Translation:Electrodynamic and Relativistic Theory of Electromagnetic Mass

By.

1. It's known that simple electrodynamic considerations lead to the value $$\tfrac{4}{3}\tfrac{U}{c^{2}}$$ for the electromagnetic mass of a spherical electricity-distribution of electrostatic energy $$U$$, when $$c$$ denotes the speed of light. On the other hand, it's known that relativistic considerations for the mass of a system containing the energy $$U$$ give the value $$\tfrac{U}{c^{2}}$$. Thus we stand before a contradiction between the two views, whose solution seems not unimportant to me, especially with respect to the great importance of the electromagnetic mass for general physics, as the foundation of the electron theory of matter.

Especially we will prove: The difference between the two values stems from the fact, that in ordinary electrodynamic theory of electromagnetic mass (though not explicitly) a relativistically forbidden concept of rigid bodies is applied. Contrary to that, the relativistically most natural and most appropriate concept of rigid bodies leads to the value $$U:c^{2}$$ for the electromagnetic mass.

We additionally notice, that relativistic dynamics of the electron was studied by, though from the standpoint only partially different from the ordinary electrodynamic one, so that the value $$\tfrac{4}{3}\tfrac{U}{c^{2}}$$ for the Electron's mass was found of course.

In this paper, 's principle will serve as a basis, being most useful for the treatment of a problem subjected to very complicated conditions - conditions of a different nature than those considered in ordinary mechanics, because our system must contract in the direction of motion according to relativity theory. However, we notice that although this contraction is of order of magnitude $$v^{2}:c^{2}$$, it changes the most important terms of electromagnetic mass, i.e, the rest mass.

2. Let us consider a system of electric charges connected by a rigid dielectric, being under the action of an electrostatic field, partially stemming from the system itself, and partially from external causes, describing a world-tube in the world. We further assume, that the motion is translatory, by which we understand: When we consider any reference frame and assume, in which at a certain moment, for example at time zero, a point of our conducting system is at rest, then all points of the conducting system must rest at time zero. From that it follows, that the world-lines of the points of our system are the orthogonal paths of a family of linear spaces; the rigidity is expressed by the condition, that the shape of the system in these spaces remains unchanged.

To be able to apply 's principle, we need a variation of motion of our system satisfying the condition of rigidity. Now we will prove, that we are led to $$\tfrac{4}{3}\tfrac{U}{c^{2}}$$ or to $$\tfrac{U}{c^{2}}$$ as the electromagnetic mass, depending on whether one chooses the first or the second of both variation systems, which in the following will defined by us, and denoted with the letters $$A$$ and $$B$$.

As we will immediately see, however, variation $$A$$ has to be excluded, because it is in contradiction with relativity theory.

Let $$T$$ be a world-tube described by the system. In the figure, we have indicated the space in a one-dimensional way by the $$X$$-axis, and replaced time $$t$$ by $$ict$$, to reach a definite metric.

Variation $$A$$: Consider as a variation satisfying the condition of rigidity, an infinitesimal displacement (parallel to space $$(x, y, z)$$ and rigid in the ordinary kinematic sense) of the parallel cross-sections of the world-tube parallel to the same space. We will obtain this variation in the figure, by displacing all cross-sections $$t$$=const of the tube by arbitrary infinitesimal lines parallel to the $$X$$-axis. If we confine ourselves to the consideration of translatory displacements, then $$\delta x,\ \delta y,\ \delta z$$ are arbitrary functions of time, and $$\delta t=0$$.



Variation $$B$$: Consider as a variation satisfying the condition of rigidity, an infinitesimal displacement (perpendicular to the world-tube and rigid in the ordinary kinematic sense) of the normal intersections of the world-tube. We will obtain such a variation in the figure, by displacing all normal-intersections of the tube parallel to itself, by arbitrary infinitesimal lines.

To be able to apply 's principle, we have to subject our variations to the additional condition, that it must vanish at the limits of the arbitrary integration field $$G$$. Due to this additional condition, the integration field contracts to $$ABEF$$ when the variation system $$A$$ is applied; so $$\delta x, \delta y, \delta z$$ must vanish in the fields $$BDE$$, $$ACF$$ because they must be constant for $$t=const$$, and must vanish at the limits of $$G$$, i.e., at the lines $$BD$$, $$AC$$. However, if we apply the variation system $$B$$, the integration field contracts to $$ABGH$$ due to the same reasons.

Now it can be immediately seen, that variation $$A$$ is in contradiction with relativity theory, because it has no invariant characteristics against the world-transformation, and is based on the arbitrary space $$xyz$$. On the other hand, variation $$B$$ has the desired invariant characteristics, and is always based on the proper-space, i.e., the space perpendicular to the world-tube, thus it is without doubt to be preferred before the previous one.

