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My physics prof. mentioned briefly, that in the framework of the relativistic Doppler effect, not only the frequency (alternatively, the wavelengh) changes when objects move with respect to each other (which I would understand), but also the intensity (which I do not understand). He said, it has to do with the fact, that the source strength, divided by the cube of the frequency, is a Lorentz invariant, but did not (want to) explain it any further. I found a publication by Johnson, M. H. and Teller, E. "Intensity changes in the Doppler effect", which is available here:

But it is the usual short version of an understandable explanation. I saw here, where it says: “A more-sophisticated method of deriving the beaming equations starts with the quantity $\displaystyle\frac{S}{\nu^3}$ This quantity is a Lorentz invariant, so the value is the same in different reference frames”. I think, that “sophisticated'' method is what I am looking for. I also have a hunch, that light gets bundled in high gravitational environments (edge of black hole) and therefore the intensity is higher, but I need a good explanation for dummies. I am Relativity level of about Schutz, Wald, Caroll and similar lecture texts.

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  • $\begingroup$ Somebody had answered my question, and I had commented on that, and it is all gone! Who does that? And why don't I get notified when somebody messes with my posting?! $\endgroup$
    – Fuzzy
    May 1, 2022 at 9:30

1 Answer 1

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$\texttt{C O N T E N T S}$

$\bl\S\texttt{ A. Lorentz boost and transformation of velocity 3-vectors}$

$\bl\S\texttt{ B. Aberration of light}$

$\bl\S\texttt{ C. Relativistic Doppler Shift}$

$\bl\S\texttt{ D. Intensity changes in the Doppler effect}$

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$\texttt{R E F E R E N C E S}$

$\texttt{Reference-01:}$ Intensity changes in the Doppler effect, M.H. Johnson and E.Teller.

$\texttt{Reference-02:}$ My answer in About de Broglie relations, what exactly is E? Its energy of what?.

$\texttt{Reference-03:}$ My answer in Deriving relativistic Doppler shift in terms of wavelength.

$\texttt{Reference-04:}$'Relativity-Special, General, and Cosmological' by W.Rindler, 2nd Ed.

$\texttt{Reference-05:}$'Modern Classical Physics. Optics, Fluids, Plasmas, Elasticity, Relativity, and Statistical Physics' by Kip S.Thorne and Roger D. Blandford, 2017.

$\texttt{Reference-06:}$High Energy Astrophysics - Lecture 3, Frank Rieger.

$\texttt{Reference-07:}$Phase space volume and relativity.

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$\bl\S $ A. Lorentz boost and transformation of velocity 3-vectors

The $\,1\p 1\m$Lorentz boost transformation with velocity $\:\bl\upsilon\e\upsilon\,\mb e_x \:$ along the common $x'\m,x\m$axis of two inertial frames $\:\mr S',\mr S\:$ expressed in differential form is \begin{align} \mr dx &\e\gamma\plr{\mr dx'\p\upsilon\mr dt'} \qquad \plr{\m c\les\upsilon\les c} \tl{A-01a}\\ \mr dy &\e\mr dy' \tl{A-01b}\\ \mr dz &\e\mr dz' \tl{A-01c}\\ \mr dt &\e\gamma\plr{\mr dt'\p\dfrac{\upsilon\mr dx'}{c^2}} \tl{A-01d}\\ \gamma &\e\plr{1\boldsymbol{-}\dfrac{\upsilon^2}{c^2}}^{\m\frac12} \tl{A-01e} \\ \beta &\e\dfrac{\,\upsilon\,}{c} \qquad \plr{\m 1\les\beta\les 1} \tl{A-01f} \end{align} see Figure-01.

