Translation:On the Thermodynamics of Moving Systems

by Friedrich Hasenöhrl

On the Thermodynamics of Moving Systems

by

Dr. Fritz Hasenöhrl

(Presented in the session of October 31, 1907.)

A specific electromagnetic momentum is connected to radiation in a moving cavity. Because the heat content of each body partly consists of radiating energy, each body has a specific electromagnetic mass that depends on its energy content, thus e.g. also on its temperature. This assertion was demonstrated by me in earlier papers.^[1] Since then, a paper by v. Mosengeil on the radiation in a moving cavity has been published, in which (besides other things) the energy content of the moving cavity has been calculated by the aid of the relation between energy and momentum.^[2] Furthermore, Planck^[3] has studied the dynamics of an arbitrarily moving system, where he presupposes the existence of the mentioned electromagnetic momentum.

Planck presupposes the validity of the so called relativity principle in the form of Einstein, then he uses the moving cavity as test body, and so he obtains theorems that are valid for each body. In the present paper I have attempted to work out a theory of an arbitrarily moving body as well. The chosen way differs essentially from the method of Planck. Only the thermodynamic theorems as well as the definition of electromagnetic momentum are presupposed. If one then requires that a co-moving observer shall not notice his motion, then the Fitzgerald-Lorentz contraction hypothesis is given.

Thus we consider an arbitrary body, whose state is given by the inner energy $U_{0}$ and the volume $v$ when at rest. If it is adiabatically brought to the velocity $q=\beta c$,^[4] then it has the specific momentum ${\mathfrak {G}}$ which must be representable as a function of $U_{0},v,\beta$. We put

${\mathfrak {G}}={\frac {1}{c}}\phi (U_{0},v,\beta ).$

There, the performed work of the translation forces is

$\int q\ dt{\frac {d{\mathfrak {G}}}{dt}}=\int _{0}^{\beta }\beta {\frac {\partial \phi }{\partial \beta }}d\beta .$

The energy of the body has increased by this amount; if we denote it by $U$, then

$U=U_{0}+\int _{0}^{\beta }\beta {\frac {\partial \phi }{\partial \beta }}d\beta =\Phi (U_{0},v,\beta ),$ ${\frac {\partial \Phi }{\partial \beta }}=\beta {\frac {\partial \phi }{\partial \beta }}.$

(1)

Furthermore, we introduce the quantity

$H=U-\beta \phi \,$,

(2)

that we can also consider as function of $U_{0},v$ and $\beta$. Then it is:

$H=U_{0}-\int _{0}^{\beta }\phi \ d\beta ,$

(3)

$\phi =-{\frac {\partial H}{\partial \beta }},$

(4)

$U=H-\beta {\frac {\partial H}{\partial \beta }}.$

(5)

In the differentiation of one of the quantities $U_{0},v,\beta$, the two others have to be kept constant. We also emphasize, that we understand $U_{0}$ as the energy which it obtains when the body is brought adiabatically and isochorically to rest; the way by which the body has acquired its momentary (moving) state, is completely irrelevant.

1. Computation of the pressure.

We denote the pressure of the resting system by $p_{0}$, that of the moving one by $p$. In order to compute the latter, we consider the following circular process:

A. The initial state shall be that of rest; $U_{0},v,p_{0}$ shall be the values of the relevant state variables. We change the volume adiabatically from $v$ to $v'=v+dv$; then the energy assumes the value $U'_{0}=U_{0}-p_{0}dv$.

B. We bring the body to the velocity $\beta c$. The energy assumes the value

$U=\Phi (U'_{0},\ v',\ \beta )$

The work of the external forces is

$U-U'_{0}=\Phi (U'_{0},\ v',\ \beta )-U'_{0}$.

C. We change (at constant velocity) the volume adiabatically by $-dv$. The external forces perform the compression work $pdv$ and the translation work $\beta d\phi$ (in order to keep the velocity constant). Thus the increase of energy is

$dU=pdv+\beta d\phi \,$.

Let $U''_{0}$ be the value now assumed by $U_{0}$; then it is

$dU=\Phi (U''_{0},\ v,\ \beta )-\Phi (U'_{0},\ v',\ \beta ),$ $d\phi =\phi (U''_{0},\ v,\ \beta )-\phi (U'_{0},\ v',\ \beta ).$

D. We bring the body to rest in an adiabatic and isochoric way. There, the work

$U''_{0}-\Phi (U''_{0},\ v,\ \beta )$

is performed. The state of the body is now given by the variables $U''_{0},\ v,\ \beta =0$. The total work of the external forces is:

$(-p_{0}dv)+(\Phi (U'_{0},\ v',\ \beta )-U'_{0})+(pdv+\beta d\phi )+(U''_{0}-\Phi (U''_{0},v,\ \beta )).$

According to the first thermodynamic main-theorem, the work must be equal to $U''_{0}-U_{0}$. The second main-theorem additionally requires, that this difference is zero. Otherwise this circular process (or the reverse one) would represent a thermal perpetual motion machine. Thus we put $U''_{0}-U_{0}$ in the previous expressions; then we notice that

