1. Introduction
Constructing a covariant theory of thermodynamics has been a long-standing problem [1–4]. For over a century, the statistical mechanics community has debated on the proper covariant definitions of fundamental thermodynamic quantities as well as how to generalize thermodynamic relations in relativistic settings [1, 4–17]. Recently, interest in relativistic thermodynamics has extended to the non-equilibrium regime [18–22]. These discussions have encompassed not only the thermodynamic relation in both Minkowski and curved spacetimes but also the question of thermal fluctuations in spacetime itself [23].
In all of the debate mentioned above, the covariant definition of temperature is one of the most fundamental issues in relativistic thermodynamics. Several transformations of temperature have been proposed [1, 5, 6, 8–11, 13]. Considering a system at the temperature T, and a moving observer whose velocity relative to the system is v, in 1907 and 1908 Plank and Einstein proposed that the temperature of the system perceived in the moving observer's frame is T′ = T/γ, where $\gamma =1/\sqrt{1-{{\boldsymbol{v}}}^{2}}$ is the Lorentz factor [5, 6] (we choose the natural unit ℏ = c = 1 hereafter). Half a century later, in 1963, Ott suggested that the temperature perceived in the moving frame of reference should be ${T}^{{\prime} }=\gamma T$ instead [8]. Shortly after Ott's proposal, Landsberg reviewed the previous work and recommended that the temperature was a Lorentz invariant, ${T}^{{\prime} }=T$ [9].
Beside the theoretical explorations, the discovery of the cosmic microwave background (CMB) provided an excellent platform for the experimental examination of these theories. Shortly after Penzias and Wilson's observation of the CMB in 1964 [24], a dipole anisotropy in CMB temperature was observed in the 1970s [25, 26] and the energy spectrum satisfied a perfect black-body spectrum (Plank's law). However, the dipole anisotropy in temperature was not predicted by any proposal of scalar temperature transformation [1, 5, 6, 8, 9]. The expression of the transformed temperature ${T}^{{\prime} }$ only contains a factor γ, which was not related to the direction of the relative motion and thus cannot explain the dipole anisotropy in CMB temperature.
The failure of the previous proposals on temperature transformations in describing the dipole anisotropy in the CMB temperature lies in the notion of equilibrium state. When promoting theories of thermodynamics to a covariant form, i.e. invariant for an arbitrary inertial observer, we need to extend the definition of equilibrium state to a general form [21, 27–29]. In the special theory of relativity, the energy and three-momentum together form the four-momentum and mix when undergoing Lorentz transformation. In thermodynamics, the inverse temperature is conjugate to energy, therefore the covariant form of the inverse temperature should be a four-vector, which includes some components being conjugate to the three-momentum. In 1968, van Kampen suggested that the temperature should be defined as a Lorentz vector [10, 11]. The Lorentz transformation depends on the direction of the relative motion, therefore the inverse temperature four-vector could be used to explain the experimental observations [21, 27, 29].
The Lorentz transformation of black-body radiation has been well studied, and the energy spectrum of a moving equilibrium electromagnetic field has been derived [30]. The scalar temperature with dipole anisotropy has been predicted theoretically [30–34]. However, we find the reduction of the inverse temperature four-vector to a scalar temperature is a consequence of the massless nature of photons. To demonstrate this result, we calculate the Lorentz transformation of massive equilibrium fields, not limited to the photons but also for massive free bosonic or fermionic fields. In the massless limit for bosonic fields, our results reproduce the main results in Ref. [30]. We demonstrate that for a massive field, the equilibrium spectrum can no longer be described by a scalar temperature with dipole anisotropy. Instead, an inverse temperature four-vector should be used to characterize the equilibrium state perceived in the moving observer's frame. Moreover, the fermionic case of our result may be applied to the study of the cosmic neutrino background (CνB) [35, 36].
The remainder of this article is organized as follows. In section 2 we use the Lorentz transformation of the equilibrium state of a massive free field to calculate the energy spectrum of the field perceived in an arbitrary inertial frame and explain the results under the framework of the inverse temperature four-vector. In section 3 we analyze some properties of the inverse temperature four-vector and demonstrate that for a massive field the moving equilibrium state can no longer be characterized by a scalar temperature with dipole anisotropy. In section 4 we summarize the main results.
2. Equilibrium spectrum of a massive field perceived in an arbitrary inertial frame of reference
2.1. Setup
Consider a massive real scalar field φ with proper mass m and Lagrangian density ${ \mathcal L }$
$\begin{eqnarray}{ \mathcal L }=\frac{1}{2}{\partial }_{\mu }\phi {\partial }^{\mu }\phi +\frac{{m}^{2}}{2}{\phi }^{2}.\end{eqnarray}$
Here we use the metric tensor ημν = diag(1, − 1, − 1, − 1). The Hamiltonian H of this field is $\begin{eqnarray}H=\int \displaystyle \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{{\boldsymbol{k}}}}\displaystyle \frac{{\omega }_{{\boldsymbol{k}}}}{2}({a}_{k}{a}_{k}^{\dagger }+{a}_{k}^{\dagger }{a}_{k}).\end{eqnarray}$
Here k denotes the three-wave vector and ωk is the angular frequency of the particle with wave vector k, which equals $\sqrt{{{\boldsymbol{k}}}^{2}+{m}^{2}}$ due to the on-shell condition. Since the on-shell four-momentum k = (ωk, k) has a one-to-one correspondence to k, we can also denote ωk as ωk without ambiguity. ak and ${a}_{k}^{\dagger }$ are the annihilation and creation operators of a particle with four-momentum k corresponding to the scalar field φ. The field operator can be expanded as $\begin{eqnarray}\phi (x)=\int \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{k}}\left[{a}_{k}{{\rm{e}}}^{-{\rm{i}}k\cdot x}+{a}_{k}^{\dagger }{{\rm{e}}}^{{\rm{i}}k\cdot x}\right],\end{eqnarray}$
in terms of the creation and annihilation operators.We may define $Dk=\frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{{\boldsymbol{k}}}}$ as a (on-shell) Lorentz invariant integral measure
$\begin{eqnarray*}\int D({\rm{\Lambda }}k)f({\rm{\Lambda }}k)=\int Dkf(k),\end{eqnarray*}$
for any Lorentz transformation Λ and any function f. Moreover, we can define a Lorentz invariant delta distribution $\begin{eqnarray}{\delta }^{* }(k-{k}^{{\prime} }):= {(2\pi )}^{3}2{\omega }_{k}{\delta }^{3}({\boldsymbol{k}}-{{\boldsymbol{k}}}^{{\prime} }),\end{eqnarray}$
which satisfies δ*(Λk) = δ*(k). The commutator of the creation and annihilation operators can then be expressed in δ* $\begin{eqnarray}[{a}_{k},{a}_{{k}^{{\prime} }}^{\dagger }]={\delta }^{* }(k-{k}^{{\prime} }).\end{eqnarray}$
The field system is in an equilibrium state and at rest in the frame of reference i, and there is a moving observer in the frame of reference ${i}^{{\prime} }$ which moves at a velocity v with respect to frame i.2.2. Spectral density at equilibrium
Suppose a system is in a thermal state at a finite temperature β and chemical potential $\tilde{\mu }$ in the frame i; the particle number operator is N. If we denote $\alpha =\beta \tilde{\mu }$, the density operator of the thermal equilibrium state is ${\rho }_{i}=\frac{{{\rm{e}}}^{-\beta H+\alpha N}}{{\rm{Tr}}({{\rm{e}}}^{-\beta H+\alpha N})}$. Then the expectation value of an operator A is ${\langle A\rangle }_{i}={\rm{Tr}}({\rho }_{i}A)$. The density operator and the expectation value are not Lorentz invariant. In frame i, we can derive the following result [30]: 6 ).