If a particle is moving with respect to the frame $\:\mr S'\:$ with velocity \begin{equation} \begin{split} &\mb u'\e\plr{\mr u_x',\mr u_y',\mr u_z'}\e\mr u'\plr{\mr n_x',\mr n_y',\mr n_z'}\e\mr u'\mb n'\\ &\m c\leseq\mr u'\leseq c \qquad \Vlr{\mb n'}\e 1 \\ \end{split} \tl{A-02} \end{equation} then to find its velocity with respect to the frame $\:\mr S\:$ \begin{equation} \begin{split} &\mb u\e\plr{\mr u_x,\mr u_y,\mr u_z}\e\mr u\,\plr{\mr n_x,\mr n_y,\mr n_z}\e\mr u\,\mb n\\ &\m c\leseq\mr u\leseq c \qquad \Vlr{\mb n}\e 1 \\ \end{split} \tl{A-03} \end{equation} we divide side-by-side each one of equations \eqref{A-01a},\eqref{A-01b},\eqref{A-01c} by equation \eqref{A-01d}, and setting \begin{equation} \begin{split} \mb u'&\e\plr{\mr u_x',\mr u_y',\mr u_z'}\e\plr{\dfrac{\mr dx'}{\mr dt'}\:,\:\dfrac{\mr dy'}{\mr dt'}\:,\:\dfrac{\mr dz'}{\mr dt'}}\\ \mb u &\e\plr{\mr u_x,\mr u_y,\mr u_z}\e\plr{\dfrac{\mr dx}{\mr dt}\:,\:\dfrac{\mr dy}{\mr dt}\:,\:\dfrac{\mr dz}{\mr dt}}\\ \end{split} \tl{A-04} \end{equation} we end up with the following Lorentz transformation of the velocity 3-vectors \begin{equation} \mr u_x\e\dfrac{\mr u_x'\p\upsilon}{1\p\dfrac{\upsilon\mr u_x'}{c^2}}\:,\quad \mr u_y\e\dfrac{\mr u_y'}{\gamma\plr{1\p\dfrac{\upsilon\mr u_x'}{c^2}}}\:,\quad \mr u_z\e\dfrac{\mr u_z'}{\gamma\plr{1\p\dfrac{\upsilon\mr u_x'}{c^2}}} \tl{A-05} \end{equation} essentially the relativistic addition of the velocities $\:\mb u'\:$ and $\:\bl\upsilon$.

$\bl\S $ B. Aberration of light

Consider that a light source at rest in frame $\:\mr S'\:$ is emitting a photon in the direction $\:\mb n'\:$ on the $x'y'\m$plane by an angle $\:\theta'\:$ with respect to $\:\bl\upsilon$. The velocity of the photon is \begin{equation} \mb u'\e c\,\mb n'\e c\plr{\cos\theta',\sin\theta',0}\e\plr{\mr u_x',\mr u_y',\mr u_z'} \tl{B-01} \end{equation} Inserting its components in equations \eqref{A-05} we find the velocity of the photon with respect to the frame $\:\mr S$ \begin{equation} \mb u\e c\,\mb n\:\e c\plr{\cos\theta\:,\sin\theta\:,0}\e\plr{\mr u_x,\mr u_y,\mr u_z} \tl{B-02} \end{equation} where \begin{equation} \boxed{\:\:\cos\theta\e\dfrac{\cos\theta'\p\beta}{1\p\beta\cos\theta'}\:, \quad \sin\theta\e\dfrac{\sin\theta'}{\gamma\plr{1\p\beta\cos\theta'}}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}\quad \texttt{(Aberration 1)} \tl{B-03} \end{equation} that is the equation for the aberration of light.

Note that due to the trigonometric identity \begin{equation} \tan\plr{\dfrac{\,\theta\,}{2}}\e\dfrac{\sin\theta}{1\p \cos\theta} \tl{B-04} \end{equation} equations \eqref{B-03} yield \begin{equation} \tan\plr{\dfrac{\,\theta\,}{2}}\e\dfrac{\sin\theta}{1\p \cos\theta}\e \dfrac{1}{\gamma\plr{1\p \beta}}\dfrac{\sin\theta'}{1\p \cos\theta'}\e\sqrt{\dfrac{1\m \beta}{1\p\beta}}\tan\plr{\dfrac{\,\theta'}{2}} \nonumber \end{equation} that is the more simple equation \begin{equation} \boxed{\:\:\tan\plr{\dfrac{\,\theta\,}{2}}\e \plr{\dfrac{c\m\upsilon}{c\p\upsilon}}^{\frac12}\tan\plr{\dfrac{\,\theta'}{2}}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:}\quad\texttt{(Aberration 2)} \tl{B-05} \end{equation} We meet above equation as (4.9) in $\texttt{Reference-04}$.

$\bl\S $ C. Relativistic Doppler Shift

According to Louis de Broglie to a massive particle moving with $''$subluminal$''$ velocity $\:\mb u\e\mr u\,\mb n, \vert\mr u\vert\les c \:$, there corresponds a $''$superluminal$''$ plane phase wave, see $\texttt{Reference-02}$.

