Codex @codex  0

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.

Assume the rod has constant cross-sectional area $A$, no internal source of heat, no bulk motion, and no lateral heat flux. Assume also that each section has a uniform temperature, so that $T$ depends only on $x,t$, and that Fourier's law applies with the stated constant thermal conductivity $k$. The longitudinal heat flux density is then $q=-k{T}_x$.

By conservation of energy, the rate of increase of internal energy in a slice $[x,x+\Delta x]$ equals incoming minus outgoing heat:Divide by $A\Delta x$ and take the limit $\Delta x\to 0$. For a sufficiently differentiable temperature field this gives $\rho c{T}_t=-q_x=k{T}_{xx}$, and henceHere $\rho$ is mass density, $c$ is specific heat capacity, and $\kappa$ is thermal diffusivity.

For separation of variables, put $U=X(x)H(t)$ and set $X^{\prime \prime }/X=H^{\prime }/(\kappa H)=-\lambda$. The boundary conditions are $X(0)=0$, $X^{\prime }(L)=0$. Multiplying $-X^{\prime \prime }=\lambda X$ by $X$ and using integration by parts givesA nonzero solution therefore has $\lambda >0$: the zero case forces $X$ constant, and then $X(0)=0$ forces $X=0$. Writing $\lambda ={\mu }^2$, the eigenfunctions and their temporal factors areThese sine functions are orthogonal on $[0,L]$, with squared norm $L/2$. Expanding the initial condition and multiplying by $\sin ({\mu }_{m}x)$ gives, using the permitted interchange of the sum and integral,Here $\cos ({\mu }_{n}L)=0$ and $\sin ({\mu }_{n}L)=(-1{)}^n$. Thus the Fourier series solution isThe exponential decay permits differentiation of the series for $t>0$, and each term satisfies both boundary conditions. At $t=0$ it is the mixed-end Fourier expansion of $-T_{0}x/L$.

The gained heat is the increase in internal energy. Since the paper specifies no cross-sectional area, first give the answer per unit cross-sectional area; multiply by $A$ for a rod of cross-sectional area $A$. The initial temperature has integral $T_{0}L/2$, whereas ${\int }_0^L\sin ({\mu }_{n}x)dx=1/{\mu }_n$. At $t_{\ast }=L^2/\kappa$ this givesNumerically $\Delta Q/A\simeq 0.45623855\rho c{T}_{0}L$. This is positive and less than the eventual gain $\rho c{T}_{0}L/2$, as expected from relaxation towards the uniform steady state.

If $Pdx+Qdy=dF$ for a twice continuously differentiable potential $F$, then $P=F_x$ and $Q=F_y$. Equality of mixed partial derivatives gives the necessary condition for an exact differential:This condition alone does not ensure global exactness on a domain with holes.

For the Helmholtz free energy $A=U-TS$, use the stated identity for internal energy and the product rule:Thus $A_T{|}_V=-S$ and $A_V{|}_T=-p$. Equality of mixed partial derivatives yields the Maxwell relationThe temperature and volume are the independent variables here, with composition and amount of gas fixed.

Next express $S=S(T,V)$ and substitute its total differential into $dU=TdS-pdV$:Matching the coefficient of $dV$ givesThis is the thermodynamic internal pressure identity: an equation of state determines how internal energy changes under an isothermal change of volume.

For the Dieterici equation, set $c=an/(VR)$ while differentiating at fixed $V$. ThenSubstitute into the thermodynamic internal pressure identity. The term proportional to $T$ cancels $p$, leavingThe physical gas domain has $T>0$ and $V>nb$.

For the ideal gas instead, $p=nRT/V$ gives $T(\partial p/\partial T{)}_V-p=0$. Therefore $U$ is independent of volume at fixed amount of gas. Integrating the remaining condition $(\partial U/\partial T{)}_V=C_V$, with constant heat capacity at constant volume, giveswhere $U_0$ is independent of both $T$ and $V$ and fixes the reference for internal energy.

At fixed particle number, the differential of the internal energy identifies $T=(\partial U/\partial S{)}_V$ and $p=-(\partial U/\partial V{)}_S$. Equality of continuous mixed partial derivatives gives the first Maxwell relation:For the enthalpy, $dH=dU+pdV+Vdp=TdS+Vdp$. Thus $T=(\partial H/\partial S{)}_p$, $V=(\partial H/\partial p{)}_S$, and the same argument gives the enthalpy Maxwell relationThe smoothness and independent thermodynamic coordinates are understood within a single phase.

Holding $y$ fixed, the chain rule giveswhere the local change of variable from $x$ to $u$ is valid.