$\begin{eqnarray}\begin{array}{l}{\langle {a}_{k}{a}_{{k}^{{\prime} }}^{\dagger }+{a}_{k}^{\dagger }{a}_{{k}^{{\prime} }}\rangle }_{i}\\ \quad =\,{\delta }^{* }(k-{k}^{{\prime} })\coth \left(\frac{\beta {\omega }_{k}-\alpha }{2}\right),\\ {\langle {a}_{k}{a}_{{k}^{{\prime} }}\rangle }_{i}={\langle {a}_{k}^{\dagger }{a}_{{k}^{{\prime} }}^{\dagger }\rangle }_{i}=0.\end{array}\end{eqnarray}$
The energy–momentum tensor of the field is $\begin{eqnarray}{T}^{\mu \nu }=\frac{1}{2}({\partial }^{\mu }\phi {\partial }^{\nu }\phi +{\partial }^{\nu }\phi {\partial }^{\mu }\phi )-{\eta }^{\mu \nu }{ \mathcal L },\end{eqnarray}$
thus T00 is the energy density operator. Now we calculate the thermal state energy density ρE $\begin{eqnarray}\begin{array}{rcl}{\rho }_{E} & = & {\langle {T}^{00}\rangle }_{i}\\ & = & {\left\langle \displaystyle \frac{1}{2}\Space{0ex}{2.5ex}{0ex}[{({\partial }^{0}\phi )}^{2}+{({\rm{\nabla }}\phi )}^{2}+{m}^{2}{\phi }^{2}\Space{0ex}{2.5ex}{0ex}]\right\rangle }_{i}\\ & = & \left\langle \displaystyle \frac{1}{2}\int \displaystyle \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{k}}\right.\int \displaystyle \frac{{{\rm{d}}}^{3}{{\boldsymbol{k}}}^{^{\prime} }}{{(2\pi )}^{3}2{\omega }_{{k}^{^{\prime} }}}\\ & & \times \,[(-{\rm{i}}{\omega }_{k}{a}_{k}{{\rm{e}}}^{-{\rm{i}}k\cdot x}+{\rm{i}}{\omega }_{k}{a}_{k}^{\dagger }{{\rm{e}}}^{{\rm{i}}k\cdot x})\\ & & \times \,(-{\rm{i}}{\omega }_{{k}^{{\prime} }}{a}_{{k}^{{\prime} }}{{\rm{e}}}^{-{\rm{i}}{k}^{^{\prime} }\cdot x}+{\rm{i}}{\omega }_{{k}^{{\prime} }}{a}_{{k}^{{\prime} }}^{\dagger }{{\rm{e}}}^{{\rm{i}}{k}^{^{\prime} }\cdot x})\\ & & +\,({\rm{i}}{\boldsymbol{k}}{a}_{k}{{\rm{e}}}^{-{\rm{i}}k\cdot x}-{\rm{i}}{\boldsymbol{k}}{a}_{k}^{\dagger }{{\rm{e}}}^{{\rm{i}}k\cdot x})\\ & & \,\cdot \,({\rm{i}}{{\boldsymbol{k}}}^{{\prime} }{a}_{{k}^{{\prime} }}{{\rm{e}}}^{-{\rm{i}}{k}^{^{\prime} }\cdot x}-{\rm{i}}{{\boldsymbol{k}}}^{{\prime} }{a}_{{k}^{{\prime} }}^{\dagger }{{\rm{e}}}^{{\rm{i}}{k}^{^{\prime} }\cdot x})\\ & & +\,{m}^{2}({a}_{k}{{\rm{e}}}^{-{\rm{i}}k\cdot x}+{a}_{k}^{\dagger }{{\rm{e}}}^{{\rm{i}}k\cdot x})\\ & & \times \,{\left.({a}_{{k}^{^{\prime} }}{{\rm{e}}}^{-{\rm{i}}{k}^{^{\prime} }\cdot x}+{a}_{{k}^{^{\prime} }}^{\dagger }{{\rm{e}}}^{{\rm{i}}{k}^{^{\prime} }\cdot x})]\Space{0ex}{1.25em}{0ex}\right\rangle }_{\,i}\\ & = & \int \displaystyle \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{k}}\int \displaystyle \frac{{{\rm{d}}}^{3}{{\boldsymbol{k}}}^{^{\prime} }}{{(2\pi )}^{3}2{\omega }_{{k}^{^{\prime} }}}\displaystyle \frac{1}{2}({\omega }_{k}{\omega }_{{k}^{{\prime} }}+{\boldsymbol{k}}\cdot {{\boldsymbol{k}}}^{{\prime} }+{m}^{2})\\ & & \left.\times \,{\Space{0ex}{2.5ex}{0ex}\langle ({a}_{k}{a}_{{k}^{^{\prime} }}^{\dagger }{{\rm{e}}}^{{\rm{i}}({k}^{^{\prime} }-k)\cdot x}+{a}_{k}^{\dagger }{a}_{{k}^{^{\prime} }}{{\rm{e}}}^{{\rm{i}}(k-{k}^{^{\prime} })\cdot x})}_{i}\right\rangle .\end{array}\end{eqnarray}$
Here we have eliminated the $\langle {a}_{k}{a}_{{k}^{{\prime} }}\rangle $ and $\langle {a}_{k}^{\dagger }{a}_{{k}^{{\prime} }}^{\dagger }\rangle $ terms whose expectation values are 0. When ${\langle ({a}_{k}{a}_{{k}^{{\prime} }}^{\dagger }{{\rm{e}}}^{{\rm{i}}({k}^{{\prime} }-k)\cdot x}+{a}_{k}^{\dagger }{a}_{{k}^{{\prime} }}{{\rm{e}}}^{{\rm{i}}(k-{k}^{{\prime} })\cdot x})\rangle }_{i}$ is non-zero, $k={k}^{{\prime} },{{\rm{e}}}^{\pm i(k-{k}^{{\prime} })\cdot x}=1$, thus the last term is exactly the thermal state expectation value in equation (Integrating out ${k}^{{\prime} }$ with the delta function, we obtain
$\begin{eqnarray}\begin{array}{rcl}{\rho }_{E} & = & \displaystyle \int \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{k}}\displaystyle \int \frac{{{\rm{d}}}^{3}{{\boldsymbol{k}}}^{{\prime} }}{{(2\pi )}^{3}2{\omega }_{{k}^{{\prime} }}}\\ & & \times \,\frac{1}{2}\left({\omega }_{k}{\omega }_{{k}^{{\prime} }}+{\boldsymbol{k}}\cdot {{\boldsymbol{k}}}^{{\prime} }+{m}^{2}\right){\delta }^{* }(k-{k}^{{\prime} })\coth \left(\frac{\beta {\omega }_{k}-\alpha }{2}\right)\\ & = & \displaystyle \int Dk\,{\omega }_{k}^{2}\coth \left(\frac{\beta {\omega }_{k}-\alpha }{2}\right)\\ & = & \displaystyle \int \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}}\frac{{\omega }_{k}}{2}\coth \left(\frac{\beta {\omega }_{k}-\alpha }{2}\right)\\ & = & \displaystyle \int | {\boldsymbol{k}}{| }^{2}{\rm{d}}{\rm{\Omega }}{\rm{d}}| {\boldsymbol{k}}| \frac{{\omega }_{k}}{2{(2\pi )}^{3}}\coth \left(\frac{\beta {\omega }_{{\boldsymbol{k}}}-\alpha }{2}\right),\end{array}\end{eqnarray}$
where dΩ is the solid angle of the wave vector k. Now we define the spectral density $\rho (\omega ,\hat{{\boldsymbol{k}}})$ to satisfy the relation ${\rho }_{E}=\int \rho (\omega ,\hat{{\boldsymbol{k}}}){\rm{d}}{\rm{\Omega }}{\rm{d}}{\rm{\Omega }}$, where $\omega ={\omega }_{{\boldsymbol{k}}}=\sqrt{{{\boldsymbol{k}}}^{2}+{m}^{2}}$ and $\hat{{\boldsymbol{k}}}$ is the unit vector along k. For the massless case we have ω = ∣k∣, then $\begin{eqnarray}\rho (\omega ,\hat{{\boldsymbol{k}}})=\frac{{\omega }^{3}}{2{(2\pi )}^{3}}\coth (\frac{\beta \omega -\alpha }{2}).\end{eqnarray}$
For black-body radiation, $\tilde{\mu }=0,\alpha =0$, the result is just Planck's law (including zero-point energy) without the two-fold spin degeneracy. If m ≠ 0, then ${\rm{d}}| {\boldsymbol{k}}| =\frac{\omega }{\sqrt{{\omega }^{2}-{m}^{2}}}{\rm{d}}{\rm{\Omega }}$, and we can see [37] $\begin{eqnarray}\rho (\omega ,\hat{{\boldsymbol{k}}})=\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\coth (\frac{\beta \omega -\alpha }{2}).\end{eqnarray}$
The spectral density is isotropic, because the observer is at rest with the equilibrium field.2.3. Spectral density perceived in an arbitrary inertial frame of reference
Consider an observable A in the frame of reference ${i}^{{\prime} }$ with a three-velocity v. The Lorentz transformation from frame i to frame ${i}^{{\prime} }$ is denoted as Λ. The density operator of the thermal state in frame ${i}^{{\prime} }$ is ${\rho }_{{i}^{{\prime} }}$. Then the expectation value of A in frame ${i}^{{\prime} }$ is ${\langle A\rangle }_{{i}^{{\prime} }}={\rm{Tr}}({\rho }_{{i}^{{\prime} }}A)$, where ${\rho }_{{i}^{{\prime} }}=U({\rm{\Lambda }}){\rho }_{i}U{({\rm{\Lambda }})}^{\dagger }$, and U(Λ) is the unitary operator corresponding to the Lorentz transformation Λ. This state can be viewed as a generalized thermal state ${\rho }_{{i}^{{\prime} }}=\frac{{{\rm{e}}}^{-\beta {u}_{\mu }{P}^{\mu }+\alpha N}}{{\rm{Tr}}({{\rm{e}}}^{-\beta {u}_{\mu }{P}^{\mu }+\alpha N})}$ [38], where uμ is the four-velocity of frame i relative to frame ${i}^{{\prime} }$ and Pμ is the four-momentum of the field, with P0 = H and ${P}^{i}=\int Dk\,{k}_{i}{a}_{k}^{\dagger }{a}_{k}$. α and βuμ are the Lagrange multipliers corresponding to the conserved quantities N and Pμ. As it is usual in relativity that the energy and momentum are treated on the same footing, the density operator ${\rho }_{{i}^{{\prime} }}$ characterizes a generalized Gibbs state with conserved particle number, energy and momentum. Using the cyclic property of trace, we have
$\begin{eqnarray}\begin{array}{rcl}{\langle A\rangle }_{{i}^{{\prime} }} & = & {\rm{Tr}}(U({\rm{\Lambda }}){\rho }_{i}U{({\rm{\Lambda }})}^{\dagger }A)\\ & = & {\rm{Tr}}({\rho }_{i}U{({\rm{\Lambda }})}^{\dagger }AU({\rm{\Lambda }}))\\ & = & {\langle U{({\rm{\Lambda }})}^{\dagger }AU({\rm{\Lambda }})\rangle }_{i}.\end{array}\end{eqnarray}$
Utilizing this relation, the (generalized) thermal state expectation value in frame ${i}^{{\prime} }$ becomes 6 ) and the invariance of δ*. We can do a similar calculation to that in equation (8 ), then we obtain 13 ). Therefore, the energy spectral distribution in frame ${i}^{{\prime} }$ is
$\begin{eqnarray}\begin{array}{rcl} & & {\langle {a}_{k}{a}_{{k}^{{\prime} }}^{\dagger }+{a}_{k}^{\dagger }{a}_{{k}^{{\prime} }}\rangle }_{{i}^{{\prime} }}\\ & = & {\rm{Tr}}\left[{\rho }_{i}U{({\rm{\Lambda }})}^{\dagger }({a}_{k}{a}_{{k}^{{\prime} }}^{\dagger }+{a}_{k}^{\dagger }{a}_{{k}^{{\prime} }})U({\rm{\Lambda }})\right]\\ & = & {\rm{Tr}}\left[{\rho }_{i}({a}_{{{\rm{\Lambda }}}^{-1}k}{a}_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}^{\dagger }+{a}_{{{\rm{\Lambda }}}^{-1}k}^{\dagger }{a}_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }})\right]\\ & = & {\langle {a}_{{{\rm{\Lambda }}}^{-1}k}{a}_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}^{\dagger }+{a}_{{{\rm{\Lambda }}}^{-1}k}^{\dagger }{a}_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}\rangle }_{i}\\ & = & {\delta }^{* }({\rm{\Lambda }}(k-{k}^{{\prime} }))\coth \left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}k}-\alpha }{2}\right)\\ & = & {\delta }^{* }(k-{k}^{{\prime} })\coth \left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}k}-\alpha }{2}\right),\end{array}\end{eqnarray}$
with equation ( $\begin{eqnarray}{\rho }_{E}^{{\prime} }=\int D{k}^{{\prime} }{\omega }_{{k}^{{\prime} }}^{2}\coth \left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2}\right),\end{eqnarray}$
with equation ( $\begin{eqnarray}{\rho }^{{\prime} }({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} })=\frac{| {{\boldsymbol{k}}}^{{\prime} }\left({\omega }^{{\prime} }\right){| }^{2}{\omega }^{{\prime} }}{2{\left(2\pi \right)}^{3}}\frac{{\rm{d}}| {{\boldsymbol{k}}}^{{\prime} }| }{{\rm{d}}{{\rm{\Omega }}}^{{\prime} }}\coth \left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2}\right).\end{eqnarray}$
This result actually contains the key result in Ref. [30] as a massless limit. If we take $| {{\boldsymbol{k}}}^{{\prime} }| ={\omega }^{{\prime} }$ and the chemical potential $\tilde{\mu }=0$, then α = 0 and equation (15 ) simplifies to Appendix ). Please note that a similar result about the mean occupation number was obtained in [39].
$\begin{eqnarray}{\rho }^{{\prime} }({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} })=\frac{{\omega {}^{{\prime} }}^{3}}{2{(2\pi )}^{3}}\coth \left(\frac{\beta \gamma (1+{\hat{{\boldsymbol{k}}}}^{{\prime} }\cdot {\boldsymbol{v}}){\omega }^{{\prime} }}{2}\right),\end{eqnarray}$
which, if multiplied by a two-fold spin degeneracy, is just the main result in Ref. [30]. In this case, in any direction ${\hat{{\boldsymbol{k}}}}^{{\prime} }$ we can see that the energy spectrum over ${\omega }^{{\prime} }$ is a perfect black-body spectrum with a modified temperature. In the more general case, m ≠ 0, we have $\begin{eqnarray}\begin{array}{rcl}{\rho }^{{\prime} }\left({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} }\right) & = & {g}_{s}\frac{{\omega {}^{{\prime} }}^{2}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\\ & & \times \coth \left(\frac{\beta \gamma \left({\omega }^{{\prime} }+{\hat{{\boldsymbol{k}}}}^{{\prime} }\cdot {\boldsymbol{v}}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}\right)-\alpha }{2}\right).\end{array}\end{eqnarray}$
For fermions, we can obtain similarly $\begin{eqnarray}\begin{array}{rcl}{\rho }^{{\prime} }({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} }) & = & -{g}_{s}\frac{{\omega {}^{{\prime} }}^{2}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\\ & & \times \tanh \left(\frac{\beta \gamma ({\omega }^{{\prime} }+{\hat{{\boldsymbol{k}}}}^{{\prime} }\cdot {\boldsymbol{v}}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}})-\alpha }{2}\right).\end{array}\end{eqnarray}$
Here gs accounts for the spin degeneracy (for details see the 2.4. Equilibrium state characterized by an inverse temperature four-vector
For a system with an inverse temperature β defined in the rest frame, and with a four-velocity uμ measured in the moving observer's frame of reference, the inverse temperature four-vector βμ of the system is defined as [10, 11, 13]
$\begin{eqnarray}{\beta }^{\mu }=\beta {u}^{\mu }.\end{eqnarray}$
We now show that the spectral density should be written as a function of βμ instead of the scalar temperature β.In the rest frame of the system, the four-velocity of the system itself is uμ = (1, 0, 0, 0), thus the inverse temperature four-vector is simply βμ = (β, 0, 0, 0). In the moving frame ${i}^{{\prime} }$, the four-velocity of the system is ${u{}^{{\prime} }}^{\mu }=\gamma (1,-{\boldsymbol{v}})$, and the contra-variant inverse temperature four-vector becomes ${\beta {}^{{\prime} }}^{\mu }=\gamma (\beta ,-\beta {v}_{x},-\beta {v}_{y},-\beta {v}_{z})$. Correspondingly, the covariant inverse temperature four-vector is ${\beta }_{\mu }^{{\prime} }={\eta }_{\mu \nu }{\beta {}^{{\prime} }}^{\nu }\,=\gamma (\beta ,\beta {v}_{x},\beta {v}_{y},\beta {v}_{z})$.
The spectral density of energy is then fully characterized by the inverse temperature four-vector of the field and the four-momentum of the particles. For bosons
$\begin{eqnarray}\begin{array}{rcl} & & {\rho }^{{\prime} }({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} })\\ & = & {g}_{s}\frac{{\omega {}^{{\prime} }}^{2}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\\ & & \times \,\coth \left(\frac{\beta \gamma ({\omega }^{{\prime} }+{\boldsymbol{v}}\cdot {\hat{{\boldsymbol{k}}}}^{{\prime} }\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}})-\alpha }{2}\right)\\ & = & {g}_{s}\frac{{\omega {}^{{\prime} }}^{2}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\coth \left(\frac{{\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }-\alpha }{2}\right),\end{array}\end{eqnarray}$
where ${P{}^{{\prime} }}^{\mu }=({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} }\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}})=({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} }| {{\boldsymbol{k}}}^{{\prime} }| )=({\omega }^{{\prime} },{{\boldsymbol{k}}}^{{\prime} })$ is the four-momentum of the particles [12].In the rest frame of the field (frame i), the only non-zero component in the inverse temperature four-vector is β0 = β, therefore βμPμ = β0 · ω + 0 · Px + 0 · Py + 0 · Pz = βω and the three-momentum does not appear in the spectral density. However, the spectrum characterized by an inverse temperature four-vector is universal in an arbitrary inertial frame. Generally, we can remove the prime notation in equation (20 ), and the expression is valid in an arbitrary inertial frame of reference. For bosons, the covariant spectral density of the generalized thermal state $\rho =\frac{{{\rm{e}}}^{-\beta {u}^{\mu }{P}_{\mu }+\alpha N}}{{\rm{Tr}}({{\rm{e}}}^{-\beta {u}^{\mu }{P}_{\mu }+\alpha N})}$ is
$\begin{eqnarray}\rho (\omega ,\hat{{\boldsymbol{k}}})={g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\coth \left(\frac{{\beta }_{\mu }{P}^{\mu }-\alpha }{2}\right),\end{eqnarray}$
and for fermions the spectral density is $\begin{eqnarray}\rho (\omega ,\hat{{\boldsymbol{k}}})=-{g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\tanh \left(\frac{{\beta }_{\mu }{P}^{\mu }-\alpha }{2}\right),\end{eqnarray}$
where βμ is the covariant inverse temperature four-vector of the field and Pμ is the contra-variant four-momentum of a particle with energy ω and moving in the direction $\hat{{\boldsymbol{k}}}$ [38]. βμ and Pμ can be defined in any inertial frame and satisfy the Lorentz transformation of four-vectors. α corresponds to the chemical potential of a particle. Since the particle number is a Lorentz scalar, the corresponding Lagrange multiplier α, which represents the conservation of particle number, should also be a Lorentz scalar [12, 39].The CνB could be approximated as an equilibrium field of neutrinos, which can be treated as a free field. The rest mass of the neutrino (or at least the heaviest flavor of the neutrino) is estimated to be in the range 10−2 eV to 10−1 eV [35]. In the early stage of the Universe, when the temperature of the Universe dropped to 1010 K (corresponding to the energy scale of 106 eV, which is 107–108 higher than the rest mass of neutrinos), the weak interaction rates began to fall far below the expansion rate of the Universe, so the neutrinos quickly fell out of equilibrium [37]. After decoupling from electrons, positrons and photons, the neutrinos underwent adiabatic evolution and ceased to thermalize. Consequently, the effect of the rest mass on the neutrino's energy spectrum had few observational consequences, remaining a relative correction of order 10−8–10−7 compared with the massless result. Currently, the effective temperature of CνB is predicted to be around 1.95 K, and the deviation of the CνB spectrum from the massless energy spectrum is estimated to be of the same order, i.e. about 10−8–10−7.