${\begin{array}{lll}d\phi &=&\phi (U_{0},v,\beta )-\phi (U'_{0},v',\beta )=\\\\&=&{\frac {\partial \phi }{\partial U_{0}}}(U_{0}-U'_{0})+{\frac {\partial \phi }{\partial v}}(v-v')=\\\\&=&{\frac {\partial \phi }{\partial U_{0}}}p_{0}dv-{\frac {\partial \phi }{\partial v}}dv\end{array}}$

and

$\Phi (U'_{0},v',\beta )-\Phi (U_{0},v,\beta )={\frac {\partial \Phi }{\partial v}}dv-{\frac {\partial \Phi }{\partial U_{0}}}p_{0}dv$

thus we obtain:

${\frac {\partial \Phi }{\partial v}}dv-{\frac {\partial \Phi }{\partial U_{0}}}p_{0}dv+\beta {\frac {\partial \phi }{\partial U_{0}}}p_{0}dv-\beta {\frac {\partial \phi }{\partial v}}dv+pdv=0$

or according to (2):

$p=p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}.$

(6)

Thus we obtain the theorem: If an arbitrary body (in the state of rest) is under the pressure $p_{0}$, then it assumes (at adiabatic-isochoric acceleration) the value $p$ as given by equation (6).

This theorem can be easier derived in the following way (even though it is less clear from the physical standpoint): At adiabatic state changes, the amount of $U$ only depends on the momentary values of the quantities $\beta$ and $v$. Thus when (at arbitrary velocity) $v$ is adiabatically changed by $dv$, then $U_{0}$ changes by $-p_{0}dv$.^[5] Then the energy increase, which is equal to the work of the external forces here, is:

$dU=-pdv+\beta d\phi \,$

and therefore also

$-pdv-\phi d\beta$

is a complete differential, thus

${\frac {\partial p}{\partial \beta }}=\left({\frac {\partial \phi }{\partial v}}\right)$

Here, $\left({\tfrac {\partial }{\partial v}}\right)$ is to be understood as a differentiation at adiabatic state change; thus if $\phi$ is given as an explicit function of $v$ and $U_{0}$, we have:

$\left({\frac {\partial \phi }{\partial v}}\right)={\frac {\partial \phi }{\partial v}}+{\frac {\partial \phi }{\partial U_{0}}}\left({\frac {\partial U_{0}}{\partial v}}\right)={\frac {\partial \phi }{\partial v}}-p_{0}{\frac {\partial \phi }{\partial U_{0}}}.$

Furthermore, since we have according to (4)

$\phi =-{\frac {\partial H}{\partial \beta }}$

the previous equation can be integrated towards $\beta$ and we obtain

$p=p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}+\mathrm {const.}$

This constant can also be a function of $U_{0}$ and $v$; it reduces to zero, since we have

$H=U_{0};\quad \left({\frac {\partial H}{\partial v}}\right)_{U_{0}}=0$

for $\beta =0,p=p_{0}$.

2. The differential of the supplied heated.

is increased (as the energy increase) by the performed work of the considered body, thus

$dQ=dU+pdv-\beta d\phi$^[6]

If we again introduce $U_{0},v$ and $\beta$ as independent variables, then it becomes:

$dQ={\frac {\partial \Phi }{\partial U_{0}}}dU_{0}+{\frac {\partial \Phi }{\partial v}}dv+{\frac {\partial \Phi }{\partial \beta }}d\beta --\beta \left({\frac {\partial \phi }{\partial U_{0}}}dU_{0}+{\frac {\partial \phi }{\partial v}}dv+{\frac {\partial \phi }{\partial \beta }}d\beta \right)+pdv.$

If we consider (1), (2) and (6), then it becomes:

$dQ={\frac {\partial H}{\partial U_{0}}}dU_{0}+{\frac {\partial H}{\partial v}}dv+\left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)dv,$

or

$dQ={\frac {\partial H}{\partial U_{0}}}(dU_{0}+p_{0}dv).$

(7)

This expression is valid in full generality.

3. The temperature of the moving body.

We first consider a system of bodies, all of them moving with the same constant velocity. Experience tells us that $dQ/T$ is then a complete differential. Equation (7) shows that this condition is satisfied, when we put

$T=T_{0}{\frac {\partial H}{\partial U_{0}}}\cdot f(\beta )$

(8a)

$T_{0}$ is the integrating denominator of $dU_{0}+p_{0}dv$, when we (analogous to the preceding) understand under $T_{0}$ the temperature assumed by the body when it is adiabatically and isochorically brought to velocity zero. The arising function of $\beta$ plays here the role of a constant, so it is irrelevant. Of course, it must have the same value for all bodies.

Since we see $\beta$ as constant in this case, it is

$dQ={\frac {\partial H}{\partial U_{0}}}(dU_{0}+p_{0}dv)=dH+pdv.$

In a system whose velocity isn't changing, $H$ plays the role of the inner energy for a co-moving observer; between the quantities $H,v,p,t$ the same relations exist, which follows from the thermodynamic main-theorems for $U_{0},v,p_{0},T_{0}$.

