Page:Zur Theorie der Strahlung in bewegten Körpern.djvu/9

moment about $B$ as having temperature O° A., then the radiation emanating from $A$ is the only one present in $R$, and the reaction principle requires (whose denial in this case would be contradicting all views expressed up to now) that radiation (12a) exerts the same pressure in $B$ as in $A$, though in opposite direction.^[1] Since the corresponding work, whose amount is again given by (14), is performed by the radiation here, then its amount is to be subtracted from (12a) to obtain the radiation absorbed in $B$, for which we again obtain the true radiation (15).

If we assume again, that $B$ has a certain temperature different from zero, then under a certain angle it is left by the total relative radiation

(12b)

$2\pi \ i\ \cos \psi \ \sin \psi \ \ d\psi$,

to which the quantity of work

$w\cdot 2\pi \ p_{2}\ \cos \psi \ \sin \psi \ \ d\psi$

is to be added in this case, to obtain the true radiation provided by $B$; as before, it provides the quantity of heat absorbed by $A$ at the other side of the cavity.

Then, the same though opposite pressure, is present at both sides:

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

We see, that the true relative radiation is solely decisive for the heat transport between $A$ and $B$; because it completely stems from the heat reservoir of one black body, and is completely transformed into the heat reservoir of the other one. The other energy present in $R$, the apparent radiation, is gained from mechanical work. At one side of the cavity, heat is steadily transformed into work, which traverses it and is retransformed at the other side as work of the same amount, so that altogether no work is performed or gained at uniform translation of our system.

1. The explanations in § 4 will confirm this.