On the Theory of Radiation of Moving Bodies

On the Theory of Radiation of Moving Bodies  (1904)  by Friedrich Hasenöhrl, translated by Wikisource

information about this edition.  related portals: Relativity. In German: Zur Theorie der Strahlung bewegter Körper, Wiener Sitzungsberichte, 113 IIa: 1039-1055

On the Theory of Radiation of Moving Bodies.

by

Dr. Fritz Hasenöhrl.

(With 2 text figures.)

(Presented in the meeting on June 23, 1904).

If a radiating surface moves with uniform velocity in the same direction as the radiation emitted by it, then work must be constantly applied to overcome the pressure exerted by the latter. Now, this work is transformed into radiation energy as well, so that more energy of that amount is emanated from a surface moving in this sense, than from a stationary one. However, if the surface is moving in the opposite direction of the emanated radiation, then the latter steadily performs work, thus less energy of that amount is emanated from the moving surface than from a stationary one.

On the other hand, if an absorbing surface is moving, so that it recedes from the incident radiation, then the latter steadily performs work; the surface is therefore only capable of absorbing – i.e. transforming into inner heat – less radiation by the equivalent of the latter. However, if the absorbing surface is approaching the incident radiation, then work from outside must be steadily performed against the pressure of radiation. This work can only occur as heat in the absorbing surface. The effect of motion is thus in this case, that the surface is absorbing more heat by the amount of performed work.

These modes of conception were already spoken out by different authors.

In the following I've tried now, to apply these theorems upon the radiation in a moving cavity. Except the radiation provided by the walls, also radiation energy must be present within it, which is gained from mechanical work and which is transformed into such work again. It is essentially determined by the motion of the cavity; its amount is, as it will be shown, proportional to the square of the system's velocity (in first approximation), thus it apparently increases the kinetic energy of the system. Circumstances are thus present, that are quite analogous to the motion of an electron. In the same way as the concept of "electromagnetic mass" is introduced there, one could also speak about an "apparent mass" here, which is caused by radiation. In the same way, as the electromagnetic mass is proportional to the static energy of the stationary electron, also the apparent mass caused by radiation is proportional to the energy content of the stationary cavity. Namely, the proportionality factor is in both cases of the same order of magnitude. Now, since the heat content of every body partly consists of radiation energy, then every body must possess such an apparent mass depending on the temperature, and which is added to the mass in the ordinary sense.

1.

Now, we consider a cylindric cavity $R$, moving with constant velocity $c$ in the direction of the arrow (Fig. 1). Let $\mathfrak{B}$ be the velocity of light and $\sigma=\tfrac{c}{\mathfrak{B}}$. Furthermore, let $A$ and $B$ be two black surfaces.

The shell surface of the cavity as well as the outer boundary of the black body shall be formed by surfaces which are perfectly reflecting into the interior. Let the cross-section of space $R$ be equal to 1, its height be equal to $D$. Let the outer space be completely free of radiation, thus at absolute temperature zero, while a certain temperature shall be attributed to surfaces $A$ and $B$.

Now we have to distinguish between absolute and relative direction of radiation; it is more convenient, to base our consideration upon the latter.

Let

$2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$

be the energy quantity, emanated by $A$ in the unit of time into the relative direction between $\phi$ and $\phi+d\phi$, where $\phi$ is thus the angle enclosed by the relative beam direction with the normal (and with the direction of velocity $c$). $i_{1}$ then must be a constant (with respect to $\phi$, as I already have emphasized in an earlier work.) This radiation now exerts a pressure upon $A$, whose component that coincides with the direction of the normal (in the sense of $-c$), shall have the value

$2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\,$

If we multiply this expression with $c$, then we obtain the work performed against this pressure in one second from the outside, which is now also transformed into radiation, so that the total radiation leaving $A$ in the given direction, has the value

$2\pi\ (i_{1}+p_{1}c)\ \cos\phi\ \sin\phi\ d\phi = 2\pi\ i\ \cos\phi\ sin\phi\ d\phi$

This expression now gives us also the amount of the radiation falling upon $B$ in the unit of time. There, one fraction of it is absorbed, and one fraction is transformed into work. If $B$ would be at the absolute temperature zero, then the radiation just considered would be the only one present in $R$. Since in this case, no force of resistance against the motion of our system is to be expected, then the same pressure into opposite directions must be effective in $A$ and $B$, so that no work is performed in the whole. Thus in $B$, the energy quantity

$2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$

is absorbed, and the energy quantity

$2\pi\ p_{1}\ \cos\phi\ \sin\phi\ d\phi\cdot c\,$

is transformed into mechanical work.