Now, let a body pass through Carnot's circular process, in which the two reservoirs have different velocities; namely, $T_{1},\beta _{1}c$ are the temperature and velocity of one reservoir; $T_{2},\beta _{2}c$ are the relevant quantities for the other one. If the theorem of the impossibility of a thermal perpetual motion machine is also valid when it assumes different velocities in its different stages, then the ratio of the heat quantities given off to the reservoirs cannot depend on the nature of the circular process. Then it must be

${\frac {dQ_{1}}{dQ_{2}}}={\frac {\left[{\frac {\partial H}{\partial U_{0}}}(dU_{0}+p_{0}dv)\right]_{1}}{\left[{\frac {\partial H}{\partial U_{0}}}(dU_{0}+p_{0}dv)\right]_{2}}}=\varphi (T_{1},T_{2},\beta _{1},\beta _{2})$

One can easily see, that this function must have the form

${\frac {\phi (T_{1},\beta _{1})}{\phi (T_{2},\beta _{2})}}$

Furthermore

${\frac {\phi (T_{1},\beta )}{\phi (T_{2},\beta )}}={\frac {T_{1}}{T_{2}}}$

must be the case as well, because the ordinary definition of temperature must hold for bodies of same velocity, so the form for $\phi$ is given by

$\phi (T,\beta )=T\cdot g(\beta ).$

We want to set this function $g(\beta )$, as well as function $f(\beta )$ in (8a), equal to one; then it becomes

$T=T_{0}{\frac {\partial H}{\partial U_{0}}}.$

(8)

Though we have to emphasize, that there is a certain arbitrariness in this. Even when we don't set these functions equal to one, we neither come into contradiction with the theorem of the impossibility of a thermal perpetual motion machine, nor with the ordinary definition of temperature which is indeed only related to bodies of same velocity. The criterion of equality of temperature is not applicable to bodies of unequal velocity, since we cannot directly bring them in reversible heat exchange, but only with the aid of an auxiliary body which assumes different velocities. Though if we don't set $g(\beta )$ equal to one, then also the entropy of the adiabatic acceleration changes.

Anyway, it is the easiest way to define $T$ by equation (8); then $dQ/T$ is a complete differential and the entropy remains constant at adiabatic acceleration.^[7]

4. The entropy of a moving body.

We arrived at the result that pressure and temperature assume the values

$p=p_{0}{\frac {\partial H}{\partial H_{0}}}-{\frac {\partial H}{\partial v}},$

(6)

$T=T_{0}{\frac {\partial H}{\partial U_{0}}}$

(8)

at isochoric-adiabatic acceleration. $H$ plays the role of the inner energy in a system moving with constant velocity.

Let the entropy of the resting system be $S_{0}(U_{0},v)$, that of the moving one can be expressed by $S(H,v)$. The relations hold

$\left({\frac {\partial S_{0}}{\partial U_{0}}}\right)_{v}={\frac {1}{T_{0}}},\quad \left({\frac {\partial S_{0}}{\partial v}}\right)_{U_{0}}={\frac {p_{0}}{T_{0}}};$

but also

$\left({\frac {\partial S}{\partial H}}\right)_{v}={\frac {1}{T}},\quad \left({\frac {\partial S}{\partial v}}\right)_{H}={\frac {p}{T}};$

since it is indeed (at constant $\beta$)

$dS={\frac {1}{T}}(dH+pdv)$.

Since the system was adiabatically brought from the state of rest to that of motion, the entropy has the same value in both cases, thus:

$S_{0}(U_{0},v)=S(H,v)\,$

therefore also

${\frac {\partial S_{0}}{\partial U_{0}}}={\frac {\partial S}{\partial U_{0}}}={\frac {\partial S}{\partial H}}{\frac {\partial H}{\partial U_{0}}},$

$\left({\frac {\partial S_{0}}{\partial v}}\right)_{U_{0}}=\left({\frac {\partial S}{\partial v}}\right)_{U_{0}}={\frac {\partial S}{\partial H}}\left({\frac {\partial H}{\partial v}}\right)_{U_{0}}+\left({\frac {\partial S}{\partial v}}\right)_{H}.$

From that, also equations (6) and (8) are immediately given.

5. Momentum.

Thus far we have presupposed the existence of momentum, without making a special assumption concerning its value. Now we want to assume in agreement with the theory of Lorentz and Abraham, that momentum is equal to the space integral of the (absolute) energy flow, divided by the square of the speed of light. If we assume that it's about the flow of the total energy, i.e. that the total inner energy is of electromagnetic nature, then momentum can be calculated by the following simple consideration.

We consider a cylindrical body of cross-section unity, moving in the direction of its axis (a differently deformed body can be imagined as divided in cylindrical parts). By an arbitrary cross-section that shares the motion, the (relative) energy flow $\pi _{1}$ shall flow in the direction of motion, the energy flow $\pi _{2}$ in the opposite direction. Since the body is imagined as homogeneous, these quantities are independent of the location of the cross-section; thus the backward (in the sense of motion) base surface will emanate the energy quantity $\pi _{1}$ in unit time, and will get the energy quantity $\pi _{2}$. The difference $\pi _{1}-\pi _{2}$ must be equal to the work performed at this surface in unit time. The force acting here is the pressure $p$; the pressure work in unit time is $pq$; thus

$pq=\pi _{1}-\pi _{2}\,$.