3. Analysis of the properties of the inverse temperature four-vector
From the study of the equilibrium state of the free fields we observe a distinct difference between massive and massless systems: for massless particles such as photons, the equilibrium state distribution can be characterized by a scalar temperature with dipole anisotropy. However, for massive particles whose momenta are no longer proportional to the energy it is impossible to characterize the equilibrium state perceived in an arbitrary inertial frame with a scalar temperature, and it is necessary to express the equilibrium state distribution in the form of inverse temperature four-vector and four-momentum.
3.1. Failure of scalar temperature description for massive fields
For massless particles such as photons, the spectral distribution in a specific direction $\hat{{{\boldsymbol{k}}}^{{\prime} }}$ satisfies the exact black-body radiation spectrum, even in a moving frame of reference (such as frame ${i}^{{\prime} }$ which has a velocity v relative to frame i). The spectral density in direction $\hat{{{\boldsymbol{k}}}^{{\prime} }}$ is $\rho ({\omega }^{{\prime} },{{\boldsymbol{k}}}^{{\prime} })={({\omega }^{{\prime} }/2\pi )}^{3}\coth [\gamma (1+{\boldsymbol{v}}\cdot \hat{{{\boldsymbol{k}}}^{{\prime} }}){\omega }^{{\prime} }/2T]$, which corresponds to a black-body spectrum at rest with an effective temperature ${T}^{{\prime} }(\hat{{{\boldsymbol{k}}}^{{\prime} }})$
$\begin{eqnarray}{T}^{{\prime} }({\hat{{\boldsymbol{k}}}}^{{\prime} })=\frac{T}{\gamma (1+{\boldsymbol{v}}\cdot {\hat{{\boldsymbol{k}}}}^{{\prime} })}=\frac{T}{\gamma (1+| {\boldsymbol{v}}| \cos \theta )}.\end{eqnarray}$
Here θ is the angle between the direction of observation ${\hat{{\boldsymbol{k}}}}^{{\prime} }$ and the direction of the relative motion $\hat{{\boldsymbol{v}}}$. This result has been studied theoretically [27, 29, 30, 32–34] and observed experimentally [25, 26, 40].This correspondence is a consequence of the massless nature of photons. In equation (21 ), for a photon with m = 0, the absolute value of three-momentum $| {{\boldsymbol{k}}}^{{\prime} }| $ is proportional to its energy ${\omega }^{{\prime} }$, therefore the contraction ${\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }\propto {\omega }^{{\prime} }$, and the ratio ${\omega }^{{\prime} }/{\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }$ could be regarded as an effective temperature ${T}_{eff}^{{\prime} }={\omega }^{{\prime} }/{\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }=T/\gamma (1+{\boldsymbol{v}}\cdot \hat{{{\boldsymbol{k}}}^{{\prime} }})$, which is independent of ${\omega }^{{\prime} }$.
However, for massive particles, the corresponding ratio is ${\omega }^{{\prime} }/{\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }=T/\gamma (1+{\boldsymbol{v}}\cdot {\hat{{\boldsymbol{k}}}}^{{\prime} }\sqrt{1-{\left(\frac{m}{{\omega }^{{\prime} }}\right)}^{2}})$, which is a function of both ${\hat{{\boldsymbol{k}}}}^{{\prime} }$ and ${\omega }^{{\prime} }$. This expression contains ${\omega }^{{\prime} }$, thus ${\omega }^{{\prime} }/{\beta }_{\mu }^{{\prime} }{P{}^{{\prime} }}^{\mu }$ can no longer be regarded as an effective temperature for massive fields[36, 41]. This explains why we can no longer use a scalar temperature to characterize the equilibrium state of massive fields.
Here we can see that it is just a coincidence that the spectrum of the CMB measured by a moving observer (the observer fixed on earth) can be characterized by a scalar temperature, because (only for massless particles) the inverse temperature four-vector can be reduced to a scalar temperature with dipole anisotropy. Generally, the moving observer needs the inverse temperature four-vector to characterize the equilibrium state in the rest frame.
3.2. Some explicit illustrations
Let us take the spectral density of fermions as an example, and recall the results in equation (22 ). We would like to highlight two situations: the zero-temperature and infinite-temperature cases.
The first illustration is the vacuum state of the field, and the corresponding chemical potential is $\tilde{\mu }=0,\alpha =0$. The factor $-\tanh ({\beta }_{\mu }{P}^{\mu }/2)=-\tanh (\beta ({u}_{\mu }{P}^{\mu })/2)$, β is the rest temperature of the equilibrium field and uμPμ is the energy measured in the rest frame of the field, which is positive definite.
For the zero-temperature state β → ∞ and $-\tanh (\beta ({u}_{\mu }{P}^{\mu })/2)\to -1$, which means that there is no particle excited in any mode. For two fermionic field systems with different velocities and zero temperatures, the difference between uμ is finite, and the difference between inverse temperature four-vectors even diverges. However, there is no difference in the spectral density. This result is intuitive: the vacuum state is invariant under Lorentz transformation.
In the infinite-temperature state β = 0, α = 0 and $-\tanh (\beta ({u}_{\mu }{P}^{\mu })/2)=0$, this temperature-dependent factor also becomes a constant, meaning that every state is equally occupied. For two fermionic field systems with different velocities and infinite temperatures, the difference between uμ is finite but the difference between the inverse temperature four-vectors is zero (β = 0, thus βμ = βuμ = 0 for arbitrary uμ, so there is no difference between βμ for infinite-temperature systems with different velocities). This result is intuitive too: the equally occupied state should also be invariant under Lorentz transformation.