If we now imagine that $B$ has the same temperature as $A$, then also $B$ emanates energy

$2\pi\ i_{0}\ \cos\phi\ \sin\phi\ d\phi\,$

in a certain direction. If this radiation exerts the pressure $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$, then it performs work by which amount the energy provided by $B$ has to be diminished, so that also $B$ is left by the radiation quantity

$2\pi(i_{0}-p_{2}c)\cos\phi\ \sin\phi\ d\phi=2\pi i'\cos\phi\ \sin\phi\ d\phi\,$

The same energy quantity also occurs in $A$, and a quite similar consideration as earlier teaches, that the energy quantity

$2\pi i_{0}\cos\phi\ \sin\phi\ d\phi$

is absorbed there, because the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\,$ must also be performed against the incident radiation from outside, which is also transformed into work.

Thus the same quantity is absorbed by both bodies $A$ and $B$, as it is emitted; upon both surfaces (in opposite direction) the same pressure

$2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$

is exerted, so that no work is performed altogether. Thus when no external force is acting, then the system maintains its velocity.

In case one of the bodies $A$ or $B$ is replaced by a perfect mirror, the radiation condition in $R$ must remain exactly the same, thus besides other things, $2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi,\,$ is the pressure upon the mirror as well.

We ask ourselves about the energy content of space $R$. The radiation traveling from $A$ to $B$, has the relative velocity:

$\mathfrak{B}(-\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{-}.$

The radiation coming from $A$ to $B$, has the velocity

$\mathfrak{B}(+\sigma\cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\phi}=\mathfrak{B}_{+}.$

both have to travel the path $\tfrac{D}{\cos\phi}$, thus the total energy content of space $R$ is

$\overset{\pi/2}{\int_{0}}\frac{D}{\cos\phi}\left[\frac{2\pi(i_{0}+p_{1}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{-}}+\frac{2\pi(i_{0}-p_{2}c)\cos\phi\sin\phi\ d\phi}{\mathfrak{B}_{+}}\right]$ $=2\pi i_{0}D\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)+$ $+2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$

(1)

The first of the two summands gives the part of the energy of space $R$, which was provided by the black bodies. My paper already cited dealt with it.

The second summand, however, represents the radiation energy gained from mechanical work. Since no work is performed at uniform motion as we have seen, this must be a work which must be performed when our system is accelerated.

One can easily determine the amount of this energy quantity in the following way. Let us imagine a system in absolute rest, and the radiation of both black surfaces $A$ and $B$ is somehow hindered: space $R$ is thus totally free of radiation. Now, in one instant the radiation from $A$ and $B$ shall be left free, and the system shall simultaneously brought to velocity $c$. From that instant on, the work $2\pi p_{1}\cos\phi\ \sin\phi\ d\phi\cdot c$ in the unit of time must be performed against the radiation emitted from $A$. Now, time $\tfrac{D}{\mathfrak{B}_{-}\cos\phi}$ is passing until this radiation arrives in $B$, and there it provides the same quantity of work itself. During this time, the work provided from outside is thus uncompensated. Also the radiation emitted from $B$ performs right from the start the work $2\pi p_{2}\cos\phi\ \sin\phi\ d\phi\cdot c$, and now the time $\tfrac{D}{\mathfrak{B}_{+}\cos\phi}$ is passing until this radiation arrives in $A$, and there the same work from outside must be performed in the opposite sense. Thus, more work was performed from the outside by

$\frac{D}{\cos\phi\cdot\mathfrak{B}_{-}}2\pi p_{1}\cos\phi\sin\phi\ d\phi\cdot c\frac{D}{\cos\phi\cdot\mathfrak{B}_{+}}2\pi p_{2}\cos\phi\sin\phi\ d\phi\cdot c$

i.e., it was gained from the radiation. If we integrate this amount over all beam directions, then we obtain the expression above.