In order to calculate the absolute energy flow, i.e. the energy flow through a cross-section imagined as at rest, we have to add the product of the energy density multiplied with the translation velocity, to the relative energy flow in the direction of motion, i.e. to the quantity $\pi _{1}-\pi _{2}$.^[8] If we denote the first one with $u$ for the moment, then the absolute energy flow through the cross-section is given by the quantity

$\pi _{1}-\pi _{2}+qu=(p+u)q\,$.

If we multiply this quantity with the volume and divide by $c^{2}$, then

${\mathfrak {G}}={\frac {1}{c^{2}}}v(p+u)q$

or, since we denote $uv$ with $U$

${\mathfrak {G}}={\frac {1}{c^{2}}}(pv+U)q.$[9]

(10)

Thus the types of inner energy don't matter, as long as they are only of electromagnetic nature (we imagine that they are composed of radiating energy and energy of arbitrarily moving electrons). Also the relative velocity and the energy flow don't matter as well; the individual energy types can of course flow with different velocities.

Of course, one comes to the same result when the individual energy flows are taken into account. For instance, let $u(\phi )\sin \phi \ d\phi$ be the density of a specific energy type, moving in a relative direction that encloses an angle between $\phi$ and $\phi +d\phi$ with the direction of motion. Then the total energy of this type is

$U=2\pi \ v\int _{0}^{\pi }\ u(\phi )\ \sin \phi \ d\phi .$

We obtain the momentum when we multiply the absolute flow, i.e. $u(\phi )\sin \phi \ d\phi \cdot \omega _{A}$ (where $\omega _{A}$ is the flow-velocity) with $\cos \varphi$, where $\varphi$ is the angle between the absolute flow direction and the direction of motion. Thus:

${\mathfrak {G}}={\frac {2\pi v}{c^{2}}}\int _{0}^{\pi }u(\phi )\cdot \omega _{A}\cdot \cos \varphi \cdot \sin \phi \ d\phi .$

However, now it is^[10]

$\omega _{A}\cos \varphi =q+\omega _{R}\cos \phi$,

where $\omega _{R}$ is the relative velocity ($\omega _{A}$ and $\omega _{R}$ are in general functions of $\phi$ or $\varphi$).

Thus

${\mathfrak {G}}={\frac {2\pi v}{c^{2}}}q\int _{0}^{\pi }u(\phi )\sin \phi \ d\phi +{\frac {2\pi v}{c^{2}}}\int _{0}^{\pi }u(\phi )\sin \phi \cos \phi \omega _{R}\ d\phi .$

The first summand is equal to ${\tfrac {1}{c^{2}}}qU$; the second one gives the surplus of the energy emanating from the basis surface over the inflowing energy, it is thus connected with the pressure work $pq$, by which we come to equation (10) again.

6. The change of volume.

A resting system in mechanical and thermal equilibrium is given, i.e. in which all bodies have the same pressure and the same temperature. If this system is adiabatically set into motion (each body adiabatically for itself), then pressure and temperature of every single body is changing, namely in different measure for every single body as we must assume from the outset. Thus the equilibrium is disturbed; if it is restored again, then the individual bodies must change their volumes. If these volume changes are different for different bodies, then they are principally observable. However, if the mechanical and thermal equilibrium is restored again by the change of dimensions of all bodies in the same way, then an influence of the common translatory motion is not observable.

This is indeed the case; first it can be shown that the pressure of a body remain unchanged when $\beta$ is adiabatically changed by $d\beta$ and $v$ by $-v{\tfrac {\beta d\beta }{1-\beta ^{2}}}$ at the same time. Thus it must be

$dp={\frac {\partial p}{\partial U_{0}}}dU_{0}+{\frac {\partial p}{\partial v}}dv+{\frac {\partial p}{\partial \beta }}d\beta =0$

(11)

when

$dv=-v{\frac {\beta d\beta }{1-\beta ^{2}}}$ and $dU_{0}=-p_{0}dv\,$

(12)

(the latter relation generally holds for adiabatic state changes, according to (7)).

We notice that according to (10) and (2)

${\begin{array}{rl}H=&U-\beta \phi =U-q{\mathfrak {G}}=U-\beta ^{2}(pv+U)\\=&U(1-\beta ^{2})-\beta ^{2}pv\end{array}}$

If we also insert for $U$ its value from (5), then it is given

$H=(1-\beta ^{2})H-(1-\beta ^{2})\beta {\frac {\partial H}{\partial \beta }}-\beta ^{2}pv$

or

$(1-\beta ^{2}){\frac {\partial H}{\partial \beta }}=-\beta (H+pv).$

(13)

Then it is

$(1-\beta ^{2}){\frac {\partial p}{\partial \beta }}=(1-\beta ^{2}){\frac {\partial }{\partial \beta }}\left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)=$ ${\begin{array}{l}=-\beta p_{0}{\frac {\partial }{\partial U_{0}}}(H+pv)+\beta {\frac {\partial }{\partial v}}(H+pv)=\\\\=-\beta \left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)-\beta v\left(p_{0}{\frac {\partial p}{\partial U_{0}}}-{\frac {\partial p}{\partial v}}\right)+\beta p=\\\\=\beta v\left({\frac {\partial p}{\partial v}}-p_{0}{\frac {\partial p}{\partial U_{0}}}\right).\end{array}}$

If we insert this value into (11) and consider (12), then we see that indeed $dp=0$.