A subluminal particle has time-like energy-momentum Lorentz 4-vector \begin{equation} \mb P\e\plr{\gamma_{\mr u}\,m\,c, \gamma_{\mr u}\,m\,\mb u}\e\plr{\dfrac{E}{c},\mb p } \tl{C-01} \end{equation} where \begin{equation} \begin{split} \!\!\!\!\!\!\!\!E&\e\gamma_{\mr u}\,m\,c^2\e \texttt{energy of the particle}\\ \!\!\!\!\!\!\!\!\mb p&\e \gamma_{\mr u}m\mb u\e \gamma_{\mr u}m\mr u\mb n\e\texttt{linear momentum 3-vector of the particle}\\ &\Vlr{\mb n}\e 1\\ \end{split} \tl{C-02} \end{equation} while a superluminal plane phase wave has time-like angular frequency Lorentz 4-vector \begin{equation} \bl\Omega\e\plr{2\pi\nu, c\dfrac{2\pi}{\lambda}\,\mb m}\e\plr{\omega, c\,\mb k} \tl{C-03} \end{equation} where \begin{equation} \begin{split} \omega &\e2\pi\,\nu\e \texttt{angular frequency of the plane phase wave}\\ \mb k&\e \dfrac{2\pi}{\lambda}\,\mb m\e\texttt{wave number 3-vector of the plane phase wave}\\ \lambda &\e \texttt{wavelength of the plane phase wave}\\ &\Vlr{\mb m}\e 1\\ \end{split} \tl{C-04} \end{equation}

This superluminal plane phase wave is moving with velocity \begin{equation} \mb w\e\dfrac{\omega}{\Vlr{\mb k}}\mb m\e\lambda\,\nu\,\mb m\e\dfrac{\omega}{\Vlr{\mb k}^2}\mb k\,,\qquad \Vlr{\mb w}\e \mr w\e\lambda\,\nu \tl{C-05} \end{equation}

The vector $\:\bl\Omega\:$ of equation \eqref{C-03} is a Lorentz 4-vector. This means that the vector $\:\plr{\omega,c\,k_x,c\,k_y,c\,k_z}\:$ is transformed as the infinitesimal displacement vector $\:\plr{c\mr dt,\mr dx,\mr dy,\mr dz}\:$ in equations \eqref{A-01a}-\eqref{A-01f}.

The de Broglie relation connects the energy-momentum Lorentz 4-vector of the particle $\:\mb P$, equation \eqref{C-01}, with the angular frequency of its accompanying plane phase wave $\:\bl\Omega$, equation \eqref{C-03} \begin{equation} \boxed{\:\:c\,\mb P\e \hbar\,\bl\Omega\:\:\vp}\quad\texttt{(de Broglie)} \tl{C-06} \end{equation}

The directions of the particle motion $\:\mb n\:$ and of propagation of the plane phase wave $\:\mb m\:$ are identical, while the product of their speeds is a Lorentz scalar invariant \begin{equation} \mb m\bl\equiv \mb n\,,\qquad \mr u\,\mr w\e \mr u'\,\mr w'\e c^2\e\texttt{ Lorentz invariant} \tl{C-07} \end{equation}

Equating time and space components in equation \eqref{C-06} we have \begin{equation} \begin{split} E&\e\hbar \omega \e h\,\nu\\ \mb p&\e\hbar\mb k\e\dfrac{\,h\,}{\lambda}\,\mb n\\ \end{split} \tl{C-08} \end{equation}

Now, all these relations are valid in the limiting case of a $''$luminal$''$ particle, that is a photon, and its accompanying $''$luminal$''$ phase wave, that is light or electromagnetic wave. In this case we have $\: \mr u\e\mr w\e c\e\lambda\,\nu\:$ and equations \eqref{C-08} give the energy-momentum 4-vector of a photon \begin{equation} \mb P\e\plr{\dfrac{E}{c},\mb p}\e \dfrac{h\,\nu}{c}\plr{1,\mb n} \tl{C-09} \end{equation} The vector $\:c\mb P\:$ of equation \eqref{C-09} is a Lorentz 4-vector. This means that the components of this vector $\:\plr{h\,\nu,h\,\nu\,n_x,h\,\nu\,n_y,h\,\nu\,n_z}\:$ are transformed as the components of the infinitesimal displacement vector $\:\plr{c\mr dt,\mr dx,\mr dy,\mr dz}\:$ in equations \eqref{A-01a}-\eqref{A-01f}.