If our system is at rest in the beginning, yet the surfaces $A$ and $B$ are not hindered of emanating radiation, then radiation energy of certain amount is present at the beginning also in $A$. The previous consideration indicates, that this energy is completely absorbed and not transformed into work. Namely this can also be directly shown, when one introduces the values for $p_{1}$ and $p_{2}$, which we will give in the following section. In order of not interrupting the progression of the investigation, I transfer the proof of this theorem into § 4 of this treatise. Thus in expression (1), the first summand is the energy provided by the radiating bodies, the second one is the energy stemming from work. We denote the latter by $L$, thus we set

$L=2\pi cD\int_{0}^{\pi/2}\sin\phi\ d\phi\left(\frac{p_{1}}{\mathfrak{B}_{-}}-\frac{p_{2}}{\mathfrak{B}_{+}}\right).$

(2)

2.

We now want to introduce the values for magnitudes $p_{1}$ and $p_{2}$, which we preliminarily left undetermined. It was surely introduced at first by Abraham in his treatises already cited. Namely, these values have been derived from Lorentz's theory of electromagnetism. Since one can arrive at the same expressions in another way (see § 3 of the present work), their correctness seem to be even more plausible. According to Abraham, the radiation pressure upon a surface is numerically equal to the incident or emanated radiation divided by the speed of light, namely this pressure acts in the absolute direction of radiation. Let $\varphi$ be the angle, which is enclosed by the absolute direction of the radiation (emanated under the relative angle $\phi$) with the direction of $c$. Then the magnitude denoted earlier as $p_1$, is

$p_{1}=\frac{1}{\mathfrak{B}}i\ \cos\varphi,$

since we understood under $p_1$ the – solely effective – normal pressure component. A relation between $\varphi$ and $\phi$ can be easily obtained by the following figure 2, which is essentially identical with Fig. 1 of my cited treatise, and which surely needs nor further explanation.

It is

$\mathfrak{B}_{-}=\sqrt{\mathfrak{B}^{2}+c^{2}-2\mathfrak{B}c\ \cos\ \varphi}=\mathfrak{B}\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}=$ $=\mathfrak{B}\left(-\sigma\ \cos\phi+\sqrt{1-\sigma^{2}\sin^{2}\ \phi}\right),$

from which the relations

$\cos\phi=\frac{1-\sigma\ \cos\varphi}{\sqrt{1+\sigma^{2}-2\sigma\ \cos\ \varphi}}$

and

$\cos\ \varphi=\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\ \phi}$

(3)

are given.

Thus it is

$p_{1}=\frac{i}{\mathfrak{B}}\left(\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right);$

(4)

if we insert this into equation $i=i_{0}+p_{1}c$, then it follows

$i=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$

(5)

and

$p_{1}=i_{0}\frac{\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{-}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$

(6)

Quite similar it is given:

$i'=i_{0}\frac{\mathfrak{B}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$

(7)

and

$p_{2}=\frac{i'}{B}\left(-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}\right)$

(8)

or

$p_{2}=i_{0}\frac{-\sigma\sin^{2}\phi+\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}_{+}\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=i_{0}\frac{\cos\phi-\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.$

(9)

If we insert this into (2), then:

$L=2\pi cDi_{0}\cdot\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\left[\cos\ \phi\left(\frac{1}{\mathfrak{B}_{-}}-\frac{1}{\mathfrak{B}_{+}}\right)+\sigma\sqrt{1-\sigma^{2}\sin^{2}\phi}\left(\frac{1}{\mathfrak{B}_{-}}+\frac{1}{\mathfrak{B}_{+}}\right)\right]=$ $=\frac{4\pi cDi_{0}\sigma}{\mathfrak{B}^{2}(1-\sigma^{2})^{2}}\int_{0}^{\pi/2}\sin\phi\ d\phi\frac{1+\cos^{2}\phi-\sigma^{2}\sin^{2}\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$

If we additionally put the energy content of the resting cavity $R$:

$\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0},$

then after execution of the integration

$L=E_{0}\frac{c^{2}}{\mathfrak{B}^{2}}\frac{1}{(1-\sigma^{2})^{2}}\left[\frac{1}{2}+\frac{1}{2\sigma^{2}}-\frac{(1-\sigma^{2})^{2}}{4\sigma^{3}}\log\frac{1+\sigma}{1-\sigma}\right].$