The simultaneous change of $T$ is:

$dT={\frac {\partial T}{\partial v}}dv+{\frac {\partial T}{\partial U_{0}}}dU_{0}+{\frac {\partial T}{\partial \beta }}d\beta =$ $=-v{\frac {\beta d\beta }{1-\beta ^{2}}}\left({\frac {\partial T}{\partial v}}-p_{0}{\frac {\partial T}{\partial U_{0}}}\right)+{\frac {\partial T}{\partial \beta }}d\beta .$

Namely, it is according to (8):

${\frac {\partial T}{\partial \beta }}=T_{0}{\frac {\partial ^{2}H}{\partial \beta \ \partial U_{0}}}=-{\frac {1}{1-\beta ^{2}}}T_{0}\beta {\frac {\partial }{\partial U_{0}}}(H+pv)=$ $=-{\frac {1}{1-\beta ^{2}}}\beta T_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {1}{1-\beta ^{2}}}\beta vT_{0}{\frac {\partial p}{\partial U_{0}}}.$

However, it is now

$T_{0}{\frac {\partial p}{\partial U_{0}}}=T_{0}\left({\frac {\partial p_{0}}{\partial U_{0}}}{\frac {\partial H}{\partial U_{0}}}+p_{0}{\frac {\partial ^{2}H}{\partial U_{0}^{2}}}-{\frac {\partial ^{2}H}{\partial U_{0}\partial v}}\right).$

If we insert herein

$T_{0}{\frac {\partial p_{0}}{\partial U_{0}}}=p_{0}{\frac {\partial T_{0}}{\partial U_{0}}}-{\frac {\partial T_{0}}{\partial v}},$

a relation known as following from the thermodynamics of resting bodies, then it becomes:

$T_{0}{\frac {\partial p}{\partial U_{0}}}=\left(p_{0}{\frac {\partial T_{0}}{\partial U_{0}}}-{\frac {\partial T_{0}}{\partial v}}\right)\cdot {\frac {\partial H}{\partial U_{0}}}+p_{0}T_{0}{\frac {\partial ^{2}H}{\partial U_{0}^{2}}}-T_{0}{\frac {\partial ^{2}H}{\partial U_{0}\partial v}}=$ $=p_{0}{\frac {\partial }{\partial U_{0}}}\left(T_{0}{\frac {\partial H}{\partial U_{0}}}\right)-{\frac {\partial }{\partial v}}\left(T_{0}{\frac {\partial H}{\partial U_{0}}}\right)=p_{0}{\frac {\partial T}{\partial U_{0}}}-{\frac {\partial T}{\partial v}}.$

Thus

${\frac {\partial T}{\partial \beta }}=-{\frac {\beta }{1-\beta ^{2}}}\cdot T-{\frac {\beta }{1-\beta ^{2}}}v\left(p_{0}{\frac {\partial T}{\partial U_{0}}}-{\frac {\partial T}{\partial v}}\right)$

and

$dT=-{\frac {\beta d\beta }{1-\beta ^{2}}}T.$

The change of temperature is thus the same for all bodies.

This expression as well as the expression for $dv$ (12) can be immediately integrated. Then we obtain:

$v=v_{0}{\sqrt {1-\beta ^{2}}},$ $T=T_{0}{\sqrt {1-\beta ^{2}}}.$

(14)

Thus we come to the result:

When the volume decreases with velocity according to the previous law, i.e. when the dimensions of matter are contracting in the direction of motion in the ratio

${\sqrt {1-\beta ^{2}}}$,

then the pressure of every body remains unchanged at adiabatic change of velocity, while the temperature of all bodies decreases in the same measure. Then, no influence of a common translatory motion is observable.

This is in agreement with the contraction hypothesis of H. A. Lorentz, as well as with the theorems derived by Planck from the so called relativity principle.

While Planck assumes the validity of the relativity principle from the outset, we arrived to a certain extent at a proof of the contraction hypothesis, by postulating the theorem that a common translatory motion is not observable for a co-moving observer; or additionally by demonstrating that a volume change must arise in the previously given way at constant pressure.

Continuation

(Presented in the session of February 6, 1908.)^[11]

7. Calculation of quantity H.

In order to express $H$ by the variables $U_{0},v,\beta$, we insert the value for $p$ from (6) into (13) and obtain

$(1-\beta ^{2}){\frac {\partial H}{\partial \beta }}+\beta H+\beta v\left(p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}\right)=0.$

(15)

This partial differential equation assumes a simpler form, when the quantities $\beta ,v,S_{0}$ are chosen instead of $\beta ,v,U_{0}$ as independent variables. Namely, $S_{0}$ shall be the value of entropy again, when the system is adiabatically brought to rest; of course $S=S_{0}$. Thus we think of $U_{0}$ as being expressed by entropy and volume; if for example

$U_{0}=F(S_{0},v)\,$.

