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For $N$ particles of a classical monatomic ideal gas, the equation of state and internal energy areOnly translational degrees of freedom contribute; the particle number is fixed throughout the thermodynamic cycle.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

Work in the shock frame, assume a calorically perfect gas with fixed specific-heat ratio $\gamma >1$, and neglect magnetic stresses and body forces across the thin shock. The internal energy per unit mass is $e=p/[(\gamma -1)\rho ]$, and the specific enthalpy is $w=\gamma p/[(\gamma -1)\rho ]$. Integrating the local conservation laws across a thin stationary control volume makes the mass flux $\rho u$, momentum flux $p+\rho u^2$, and energy flux $u(\rho u^2/2+\gamma p/(\gamma -1))$ continuous. Hence the Rankine-Hugoniot conditions for a perfect gas areMultiplication of the last equality by $m$ gives continuity of the stated energy flux. These express that mass, momentum and energy cannot accumulate in an infinitesimally thin steady shock, although entropy production occurs on the physical compression branch. The same constant in the polytropic equation of state $p=K{\rho }^{\gamma }$ must therefore not be imposed on both sides. For flow from $z>0$ to $z<0$, both signed velocities are negative; the flux equations and ratios remain valid.

In the Strong-shock Rankine-Hugoniot conditions, $p_1/({\rho }_1{u}_1^2)=1/(\gamma M_1^2)$ tends to zero. Put $r={\rho }_2/{\rho }_1$. The first two flux balances give $u_2=u_1/r$ and $p_2={\rho }_1{u}_1^2(1-1/r)$ to leading order. The energy balance becomesDiscarding the unchanged-state root $r=1$ gives the compressive shock compression ratio and the required limits:

The conservation of energy to use here is the total kinetic energy plus internal energy inside the shock, not just the internal energy of a uniform-pressure core. With no radiative loss and negligible initial ambient energy,Substitute the adiabatic superbubble similarity solution and define the finite positive dimensionless integralThen $\dot{E}t=4\pi {\rho }_1{\dot{R}}^2{R}^{3}I$. Since $\dot{R}=3R/(5t)$ and $R^5={\alpha }^5(\dot{E}/{\rho }_1)t^3$,This is the energy normalization of a continuously driven spherical shock. It fixes $\alpha$ once the similarity profiles are determined. If the profiles describe only the shocked ambient layer, the energy of the interior hot bubble must also be included before using this normalization. A radiatively cooled shell has a different energy budget; its familiar numerical coefficient should not be substituted into this loss-free model.

Consider a spherical stellar polytrope with $P=K{\rho }^{1+1/n}$, constant $K$, $0<n<5$, zero surface pressure and finite radius $R$. Its specific enthalpy, measured from zero density, is $h={\int }_0^{P}d{P}^{\prime }/\rho =(n+1)P/\rho$. Hydrostatic equilibrium gives $h+\Phi =\Phi (R)=-GM/R$. Integrating over mass therefore yieldsThe static stellar virial theorem gives $3\int PdV=-\Omega$, and eliminating the pressure integral proves the gravitational energy of a stellar polytrope:For an ideal gas with constant specific-heat ratio, the accompanying internal energy isTo express both energies using $\gamma$ alone one additionally identifies the model as an adiabatic stellar polytrope, so $n=1/(\gamma -1)$. ThenFor $\gamma =5/3$ these become $\Omega =-6G{M}^2/(7R)$ and $U=3G{M}^2/(7R)$. A general polytropic index need not be determined by the thermodynamic specific-heat ratio; without the adiabatic identification both $n$ and $\gamma$ remain.

For an ideal gas with constant specific-heat ratio $\gamma$, the specific internal energy is $u=c_{V}T$ and $P=\rho (c_P-c_V)T=(\gamma -1)\rho u$. HenceThe stellar virial theorem consequently reads $\ddot{I}/2=2\mathcal{T}+3(\gamma -1)U-3P_{S}V+\Omega$. For a static nonrotating star with negligible surface pressure, $\mathcal{T}=\ddot{I}=P_S=0$, so $\Omega =-3(\gamma -1)U$.

Fully ionized nonrelativistic ions and electrons form a monatomic ideal gas, with $\gamma =5/3$, provided electron degeneracy pressure and radiation pressure are negligible. ThusBeing steady alone does not eliminate rotation or other macroscopic kinetic energy; those are additional assumptions needed for this relation.

Work in an inertial frame centred on the centre of mass, and follow a fixed material mass, with no stellar wind or accretion through its boundary. Assume nonrelativistic motion, Newtonian self-gravity, no external gravitational field, and sufficiently regular integrable fields. DefineHere $I$ is the scalar second mass moment, rather than a moment of inertia about one axis. The kinetic energy $\mathcal{T}$ measures resolved internal motions relative to the centre of mass; random microscopic motion contributes to pressure and internal energy, and must not be counted again in $\mathcal{T}$.