If we neglect herein magnitudes of order $\sigma^3$, then it becomes

$L=E_{0}\frac{c^{2}}{B^{2}}\cdot\frac{4}{3}.$

This expression has now the form and the dimension of a kinetic energy. Thus one can say, that the kinetic energy of our system was apparently increased by $L$, or that to the mechanical mass of our system, also an apparent mass

$\mu=\frac{8}{3}\frac{E_{0}}{\mathfrak{B}^{2}}$

has been added. Namely, the introduction of that mass is quite similar to that of the electromagnetic mass. If we (for the moment) denote the energy of a resting electron with $\epsilon_{0}$, then the electromagnetic mass of it is equal to $\tfrac{4}{3}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$ or $\tfrac{8}{5}\tfrac{\epsilon_{0}}{\mathfrak{B}^{2}}$, depending on whether one deals with surface or volume charge. The relation is thus the same in terms of order of magnitude. Also the exact expressions have a certain similarity, since the quantity $\log\tfrac{1+\sigma}{1-\sigma}$ plays a role in both of them.

3.

In this section, I still want to make same remarks concerning the value of the radiation pressure.

The total pressure upon one of the surfaces $A$ or $B$ – it is irrelevant as to whether it is imagined as black of reflecting – has the value

$\begin{array}{ll} P & =2\pi(p_{1}+p_{2})\cos\phi\ \sin\phi\ d\phi\\ \\ & =4\pi i_{0}\frac{\cos^{2}\phi\ \sin\phi\ d\phi}{\mathfrak{B}(1-\sigma^{2})\sqrt{1-\sigma^{2}\sin^{2}\phi}}.\end{array}$

Herein, we want to introduce the radiation density $\rho$, which for example falls upon $B$. It is

$\rho=\frac{2\pi i\ \sin\phi\ d\phi}{\mathfrak{B}_{-}}=\frac{2\pi i_{0}\ \sin\phi\ d\phi\cdot\mathfrak{B}}{\mathfrak{B}_{-}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}$

(this is given from (5)).

Thus it is

$P=2\rho\frac{\mathfrak{B}_{-}^{2}\cos^{2}\phi}{\mathfrak{B}^{2}(1-\sigma^{2})}.$

If we want to introduce herein the absolute beam direction again, then we use the relation immediately given from Fig. 2

$\mathfrak{B}_{-}cos\phi=\mathfrak{B}\ cos\varphi-c=\mathfrak{B}(\cos\ \varphi-\sigma).$

And it becomes:

$P=2\rho\frac{(\cos\ \varphi-\sigma)^{2}}{1-\sigma^{2}},$

an expression already given by Abraham as well.

Now, by generalization of the thought that was first spoken out by Larmor, the same result can be derived from the standpoint of the elastic theory of light, which I would like to show soon.

We consider a light source, moving under the (absolute) angle $i$ against the $X$-axis. It is given by

$\zeta=A\cos\ m(x\ \cos\ i+y\ \sin\ i+\mathfrak{B}t).$

If this wave falls upon a mirror lying perpendicular to the $X$-axis, then a reflected wave is formed, which is given by

$\zeta'=A'\cos\ m'(x\ \cos\ r-y\ \sin\ r-\mathfrak{B}t).$

Herein, $r$ is the angle of reflection.

At the surface of the mirror it must be $\zeta+\zeta'=0$. If it is moving with velocity $c$ in the direction of the normal, i.e., in the direction of the positive $X$-axis, then it must be

$x=ct,\ \zeta+\zeta'=0$

This gives

$A\ cos\ m(t(\mathfrak{B}+c\ \cos\ i)+y\ \sin\ i]+$ $+A'\cos\ m'[t(c\ \cos\ r-\mathfrak{B})-y\ \sin\ r]=0.$

This equation can only then be satisfied for all values of $y$ and $t$, when

$\begin{array}{rl} A= & -A',\\ m(\mathfrak{B}+c\ \cos\ i)= & m'(\mathfrak{B}-c\ \cos\ r),\\ m\ \sin i= & m'\ \sin r.),\end{array}.$

These equations give us the law of reflection

$\frac{\sin\ i}{1+\sigma\ \cos\ i}=\frac{\sin\ r}{1-\sigma\ \cos\ r}$

(10)

and the Doppler effect

$\frac{m'}{m}=\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}$

in full agreement with Abraham. Incidentally, equation (10) is also easily given from equation (12) of my earlier treatise.

