Page:Zur Thermodynamik bewegter Systeme.djvu/10

 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 $$\pi_{1} - \pi_{2}$$. 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

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