Then it is:

${\frac {\partial }{\partial v}}-p_{0}{\frac {\partial }{\partial U_{0}}}=\left({\frac {\partial }{\partial v}}\right)_{S_{0}}$,

because according to (7), $U_{0}$ is changed by $-p_{0}dv$ at adiabatic volume change. If we furthermore introduce the variable

$\varkappa ={\sqrt {1-\beta ^{2}}}$

instead of $\beta$, then (15) becomes

$-\beta \varkappa {\frac {\partial H}{\partial \varkappa }}+\beta H-\beta v\left({\frac {\partial H}{\partial v}}\right)_{S_{0}}=0,$

or when $\beta$ is different from zero:

$\varkappa {\frac {\partial }{\partial \varkappa }}\left({\frac {H}{\varkappa }}\right)+v\left({\frac {\partial }{\partial v}}\right)_{S_{0}}\left({\frac {H}{\varkappa }}\right)=0.$

It follows from this equation, that $H/\varkappa$ must be a function of $v/\varkappa$, which of course must also depend on $S_{0}$. Furthermore, $H$ must be identical with $U_{0}$ for $\beta =0,\ \varkappa =1$. We satisfy these requirements when we put

${\frac {H}{\varkappa }}=F\left(S_{0},{\frac {v}{\varkappa }}\right)$.

$F\left(S_{0},{\tfrac {v}{\varkappa }}\right)$ is evidently the energy amount of the resting system, when it is adiabatically expanded from $v$ to $v/\varkappa$; if we denote this energy value with $U'_{0}$, then

$H={\sqrt {1-\beta ^{2}}}\cdot U'_{0}.$

Now, if we assume in accordance with Lorentz's hypothesis, that the velocity change is accompanied with a volume change proportional to ${\sqrt {1-\beta ^{2}}}$, then $U'_{0}$ is the energy of the resting body; then we remove the prime and thus put:

$H={\sqrt {1-\beta ^{2}}}\cdot U{}_{0}.$

(16)

8. Summary of results.

With the aid of equations (2) and (10), momentum and total energy ($U$) can be expressed by the state variables of the resting system. (We have to consider here, that equations (1), (3), (4) and (5) may not be applied now; they only hold for velocity changes at constant volume.) We obtain

$U=H+\beta \phi ={\sqrt {1-\beta ^{2}}}\,U_{0}+\beta ^{2}(U+pv),$

from which it is given under consideration of the first equation (14):

$U={\frac {1}{\sqrt {1-\beta ^{2}}}}(U_{0}+\beta ^{2}p_{0}v_{0}).$

(17)

Finally, the momentum is according to (10):

${\mathfrak {G}}={\frac {\beta }{c}}(pv+U)={\frac {\beta }{c{\sqrt {1-\beta ^{2}}}}}(U_{0}+p_{0}v_{0}).$

(18)

If we summarize everything, we come to the result:

If a body whose state at rest is given by the variables $v_{0},U_{0},p_{0},T_{0},S_{0}$, is adiabatically brought to velocity $\beta c$, then the state variables assume the value:

$v=v_{0}{\sqrt {1-\beta ^{2}}}$

(14)

$p=p_{0}\,$

$T=T_{0}{\sqrt {1-\beta ^{2}}}$

(14)

$U={\frac {1}{\sqrt {1-\beta ^{2}}}}(U_{0}+\beta ^{2}p_{0}v_{0})$

(17)

$H={\sqrt {1-\beta ^{2}}}\cdot U{}_{0}$

(16)

$S=S_{0}\,$

${\mathfrak {G}}={\frac {\beta }{c}}{\frac {1}{\sqrt {1-\beta ^{2}}}}(U_{0}+p_{0}v_{0}).$

(18)

These equations are in agreement with the results of the paper of Planck^[12]. Besides thermodynamics, Planck used the relativity principle, while stating equation (10) for momentum is essential in our work.

9. Application to cavity radiation.

We base our calculation on the relative ray path. We consider radiation that encloses angles between $\phi$ and $\phi +d\phi$ with the direction of motion; it carries – in unit volume through the unit surface of a perpendicular (co-moving) plane – the energy amount:

$2\pi J\ \sin \phi \ \cos \phi \ d\phi .$

We call $J$ the intensity of the total (relative) radiation. If this radiation is incident upon an absorbing surface, it performs the pressure work:^[13]

$q\cdot {\frac {2\pi J\sin \phi \cos \phi \ d\phi }{c}}\cdot \cos \varphi =2\pi J\sin \phi \cos \phi \ d\phi \ \beta \cos \varphi ,$

where $\varphi$ is the angle between the absolute radiation direction and the direction of motion. The difference:

$2\pi J\ \sin \phi \ \cos \phi \ d\phi (1-\beta \ \cos \varphi )=2\pi i\ \sin \phi \ \cos \phi \ d\phi$

we call the true (relative) radiation. The true radiation intensity

$i=J(1-\beta \cos \varphi )$

(19)

is crucial for the heat transport between bodies of equal velocity.^[14]

We employ the standpoint of Lorentz's contraction hypothesis and introduce the angle $\phi '$ by the equation

${\begin{array}{rl}\operatorname {tg} \ \phi '=&\varkappa \ \operatorname {tg} \ \phi \\\varkappa {}^{2}=&1-\beta ^{2}\end{array}}$

(20)

Then instead of $\phi$, $\phi '$ is the angle read by a co-moving observer at a transporter, whose dimensions are contracted in the direction of motion in the ratio $1:\varkappa$. If we set

$i\ \sin \phi \ \cos \phi \ d\phi =i'\ \sin \phi '\ \cos \phi '\ d\phi ',$

then $i'$ is the true radiation intensity observed by the co-moving observer, whose measuring rods have experienced the mentioned contraction.