Now, to derive the value of the pressure from it, we have to assume that the energy density of a light wave is proportional (at equal amplitude) to the -2. power of the wave length, thus to the quantity $m^2$. Thus when we denote by $\rho$ the density of the incident wave, by $\rho'$ that of the reflecting one, then it is

$\frac{\rho'}{\rho}=\left(\frac{m'}{m}\right)^{2}=\left(\frac{1+\sigma\ \cos\ i}{1-\sigma\ \cos\ r}\right)^{2}.$

Now, the energy falling upon the mirror in unit time, is contained in space $\mathfrak{B}\ cos\ i+c$; the reflected energy in space $\mathfrak{B}\ cos\ r-c$. Thus the amount the incident energy per second is:

$\rho(\mathfrak{B}\ \cos\ i+c)=\rho\mathfrak{B}(\cos\ i+\sigma)$

and the amount of the energy reflected in the same time

$\rho'(\mathfrak{B}\ \cos\ r-c)=\rho'\mathfrak{B}(\cos\ r-\sigma).$

The difference of these two expressions must be equal to the work of the radiation pressure per unit time. Thus it must be

$Pc=\rho'\mathfrak{B}(\cos\ r-\sigma)-\rho\mathfrak{B}(\cos\ i+\sigma)$

Or

$cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\left(\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}\right)^{2}\frac{\cos\ r-\sigma}{\cos\ i+\sigma}-1\right].$

With the aid of the easily derivable relation, already given by Abraham:

$\frac{1+\sigma\cos\ i}{1-\sigma\cos\ r}=\frac{\sigma+\cos\ i}{\sigma-\cos\ r}=\frac{\sin\ i}{\sin\ r}$

and the equation:

$\sin\ r=\frac{\sin\ i(1-\sigma^{2})}{1+\sigma^{2}+2\sigma\ \cos\ i}$

it becomes

$cP=\mathfrak{B}\rho(\cos\ i+\sigma)\left[\frac{1+\sigma^{2}+2\sigma\ \cos\ i}{\ i-\sigma^{2}}-1\right].$

and

$P=2\rho\frac{(\cos\ i+\sigma)^{2}}{1-\sigma^{2}}.$

The agreement with the value given by Abraham is thus a complete one (here, $\sigma$ has the opposite sign as earlier.)

Of course, upon this foundation one can calculate the values $p_{1}$ and $p_{2}$, when one assumes that a moving body is emanating waves, whose amplitude is the same as in the stationary state, while the density of the radiation energy is inversely proportional to the square of the wavelength.

Incidentally, the values of $p_{1}$ and $p_{2}$ have been derived by Poynting by a peculiar consideration. However, Poynting confines himself to the case of perpendicular emission. If one accordingly put $\cos\varphi = \cos\phi = 1$ in the expressions given here, then the agreement is complete.

4.

Now, we have to provide the proof of the assertion made in § 1.

Our system given by Fig. 1 shall be at rest at the beginning. Then the energy quantity

$E_{0}=\frac{4\pi i_{0}}{\mathfrak{B}}\cdot D.$

is in space $R$. Now it is the question, what is happening with the energy, when the system is suddenly brought to velocity $c$. We can assume from the outset, that only a fraction of it becomes absorbed, while another fraction is transformed into mechanical work.

Now we have to notice in particular, that this energy is uniformly distributed into all absolute directions. Thus when we maintain our earlier mode of expression, then the density of energy whose absolute direction of propagation encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is:

$\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ d\varphi.$

As soon as the system is in motion, it is about the distribution in terms of the relative propagation direction. Then the density of energy, whose relative propagation direction encloses an angle between $\phi\,$ and $\phi+d\phi\,$ with the normal, is evidently:

$\frac{2\pi i_{0}}{\mathfrak{B}}\sin\varphi\ \frac{d\varphi}{d\phi}d\phi=-\frac{2\pi i_{0}}{\mathfrak{B}}\ \frac{d\ \cos\varphi}{d\phi}d\phi.$

This becomes by differentiation of equation (3):

