Page:Die Kaufmannschen Messungen.djvu/2

 of valuable data, and he, what must be recognized particularly grateful, has also given numerical data to the public, which is rich and reliable so that everyone is in a position to independently verify and complete the conclusions reached by.

I even more liked it to make use from this suggestion, as indeed the question at which 's experiments was aimed for, is almost vital for various electrodynamic theories. It is known that there exist already a number of excellent mathematical investigations from several of these theories, and their physical meaning, of course, would be repealed at once if that theory would be defeated in the resulting competition.

§ 1. Equations of motion.
We can presuppose that the method by which has examined the contents of the various theories by his measurements is known. At first I was interested to see how far each of those measured deflections are removed from those, which can be calculated from the various theories on the basis of the measured "constants of apparatus" from the outset. Since I preferred not to reduce the measured deflections $$(\overline{y},\overline{z})$$ from the start "to infinitely small deflections (y', z'), the equations of motion of the electrons had to be fully integrated. This gives for all compared theories:

$$\begin{matrix} \frac{d}{dt}\left(\frac{\partial H}{\partial\dot{x}}\right) & = & -\frac{e}{c}\dot{z}\mathfrak{H}\\ \\\frac{d}{dt}\left(\frac{\partial H}{\partial\dot{y}}\right) & = & e\mathfrak{E}\\ \\\frac{d}{dt}\left(\frac{\partial H}{\partial\dot{z}}\right) & = & \frac{e}{c}\dot{x}\mathfrak{H}.\end{matrix}$$

Here, H is the kinetic potential (the Lagrangian function) of a moving electron as a function of the velocity

$$q=\sqrt{\dot{x}^{2}+\dot{y}^{2}+\dot{z}^{2}}$$,

$$\mathfrak(E)$$ and $$\mathfrak(H)$$ is the electric and magnetic field strength, both acting in the y-direction as known functions of x, e is the electrical elementary quantum, c is the speed of light. The electrical quantities are measured in electrostatic units.

The electron moves from the radiation source:

x = x0 = 0 y = 0 z = 0

through the diaphragm opening:

x=x1=1,994 y=0 z=0

to the point of the photographic plate:

x=x2=3,963 y=$$\overline{y}$$ z=$$\overline{z}$$.

To hit the diaphragm opening straight, one electron has to be emanated with a certain initial velocity in a certain direction from the radiation source. On that occasion, a certain curve ($$\overline{y}, \overline{z}$$) emerges on the photographic plate (x = x2), whose points depend on a single parameter, for example on the initial velocity.

§ 2. Determination of the external field components.
The integration of the equations of motion still requires knowledge of $$\mathfrak{H}$$ and $$\mathfrak{E}$$ as functions of x. The magnetic field strength $$\mathfrak{H}$$ I assumed to be constant, and so great that the value of the "magnetic field integral" is the same as 's. Its value is:

$$\int_{x_{0}}^{x_{2}}dx\int_{x_{0}}^{x}\mathfrak{H}dx-\frac{x_{2}-x_{0}}{x_{1}-x_{0}}\int_{x_{0}}^{x_{1}}dx\int_{x_{0}}^{x}\mathfrak{H}dx=557,1$$.

Herein we set $$\mathfrak{H}$$ constant, so it follows:

$$\frac{1}{2}(x_{2}-x_{1})(x_{2}-x_{0})\mathfrak{H}=557,1$$

and from the values given by x0, x1 and x2:

$$\mathfrak{H}=142,8$$.

The electric field strength $$\mathfrak{E}$$ is zero between the diaphragm and the photographic plate, and constant between the capacitor plates at a proper distance from the edges. As, we connect the field strength to its value in the homogeneous part of the field and take it as unity, calling $$\mathfrak{E}_{1}$$ the field strength measured in this way. Then we have:

for $$x_{1}<x<x_{2}:\mathfrak{E}_{1}=0$$.

Between the radiation source and the diaphragm I have $$\mathfrak{E}_{1}$$ assumed to be symmetrical with respect to the center of that distance: $$x=\frac{x_{1}}{2}$$, so that if one puts:

The increase of the electric field strength $$\mathfrak{E}_{1}$$ from the radiation source to its constant value 1 is assumed to be linear, as well as the decrease to the value 0 at the diaphragm. That is: