Black-Body Radiation in an Accelerated Frame

Let $\left(\mathcal{M},\mathbf{x}\right)$ be a Minkowski space equipped with a metric $\eta$. The (global) coordinate $\mathbf{x}$ =$\left(x,y,z,t\right)$, with $t:=c\mathfrak{t}$ is the time coordinate and $\left(x,y,z\right)$ is the spatial part, is used to parametrize $\mathcal{M}$. In this coordinate, the metric $\eta$ is a diagonal metric with signature $\left(-1,1,1,1\right)$. Naturally, one could attach an inertial reference of frame to the coordinate $\mathbf{x}$, and let $\mathcal{O}$ be an observer at rest with respect to this inertial frame. Let us call this frame/observer as $\mathcal{O}.$

Suppose we have another observer $\mathcal{A}$ that moves with a constant acceleration of magnitude $a$ in the direction of the $x$-axis, with respect to $\mathcal{O}$. The trajectory of $\mathcal{A}$ in $\mathcal{M}$ is $\mathbf{x}\left[\lambda\right]=x^{\mu}\left[\lambda\right]$, with $\lambda$ is a parameter (a proper time with respect to $\mathcal{A}$), satisfying:

	$\displaystyle t\left[\lambda\right]$	$\displaystyle=\frac{1}{a}\sinh a\lambda,$		(1)
	$\displaystyle x\left[\lambda\right]$	$\displaystyle=\frac{1}{a}\cosh a\lambda,$		(2)

and $y,z$ being constant. The 4-acceleration is $a^{\mu}=\frac{d^{2}x^{\mu}}{d\lambda^{2}},$ where the non-zero components are:

	$\displaystyle a^{t}$	$\displaystyle=a\sinh a\lambda,$
	$\displaystyle a^{x}$	$\displaystyle=a\cosh a\lambda,$

such that the magnitude of $a^{\mu}$ is $\sqrt{a^{\mu}a_{\mu}}=a$. Note here that $a$ is positive definite: $a\geq 0$. From (1) and (2), the trajectory of $\mathcal{A}$ satisfies $x^{2}-t^{2}=a^{-2}$, describing a collection of hyperboloids with asymptotes at the lines $x=-t$ and $x=t$. The asymptotes divide $\mathcal{M}$ into 4 regions, and let us focus on one of the region where $0<x<\infty$ and $-x<t<x$, usually labeled as the (right) Rindler wedge, see FIG. 1.

One could choose a new coordinate patch $\bar{\mathbf{x}}$ =$\left(\xi,\bar{y},\bar{z},\tau\right)$, to parametrize the Rindler wedge as follows:

$$\begin{array}[]{cc}\tau=&\frac{1}{2\alpha}\left(\ln\alpha\left(x+t\right)-\ln\alpha\left(x-t\right)-2\varsigma\right),\\ \xi=&\frac{1}{2\alpha}\left(\ln\alpha\left(x+t\right)+\ln\alpha\left(x-t\right)\right),\end{array}$$

(3)

and their inverse:

$$\begin{array}[]{cc}t&=\frac{1}{\alpha}e^{\alpha\xi}\sinh\left(\alpha\tau+\varsigma\right),\\ x&=\frac{1}{\alpha}e^{\alpha\xi}\cosh\left(\alpha\tau+\varsigma\right),\end{array}$$

(4)

with $\bar{y}=y$ , $\bar{z}=z$, and $\alpha,$ $\varsigma$ are constant parameters. The (right) Rindler wedge is covered by the coordinate patch with $-\infty<\xi,\tau<\infty$. For $\varsigma=0$, the coordinate is usually known as the Rindler coordinates in the Lass/radar representation (Carrol, ). In this coordinate, (1)-(2) becomes:

	$\displaystyle\tau\left[\lambda\right]$	$\displaystyle=\frac{1}{\alpha}\left(a\lambda-\varsigma\right),$
	$\displaystyle\xi\left[\lambda\right]$	$\displaystyle=\frac{1}{\alpha}\ln\frac{\alpha}{a}.$

Notice that along the trajectory $\mathbf{x}\left[\lambda\right]$, $\xi$ is constant and $\tau$ is proportional to the proper time $\lambda$, although it is shifted by a constant amount $-\frac{\varsigma}{\alpha}$. An observer moving along constant $\xi$ will experience an acceleration of magnitude:

$$a=\alpha e^{-\alpha\xi}.$$

(5)

Since $a$ is positive definite, then so does $\alpha$: $\alpha\geq 0$. Naturally, the coordinate $\bar{\mathbf{x}}$ is adopted by the observer/frame $\mathcal{A}$.

One can obtain the infinitesimal transformation of (3), which is:

	$\displaystyle d\tau$	$\displaystyle=\frac{xdt-tdx}{\alpha\left(x^{2}-t^{2}\right)}=e^{-\alpha\xi}\left(\cosh\bar{\varsigma}dt-\sinh\bar{\varsigma}dx\right),$		(6)
	$\displaystyle d\xi$	$\displaystyle=\frac{xdx-tdt}{\alpha\left(x^{2}-t^{2}\right)}=e^{-\alpha\xi}\left(\cosh\bar{\varsigma}dx-\sinh\bar{\varsigma}dt\right),$

where we write $\bar{\varsigma}=\bar{\varsigma}\left[\tau\right]=\alpha\tau+\varsigma$, and their inverses:

	$\displaystyle dt$	$\displaystyle=e^{\alpha\xi}\left(\cosh\bar{\varsigma}d\tau+\sinh\bar{\varsigma}d\xi\right)=\alpha\left(xd\tau+td\xi\right),$
	$\displaystyle dx$	$\displaystyle=e^{\alpha\xi}\left(\cosh\bar{\varsigma}d\xi+\sinh\bar{\varsigma}d\tau\right)=\alpha\left(xd\xi+td\tau\right),$		(7)

with $d\bar{y}=dy$ and $d\bar{z}=dz$. $\bar{\varsigma}$ is the time-dependent rapidity that is linear to the “proper time” $\tau$. For $\alpha=0$, the acceleration $a$ vanishes, and (6)-(7) will return to the standard Lorentz transformation (81)-(82).

The line element in Minkowksi space could be written in Rindler coordinate using (7) as follows:

$$ds^{2}=g_{\mu\nu}d\bar{x}^{\mu}d\bar{x}^{\nu}=e^{2\alpha\xi}\left(-d\tau^{2}+d\xi^{2}\right)+d\bar{y}^{2}+d\bar{z}^{2};$$

(8)

The coefficient of the line element gives the Rindler metric $g_{\mu\nu}$ in Lass representation. One could also consider the left Rindler wedge (the region IV in FIG. 1), where it could be covered by a coordinate patch similar to the RRW, but with the flip in the sign of (4) as follows:

$$\begin{array}[]{cc}t&=-\frac{1}{\alpha}e^{\alpha\xi}\sinh\left(\alpha\tau+\varsigma\right),\\ x&=-\frac{1}{\alpha}e^{\alpha\xi}\cosh\left(\alpha\tau+\varsigma\right).\end{array}$$

(9)

In this region, the future-directed time-like Killing vector is $-\partial\tau$, instead of $\partial\tau$ in region I.

Suppose with respect to an inertial frame $\mathcal{O}$ we have an object moving with a constant velocity $\mathbf{v}$ as follows:

$$\mathbf{v}=\left(v_{x},v_{y},v_{z}\right)=\left(\frac{dx}{dt},\frac{dy}{dt},\frac{dz}{dt}\right).$$

According to the accelerated frame $\mathcal{A}$, the velocity of the object is:

$$\bar{\mathbf{v}}\left[\tau,\xi\right]=\left(v_{\xi},\bar{v}_{y},\bar{v}_{z}\right)=\left(\frac{d\xi}{d\tau},\frac{d\bar{y}}{d\tau},\frac{d\bar{z}}{d\tau}\right).$$

Notice that the components of $\bar{\mathbf{v}}\left[\tau,\xi\right]$ are functions of the Rindler coordinate $\left(\tau,\xi\right)$. Using the transformation (6)-(7), one can obtain the relation between the instantaneous $\bar{\mathbf{v}}\left[\tau,\xi\right]$ and $\mathbf{v}$ as follows:

$$v_{\xi}=\frac{v_{x}-\tanh\bar{\varsigma}}{1-v_{x}\tanh\bar{\varsigma}}=\frac{\left(xv_{x}-t\right)}{\left(x-tv_{x}\right)},$$

(10)

$$\bar{v}_{y,z}=\frac{e^{\alpha\xi}v_{y,z}}{\cosh\bar{\varsigma}\left(1-v_{x}\tanh\bar{\varsigma}\right)}=\frac{\alpha\left(x^{2}-t^{2}\right)v_{y,z}}{\left(x-tv_{x}\right)}.$$

(11)

Relation (10)-(11) could be derived from the standard velocity addition formula (or, similarly, from the relativistic acceleration formula). Their inverses could also be obtained as follows:

$$v_{x}=\frac{v_{\xi}+\tanh\bar{\varsigma}}{1+v_{\xi}\tanh\bar{\varsigma}},$$

$$v_{y,z}=\frac{\bar{v}_{y,z}}{e^{\alpha\xi}\cosh\bar{\varsigma}\left(1+v_{\xi}\tanh\bar{\varsigma}\right)}.$$

In the following paragraphs, we will derive the light aberration formula for an accelerated observer $\mathcal{A}$. For a brief review on the light aberration formula for an inertial, moving observer, one could refer to Appendix B, or consult (Johnson, ). Let us consider a black-body source (photons inside a cavity) at rest with respect to an inertial frame $\mathcal{O}.$ Electromagnetic radiations (photons) are emitted by the black-body with propagation velocity $c$ in the direction of $\hat{n}=\left(\cos\theta,\sin\theta,0\right)$, with $\theta$ is the (polar) observation (or inclination) angle between the axis $x$ and $\hat{n}$ on plane $dx\wedge dy$. The velocity of the photon with respect to frame $\mathcal{O}$ is:

$$\mathbf{v}=\left(v_{x},v_{y},v_{z}\right)=c\hat{n}=\left(c\cos\theta,c\sin\theta,0\right).$$