So, let again the photon emitted by the light source in its rest frame $\:\mr S'\:$ as shown in Figure-01. For its energy-momentum 4-vector we have \begin{equation} c\mb P'\e h\,\nu'\plr{1,\mr n_x',\mr n_y',\mr n_z'}\e\plr{h\,\nu',h\,\nu'\cos\theta',\,h\,\nu'\sin\theta',0} \tl{C-10} \end{equation} Its energy-momentum 4-vector in the frame $\:\mr S\:$ is \begin{equation} c\mb P\:\e h\,\nu\:\plr{1,\mr n_x,\mr n_y,\mr n_z}\e\plr{h\,\nu,h\,\nu\cos\theta,\,h\,\nu\sin\theta,0} \tl{C-11} \end{equation} Inserting these vectors in equations \eqref{A-01a}-\eqref{A-01d} in place of $\:\plr{c\,\mr dt',\mr d\mb r' }\:$ and $\:\plr{c\,\mr dt,\mr d\mb r}\:$ respectively we have in details \begin{align} h\,\nu\cos\theta &\e\gamma\plr{h\,\nu'\cos\theta'\p\beta\,h\,\nu'} \tl{C-12a}\\ h\,\nu\sin\theta &\e h\,\nu'\sin\theta' \tl{C-12b}\\ 0 &\e 0 \tl{C-12c}\\ h\,\nu &\e\gamma\plr{h\,\nu'\p \beta\,h\,\nu'\cos\theta'} \tl{C-12d} \end{align} that is \begin{align} \dfrac{\nu}{\nu'} &\e\dfrac{\gamma\,\plr{\cos\theta'\p\beta}}{\cos\theta} \tl{C-13a}\\ \dfrac{\nu}{\nu'} &\e\dfrac{\sin\theta'}{\sin\theta} \tl{C-13b}\\ \dfrac{\nu}{\nu'} &\e\gamma\plr{1\p\beta\cos\theta'} \tl{C-13c} \end{align} Equating the right hand sides firstly of equations \eqref{C-13a}, \eqref{C-13c} and secondly of equations \eqref{C-13b}, \eqref{C-13c} we get the following equations respectively \begin{equation} \cos\theta\e\dfrac{\cos\theta'\p\beta}{1\p\beta\cos\theta'}\:, \quad \sin\theta\e\dfrac{\sin\theta'}{\gamma\plr{1\p\beta\cos\theta'}} \tl{C-14} \end{equation} identical to equations \eqref{B-03}. We meet again the aberration of light as discussed in $\:\bl\S\texttt{B}$.

From the first of the aberration equations \eqref{C-14} we have \begin{equation} \cos\theta'\e\dfrac{\cos\theta\m\beta}{1\m\beta\cos\theta} \tl{C-15} \end{equation}

Inserting this expression of $\:\cos\theta'\:$ firstly in the second of the aberration equations \eqref{C-14} we get \begin{equation} \sin\theta'\e\dfrac{\sin\theta}{\gamma\plr{1\m\beta\cos\theta}} \tl{C-16} \end{equation} and secondly by insertion in equation \eqref{C-13c} we have \begin{equation} \dfrac{\nu}{\nu'}\e\dfrac{1}{\gamma\plr{1\m\beta\cos\theta}} \tl{C-17} \end{equation} Finally differentiating any of the equations \eqref{C-14},\eqref{C-15} or \eqref{C-16} yields \begin{equation} \dfrac{\mr d\theta'}{\mr d\theta}\e\dfrac{\nu}{\nu'} \tl{C-18} \end{equation}

Defining the Doppler factor \begin{equation} \mr D\bl\equiv\dfrac{\nu}{\nu'} \tl{C-19} \end{equation} all these relations are given in one stroke below \begin{equation} \!\!\!\!\!\!\!\!\!\!\boxed{\:\mr D\e\dfrac{\nu\texttt{(shifted)}}{\nu'\texttt{(unshifted)}}\e\gamma\plr{1\p\beta\cos\theta'}\e\dfrac{1}{\gamma\plr{1\m\beta\cos\theta}}\e\dfrac{\sin\theta'}{\sin\theta}\e\dfrac{\mr d\theta'}{\mr d\theta}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:} \tl{C-20} \end{equation}