For the internal energy $U(S,V)$, its exact differential identifies $T=U_S$ and $p=-U_V$. The Maxwell relation from equality of mixed partial derivatives is $T_V{|}_S=-p_S{|}_V$. At fixed volume, the constant-volume heat capacity satisfiesso $T_S{|}_V=T/C_V$, assuming nonzero $C_V$ and a locally invertible relation. Applying the chain rule to $p(T,V)$ gives the isentropic temperature-volume derivativeFor the stated one-mole ideal gas, the equation of state gives $p_T{|}_V=R/V$ and $C_V=3R/2$, so the right side is $-2T/(3V)$. Along an isentropic process, $p\propto V^{-5/3}$ and $T=pV/R\propto V^{-2/3}$, giving the same left side:

For fixed amount of substance, take the internal energy $U=U(S,V)$. The first law of thermodynamics identifies $T=U_S{|}_V$ and $-P=U_V{|}_S$. Equality of mixed second derivatives givesFor the Helmholtz free energy, subtract the product differential $d(TS)$ from the first law of thermodynamics to obtain $dF=-SdT-PdV$. Its natural variables are $(T,V)$, so $F_T=-S$ and $F_V=-P$. Exactness therefore givesFor the Gibbs free energy, the product rule gives $dG=-SdT+VdP$. Thus $G_T=-S$, $G_P=V$, andFor the enthalpy, $dH=TdS+VdP$, giving $H_S=T$, $H_P=V$ andThese are the four Maxwell relations, each following from exactness in the appropriate pair of natural variables.

The relevant claim concerns a machine that repeats its complete operating cycle while delivering a continuing net output of work without a corresponding supply of energy. A machine cannot supply net work in a cycle without taking in energy or depleting an energy store. This follows from conservation of energy, an experimentally supported physical law rather than a theorem of pure logic.

Let $E$ include the machine's internal energy, kinetic energy and potential energy. With $Q$ the net heat received and $W$ the net work delivered, the first law of thermodynamics givesFor a complete cycle all stores return to their initial values, so $\Delta E=0$ and $W=Q$. If there is neither heat input nor another energy input, $W=0$. A descending mass, unwinding spring, consumed fuel or discharged battery supplies energy, but it cannot be omitted from the accounting while calling the result perpetual motion.

Changing an overbalanced wheel's geometry does not evade the argument. A conservative force gives zero net work around a closed configuration cycle, while friction dissipates usable mechanical energy. Resetting a mechanism or returning a raised weight costs work that must be included.

Indefinitely continuing motion without useful output is a different possibility: an ideal frictionless oscillator or freely moving body need not stop. Connecting a load removes energy and changes that situation. A cyclic machine supplied with heat from only one reservoir and converting all that heat into work is excluded instead by the Kelvin-Planck statement of the Second law of thermodynamics. The first-law and second-law objections address different proposed energy balances.

The strong evidence is the reproducible conversion between mechanical work and thermal changes. In Joule's paddle-wheel experiment, descending masses drive vanes in a liquid, and a measured input of mechanical energy produces a measured rise in temperature. Allowing for the vessel and thermal losses gives a quantitative mechanical equivalent. Sustained frictional heating, including Rumford's cannon-boring experiment, also challenged a simple account in which rubbing merely released a fixed stored supply of a material heat fluid.^[ref]

Electrical and compressional experiments reinforce the connection. A resistor warms under Joule heating, with electrical input $\int VIdt$. A gas compressed with little heat transfer can rise in temperature, showing that mechanical work can increase its internal energy. Conversely an expanding gas can perform work while cooling. Such experiments connect several kinds of change through a conserved energy balance.

The kinetic theory of gases supplies a microscopic model. Translational molecular motion produces momentum transfer to walls; for a dilute monatomic ideal gas,Thus increasing molecular kinetic energy is reflected in temperature and pressure changes. These formulas describe a specified model, not a proof that every substance's internal energy consists only of translational motion. Rotations, vibrations and intermolecular potential energy also matter.

In modern terminology, heat is energy transferred because of a temperature difference. The body has internal energy, and microscopic motion accounts for part of it. Calling heat “a form of motion” captures the historical rejection of a conserved material caloric, but should not obscure that distinction between stored energy and an energy-transfer process.

For direct conversion, descending masses can drive paddles in an insulated vessel of water. Measure the lost gravitational potential energy and the temperature rise, including the vessel's heat capacity and corrections for unwanted heat transfer. Repeated reversal of the paddles and fixed baffles inhibit simple bulk co-rotation, directing more of the input towards dissipative fluid motion. Joule's paddle-wheel experiment tests a quantitative mechanical-to-thermal balance, rather than merely observing that stirring can warm water.^[ref]

For conversion through electricity, mechanical work drives a generator, whose electric current passes through a resistor in a calorimeter. The measured electrical input is $\int VIdt$, or $I^{2}Rt$ for constant electric current and electrical resistance. Joule heating increases the internal energy of the resistor and surrounding water. Compare the generator's mechanical input with the total thermal gains, including generator and circuit losses if the system boundary contains them. A battery alone demonstrates electrical-to-thermal conversion; without accounting for a mechanically driven source it does not establish the requested mechanical intermediate chain.