(12)

The 4-velocity of the photon satisfies the null-vector condition, see Appendix X. To derive the 3-velocity in $\mathcal{A}$, let us first write the 4-velocity in Rindler coordinate:

$$\mathbf{u}=\bar{\mathbf{u}}=\frac{\partial\tau}{\partial\lambda}\left(1,\bar{\mathbf{v}}\right),$$

where $\lambda$ is a real parameter (usually the proper time with respect to the moving object) and $\bar{\mathbf{v}}$ is the 3-velocity in $\mathcal{A}$:

$$\bar{\mathbf{v}}=\left(v_{\xi},\bar{v}_{y},\bar{v}_{z}\right)=\left(\frac{\partial\xi}{\partial\tau},\frac{\partial\bar{y}}{\partial\tau},\frac{\partial\bar{z}}{\partial\tau}\right).$$

Using the null-vector condition for light, we have:

$$\bar{u}^{\mu}\bar{u}_{\mu}=g_{\mu\nu}\bar{u}^{\mu}\bar{u}^{\nu}=\left(\frac{\partial\tau}{\partial\lambda}\right)^{2}\left(-e^{2\alpha\xi}+e^{2\alpha\xi}v_{\xi}^{2}+\bar{v}_{y}^{2}+\bar{v}_{z}^{2}\right)=0.$$

In $\mathcal{O}$, the light only propagates in the direction $\hat{n}$ on plane $dx\wedge dy$, so even in $\mathcal{A}$, the $\bar{z}$ component of the velocity of light is zero. The null condition then becomes:

$$v_{\xi}^{2}+\left(e^{-\alpha\xi}\bar{v}_{y}\right)^{2}=1.$$

Writing $v_{\xi}^{2}=\cos^{2}\bar{\theta}$ and $\left(e^{-\alpha\xi}\bar{v}_{y}\right)^{2}=\sin^{2}\bar{\theta},$ the 3-velocity of the photon in $\mathcal{A}$ is:

$$\bar{\mathbf{v}}=\left(v_{\xi},\bar{v}_{y},\bar{v}_{z}\right)=c\hat{\bar{n}}=\left(c\cos\bar{\theta},ce^{\alpha\xi}\sin\bar{\theta},0\right),$$

(13)

with $\hat{\bar{n}}$ and $\bar{\theta}$ are the propagation direction and the inclination angle of the photon at $\mathcal{A}$, respectively. Notice that since the observer $\mathcal{A}$ is accelerated in the direction $+x$, it will receive only photons that are moving towards the observer, namely, the ones that have the $+x$ component in the velocity. The relation between $\bar{\mathbf{v}}$ and $\mathbf{v}$ can be obtained using the velocity addition formula in the accelerated frame (10)-(11). Inserting (12)-(13) to (10)-(11) will result in the light aberration for an accelerated frame $\mathcal{A}$:

	$\displaystyle\cos\bar{\theta}$	$\displaystyle=\frac{\cos\theta-\tanh\bar{\varsigma}}{1-\cos\theta\tanh\bar{\varsigma}},$		(14)
	$\displaystyle\sin\bar{\theta}$	$\displaystyle=\frac{\sin\theta}{\cosh\bar{\varsigma}\left(1-\cos\theta\tanh\bar{\varsigma}\right)},$		(15)
	$\displaystyle\tan\frac{\bar{\theta}}{2}$	$\displaystyle=\frac{1}{\cosh\bar{\varsigma}\left(1-\tanh\bar{\varsigma}\right)}\tan\frac{\theta}{2},$		(16)

where the last equation is obtained from the trigonometry identity $\tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}$. One could also obtained their inverses:

	$\displaystyle\cos\theta$	$\displaystyle=\frac{\cos\bar{\theta}+\tanh\bar{\varsigma}}{1+\cos\bar{\theta}\tanh\bar{\varsigma}},$		(17)
	$\displaystyle\sin\theta$	$\displaystyle=\frac{\sin\bar{\theta}}{\cosh\bar{\varsigma}\left(1+\cos\bar{\theta}\tanh\bar{\varsigma}\right)},$		(18)
	$\displaystyle\tan\frac{\theta}{2}$	$\displaystyle=\frac{1}{\cosh\bar{\varsigma}\left(1+\tanh\bar{\varsigma}\right)}\tan\frac{\bar{\theta}}{2}.$		(19)

The aberration formula in $\mathcal{A}$ has exactly the same form with the ones in a constant moving frame $\mathcal{O}^{\prime}$ (see relation (89)-(91)), however, it should be kept in mind that $\bar{\theta}$ is time-dependent, namely $\bar{\theta}=\bar{\theta}\left[\tau\right]$, while $\theta^{\prime}$ is constant.