The paper's symmetric compressive tensor $\mathbb{P}$ is minus the tensile stress tensor, so its force per unit volume is $-{\partial }_j{P}_{ij}$. Differentiating the material integral twice and substituting the equation of motion givesIntegration by parts and the divergence theorem turn the stress contribution intoFor the requested scalar stellar virial theorem, assume isotropic pressure, $P_{ij}=P{\delta }_{ij}$, with uniform surface pressure $P_S$. Then $\mathrm{tr}\mathbb{P}=3P$ and the surface integral is $P_S\oint \mathbf{r}\cdot \mathbf{n}dS=3P_{S}V$. No spherical shape is needed for this step.

Finally, symmetrizing the pair integral for the Newtonian gravitational potential givesThereforeAnisotropic magnetic or viscous stresses retain their full trace and boundary terms; a changing integration mass adds transport terms.

The chemical potential is the change in internal energy when one particle is added at fixed entropy and volume:Equivalently, $\mu =(\partial F/\partial N{)}_{T,V}$ for the Helmholtz free energy.

Let a microstate $s$ have energy $E_s$ and particle number $N_s$. Maximizing the Gibbs entropy subject to normalization and fixed mean values of $E$ and $N$ is the maximum-entropy derivation of equilibrium ensembles. With Lagrange multipliers $\alpha$, $\beta$, and $-\beta \mu$, varyThe stationarity equation is $-\log p_s-1-\alpha -\beta (E_s-\mu N_s)=0$. Normalization therefore gives the grand canonical ensembleHere $\mathcal{Z}$ is the grand canonical partition function and $k_B$ is the Boltzmann constant.

For one fermionic quantum state of energy $\epsilon$, the Pauli exclusion principle permits occupation numbers only $n=0,1$. Its two grand-canonical weights are $1$ and $e^{-\beta (\epsilon -\mu )}$, so its mean occupation is the Fermi-Dirac distribution

For a free nonrelativistic particle, $\epsilon ={\hslash }^2{k}^2/(2m)$. In a region of area $A$, the number of wave-vector states in the annulus from $k$ to $k+dk$, including the two spin angular momentum states, isSince $kdk=(m/{\hslash }^2)d\epsilon$, the two-dimensional free-electron density of states is constant:

We now use units in which $k_B=1$, as in the question. At zero temperature, the Fermi-Dirac distribution is a step function, and henceAt positive temperature the same fixed particle number satisfies the exact relationThusThis is the low-temperature particle-number cancellation for constant density of states. Therefore, for $T\ll {\epsilon }_F$, the mean number of particles in the energy interval $[\epsilon ,\epsilon +d\epsilon ]$ is

Finally, compare the internal energy with its zero-temperature value. The thermally excited particles above ${\epsilon }_F$ and holes below ${\epsilon }_F$ have equal leading particle numbers, so their terms proportional to ${\epsilon }_F$ cancel. With $x=|\epsilon -{\epsilon }_F|/T$,where the integral follows from the fermion-to-boson thermal integral ratio. Differentiation gives the linear low-temperature heat capacity of a two-dimensional Fermi gas:The requested power law is therefore linear in $T$, with exponent one.

The zero-temperature internal energy isUsing part (i),Consequently the adiabatic equation of state isThe pressure is degeneracy pressure: it remains nonzero at zero temperature because of the Pauli exclusion principle. By comparison, a classical ultrarelativistic ideal gas at fixed temperature obeys $pV=N{k}_{B}T$, so its corresponding volume exponent is one and its pressure vanishes as $T\to 0$.

A phonon has zero chemical potential and energy $\hslash \omega$. The total internal energy is consequentlyWith the change of variable $x=\beta \hslash \omega$ and $\beta \hslash {\omega }_D=T_D/T$, this becomes

The one-dimensional mass flux and momentum flux are respectivelyIn a steady flow, the conservation laws make both the mass flux and the given energy fluxconstant. Wherever $J_m\ne 0$, their ratio is therefore constant:where $W=p/(\gamma -1)$ is the internal energy per unit volume of the ideal gas. This is the required steady-flow energy integral.

A body loses internal energy by emitting thermal radiation faster than it absorbs radiation or receives other heating. In a gravitating gas this can remove pressure support and allow further contraction.

For collinear motion without external forces, a rocket ejecting material backwards at constant speed $u$ in its instantaneous rest frame obeys $m{\gamma }^{2}dv+udm=0$. Here $m$ is the rocket's rest mass, including stored internal energy, and $\gamma$ is its Lorentz factor in the chosen inertial frame. The equation follows from momentum conservation, conservation of relativistic energy, and the relativistic velocity-addition formula; expelled rest mass is not simply the decrease in rocket rest mass.

Specific internal energy is internal energy per unit mass. For a perfect gas with constant specific-heat ratio, $e=p/[(\gamma -1)\rho ]$. Its material balance is $\rho De/Dt=-p\nabla \cdot \mathbf{u}+Q$, where $Q$ is heat supplied per unit volume per unit time.