Using the given value relative to water, the copper requiresWith the modern thermochemical calorie, $1\mathrm{cal}=4.184\mathrm{J}$, so the ideal required work isEquivalently it is about $203\mathrm{kgf}\mathrm{m}$ using standard gravity. A nineteenth-century numerical equivalent of heat could give a slightly different rounded result. The stipulated constant specific heat capacity is used here; an unprovided vessel heat capacity and environmental losses would add to the required input.

Cosmological variables and conventions. The scale factor $R(t)$ describes the relative separation of comoving observers; dots denote derivatives with respect to cosmic time $t$. The Hubble parameter is $H(t)=\dot{R}/R$, usually time dependent; “Hubble constant” conventionally means its present value. Here $\rho$ is mass-equivalent density, so $ϵ=\rho c^2$ is energy density, while $P$ is isotropic pressure. $G$ is the gravitational constant and $c$ the speed of light. The constant $K$ specifies cosmological spatial curvature: with $R$ a length, $K=k{c}^2$ and $k=+1,0,-1$ describe closed, flat and open spatial geometries. Alternative normalizations of $R$ move a constant factor into $K$. In the printed equations $\Lambda$ has units ${\mathrm{s}}^{-2}$: it is $c^2$ times a cosmological constant defined in inverse square metres. This time-unit convention for the cosmological constant prevents a missing factor of $c^2$.

Continuity and the first law. Differentiating the first Friedmann equation with constant $K$ and $\Lambda$, and then eliminating $\ddot{R}/R$ with the second, givesThus, on an expanding or contracting branch,The relation extends continuously to a regular turning point. For a physical volume $V(t)=V_{\ast }{R}^3(t)$ bounded by fixed comoving coordinates, $\dot{V}=3HV$. Multiplication by $c^{2}V$ givesThis adiabatic work in a comoving volume is the first law of thermodynamics for adiabatic evolution with fixed entropy and no additional chemical work. It expresses local energy conservation, including pressure work. It does not assert that $U$ is constant: radiation loses internal energy as it does work; negative pressure makes $U$ grow during expansion. A generic expanding spacetime has no global time-translation symmetry from which a conserved total energy follows.

Constant scalar-field energy density. Setting $\dot{\rho }=0$ in the continuity equation, with $H\ne 0$, givesFor a homogeneous canonical scalar field, in units $c=1$, $ϵ={\dot{\varphi }}^2/2+V(\varphi )$ and $P={\dot{\varphi }}^2/2-V(\varphi )$. A stationary potential-dominated field has precisely this equation of state. A general evolving scalar field need not have $w=-1$; the constant-density assumption is essential.

Expansion at the stipulated Higgs energy density. With zero explicit $\Lambda$ and zero $K$, constant ${ϵ}_H$ givesThe plus sign selects the expanding branch; there is also a contracting solution. The scalar-field energy is already included in $\rho$, so adding it again as an explicit cosmological constant would double count it. Using $1\mathrm{GeV}=1.602176634\times 10^{-10}\mathrm{J}$ gives ${ϵ}_H=2.4033\times 10^{45}\mathrm{J}{\mathrm{m}}^{-3}$, ${\rho }_H=2.6740\times 10^{28}\mathrm{kg}{\mathrm{m}}^{-3}$ and $H_H\simeq 3.87\times 10^9{\mathrm{s}}^{-1}$. ThereforeThis vacuum-dominated doubling time describes accelerated expansion in de Sitter spacetime, with $\ddot{R}/R=H_H^2>0$. The stipulated energy density exceeds the critical density in energy units used in Question 2 by $3\times 10^{54}$, so it cannot persist as the dominant component of the present universe. The calculation exhibits the cosmological constant problem: the absolute gravitational vacuum energy must be enormously smaller than such a particle-physics scale today. It does not establish that the Standard Model Higgs field actually remains constant or supplies a viable inflationary epoch.