In this paper we will only consider the longitudinal relativistic Doppler effect (LongitudDoppler, ), where the observer/source velocity has component parallel to the wave propagation. To derive this, let us first consider the 4-momentum constraint (107) which is valid for any reference frame. The photon is massless, so using the de Broglie postulate $E=\hbar\omega$ and $\mathbf{p}=\hbar\mathbf{k}$, (107) becomes the dispersion relation:

$$\omega^{2}=\left|\mathbf{k}\right|^{2},$$

(20)

for the wave vector of the photon satisfies $\mathbf{k}=\frac{2\pi\hat{n}}{\lambda}.$ In $\mathcal{O},$ the dispersion relation becomes:

$$\omega^{2}=k_{x}^{2}+k_{y}^{2}.$$

Since the coordinate axis $x$ is perpendicular to $y$, one could write $k_{x}=\omega\cos\theta$ and $k_{y}=\omega\sin\theta$, with $\theta$ is the inclination angle at $\mathcal{O}.$ Therefore, the 4-momentum of a photon with frequency $\omega$ according to the rest frame $\mathcal{O}$ (with respect to the black-body source) is $\mathbf{p}=\left(p_{t},p_{x},p_{y},p_{z}\right)$, satisfying:

$$\mathbf{p}=\frac{\hbar\omega}{c}\left(1,\hat{n}\right)=\frac{\hbar\omega}{c}\left(1,\cos\theta,\sin\theta,0\right),$$

(21)

Now, let us obtain the photon’s 4-momentum according to the accelerated frame $\mathcal{A}$, that is $\bar{\mathbf{p}}=\left(p_{\tau},p_{\xi},\bar{p}_{y},\bar{p}_{z}\right).$ The dipersion relation (20) is also satisfied in $\mathcal{A}$, however, written in different coordinate, it becomes:

$$e^{2\alpha\xi}\bar{\omega}^{2}=e^{2\alpha\xi}k_{\xi}^{2}+k_{\bar{y}}^{2},$$

(22)

(notice that the dispersion relation comes from $p_{\mu}p^{\mu}=0,$ hence the metric components needs to be taken accounted). Since the coordinate axis $\xi$ and $\bar{y}$ are also perpendicular to one another, one could defined that $k_{\xi}=\bar{\omega}\cos\bar{\theta}$ and $k_{\bar{y}}=\bar{\omega}e^{\alpha\xi}\sin\bar{\theta}$. Notice that $\bar{\theta}$ is the inclination angle according to $\mathcal{A}$. With this, then the 4-momentum at the accelerated frame is:

$$\bar{\mathbf{p}}=\hbar\bar{\omega}\left(1,\hat{\bar{n}}\left[\xi,\tau\right]\right)=\hbar\bar{\omega}\left(1,\cos\bar{\theta},e^{\alpha\xi}\sin\bar{\theta},0\right).$$

(23)

$\bar{\omega}$ is the frequency of the photon, observed by $\mathcal{A}$. Here, we assume that de Broglie postulate is still valid in $\mathcal{A},$ namely the relation $\bar{\mathbf{p}}\left[\tau,\xi\right]=\hbar\bar{\mathbf{k}}\left[\tau,\xi\right]$ is satisfied for every position in space and each instantaneous time. One could refer to Appendix B for a more detailed explanation on the photon’s momentum.

It needs to be kept in mind that there exists another solution to (20), namely: $\omega=-k,$ giving another possible value for $k_{x}=-\omega\cos\theta$ and $k_{y}=-\omega\sin\theta$ in $\mathcal{O}$ and $k_{\xi}=-\bar{\omega}\cos\bar{\theta}$ and $k_{\bar{y}}=-\bar{\omega}e^{\alpha\xi}\sin\bar{\theta}$ in $\mathcal{A}$. However, since we are considering only the right Rindler wedge (RRW), the reasonable case for an observer in RRW to receive the signal is by the waveform that moves to the right (in $+x$ direction). For an observer in the left Rindler wedge (LRW, which we will consider in Section IV), we should consider the case where $\omega=-k.$

The (generalized) momentum is naturally a covector, however, in this paper, we use its contravariant counterpart to be consistent with the 4-velocity (12)-(13). The entire results do not depends on the choice of vector/covector for the derivation. Since the 4-momentum is a 4-vector, the transformation between $\mathbf{p}$ and $\bar{\mathbf{p}}$ follows (6)-(7), where we write the cofficient of the transformation in terms of hyperbolic functions:

$$\begin{array}[]{cc}p_{\tau}&=e^{-\alpha\xi}\left(p_{t}\cosh\bar{\varsigma}-p_{x}\sinh\bar{\varsigma}\right),\\ p_{\xi}&=e^{-\alpha\xi}\left(p_{x}\cosh\bar{\varsigma}-p_{t}\sinh\bar{\varsigma}\right),\end{array}$$