In intermediate cases, when temperatures are finite, the necessary and sufficient condition for two field systems, S and ${S}^{{\prime} }$ (with the same m), to have the same spectral density is
$\begin{eqnarray*}\begin{array}{rcl}{\beta }_{\mu } & = & {\beta }_{\mu }^{{\prime} },\\ \tilde{\mu } & = & \tilde{{\mu }^{{\prime} }}.\end{array}\end{eqnarray*}$
Here ${\beta }_{\mu }/{\beta }_{\mu }^{{\prime} }$ and $\tilde{\mu }/\tilde{{\mu }^{{\prime} }}$ represent the inverse temperature four-vector and the chemical potential of $S/{S}^{{\prime} }$. Under this circumstance $\begin{eqnarray}\begin{array}{rcl}\beta & = & \sqrt{{\beta }_{\mu }{\beta }^{\mu }}=\sqrt{{\beta }_{\mu }^{{\prime} }{\beta {}^{{\prime} }}^{\mu }}={\beta }^{{\prime} },\\ \alpha & = & \beta \tilde{\mu }={\beta }^{{\prime} }\tilde{{\mu }^{{\prime} }}={\alpha }^{{\prime} },\\ {u}_{\mu } & = & \frac{{\beta }_{\mu }}{\beta }=\frac{{\beta }_{\mu }^{{\prime} }}{{\beta }^{{\prime} }}={u}_{\mu }^{{\prime} }.\end{array}\end{eqnarray}$
The equivalence of the inverse temperature four-vector implies that the two systems move at the same velocity (the same uμ) and have the same rest temperature in each co-moving frame. In this case, the thermal equilibrium relation becomes straightforward.It is worthwhile pointing out that a single Lorentz scalar β is insufficient to characterize the thermal equilibrium relation. We can easily construct two thermal equilibrium systems R1 and R2 with the same rest temperature β, the same chemical potential $\tilde{\mu }$ and a relative velocity v. In the rest frame of R1, the energy spectra of R1 and R2 are
$\begin{eqnarray*}\begin{array}{rcl} & & {\rho }_{1}(\omega ,\hat{{\boldsymbol{k}}})\\ & = & -{g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\tanh \left(\frac{\beta \omega -\alpha }{2}\right),\\ & & {\rho }_{2}(\omega ,\hat{{\boldsymbol{k}}})\\ & = & -{g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\tanh \left(\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot \hat{{\boldsymbol{k}}}\sqrt{{\omega }^{2}-{m}^{2}})-\alpha }{2}\right).\end{array}\end{eqnarray*}$
If we let the two fields couple to each other and allow particle flux in a specific direction and a specific frequency (via a filter, collimator or a resonant cavity), then the average net energy flow is usually non-zero, indicating that the two field systems are not in equilibrium with each other.The discussions above demonstrate that the spectral density characterized by an inverse temperature four-vector is sufficient to characterize the thermal equilibrium relation. The inverse temperature four-vector provides a covariant description of the thermal equilibrium states.
3.3. Classical and non-relativistic limit of the spectrum
In the classical limit (quantum statistical effects can be neglected) we expect that the bosonic spectrum recovers a Jüttner–Synge distribution. Furthermore, in the non-relativistic limit we expect that the spectrum recovers a shifted Maxwellian distribution.
To obtain the classical limit, we should firstly subtract the influence of zero-point fluctuation. Denote the zero-point spectral density as ρ0(ω). For bosons
$\begin{eqnarray}\begin{array}{rcl} & & \rho (\omega ,\hat{{\boldsymbol{k}}})-{\rho }_{0}(\omega )\\ & = & {g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\left[\frac{1}{2}\coth \left(\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})-\alpha }{2}\right)-\frac{1}{2}\right]\\ & = & {g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\frac{\exp (\frac{-\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})+\alpha }{2})}{2\sinh (\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})-\alpha }{2})}\\ & \approx & {g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\exp (-\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})+\alpha )\\ & = & {g}_{s}{{\rm{e}}}^{\alpha }\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\exp (-{\beta }_{\mu }{P}^{\mu }).\end{array}\end{eqnarray}$
Here we have used the approximation $\begin{eqnarray*}2\sinh \left(\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})-\alpha }{2}\right)\approx \exp \left(\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})-\alpha }{2}\right),\end{eqnarray*}$
because $\begin{eqnarray*}\exp \left(\frac{-\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})+\alpha }{2}\right)\approx 0.\end{eqnarray*}$
This classical approximation requires that the density of particles is low enough that we can ignore the exchange interactions of identical bosons. In this limit, the Jüttner–Synge distribution is recovered.The non-relativistic condition requires that ∣v∣ ≪ 1, ∣k∣ ≪ m and β ≫ 1/m. We can then perform the Taylor expansion
$\begin{eqnarray}\begin{array}{rcl}\omega & = & \sqrt{{{\boldsymbol{k}}}^{2}+{m}^{2}}\approx m+\frac{{{\boldsymbol{k}}}^{2}}{2m},\\ \gamma & = & \frac{1}{\sqrt{1-{{\boldsymbol{v}}}^{2}}}\approx 1+\frac{{{\boldsymbol{v}}}^{2}}{2}.\end{array}\end{eqnarray}$
Notice that ${\boldsymbol{k}}=\hat{{\boldsymbol{k}}}\sqrt{{\omega }^{2}-{m}^{2}}$ is an exact result. We obtain $\begin{eqnarray}\begin{array}{rcl}\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}}) & \approx & \beta m+\beta \left(\frac{1}{2}m{{\boldsymbol{v}}}^{2}+\frac{{{\boldsymbol{k}}}^{2}}{2m}+{\boldsymbol{v}}\cdot {\boldsymbol{k}}\right)\\ & = & \beta \left(m+\frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}\right).\end{array}\end{eqnarray}$
With both the classical limit and the non-relativistic expansion, we have $\begin{eqnarray}\begin{array}{rcl} & & {g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\exp (-\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})+\alpha )\\ & \approx & {g}_{s}{{\rm{e}}}^{-\beta m+\alpha }\frac{({m}^{2}+{{\boldsymbol{k}}}^{2})| {\boldsymbol{k}}| }{{(2\pi )}^{3}}{{\rm{e}}}^{-\beta \frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}}.\end{array}\end{eqnarray}$
The relativistic chemical potential already includes the rest energy mc2. If we hope to recover the non-relativistic results, we may redefine the non-relativistic chemical potential as ${\tilde{\mu }}_{nr}=\tilde{\mu }-m$, then $\begin{eqnarray*}{{\rm{e}}}^{-\beta m+\alpha }={{\rm{e}}}^{\beta (\tilde{\mu }-m)}={{\rm{e}}}^{\beta {\tilde{\mu }}_{nr}}.\end{eqnarray*}$