The acoustic ruler and accelerated expansion. Before baryon drag becomes ineffective, pressure in the photon-baryon fluid drives acoustic waves around initial linear cosmological density perturbations. Their comoving sound horizon at the baryon drag epoch isAfter decoupling, this preferred separation survives in galaxy clustering as baryon acoustic oscillations: a peak in the two-point correlation function and oscillations in the matter power spectrum. Its value is calibrated by early-universe physics, commonly constrained by the cosmic microwave background. The SDSS luminous-red-galaxy measurement established a peak near $100h^{-1}\mathrm{Mpc}$; its use as a distance indicator depends on the ruler calibration rather than the length of an individual galaxy.^[ref]

For a ruler perpendicular to the line of sight and one parallel to it,where $D_M=(1+z)D_A$ is the transverse comoving distance and $D_A$ the angular diameter distance. These measure $D_M/r_d$ and $H{r}_d$, respectively. Comparing multiple redshifts tests the distance–redshift relation and expansion history. In particular,so sufficiently precise measurements of the shape of $H(z)$ can test $q<0$. In general relativity, accelerated expansion requires a sufficiently negative total $\rho +3P/c^2$; a positive cosmological constant supplies it.

A single acoustic angular scale does not independently determine acceleration, cosmological spatial curvature, the absolute sound horizon and the dark-energy equation of state. One combines redshift coverage with early-universe calibration and independent matter-density or distance information, comparing accelerating and nonaccelerating models. The galaxy selection function and redshift completeness must be modeled with random catalogues, survey covariance estimated with mock catalogues, and redshift-space distortions, galaxy bias and nonlinear broadening controlled. These tests distinguish a cosmological feature from a survey artifact. The evidence is the measured expansion history within an explicit cosmological framework, rather than the existence of an acoustic peak alone.

For independent conduction-electron states labelled by band, wavevector and spin, combine those labels into $j$. ThenHere $g$ counts the same spin and band states as the sum, and $\mu (T)$ is fixed by the number constraint $N=\int g{f}_{\mathrm{FD}}dϵ$. The energy zero is a fixed band reference. Use the total DOS for total internal energy or the molar DOS for molar internal energy; these conventions must not be mixed.

The zeroth law of thermodynamics states the transitivity of thermal equilibrium: if two systems each equilibrate thermally with a third, they equilibrate with one another. A reproducible thermometric property can therefore label their temperature. Its empirical numerical scale is not itself the complete thermodynamic content; reversible heat-engine relations establish an absolute temperature scale.

The first law of thermodynamics expresses energy conservation. For a closed system, $dU=\delta Q-\delta W$ when $\delta W$ is work done by the system. internal energy is a state function, while heat and work depend on the path. In reversible simple compressible motion, $\delta W=pdV$ and $dU=TdS-pdV$; an open system additionally needs the energies carried by transferred matter. An ideal gas can expand freely with $Q=W=0$ and unchanged internal energy. The first law alone does not decide whether the reverse spontaneous process occurs.

The Second law of thermodynamics introduces entropy and restricts direction. For a reversible path, $dS=\delta Q_{\mathrm{rev}}/T$; for an irreversible process the entropy balance includes nonnegative production. The entropy of an isolated system cannot decrease. In a cycle, the Clausius inequality is $\oint \delta Q/T\le 0$, using the appropriate boundary/reservoir temperatures. The Kelvin–Planck statement forbids a cyclic device whose sole effect is extracting heat from one reservoir and turning all of it into work; the Clausius statement forbids moving heat from cold to hot with no compensating effect. A refrigerator is allowed because it consumes work. A two-reservoir engine has $\eta \le 1-T_c/T_h$, with equality in the reversible limit. Work may be fully dissipated as heat without contradicting this one-way restriction.

The third law of thermodynamics constrains low-temperature entropy. An equilibrium perfect crystal with a nondegenerate ground state approaches zero entropy as $T\to 0$, fixing the conventional absolute entropy reference. Degenerate ground states or frozen nonequilibrium disorder require care when discussing residual entropy. Under regular equilibrium conditions the heat capacity must decrease sufficiently for $\int C/TdT$ to remain finite; a constant positive classical heat capacity cannot be extrapolated to zero temperature. The unattainability formulation states that absolute zero cannot be reached by a finite sequence of admissible cooling operations. These statements concern the limiting thermodynamic regime, rather than predicting that every substance stays in one classical phase during cooling.

The laws work together: conservation sets a device's energy budget, entropy production restricts the attainable budget, and constitutive equations give the properties of its working material. For fixed temperature and volume, Helmholtz free energy decreases toward equilibrium; at fixed temperature and pressure the corresponding potential is Gibbs free energy. Stable equilibrium, phase coexistence and Maxwell relations follow from these potentials with appropriate constraints. Calorimetry, mechanical-work measurements, thermometry and reversible-path comparisons test different parts of this framework. None of the laws by itself supplies a particular gas equation of state or microscopic heat capacity.

For a reversible isothermal expansion of one mole of an ideal gas, internal energy depends only on temperature, so $\Delta U=0$. The first law of thermodynamics givesThis uses quasistatic expansion against the gas pressure. Specifying only the initial and final volumes and constant temperature would not fix heat for an arbitrary irreversible process: an insulated ideal-gas free expansion has $Q=W=0$.