(24)

with their inverses:

$$\begin{array}[]{cc}p_{t}&=e^{\alpha\xi}\left(p_{\tau}\cosh\bar{\varsigma}+p_{\xi}\sinh\bar{\varsigma}\right),\\ p_{x}&=e^{\alpha\xi}\left(p_{\xi}\cosh\bar{\varsigma}+p_{\tau}\sinh\bar{\varsigma}\right),\end{array}$$

(25)

$\bar{p}_{y}=p_{y}$ and $\bar{p}_{z}=p_{z}.$ Moreover, by inserting (21)-(23) to the transformation (24), we can obtain the 3 equivalent forms of the relativistic Doppler effect in the accelerated frame as follows:

	$\displaystyle\frac{\bar{\omega}}{\omega}$	$\displaystyle=e^{-\alpha\xi}\left(\cosh\bar{\varsigma}-\cos\theta\sinh\bar{\varsigma}\right),$		(26)
		$\displaystyle=e^{-\alpha\xi}\frac{\left(\cos\theta\cosh\bar{\varsigma}-\sinh\bar{\varsigma}\right)}{\cos\bar{\theta}},$		(27)
		$\displaystyle=\frac{\sin\theta}{\sin\bar{\theta}}\,e^{-\alpha\xi},$		(28)

together with their inverses:

	$\displaystyle\frac{\omega}{\bar{\omega}}$	$\displaystyle=e^{\alpha\xi}\left(\cosh\bar{\varsigma}+\cos\bar{\theta}\sinh\bar{\varsigma}\right),$		(29)
		$\displaystyle=e^{\alpha\xi}\frac{\left(\cos\bar{\theta}\cosh\bar{\varsigma}+\sinh\bar{\varsigma}\right)}{\cos\theta}.$

Notice that $\bar{\omega}=\bar{\omega}\left[\tau,\xi\right]$ is a function of time and position in the Rindler coordinate. From (26) and (29), one could retrieve the aberration formula (14)-(18).

Let us consider one of the light aberation formula (17). Differentiating (17) with respect to any parameter $\lambda$ give:

$$-\sin\theta\frac{d\theta}{d\lambda}=\frac{\sin\bar{\theta}\left(-\frac{d\bar{\theta}}{d\lambda}+\alpha\sin\bar{\theta}\frac{d\tau}{d\lambda}\right)}{\left(\cosh\bar{\varsigma}+\cos\bar{\theta}\sinh\bar{\varsigma}\right)^{2}},$$

while using (18), (28), and the chain rule $\frac{d\tau}{d\lambda}=\frac{d\tau}{d\bar{\theta}}\frac{d\bar{\theta}}{d\lambda}$, gives:

$$\frac{d\theta}{d\bar{\theta}}=\frac{\bar{\omega}}{\omega}e^{\alpha\xi}\left(1-\frac{\alpha}{\dot{\bar{\theta}}}\sin\bar{\theta}\right),$$

(30)

with $\dot{\bar{\theta}}=\frac{d\bar{\theta}}{d\tau}$. $\dot{\bar{\theta}}$ is the rate of change of the inclination angle $\bar{\theta}$ as observed by the accerelated observer $\mathcal{A}$. In $\mathcal{A}$, the inclination angle $\bar{\theta}$ is time-dependent, satisfying equation (14); differentiating (14) with respect to $\tau$, we obtain:

$$\dot{\bar{\theta}}=\frac{d\bar{\theta}}{d\tau}=\sin\bar{\theta},$$

(31)

and hence (30) can be simplified into:

$$\frac{d\theta}{d\bar{\theta}}=\frac{\bar{\omega}}{\omega}e^{\alpha\xi}\left(1-\alpha\right).$$

(32)

Using the transformation of the inclination (polar) angle (30), one could obtain the transformation for the solid angle $d\Omega=\sin\theta d\theta d\phi$ as follows:

$$\frac{d\Omega}{d\bar{\Omega}}=\left(\frac{\bar{\omega}}{\omega}\right)^{2}e^{2\alpha\xi}\left(1-\alpha\right),$$

(33)

where we use the Doppler effect (28) and the fact that the transformation of the azimuth angle satisfies $d\phi=d\bar{\phi}$.

As explained in the preceeding subsection, observer $\mathcal{A}$ is accelerated along the $+x$ direction; causing it to receive only photons approaching from that direction, namely, those with the $+x$ component in their velocity. Moreover, $\mathcal{A}$ is moving away from the black-body source at rest with respect to $\mathcal{O}.$ As a consequence, the frequency of the photons received by the accelerated frame $\mathcal{A}$ is redshifted by equation (26). However, in constrast with the inertial case presented in Appendix B where the redshift remains constant, the redshift in the accelerated frame $\mathcal{A}$ increases in time, with the wavelength of the photons shift progressively towards the infrared range. To obtain the blueshift case, one could invert the situation by flipping the sign of $\tau$ (or $t$), effectively moving backward in time.