The quantity $i'$ must be constant, i.e. independent of angle $\phi '$, i.e. the true radiation is uniformly distributed into all directions in the contracted system; it obeys Lambert's $\cos$-law. Then two arbitrary oriented, equally moving surface elements are radiating the same amount of heat to each other. A mirror brought into a cavity doesn't change the distribution of radiation, since the ordinary reflection laws hold for the relative ray path in the contracted system. (This was shown in the most general way by H. A. Lorentz^[15] and can be directly proven in this case.)

From (20) it easily follows:

$\sin \phi '\cos \phi '\ d\phi '=\sin \phi \cos \phi \ d\phi {\frac {\varkappa ^{2}}{(1-\beta ^{2}\sin ^{2}\phi )^{2}}},$

thus it is:

$i=i'{\frac {\varkappa ^{2}}{(1-\beta ^{2}\sin ^{2}\phi )^{2}}}=i'{\frac {(1-\beta ^{2}\cos ^{2}\phi ')^{2}}{\varkappa ^{2}}}$.[16]

The energy content of the cavity is:

$U=2\pi v\int _{0}^{\pi }{\frac {J\sin \phi \ d\phi }{c'}},$

where $c'$ means the relative velocity:

$c'=c\left(-\beta \cos \phi +{\sqrt {1-\beta ^{2}\sin ^{2}\phi }}\right).$

If we set in the previous integral according to (19):

$J=i+J\beta \ \cos \varphi ,$

then it becomes:

$U=2\pi v\int _{0}^{\pi }{\frac {i\ \sin \phi \ d\phi }{c'}}+2\pi v\beta \int _{0}^{\pi }{\frac {J\cos \phi \ \sin \phi \ d\phi }{c'}}.$

The second summand is equal to:

$q\cdot {\frac {2\pi v}{c^{2}}}\int _{0}^{\pi }{\frac {J}{c'}}\cdot c\ \cos \varphi \cdot \sin \phi \ d\phi =q{\mathfrak {G}},$

as one can most simply recognize by comparison with the penultimate equation of p. 11 of my first report. In consequence of (2) it is therefore:

$H=2\pi v\int _{0}^{\pi }{\frac {i\ \sin \phi \ d\phi }{c'}}.$

The quantity $H$ is thus identical with the energy of the true radiation, which was indeed to be expected. If we introduce $i'$ and $\phi '$ by means of (20) and (21), then

${\begin{array}{rl}H&={\frac {2\pi vi'}{c\varkappa ^{2}}}\int _{0}^{\pi }\sin \phi 'd\phi '(1+\beta \ \cos \ \phi ')\\\\&={\frac {4\pi v}{\varkappa ^{2}}}\cdot {\frac {i'}{c}}.\end{array}}$

However, according to (16) it is:

$H={\sqrt {1-\beta ^{2}}}\cdot U_{0}=\varkappa \cdot {\frac {4\pi v_{0}}{c}}i_{0},$

where $i_{0}$ is the radiation intensity in the resting cavity. Thus it must be

$\varkappa v_{0}i_{0}={\frac {1}{\varkappa ^{2}}}vi'$;

or, since $v=\varkappa v_{0}$:

$i'=\varkappa ^{2}i_{0}.$

This is in agreement with the generally valid theorems of the theory of H. A. Lorentz.^[17]

If we set in accordance with the Stefan-Boltmann law

$i_{0}=\sigma T_{0}^{4},$

then it follows (see. (14)):

$i'=\varkappa ^{2}\sigma T_{0}^{4}={\frac {\sigma }{\varkappa ^{2}}}T^{4}.$

The constant of the Stefan-Boltzmann law is thus to be divided by $\varkappa ^{2}$.

The energy density of the true radiation is:

${\frac {H}{v}}={\frac {4\pi }{c}}\cdot {\frac {i'}{\varkappa ^{2}}}={\frac {4\pi }{c}}i_{0};$

thus it has the same value as in the resting cavity.

The total energy follows from (17): it has the value:

$U={\frac {1}{\sqrt {1-\beta ^{2}}}}\left(1+{\frac {1}{3}}\beta ^{2}\right)U_{0}={\frac {4\pi v_{0}i_{0}}{c}}\cdot {\frac {1+{\frac {1}{3}}\beta ^{2}}{\sqrt {1-\beta ^{2}}}}.$