$=\frac{2\pi i_{0}}{\mathfrak{B}}\sin\phi\ d\phi\frac{\left(\sqrt{1-\sigma^{2}\sin^{2}\phi}-\sigma\ \cos\phi\right)^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}=$ $=\frac{2\pi i_{0}}{\mathfrak{B}}\frac{\mathfrak{B}_{-}^{2}}{\mathfrak{B}^{2}\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$

Since this energy is located in a cylinder of height $D$, then the energy quantity falling upon $B$ in this direction is:

$\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$

Previously (§ 1) we saw, that from the incident radiation $2\pi\ i\ cos\phi\ sin\phi\ d\phi$, the fraction $2\pi\ i_{0}\ \cos\phi \sin\phi\ d\phi$ is absorbed, while the fraction $2\pi\ p_{1}\ \cos\phi \sin\phi d\phi$ is transformed into work. Thus from the totality of the incident energy quantity, the fraction $\tfrac{i_{0}}{i}$ is absorbed and the fraction $\tfrac{p_{1}c}{i}$ is transformed into work.

Thus from the just incident energy quantity, the fraction

$\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{i_{0}}{i}$

is absorbed; the fraction

$\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{-}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi\cdot\frac{p_{1}c}{i}$

is transformed into work.

In a similar way one recognizes, that the energy quantity

$\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\cdot\frac{\mathfrak{B}_{+}^{2}}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\sin\phi\ d\phi$

falls upon $A$, and that we have to multiply this energy quantity with $\frac{i_{0}}{i}$ and $-\frac{p_{2}c}{i}$ respectively, to obtain the fraction of them which becomes absorbed or transformed into work.

Altogether, the energy quantity

$\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{i_{0}}{i}+\mathfrak{B}_{+}^{2}\frac{i_{0}}{i'}\right)=M$

is absorbed, while the energy quantity

$\int_{0}^{\pi/2}\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}\left(\mathfrak{B}_{-}^{2}\frac{p_{1}c}{i}-\mathfrak{B}_{+}^{2}\frac{p_{2}c}{i'}\right)=N$

is transformed into work. One can convince oneself, that exactly the same expressions are valid, when for example the black surface $B$ is replaced by a mirror. Namely, the radiation coming from $A$ to $B$ is not absorbed in $B$, but is partly reflected, i.e., from the totality of the energy incident in $B$, the fraction $\tfrac{i'}{i}$ is reflected; it comes back to $A$, and the fraction $\frac{i_{0}}{i'}$ is absorbed there, so that eventually the fraction $\tfrac{i'}{i}\cdot\tfrac{i_{0}}{i'}=\tfrac{i_{0}}{i}$ of the radiation propagating at the beginning in the direction from $A$ to $B$, is absorbed again. Thus $M$ and therefore also $N$ remain unchanged.

Now, if one uses equations (4), (5), (7) and (8), then

$M=\frac{2\pi i_{0}D}{\mathfrak{B}^{4}}\int_{0}^{\pi/2}\sin\phi\ d\phi(\mathfrak{B}_{-}^{3}+\mathfrak{B}_{+}^{3})$

and

$N=\frac{2\pi i_{0}D}{\mathfrak{B}^{3}}\int_{0}^{\pi/2}\frac{\sin\phi\ d\phi}{\sqrt{1-\sigma^{2}\sin^{2}\phi}}$ $\cdot\left[\sigma^{2}\sin^{2}\phi(\mathfrak{B}_{-}^{2}+\mathfrak{B}_{+}^{2})+\sigma\cos\phi\sqrt{1-\sigma^{2}\sin^{2}\phi}(\mathfrak{B}_{-}^{2}-\mathfrak{B}_{+}^{2})\right]$

By execution of this integration one can convince oneself, that it is quite exactly

$M=\frac{4\pi i_{0}D}{\mathfrak{B}}=E_{0};\ N=0;$

Thus the total energy initially present in $R$, is absorbed: Thus the earlier assertions concerning the work necessary for the acceleration, are confirmed. On the other hand we also see, that from the energy content of the cavity, i.e. in expression (1), only the first part is provided by the heat content of the radiating body. This is an assumption, at which I started in my cited treatise concerning the dimension changes of matter due to the motion through the aether, and which was perhaps not sufficiently justified at that place.

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