Another important parameter we must determined in the accelerated frame $\mathcal{A}$ is the modes distribution of the photons in the cavity. To determine how this quantity transform from the inertial to accelerated frame, first we need to determine transformation of the phase-space volume. In the inertial frame $\mathcal{O}$ and the accelerated frame $\mathcal{A}$, the infinitesimal 3-volume elements are defined as, respectively:

	$\displaystyle d^{3}\mathbf{x}$	$\displaystyle=dx\wedge dy\wedge dz,$		(34)
	$\displaystyle d^{3}\mathbf{\bar{x}}$	$\displaystyle=d\xi\wedge d\bar{y}\wedge d\bar{z}.$

To obtain the density of states of our black-body system, we need to calculate how much modes are inside a volume element. Let the finite volume element in $\mathcal{O}$ be $\Delta x\Delta y\Delta z$. The spatial length $\Delta x$ is defined by 2 simultaneous events $\mathbf{p}=\left(t_{i},x_{i}\right)$ and $\mathbf{q}=\left(t_{f},x_{f}\right)$ along the $x$-axis where $t_{i}=t_{f}=0$, $x_{i}=0$ and $x_{f}=\ell$. According to $\mathcal{O},$ the length between these 2 event is $\Delta x=x_{f}-x_{i}=\ell.$ For an observer at $\mathcal{A},$ whose accelerated in the $x$-direction with respect to $\mathcal{O},$ the 2 events $\mathbf{p}=\left(\tau_{i},\xi_{i}\right)$ and $\mathbf{q}=\left(\tau_{f},\xi_{f}\right)$ are not simultaneous, but are separated by a time interval $\Delta\tau=\tau_{f}-\tau_{i}$. Let us labeled the spatial length of these 2 events in $\mathcal{A}$ as $\Delta\xi=\xi_{f}-\xi{}_{i},$where $\tau_{f},\tau_{i},\xi{}_{i},\xi_{f},$ could be obtained from (3). Taking the infinitesimal limit $\Delta\xi\rightarrow d\xi$, the transformation of the infinitesimal spatial length satisfies (6). Notice that the measurement of length in $\mathcal{O}$ is obtained from 2 simultaneous event, hence in $\mathcal{O}$, $dt=0;$ this is not the case in $\mathcal{A}.$ Therefore, the (infinitesimal) spatial length of $\mathbf{p},\mathbf{q}$ in $\mathcal{A}$ is:

$$d\xi=e^{-\alpha\xi}\cosh\bar{\varsigma}\,dx=\frac{x}{\alpha\left(x^{2}-t^{2}\right)}\,dx.$$

(35)

Furthermore, the infinitesimal volume element $dx\wedge dy\wedge dz$ is perceived by an observer in $\mathcal{A}$ as the volume swept by the plane $d\bar{y}\wedge d\bar{z}=dy\wedge dz$ from $\xi{}_{i}$ to $\xi_{f},$ along an infinitesimal time interval $d\tau.$ This is the physical interpretation of $d^{3}\mathbf{\bar{x}}=d\xi\wedge d\bar{y}\wedge d\bar{z},$ see also discussion on Appendix B. Inserting (35) to (34) gives the volume transformation:

$$d^{3}\mathbf{\bar{x}}=e^{-\alpha\xi}\cosh\bar{\varsigma}\,d^{3}\mathbf{x}=\frac{x}{\alpha\left(x^{2}-t^{2}\right)}\,d^{3}\mathbf{x}.$$

(36)

For the next step, we need to determine the infinitesimal volume element in the momentum space, this could be obtained from (21)-(23); in the inertial frame $\mathcal{O}$ and the accelerated frame $\mathcal{A}$, they are, respectively:

	$\displaystyle d^{3}\mathbf{p}$	$\displaystyle=dp_{x}\wedge dp_{y}\wedge dp_{z},$		(37)
	$\displaystyle d^{3}\mathbf{\bar{p}}$	$\displaystyle=dp_{\xi}\wedge d\bar{p}_{y}\wedge d\bar{p}_{z}.$

The 4-momentum is subjected to the energy-momentum constraint, see Appendix B:

$$\left|{}^{4}\mathbf{p}\right|=p_{\mu}p^{\mu}=-p_{t}^{2}+p_{x}^{2}+p_{y}^{2}+p_{z}^{2}=m^{2}c^{2}.$$

(38)

Taking the differential of (38) gives:

$$dp_{t}=\frac{1}{p_{t}}\left(p_{x}dp_{x}+p_{y}dp_{y}+p_{z}dp_{z}\right).$$

(39)

Furthermore, using (4), one could show that:

$$x^{2}-t^{2}=\frac{1}{\alpha^{2}}e^{2\alpha\xi},$$

(40)

so that the transformation of the 4-momentum (24) can be written as follows:

	$\displaystyle p_{\tau}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left(xp_{t}-tp_{x}\right),$		(41)
	$\displaystyle p_{\xi}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left(xp_{x}-tp_{t}\right),$

with $\bar{p}_{y}=p_{y}$ and $\bar{p}_{z}=p_{z}.$ Differentiating (41) with any parameter will give the infinitesimal version of 4-momentum transformation as follows:

	$\displaystyle dp_{\tau}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left[\left(2t\left[\frac{xp_{t}-tp_{x}}{x^{2}-t^{2}}\right]-p_{x}\right)dt+\left(p_{t}-2x\left[\frac{xp_{t}-tp_{x}}{x^{2}-t^{2}}\right]\right)dx+xdp_{t}-tdp_{x}\right],$		(42)
	$\displaystyle dp_{\xi}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left[\left(2t\left[\frac{xp_{x}-tp_{t}}{x^{2}-t^{2}}\right]-p_{t}\right)dt+\left(p_{x}-2x\left[\frac{xp_{x}-tp_{t}}{x^{2}-t^{2}}\right]\right)dx+xdp_{x}-tdp_{t}\right].$

Note that in equation (42), there exist infinitesimal coordinates components, namely $dt$ and $dx$. This is due to the fact that the 4-momentum transformation from inertial to accelerated frame (41) varies with coordinates $\left(x,t\right)$, in constrast to the transformation between inertial frames (98) in Appendix B, which is independent from the temporal and spatial coordinates.

Similar to the inertial case in Appendix B, using the hypersurface constraint $dt=0$ and inserting the infinitesimal momentum constraint (39), (42) becomes:

	$\displaystyle dp_{\tau}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left[\left(p_{t}-2x\left[\frac{xp_{t}-tp_{x}}{x^{2}-t^{2}}\right]\right)dx+\left(x\frac{p_{x}}{p_{t}}-t\right)dp_{x}+x\frac{1}{p_{t}}\left(p_{y}dp_{y}+p_{z}dp_{z}\right)\right],$		(43)
	$\displaystyle dp_{\xi}$	$\displaystyle=\frac{1}{\alpha\left(x^{2}-t^{2}\right)}\left[\left(p_{x}-2x\left[\frac{xp_{x}-tp_{t}}{x^{2}-t^{2}}\right]\right)dx+\left(x-t\frac{p_{x}}{p_{t}}\right)dp_{x}-t\frac{1}{p_{t}}\left(p_{y}dp_{y}+p_{z}dp_{z}\right)\right],$

$d\bar{p}_{y}=dp_{y},$ and $d\bar{p}_{z}=dp_{z}$. Now, one can construct altogether the infinitesimal phase-space volume element in frame $\mathcal{A}$ as follows:

$$d^{3}\bar{\mathbf{x}}\wedge d^{3}\bar{\mathbf{p}}=d\xi\wedge d\bar{y}\wedge d\bar{z}\wedge dp_{\xi}\wedge d\bar{p}_{y}\wedge d\bar{p}_{z}.$$

Inserting (36) and (43) into the equation above, we obtain the transformation of the phase-space volume element from the inertial frame $\mathcal{O}$ to the accelerated frame $\mathcal{A}$:

$$d^{3}\bar{\mathbf{x}}\wedge d^{3}\bar{\mathbf{p}}=\frac{x\left(x-t\frac{p_{x}}{p_{t}}\right)}{\alpha^{2}\left(x^{2}-t^{2}\right)^{2}}\,d^{3}\mathbf{x}\wedge d^{3}\mathbf{p}.$$

(44)

Note that by the definition of the wedge product $\wedge,$ the terms containing equivalent components of differential forms, i.e. $dx\wedge dx$, vanishes. Using (40) and then (21), one could simplify (44) into:

$$\frac{d^{3}\bar{\mathbf{x}}d^{3}\bar{\mathbf{p}}}{d^{3}\mathbf{x}d^{3}\mathbf{p}}=\left(1-\frac{p_{x}}{p_{t}}\tanh\bar{\varsigma}\right)e^{-2\alpha\xi}\cosh^{2}\bar{\varsigma}.$$

(45)

One the other hand, we have, from (24):

$$\frac{p_{\tau}}{p_{t}}=\left(1-\frac{p_{x}}{p_{t}}\tanh\bar{\varsigma}\right)e^{-\alpha\xi}\cosh\bar{\varsigma},$$

and therefore, using (21)-(23):

$$\frac{d^{3}\bar{\mathbf{x}}d^{3}\bar{\mathbf{p}}}{d^{3}\mathbf{x}d^{3}\mathbf{p}}=\frac{p_{\tau}}{p_{t}}e^{-\alpha\xi}\cosh\bar{\varsigma}=\frac{\bar{\omega}}{\omega}e^{-\alpha\xi}\cosh\bar{\varsigma},$$

(46)

where we write the transformation (45) in terms of the Doppler factor $\frac{\bar{\omega}}{\omega}$. (46) is the relation between the phase-space volume element in $\mathcal{O}$ and $\mathcal{A}.$

Let us define the relativistic distribution function, or density of state $f\left(\mathbf{x},\mathbf{p}\right)$ as the number of world-lines that cross the phase-space, i.e, the states $dN$, per phase-space volume element (Liboff, ):

$$f\left(\mathbf{x},\mathbf{p}\right)=\frac{dN}{d^{3}\mathbf{x}d^{3}\mathbf{p}}.$$

(47)