3. Let us denote for the sake of convenience, $$(t, x, y, z)$$ and $$\left(x_{0},x_{1},x_{2},x_{3}\right)$$ respectively as space-time coordinates, and $$F_{ik}$$ as the electromagnetic field.

's principle, that combines the laws of, of , and of mechanics, reads: The total action, i.e., the sum of actions of the electromagnetic field, of the electric charges, of the material masses, and in the case of general relativity, of the metric field, remains stationary when a variation (which satisfies the problem-conditions, and which vanishes at the limits of the arbitrary integration field "$$G$$") of the following things takes place: of the components of the four-potential, of the coordinates of the points of the world-lines followed by the charges and the masses, and of the components of the metric tensor.

In our case, however, the metric field remains unchanged, because we assumed it to be Euclidean since the material masses are missing, and because the only magnitudes to be varied, are the coordinates of the points of the world-lines followed by the charges. Thus it suffices to set the variation of the charges $$=0$$, i.e.:

where the first integration is to be extended over the charge element $$de$$ of the system, the second over the distance contained in the integration field $$G$$ of the world-line followed by $$de$$.

Now, we will separately derive the conclusions from the two variation systems $$A$$ and $$B$$.

4. Conclusions from variation $$A$$. In this case, the integration field contracts to $$ABEF$$. If $$t_1$$ and $$t_1$$ mean the times of $$A$$ and $$B$$, and if we consider that $$\delta x_i$$ only depend on the time, and that $$dt=0$$, then we can (1) write:

$\begin{array}{c} \sum\limits _{ik}\int\limits _{t_{1}}^{t_{2}}dt\ \delta x_{i}\int\ de\ F_{ik}\frac{dx_{\mu}}{dt}=0\\ \\(i=1,2,3)\ (k=0,1,2,3).\end{array}$|undefined

However, because $$\delta x_i$$ are arbitrary functions of time, we obtain from it the three equations:

$\sum\limits _{k}\int\ de\ F_{ik}\frac{dx_{k}}{dt}=0$|undefined

i.e. when $$E$$ and $$H$$ mean the electric and magnetic force:

$\int\left\{ E_{x}+\frac{1}{2}\left(\frac{dy}{dt}H_{z}-\frac{dz}{dt}H_{y}\right)\right\} de=0$

and the two corresponding equations for $$y$$ and $$z$$.

If the velocity in the reference system $$(t, x, y, z$$ is equal to zero in the relevant instant, then the three previous equations contract to a single vectorial equation:

We would have arrived at these equations without further ado, when we (as it ordinarily happens in the derivation of the electromagnetic mass and as it was essentially done by as well) would have assumed from the outset, that the total force of the systems is equal to zero. However, we have derived eq. (2) from 's principle, to demonstrate the source of the error. From (2), $$\tfrac{4}{3}\tfrac{U}{c^{2}}$$ immediately follows as electromagnetic mass.

Namely, if we notice that $$E$$ is the sum of a portion $$E^i$$ stemming from the system itself, and of a portion $$E^e$$ stemming from external causes, then we obtain from eq. (2):

$\int E^{i}de+\int E^{e}de=0$

On the other hand, either direct observation or in a familiar manner the consideration of electromagnetic momentum shows, that:

$\int E^{i}de=-\frac{4}{3}\frac{U}{c^{2}}\Gamma$|undefined

when $$\Gamma$$ means the acceleration. If we also notice, that $$\int E^{e}de$$ represents the total external force $$F$$, then we find:

$F=\frac{4}{3}\frac{U}{c^{2}}\Gamma$|undefined

A comparison of this equation with the basic law of point-dynamics $$F=m\Gamma$$ eventually gives us:

$m=\frac{4}{3}\frac{U}{c^{2}}$|undefined

5. Conclusions from variation $$B$$: In this case, the integration field contracts to $$ABGH$$. Let us imagine, that it is cut by infinitely many spaces perpendicular to the world-tube, into infinitely thin layers. With respect to any layer we also assume, that $$(t, x, y, z)$$ is the rest-system. Then $$dt=0,\ \delta x,\ \delta y,\ \delta z$$ will be arbitrary constants for our layer.