In classical mechanics, the spin degrees of freedom can be ignored so gs = 1, and the classical non-relativistic spectrum finally becomes $\begin{eqnarray}\begin{array}{rcl} & & \rho (\omega ,\hat{{\boldsymbol{k}}})-{\rho }_{0}(\omega )\\ & \approx & {g}_{s}{{\rm{e}}}^{-\beta m+\alpha }\frac{({m}^{2}+{{\boldsymbol{k}}}^{2})| {\boldsymbol{k}}| }{{(2\pi )}^{3}}{{\rm{e}}}^{-\beta \frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}}\\ & = & \frac{({m}^{2}+{{\boldsymbol{k}}}^{2})| {\boldsymbol{k}}| }{{(2\pi )}^{3}}{{\rm{e}}}^{\beta \left({\tilde{\mu }}_{nr}-\frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}\right)},\end{array}\end{eqnarray}$
which is exactly a shifted Maxwellian distribution with a center of mass velocity −v.For the fermions, we observe the shift of the Fermi surface in the zero-temperature and non-relativistic limit. Denoting the Heaviside step function as θ(x), then
$\begin{eqnarray}\begin{array}{rcl} & & \rho (\omega ,\hat{{\boldsymbol{k}}})-{\rho }_{0}(\omega )\\ & = & {g}_{s}\frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{{(2\pi )}^{3}}\left[-\frac{1}{2}\tanh \left(\frac{\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}})-\alpha }{2}\right)+\frac{1}{2}\right]\\ & \approx & {g}_{s}\frac{({m}^{2}+{{\boldsymbol{k}}}^{2})| {\boldsymbol{k}}| }{{(2\pi )}^{3}}\theta (\alpha -\gamma \beta (\omega +{\boldsymbol{v}}\cdot {\boldsymbol{k}}))\\ & \approx & {g}_{s}\frac{({m}^{2}+{{\boldsymbol{k}}}^{2})| {\boldsymbol{k}}| }{{(2\pi )}^{3}}\theta \left(\alpha -\beta m-\beta \frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}\right).\end{array}\end{eqnarray}$
Using the non-relativistic chemical potential ${\tilde{\mu }}_{nr}=\tilde{\mu }-m$, we express the step function as $\begin{eqnarray}\begin{array}{rcl}\theta (\alpha -\beta m-\beta \frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}) & = & \theta \left[\beta \left({\tilde{\mu }}_{nr}-\frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}\right)\right]\\ & = & \theta \left({\tilde{\mu }}_{nr}-\frac{{({\boldsymbol{k}}+m{\boldsymbol{v}})}^{2}}{2m}\right),\end{array}\end{eqnarray}$
from which we find that the non-relativistic chemical potential in the low-temperature limit is exactly the Fermi energy ϵF, and the Fermi surface is shifted by a constant −mv in the momentum space due to the relative motion between the observer and the system. The non-relativistic analysis proves that the inverse temperature four-vector is consistent with non-relativistic statistical mechanics.3.4. Equilibrium state with conserved charge
The field system of interest may have some conserved charges. In this circumstance, the field will not thermalize to the full Hilbert space but within a small section with a certain value of charge. In this case, we should introduce a generalized chemical potential $\tilde{\alpha }$ as a Lagrange multiplier for the charge restriction. For example, the complex scalar field φ, with Lagrangian density
$\begin{eqnarray}{ \mathcal L }={\partial }_{\mu }{\phi }^{\dagger }{\partial }^{\mu }\phi +{m}^{2}{\phi }^{\dagger }\phi ,\end{eqnarray}$
has a globally conserved U(1) charge Q = ∫d3xj0(x), where $\begin{eqnarray}{j}^{\mu }={\rm{Im}}({\phi }^{\dagger }{\partial }^{\mu }\phi -\phi {\partial }^{\mu }{\phi }^{\dagger }),\end{eqnarray}$
so we may refine our definition of thermal state to be a generalized Gibbs form $\begin{eqnarray}\rho ({\beta }^{\mu },\tilde{\alpha })=\frac{{{\rm{e}}}^{-{\beta }^{\mu }{P}_{\mu }-\tilde{\alpha }Q}}{{\rm{t}}{\rm{r}}({{\rm{e}}}^{-{\beta }^{\mu }{P}_{\mu }-\tilde{\alpha }Q})}.\end{eqnarray}$
Obviously Q is invariant under the action of a Poincaré group. So to calculate the energy spectral density, we simply repeat the previous process and the result for a complex scalar field is $\begin{eqnarray}\begin{array}{rcl}\rho (\omega ,\hat{{\boldsymbol{k}}}) & = & \frac{{\omega }^{2}\sqrt{{\omega }^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\left[\coth \left(\frac{{\beta }_{\mu }{P}^{\mu }+\tilde{\alpha }}{2}\right)\right.\\ & & \left.+\,\coth \left(\frac{{\beta }_{\mu }{P}^{\mu }-\tilde{\alpha }}{2}\right)\right].\end{array}\end{eqnarray}$
The physics behind this result is clear: it is just the total energy of particles with the chemical potential $\tilde{\alpha }/\beta $ and antiparticles with the chemical potential $-\tilde{\alpha }/\beta $. This simple result is also comparable with the discussion of electron–positron equilibrium in [37].The particle number N and the four-momentum Pμ are just some of the conserved charges. For systems with higher symmetry we can simply introduce the corresponding Lagrange multipliers and express the thermal equilibrium state in the generalized Gibbs form. It is worth pointing out that the inverse temperature four-vector reflects the energy–momentum conservation of a relativistic system, which is valid whenever there is Poincaré group symmetry. Therefore the inverse temperature four-vector is always necessary in relativistic thermodynamics.
4. Summary
In this article, we derive the equilibrium spectrum distribution of either bosonic or fermionic massive free field perceived in an arbitrary inertial frame of reference. In the massless limit, our result recovers the main result of Ref. [30], and is consistent with observations [25, 26, 40]. However, our study is generalized to both bosonic and fermionic massive fields. Furthermore, the analytical expression of the spectrum proves that for a massive field system an effective scalar temperature description is insufficient, and the perfect black-body spectral density of CMB is nothing more than a coincidence due to the massless nature of photons. Therefore, it is necessary to introduce the inverse temperature four-vector to characterize the thermal equilibrium state, and the spectral density based on the inverse temperature four-vector is sufficient to provide a covariant description of the equilibrium state perceived from an arbitrary inertial frame of reference. Our results might be applied to realistic systems such as the CνB [35], which is typically considered as a relativistic massive free field. If the rest mass of neutrinos is non-negligible, the spectrum of CνB would be quite different from the CMB's: the spectrum cannot be characterized by any scalar temperature with dipole anisotropy, and the inverse temperature four-vector description of the CνB spectrum would be a necessity.