The relation between the infinitesimal solid angles $\:\mr d\Theta'\e\sin\theta'\mr d\theta'\mr d\phi'\:$ and $\:\mr d\Theta\e\sin\theta\mr d\theta\mr d\phi\:$ could be derived from equations \eqref{C-20}, the Figure-02 and the fact that for the azimuth around $\:\bl\upsilon\:$ angle $\:\phi\:$ we have $\:\mr d\phi\e\mr d\phi'\:$ \begin{equation} \dfrac{\mr d\Theta'}{\mr d\Theta}\e\dfrac{\sin\theta'\mr d\theta'\mr d\phi'}{\sin\theta\,\mr d\theta\,\mr d\phi}\e\underbrace{\plr{\dfrac{\sin\theta'}{\sin\theta}}}_{\mr D}\underbrace{\plr{\dfrac{\mr d\theta'}{\mr d\theta}}}_{\mr D}\underbrace{\plr{\dfrac{\mr d\phi'}{\mr d\phi}}}_{1}\e\mr D^2 \tl{C-21} \end{equation} that is \begin{equation} \boxed{\:\dfrac{\mr d\Theta'}{\mr d\Theta}\e \mr D^2\e\plr{\dfrac{\nu}{\nu'}}^2\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:} \tl{C-22} \end{equation}

$\bl\S $ D. Intensity changes in the Doppler effect

The proof of the Lorentz invariance of the scalar $\:I_\nu/\nu^3\:$ is based on the Lorentz invariance of the infinitesimal volume element in phase space. But at first we must define the Specific Intensity $\:I_\nu$. So consider a set of rays and construct the infinitesimal area $\:\mr dA\:$ normal to a given ray and look at all rays passing through area element within solid angle $\:\mr d\Omega\:$ of the given ray as shown in the Figure(1) below. enter image description here

The photon's Specific Intensity $\:I_\nu\:$ is defined to be the total energy \begin{equation} \mr dE\e h\nu\mr dN \tl{D-01} \end{equation} (where $\:\mr dN\:$ is the number of photons) that crosses this area, per unit area $\:\mr dA$, per unit time $\:\mr dt$, per unit frequency $\:\mr d\nu$, and per unit solid angle $\:\mr d\Omega$ \begin{equation} I_\nu\bl\equiv\dfrac{\mr dE}{\mr dA\,\mr dt\,\mr d\nu\,\mr d\Omega} \tl{D-02} \end{equation} (i.e., per unit everything)

As respect to the invariance of the infinitesimal volume element in phase space we note in summary the following (see $\texttt{Reference-05}$, Chapter 3): As tools for the study of a collection of a very large number of identical particles (all with the same rest mass $\:m$) consider a tiny 3-dimensional volume $\:\mr d\mc V_x\:$ centered on some location $\:\mb x \:$ in physical space and a tiny 3-dimensional volume $\:\mr d\mc V_p\:$ centered on location $\:\mb p\:$ in momentum space. Together these make up a tiny 6-dimensional volume (in Newtonian theory) \begin{equation} \mr d^2\mc V\bl\equiv\mr d\mc V_x\mr d\mc V_p \tl{D-03} \end{equation} In any Cartesian coordinate system, we can think of $\:\mr d\mc V_x\:$ as being a tiny cube located at $\:\plr{x,y,z}\:$ and having edge lengths $\:\mr dx,\mr dy,\mr dz$, and similarly for $\:\mr d\mc V_p$. Then, as computed in this coordinate system, these tiny volumes are \begin{equation} \mr d\mc V_x\e\mr dx\,\mr dy\,\mr dz\,,\quad \mr d\mc V_p\e\mr dp_x\,\mr dp_y\,\mr dp_z\,,\quad \mr d^2\mc V\e\mr dx\,\mr dy\,\mr dz\,\mr dp_x\,\mr dp_y\,\mr dp_z \tl{D-04} \end{equation} Denote by $\:\mr dN\:$ the number of particles (all with rest mass $\:m$) that reside inside $\:\mr dN\:$ in phase space (at some moment of time $\:t$). Stated more fully: $\mr dN\:$ is the number of particles that, at time $\:t$, are located in the 3-volume $\:\mr d\mc V_x\:$ centered on the location $\:\mb x\:$ in physical space and that also have momentum vectors whose tips at time $\:t\:$ lie in the 3-volume $\:\mr d\mc V_p\:$ centered on location $\:\mb p\:$ in momentum space. Denote by \begin{equation} \boxed{\:\:\mc N\plr{\mb x,\mb p,t}\bl\equiv\dfrac{\mr dN}{\mr d^2\mc V}\e\dfrac{\mr dN}{\mr d\mc V_x\mr d\mc V_p}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tl{D-05} \end{equation} the number density of particles at location $\:\plr{\mb x,\mb p}\:$ in phase space at time $\:t$. This is also called the *distribution function. In Newtonian theory, the volumes $\:\mr d\mc V_x\:$ and $\:\mr d\mc V_p\:$ occupied by our collection of $\:\mr dN\:$ particles are independent of the reference frame that we use to view them. Not so in relativity theory: $\mr d\mc V_x\:$ undergoes a Lorentz contraction when one views it from a moving frame, $\:\mr d\mc V_p\:$ also changes; but their product $\:\mr d^2\mc V\e\mr d\mc V_x\mr d\mc V_p\:$ is the same in all frames. More precisely in relativity theory it has been proved on one hand that \begin{equation} \boxed{\:\:E\,\mr d\mc V_x\e p_0\,\mr d\mc V_x\e\texttt{Lorentz invariant} \Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tl{D-06} \end{equation} and on the other hand that \begin{equation} \boxed{\:\:\dfrac{\mr d\mc V_p}{E}\e \dfrac{\mr d\mc V_p}{p_0}\e\texttt{Lorentz invariant} \Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tl{D-07} \end{equation} Therefore, in relativity theory for the distribution function we have also \begin{equation} \boxed{\:\:\mc N\bl\equiv\dfrac{\mr dN}{\mr d^2\mc V}\e\dfrac{\mr dN}{\mr d\mc V_x\mr d\mc V_p}\e\texttt{Lorentz invariant}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tl{D-08} \end{equation} Although in derivations and proofs it is assumed nonzero rest mass ($m\bl\ne 0$), the conclusions \eqref{D-06} and \eqref{D-07} continue to hold if we take the limit as $m\rightarrow 0$ and the 4-momenta become null. Correspondingly, \eqref{D-06} to \eqref{D-08} are valid for particles with zero mass also, like photons.