In earlier papers – already cited multiple times – I have tried to calculate the energy content and apparent mass of a moving cavity. There I assumed isotropic distribution (in all directions) of true radiation. A contradiction with the second thermodynamic main-theorem arising at that occasion, could be solved by two different hypotheses. Either by the assumption of a change of the emission capability of the black body, or by the hypothesis of a change of the dimensions of matter due to motion. I have confined myself to the study of the latter hypothesis, though I have explicitly emphasized the possibility of the first assumption.^[18] The isotropy of the distribution of true radiation was assumed by me, so that the mutual radiation of two elements is equal. If one employs Lorentz's contraction hypothesis, as it was done by me, then this assumption must of course be modified, so that the isotropy of the true radiation is related to the contracted system. This modification was indicated by me,^[19] though I have confined myself to note, that the term of the apparent mass which is independent of velocity is not affected by this. The reason for this was, that a way of calculating the emission capability of the black body as a function of velocity was unknown to me at this time, thus I had to confine myself to the first term of the corresponding expansions. Since then, this gap was filled by the work of v. Mosengeil. By that and the fortunate thought of this author, to assume a change of temperature with velocity, it became possible to definitely determine the processes in a moving cavity. Based on the concept of true radiation, I did this in § 9 of the present paper; then one comes to the same result as v. Mosengeil. The deprecatory criticism of my earlier work by v. Mosengeil is thus not justified. I have used the opportunity to discuss this in more detail, since my first reply^[20] remained unconsidered in the new publication of the paper of v. Mosengeil.

In any case, the concept of a mass depending on inner radiation and thus on temperature was first stated in my earlier papers, and also the term of the mass which is independent of velocity was calculated for the cavity. Thus it is incomprehensible to me, why Planck – who discussed those things in detail in the introduction of his last publication^[21] – doesn't mention my papers at all.

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1. F. Hasenöhrl, these proceedings, CXIII, p. 1039, 1904; Ann. d. Phys. (4), 15, p. 344, 1904, and 16, p. 589, 1905.
2. K. v. Mosengeil, Berlin Dissertation 1906; Ann. d. Phys. (4), 22, p. 867, 1906. – Later, I will discuss v. Mosengeil’s criticism of my papers.
3. M. Planck, Berliner Berichte, 1907, p. 542.
4. Only reversible processes shall be discussed, $c$ is the speed of light in the aether.
5. Compare the following section 2.
6. Planck was the first who alluded to the fact, that the translation work $\beta d\phi$ must be considered in the determination of the temperature of a moving cavity.
7. This was also concluded in the papers of v. Mosengeil and Planck at the determination of the temperature of a moving cavity.
8. Compare for instance M. Abraham, Theorie der Elektrizität, II., p. 108.
9. The method given here is based on the consideration already given by me in an earlier paper (these proceedings, CXIII., p. 1039, 1904). Equation (10) was already derived by Planck. The method of Planck, however, has nothing at all to do with the one used here.
10. For instance, compare F. Hasenöhrl, Ann. d. Phys., 15, p. 347, 1904 (however, only radiating energy is considered there. We now have to replace the quantities denoted as $c$ and $c'$ by $\omega _{A}$ and $\omega _{R}$. Since the relations are purely geometrical, this replacement is allowed without further ado.
11. Compare these proceedings, CXVI, p. 1391 (1907).
12. Berliner Berichte, 1907, p. 542.
13. M. Abraham, Boltzmann-Festschrift, p. 90, 1904. Compare for instance F. Hasenöhrl, Jahrb. d. Radioaktivität, 2, p. 281 (1905).
14. This terminology agrees with the one used in an earlier paper (Ann. d. Phys., 15 [1904]). There, $i$ and $i_{0}$ was written instead of $J$ and $i$. See also Jahrb. d. Radioaktivität und Elektronik, 2, p. 283 (1905).
15. H. A. Lorentz, Versl. kon. Akad. v. Wetensch. Amsterdam, 7, p. 507 (1899) and 12, p. 886 (1904). – See also M. Abraham, Theorie der Elektrizität, II, p. 282 (1905).
16. The absolute radiation intensity can be calculated from this equation. It is equal to

    $J_{abs}=i\left({\frac {c}{c'}}\right)^{4}$

    (see. F. Hasenöhrl, Ann. d. Phys., 16, p. 589 [1905]), where

    ${\frac {c'}{c}}={\sqrt {1+\beta ^{2}-2\beta \cos \varphi }}$

    and $\varphi$ is again the direction of the absolute rays. Thus it is

    $J_{abs}=i'{\frac {\varkappa ^{2}}{(1-\beta ^{2}\sin ^{2}\phi )^{2}\left({\frac {c}{c'}}\right)^{4}}}=i'{\frac {\varkappa ^{2}}{(1-\beta \cos \varphi )^{4}}},$

    because $c'{\sqrt {1-\beta ^{2}\sin ^{2}\phi }}=c(1-\beta \cos \varphi )$ (see F. Hasenöhrl, Ann. d. Phys., 15, p. 347, Gl. 7 [1904]). That the absolute radiation intensity is changing with direction proportional to $(1-\beta \cos \varphi )^{4}$, was already demonstrated by v. Mosengeil in another way (Ann. d. Phys., 22, p. 875, eq. 11 [1907]).

17. See for instance, M. Abraham, Theorie der Elektrizität, II, p. 282 (1905)
18. So when v. Mosengeil says at the end of his paper, that I considered the dimension change as necessary, then this is based on a misunderstanding.
19. Ann. d. Phys. (4), 15, p. 350 (1904).
20. Ann. d. Phys., 22, p. 791 (1907).
21. Berl. Ber., 1907, p. 542.