System

Now, consider the gravity. We know the Hilbert action $\int R \sqrt{-g}d^4x$. We will see if the same action above is given to the gravity theory:$$\int (R+\frac{4\pi G}{{\mu }_0}{F}^2)\sqrt{-g}{d}^{4}x.$$First, the Einstein-Hilbert action and its variation is well known, and you might know the details.$${\delta }_g\int R\sqrt{-g}{d}^{4}x=\int G_{\mu \nu }\sqrt{-g}\delta g^{\mu \nu }{d}^{4}x$$Second, if we take the variation on the electromagnetic terms, surprisingly it may gives the energy-momentum tensor of the field (you might vary the coefficient).$$\begin{array}{rl}& {\delta }_g\int F^2\sqrt{-g}{d}^{4}x\\ & =\int (2F_{\mu \sigma }{F_{\nu }}^{\sigma }-\frac{1}{2}{F}^2{g}_{\mu \nu })\sqrt{-g}\delta g^{\mu \nu }{d}^{4}x\end{array}$$Thus, in total, the equation of motion gives the energy-momentum tensor of the electromagnetic field as a source of the gravity; just right fit into the Einstein's gravitational theory.$$\begin{array}{rl}& {\delta }_g\int (R+\frac{4\pi G}{{\mu }_0}{F}^2)\sqrt{-g}{d}^{4}x\\ & =\int (G_{\mu \nu }-8\pi G{T}_{\mu \nu }^{(\mathrm{em})})\sqrt{-g}\delta g^{\mu \nu }{d}^{4}x.\end{array}$$
     In conclusion, we might see that the electromagnetic action which gives the maxwell equations and right energy-momentum tensor also gives the gravitational equations with the same energy-momentum tensor as the source if the action is added to the gravitational theory. In other word, the total action gives the gravitational theory if it is variated by gravitational field while the same action gives electromagnetic theory when it is variated by electromagnetic potential.

Conclusion

We have defined energy-momentum tensor of non-interacting fluid and perfect fluid. Then, we could define the energy-momentum tensor of interaction itself, in way of conserving the total energy-momentum, as we defined stress tensor of electromagnetic field in classical electrodynamics. Then, surprisingly, regardless of its origin, the meaning of each component of the tensor corresponded to the meaning in the original non-interacting fluid case. We have confirmed that perfect fluid of random gas of classical particles and photons gives its pressure naturally into the energy-momentum tensor as statistical mechanics predicted.

     Furthermore, we have seen that the energy-momentum tensor can be derived canonically from the Lagrangian which is the same defined above for some field. In case of electromagnetism, we also have found that variated by metric the same action gives the Einstein's field equation i.e. the energy-momentum tensor. In conclusion, the G-EM system action may gives the equation of motion of G-EM and the conservative current from one action by varying the variation variables.

References

[1] Wikipedia: Navier-Stokes equations
[2] P. Tourrenc, Relativity and Gravitation (Cambridge Univ. Press, UK, 1997), pp. 72.
[3] ion.uwinnipeg.ca/~vincent/4500.6-001/Cosmology/EnergyMomentum\Tensors.htm
[4] Wikipedia: Maxwell-Boltzmann distribution
[5] D. Griffiths, Introduction to Electrodynamics (Prentice-Hall, NJ, 1999), pp. 347-352.
[6] Wikipedia: Photon gas
[7] L. Ryder, Quantum Field Theory (2nd ed., Cambridge Univ. Press, UK, 1996), pp. 83-90.

Appendix

During the presentation, there was an interesting question about the features of the (relativistically) exact Boltzmann gas so here the calculation goes below.

     First, we may define the partition function (considered only in momentum space as there is no interaction) as $$Z=\int d^{3}p{e}^{-\beta ϵ}=4\pi \int dp{p}^2{e}^{-\beta ϵ}$$for a single particle. As every function we consider depends on only the magnitude of the momentum, the angular part was integrated up. We know the relation between energy and momentum: $p^2+m^2={ϵ}^2$, so using it, the integrating variable can be changed. $$\frac{1}{4\pi }Z={\int }_m^{\mathrm{\infty }}{e}^{-\beta ϵ}ϵ\sqrt{{ϵ}^2-m^2}dϵ=\frac{m^2}{\beta }{K}_2(\beta m)$$Then, other quantities we need are obtained as below: $$\begin{gathered} \begin{array}{rl}\frac{1}{4\pi }⟨ϵ⟩Z& ={\int }_m^{\mathrm{\infty }}{e}^{-\beta ϵ}{ϵ}^2\sqrt{{ϵ}^2-m^2}dϵ\\ & =\frac{m^2}{{\beta }^2}(\beta m{K}_1(\beta m)+3K_2(\beta m))\end{array} \\ \frac{1}{4\pi }⟨{ϵ}^{-1}⟩Z={\int }_m^{\mathrm{\infty }}{e}^{-\beta ϵ}\sqrt{{ϵ}^2-m^2}dϵ=\frac{m}{\beta }{K}_1(\beta m) \end{gathered}$$Divided by the partition function, the expectation values are: $$\begin{gathered} ⟨ϵ⟩=\frac{3}{\beta }+m\frac{K_1(\beta m)}{K_2(\beta m)} \\ ⟨{ϵ}^{-1}⟩=\frac{K_1(\beta m)}{m{K}_2(\beta m)} \end{gathered}$$Then, the energy density may be easily obtained as: $$⟨\rho ⟩=n⟨ϵ⟩=mn\frac{K_1(\beta m)}{K_2(\beta m)}+3nkT$$with the number density $n=\frac{N}{V}$.

     Now, you may remember the pressure $p=\frac{1}{3}⟨\rho v^2⟩=\frac{1}{3}n⟨ϵv^2⟩$ in $(9)$ (From here, $p$ means the pressure) and remind the relation between the relativistic factor $\gamma$ and the velocity $v$ of a single particle: $v^2={\gamma }^2-1$. Then the pressure is $\frac{1}{3}n⟨ϵv^2⟩=\frac{1}{3}mn⟨\gamma v^2⟩=\frac{1}{3}mn⟨\gamma -{\gamma }^{-1}⟩$. $$p=\frac{mn}{3}⟨\gamma -{\gamma }^{-1}⟩=\frac{n}{3}(⟨ϵ⟩+m^2⟨{ϵ}^{-1}⟩)=nkT$$This re-confirms the equation of states again. Also, here we can find $T_{\mu }^{\mu }=mn⟨{\gamma }^{-1}⟩$, which can be estimated by thinking with the single particle models.

     In case of considering Bose or Fermi statistics, it is hard to calculate analytically. As we know that the difference between two appears in low temperature which matches to the non-relativistic and not-so-dense limit, the calculation above may fit to most case above room temperature and non-relativistic calculations may fit to the area where the two statistics make difference; besides, the fermi statistics with relativistic Fermi energy may be thought to make some other interesting feature.