It is also: $$\delta x=\delta y=\delta z=0$$, because the velocity relative to the rest-system vanishes at time $$0$$, and $$dt=$$ the height of the layer $$=d\tau(1-\varkappa\times P-O)$$. Here, $$P-O$$ means the vector whose origin is located at point $$O$$, in which any arbitrary but fixed world-line $$L_0$$ hits space $$t=0$$, and whose endpoint is located in point $$P_0$$, in which the world-line $$L_0$$ followed by the charge element $$de$$ hits the space $$t=0$$: $$\varkappa$$ and $$d\tau$$ mean the curvature and the element of world-line $$L_0$$ contained in the layer, $$\times$$ eventually means the scalar product. The contribution to integral (1) stemming from our layer, thus becomes:

$\begin{array}{c} \int de\left\{ F_{10}\delta x+F_{20}\delta y+F_{30}\delta z\right\} \\ (1-\varkappa\times P-O)d\tau.\end{array}$

Now, if $$\Gamma$$ is the acceleration, then it is:

$\varkappa=-\frac{\Gamma}{c^{2}}$|undefined

If we also consider, that in the integration with respect to $$de$$, $$\delta x,\ \delta y,\ \delta z$$ and $$\delta\tau$$ can be seen as constant, then we find for the previous integral:

$\begin{array}{c} -d\tau\left[\delta x\int E_{x}\left(1+\frac{\Gamma\times P-O}{c^{2}}\right)de\right.\\ \\+\delta y\int E_{y}\left(1+\frac{\Gamma\times P-O}{c^{2}}\right)de\\ \\\left.+\delta z\int E_{z}\left(1+\frac{\Gamma\times P-O}{c^{2}}\right)de\right]\end{array}$|undefined

This expression must vanish for all possible values of $$\delta x,\ \delta y,\ \delta z$$. Therefore we obtain three equations, that can be combined into a single vectorial equation:

This equation takes the place of (2), and gives the value $$\tfrac{U}{c^{2}}$$ for the electromagnetic mass. Namely, if we put $$E^{(i)}+E^{(e)}$$ in (3) instead of $$E$$, then we find:

$\int E^{(e)}de+\int E^{(e)}\frac{\Gamma\times P-O}{c^{2}}de+\int E^{(i)}de+\int E^{(i)}\frac{\Gamma\times P-O}{c^{2}}de$|undefined

or as it was previously found:

$\begin{array}{c} \int E^{(e)}de=-\frac{4}{3}\frac{U}{c^{2}}\Gamma\\ \\\int E^{(e)}de+\int E^{(e)}\frac{\Gamma\times P-O}{c^{2}}de-\frac{4}{3}\frac{U}{c^{2}}\Gamma+\int E^{(i)}\frac{\Gamma\times P-O}{c^{2}}de=0\end{array}$|undefined

It follows first, that $$E^{(e)}$$ is of order of magnitude $$\Gamma$$. If we neglect terms containing $$\Gamma^2$$, we can neglect the second integral and find, when we set $$\int E^{(e)}de=F$$ as before:

In order to calculate the last integral, we consider that $$E^{(i)}$$ is equal to the sum of the force

$\int\frac{P-P'}{r^{3}}de$|undefined

(where $$P$$ means the up-point, $$P'$$ the source-point of charge $$de'$$, and $$r$$ the distance $$PP'$$), and a force of order $$\Gamma$$. The last-mentioned one would only give terms of order $$\Gamma^2$$, which are neglected by us. By that, the last integral (4) becomes:

$\frac{1}{c^{2}}\iint\frac{P-P'}{r^{3}}(\Gamma\times P-O)de\ de'$|undefined

If we interchange $$P$$ and $$P'$$, by which the integral is not changed, and take the arithmetic average of the two values thus obtained, then we find:

The $$x$$-component of this expression is:

$\begin{array}{c} \frac{1}{2c^{2}}\iint\frac{P-P'}{r^{3}}\\ \\\left\{ \Gamma_{x}(x-x')+\Gamma_{y}(y-y')+\Gamma_{z}(z-z')\right\} de\ de'\end{array}$|undefined

In the case of spherical symmetry, the integrals

$\begin{array}{c} \iint\frac{(x-x')^{2}}{r^{3}}de\ de',\ \iint\frac{(x-x')(y-y')}{r^{3}}de\ de',\\ \\\iint\frac{(x-x')(z-z')}{r^{3}}de\ de'.\end{array}$|undefined

still can be calculated without any ado, if one considers that the expressions

$(x-x')^{2},\ (x-x')(y-y),\ (x-x')(z-z')\,$

can be replaced by their mean value for all directions of $$PP'$$; these mean values are $$\tfrac{1}{3}r^{2},0,0$$. By that, the three integrals become:

$\frac{1}{3}\int\frac{de\ de'}{r}=\frac{2}{3}U,\ 0,\ 0$

and the $$x$$-component of (5):

$\frac{1}{3}\frac{U}{c^{2}}\Gamma_{x},$|undefined

thus the integral (5) becomes:

$\frac{1}{3}\frac{U}{c^{2}}\Gamma$.|undefined

If one substitutes this value into (4), then one finds:

$F=\frac{U}{c^{2}}\Gamma$,|undefined

thus the electromagnetic mass $$\tfrac{U}{c^{2}}$$.


 * Pisa, January 1922.

(Received May 9, 1922.)