Appendix Calculation of a general free field thermal spectrum
Here we consider a general free field ψ, either bosonic or fermionic, with the quadratic Hamiltonian
$\begin{eqnarray}H=\displaystyle \sum _{s}\int \frac{{{\rm{d}}}^{3}{\boldsymbol{k}}}{{(2\pi )}^{3}2{\omega }_{{\boldsymbol{k}}}}\frac{{\omega }_{{\boldsymbol{k}}}}{2}(\sigma {b}_{ks}{b}_{ks}^{\dagger }+{b}_{ks}^{\dagger }{b}_{ks}).\end{eqnarray}$
Here bks and ${b}_{ks}^{\dagger }$ denote the creation and annihilation operators for a particle with four-momentum k and spin s, and σ denotes the fermionic parity of bks. The (anti-)commutation relation is ${[{b}_{ks},{b}_{{k}^{{\prime} }{s}^{{\prime} }}^{\dagger }]}_{\sigma }={(2\pi )}^{3}2{\omega }_{k}\delta (k-{k}^{{\prime} }){\delta }_{s{s}^{{\prime} }}$.If we still define the density operator of the thermal state in frame i to be ${\rho }_{i}=\frac{{{\rm{e}}}^{-\beta H+\alpha N}}{{\rm{Tr}}({{\rm{e}}}^{-\beta H+\alpha N})}$ then the expectation values of the following operators in frame i are 9 )
$\begin{eqnarray}\begin{array}{l}{\langle \sigma {b}_{ks}{b}_{{k}^{{\prime} }{s}^{{\prime} }}^{\dagger }+{b}_{ks}^{\dagger }{b}_{{k}^{{\prime} }{s}^{{\prime} }}\rangle }_{i}={\delta }^{* }(k-{k}^{{\prime} }){f}_{\sigma }(\frac{\beta {\omega }_{k}-\alpha }{2}){\delta }_{s{s}^{{\prime} }},\\ \qquad {\langle {b}_{ks}{b}_{{k}^{{\prime} }{s}^{{\prime} }}\rangle }_{i}={\langle {b}_{ks}^{\dagger }{b}_{{k}^{{\prime} }{s}^{{\prime} }}^{\dagger }\rangle }_{i}=0,\end{array}\end{eqnarray}$
where ${f}_{+1}(x):= \coth (x)$ and ${f}_{-1}(x):= -\tanh (x)$. Correspondingly, if we perform the Lorentz transformation from frame i to frame ${i}^{{\prime} }$, then the thermal state expectation value becomes $\begin{eqnarray}\begin{array}{rcl} & & {\langle \sigma {b}_{ks}{b}_{k^{\prime} s^{\prime} }^{\dagger }+{b}_{ks}^{\dagger }{b}_{k^{\prime} s^{\prime} }\rangle }_{i^{\prime} }\\ & = & \displaystyle \sum _{s^{\prime\prime} }\langle \sigma {D}_{ss^{\prime\prime} }^{* }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){b}_{{{\rm{\Lambda }}}^{-1}ks^{\prime\prime} }{D}_{s^{\prime} s^{\prime\prime} }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){b}_{k^{\prime} s^{\prime\prime} }^{\dagger }\\ & & +\,{D}_{ss^{\prime\prime} }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){b}_{ks}^{\dagger }{D}_{s^{\prime} s^{\prime\prime} }^{* }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){b}_{k^{\prime} s^{\prime} }{\rangle }_{i}\\ & = & {\delta }^{* }(k-k^{\prime} ){f}_{\sigma }(\frac{\beta {\omega }_{k}-\alpha }{2}){\delta }_{ss^{\prime} },\\ & & {\langle {b}_{ks}{b}_{k^{\prime} s^{\prime} }\rangle }_{i^{\prime} }={\langle {b}_{ks}^{\dagger }{b}_{k^{\prime} s^{\prime} }^{\dagger }\rangle }_{i^{\prime} }=0,\end{array}\end{eqnarray}$
where Dss″(Λ, Λ−1k) is the representation of a Wigner rotation of spin, satisfying $\begin{eqnarray*}\begin{array}{rcl} & & \displaystyle \sum _{s^{\prime\prime} }{D}_{ss^{\prime\prime} }^{* }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){D}_{s^{\prime} s^{\prime\prime} }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k)\\ & = & \displaystyle \sum _{s^{\prime\prime} }{D}_{ss^{\prime\prime} }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k){D}_{s^{\prime} s^{\prime\prime} }^{* }({\rm{\Lambda }},{{\rm{\Lambda }}}^{-1}k)={\delta }_{ss^{\prime} }.\end{array}\end{eqnarray*}$
Then we can calculate the energy density in a similar way to equation ($\begin{eqnarray}\begin{array}{rcl}{\rho }_{E}^{{\prime} } & = & \displaystyle \sum _{s}\displaystyle \int D{k}^{{\prime} }\displaystyle \int D{k}^{{\prime\prime} }\frac{1}{2}({\omega }_{{k}^{{\prime} }}{\omega }_{{k}^{{\prime\prime} }}+{{\boldsymbol{k}}}^{{\prime} }\cdot {{\boldsymbol{k}}}^{{\prime\prime} }+{m}^{2})\\ & & \times \,\left[{\delta }^{* }({{\rm{\Lambda }}}^{-1}{k}^{{\prime} }-{{\rm{\Lambda }}}^{-1}{k}^{{\prime\prime} }){f}_{\sigma }\left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2}\right)\right]\\ & = & {g}_{s}\displaystyle \int D{k}^{{\prime} }\displaystyle \int D{k}^{{\prime\prime} }\frac{1}{2}({\omega }_{{k}^{{\prime} }}{\omega }_{{k}^{{\prime\prime} }}+{{\boldsymbol{k}}}^{{\prime} }\cdot {{\boldsymbol{k}}}^{{\prime\prime} }+{m}^{2})\\ & & \times \,[{\delta }^{* }({k}^{{\prime} }-{k}^{{\prime} }){f}_{\sigma }(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2})]\\ & = & {g}_{s}\displaystyle \int D{k}^{{\prime} }{\omega }_{{k}^{{\prime} }}^{2}{f}_{\sigma }\left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2}\right)\\ & = & {g}_{s}\displaystyle \int \frac{{{\rm{d}}}^{3}{{\boldsymbol{k}}}^{{\prime} }}{(2{\pi }^{3})}\frac{{\omega }_{{k}^{{\prime} }}}{2}{f}_{\sigma }\left(\frac{\beta {\omega }_{{{\rm{\Lambda }}}^{-1}{k}^{{\prime} }}-\alpha }{2}\right).\end{array}\end{eqnarray}$
Therefore, the spectral distribution in frame ${i}^{{\prime} }$ is $\begin{eqnarray}\begin{array}{rcl}{\rho }^{{\prime} }({\omega }^{{\prime} },{\hat{{\boldsymbol{k}}}}^{{\prime} }) & = & {g}_{s}\frac{{\omega {}^{{\prime} }}^{2}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}}}{2{(2\pi )}^{3}}\\ & & \cdot {f}_{\sigma }\left(\frac{\beta \gamma ({\omega }^{{\prime} }+{\hat{{\boldsymbol{k}}}}^{{\prime} }\cdot {\boldsymbol{v}}\sqrt{{\omega {}^{{\prime} }}^{2}-{m}^{2}})-\alpha }{2}\right),\end{array}\end{eqnarray}$
where gs accounts for the spin degeneracy.