Consider that photons hit the surface area $\:\mr dA\:$ in time interval $\:\mr dt\:$ as shown in the Figure(2) below. enter image description here

Since the photons move at the speed of light $\:c$, the product of that surface area with $\:c\:$ times the time $\:\mr dt\:$ is equal to the volume they occupy at a specific moment of time: \begin{equation} \mr d\mc V_x\e\mr dA\,\mr dt \tl{D-09} \end{equation} Focus attention on a set $\:S\:$ of photons in this volume that all have nearly the same frequency $\:\nu\:$ and propagation direction $\:\mb n$. Their energies $\:E\:$ and momenta $\:\mb p\:$ are related to $\:\nu\:$ and $\:\mb n\:$ by equation \eqref{C-09}, that is \begin{equation} E\e h\nu\,,\quad \mb p\e\plr{h\nu/c}\mb n \tl{D-10} \end{equation} Their frequencies lie in a range $\:\mr d\nu\:$ centered on $\:\nu\:$, and they come from a small solid angle $\:\mr d\Omega\:$ centered on $\m\mb n$; the volume they occupy in momentum space is related to these quantities by \begin{equation} \mr d\mc V_p\e \Vlr{\mb p}^2\mr d\Omega\,\mr d \Vlr{\mb p}\e\plr{h\nu/c}^2\mr d\Omega\,\plr{h\mr d\nu/c}\e\plr{h/c}^3\nu^2\mr d\Omega\,\mr d\nu \tl{D-11} \end{equation} From the definition of specific intensity, equation \eqref{D-02}, using equations \eqref{D-01}, \eqref{D-09} and \eqref{D-11} we have \begin{equation} \boxed{\:\:\mc N\bl\equiv\dfrac{\mr dN}{\mr d^2\mc V}\e\dfrac{\mr dN}{\mr d\mc V_x\mr d\mc V_p}\e\dfrac{c^2}{h^4}\dfrac{I_\nu}{\nu^3}\Vp{\dfrac{\dfrac{a}{b}}{\dfrac{a}{b}}}\:\:} \tl{D-12} \end{equation} that is the scalar $\:I_\nu/\nu^3\:$ is except a constant identical to the distribution function $\:\mc N\:$ so by equation \eqref{D-08} a Lorentz invariant scalar.

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(1) Figure extracted from $\texttt{Reference-06}$

(2) Figure extracted from $\texttt{Reference-05}$

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enter image description here

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enter image description here enter image description here

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    $\begingroup$ Thanks, but I don't see in your derivation, that the intensity canges with the cube of the frequency. But Planck's law of blackbody radiation (of a star) might do the job: If the spectral intensity is proportional to $\nu^3/\left(e^{h\nu/kT}-1\right)$, and the frequency $\nu$ changes because of the Doppler effect, then of course also the spectral intensity changes with $\nu^3$. Is that acceptable, or am I on the wrong track? $\endgroup$
    – Fuzzy
    Apr 30, 2022 at 10:32

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