We assume that the density of state $f\left(\mathbf{x},\mathbf{p}\right)$ is invariant under coordinate transformation, namely:

$$\frac{dN}{d^{3}\mathbf{x}d^{3}\mathbf{p}}=\frac{d\bar{N}}{d^{3}\mathbf{\bar{x}}d^{3}\mathbf{\bar{p}}},$$

with $d\bar{N}$ is the number of states per phase-space volume $d^{3}\mathbf{\bar{x}}\,d^{3}\mathbf{\bar{p}}$ of the accelerated frame $\mathcal{A}$. This is a reasonable assumption, and for a detailed explanation, one could refer to Appendix B. Some previous works in the existing literature have assume the invariance of the number of states $dN$ instead of $f\left(\mathbf{x},\mathbf{p}\right)$, however, this assumption is not suitable for our case because our treatment in this works relies only on a coordinate transformations at some part of the Minkowski space, i.e., the Rindler wedge. While these transformations leave the world-lines invariant, they alter the unit volume due to the coordinate transformation, leading to a different count of world-lines crossing the unit volume. Similar arguments are implicitly employed in (Peebles, ; Heer, ; Henry, ) as well.

By the invariance of the density of state under (infinitesimal) coordinate transformation, one could obtain the relation between the number of states/modes in inertial frame $\mathcal{O}$ and the accelerated frame $\mathcal{A}$ as follows:

$$\frac{d\bar{N}}{dN}=\frac{d^{3}\bar{\mathbf{x}}d^{3}\bar{\mathbf{p}}}{d^{3}\mathbf{x}d^{3}\mathbf{p}}=\frac{\bar{\omega}}{\omega}e^{-\alpha\xi}\cosh\bar{\varsigma}.$$

(48)

We already had all the Rindler-transformed variables to derive the Planck’s law in a uniformly-accelerated frame. Similar to the inertial frame case that we derived in Appendix B, in the rest frame $\mathcal{O}$, we use the Planck’s law version of equation (77) as follows:

$$dN=\frac{\omega^{2}}{4\pi^{3}c^{3}}\frac{1}{e^{\nicefrac{{w\omega}}{{\omega_{\mathrm{max}}}}}-1}d\omega d\Omega dV,$$

(49)

where the term containing the temperature of the black-body, i.e., $k_{B}T$, is replaced by $\omega_{\mathrm{max}}$, the angular frequency that maximize the radiation energy $E$ of the black-body. This is possible by the Wien’s displacement law, see a detailed explanation on the Appendix A. The reason for this substitution is to avoid the problem of the temperature in moving bodies. As explained in the Introduction, currently, there is no consensus regarding the transformation of temperature in moving bodies (Farias, ). However, we convincingly agree on the transformation of the black-body’s frequency, hence, if $\omega$ satisfies the relativistic Doppler effect, then so does the $\omega_{\mathrm{max}}$. The dimensionless constant $w$ in (49) is the Wien coefficient, obtained from solving the non-linear equation that arise from the maximization of energy with respect to $\omega$. One could refer to Appendix B for a detailed derivation.

In this section, our objective is to ascertain the validity of the distribution (49) in the accelerated frame $\mathcal{A}$. Inserting the relativistic Doppler effect (29), the transformation of solid angle (33), the volume contraction (36), and the transformation of the number of modes distribution (48) to (49), we obtain:

$$d\bar{N}=\left(1-\alpha\right)\frac{e^{2\alpha\xi}}{4\pi^{3}c^{3}}\frac{\bar{\omega}^{2}}{e^{\nicefrac{{w\bar{\omega}}}{{\overline{\omega_{\mathrm{max}}}}}}-1}d\bar{\omega}d\bar{\Omega}d\bar{V}.$$

(50)

One can observe that in the right hand side of (50) there is a scale factor:

$$\varepsilon\left[\alpha,\xi\right]=e^{2\alpha\xi}\left(1-\alpha\right),$$

(51)

that prevents the distribution (50) to be the original Planckian (49). The scale factor depends on the acceleration $\alpha$ and the spatial coordinate $\xi$. For the case with $\alpha=0$ where the acceleration vanishes, the distribution becomes Planckian as in (49). The fundamental distinction between the distribution (50) and the distribution (52) in the moving inertial frame lies in the fact that all the variables in the (50) are coordinate-dependent, namely, they vary with $\tau$ and $\xi$, which represent the proper time and position in $\mathcal{A}$. This is due to the fact that the transformations of these variables, namely (29), (33), (36), and (48) are coordinate-dependent. However, since we are considering all the possible value of frequencies from $0\leq\omega<\infty$, the time dependence of $\bar{\omega}$ does not explicitly shown in the law. Consequently, the ’black-body’ spectrum (50) in the accelerated frame $\mathcal{A}$ experiences a redshift that increase over time and varying with position, where the wavelength of the photon increases toward the infrared direction. This behavior distinguishes (50) from the black-body spectrum in the inertial frames, where the spectrum shifts similarly occur, but the magnitudes of these shifts remain constant, as indicated by equation (100).