Page:Prinzipien der Dynamik des Elektrons.djvu/7

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to investigate, whether Cohn's theory explains the facts, but this would be the task of the originator of the theory. I see the advantage of Lorentz's theory mainly in the fact, that one can derive the electromagnetic momentum from it. This simplifies the mathematical treatment and makes the analogy to ordinary mechanics possible. With respect to the second question I have made the following consideration. At rapid current fluctuations, Hertz's oscillations for instance, emitted energy comes into consideration; then it is not allowed anymore to calculate with a quasi-stationary current. The matter is similar with respect to rapidly accelerated motion of electrons. The radiation of energy and momentum can be calculated by the aid of the Wiechert-Des Coudres point theorem in the Lorentz-Festschrift. I have executed the calculation, and from that I found the basis for the determination of the area of applicability of quasi-stationary motion.

Meyer (Königsberg): I have a concern that you can maybe solve without further ado. These double integrals indeed occur very often. There, you have an ideal surface, that is moved into infinity by you, and then you say that the integral becomes equal to zero.

Now I would like to ask, whether the proof, that the integral vanishes when passing the limit, was quite easy. It was alluded to the fact, that at this place, physicists sometimes calculate not precisely and make things easy for them. Neumann has considered it necessary, to make a development with respect to spherical functions, in the cases where he wanted to prove it. Now, is the proof so simple in your case, that the integral vanishes when passing the limit?

Abraham: The assumption, that the boundary surface of the field lies in infinity, becomes irrelevant, when, for instance, the motion takes place very close to a conducting surface, as for example when cathode rays are reflected. If such cases are excluded, then it is allowed to treat the problem, as if the electron would be located alone in space. The proof, that the considered surface integrals then vanish, is not very hard, as long as one only offers a certain physical understanding for those things.