Translation:On the spacetime lines of a Minkowski world/Paragraph 5

Ponderomotive force at the reference-point.
As such we take the force

$F_{\alpha}(y)=\underset{\beta}{\sum}\frac{F_{\alpha\beta}(y)\mathrm{P}^{(\alpha)}(y)}{\sqrt{1-\mathfrak{w}^{2}/c^{2}}},\quad(\alpha=1,2,3,4)$|undefined

where $$\mathfrak{w}$$ is the velocity of the charge at the reference-point, $$\rho$$ its density, and

$\mathrm{P}^{(1)}(y)=\rho\frac{\mathfrak{w}_{x}}{c},\ \mathrm{P}^{(2)}(y)=\rho\frac{\mathfrak{w}_{y}}{c},\ \mathrm{P}^{(3)}(y)=\rho\frac{\mathfrak{w}_{z}}{c},\ \mathrm{P}^{(4)}(y)=i\rho$.|undefined

It is also

$F_{1}=\mathfrak{F}_{x},\ F_{2}=\mathfrak{F}_{y},\ F_{3}=\mathfrak{F}_{z},\ F_{4}=\frac{i}{c}(\mathfrak{Fw})$,

furthermore

$\underset{\beta}{\sum}F_{\alpha\beta}(y)\mathrm{P}^{(\beta)}(y)=\underset{\beta}{\sum}F_{\alpha\beta}(y)\underset{\gamma}{\sum}\frac{\partial}{\partial y^{(\gamma)}}F_{\beta\gamma}(y)=\underset{\beta}{\sum}\frac{\partial}{\partial y^{(\beta)}}S_{\alpha\beta}$,|undefined

in which the $$S_{\alpha\beta}=S_{\beta\alpha}$$ are the stresses, thus

$\begin{aligned}S_{11} & =\frac{1}{2}\left(\mathfrak{E}_{x}^{2}-\mathfrak{E}_{y}^{2}-\mathfrak{E}_{z}^{2}\right)+\frac{1}{2}\left(\mathfrak{H}_{x}^{2}-\mathfrak{H}_{y}^{2}-\mathfrak{H}_{z}^{2}\right),\\ S_{12} & =\mathfrak{E}_{x}\mathfrak{E}_{y}+\mathfrak{H}_{x}\mathfrak{H}_{y}\ \mathrm{etc.,}\\ S_{14} & =-i\left(\mathfrak{E}_{y}\mathfrak{H}_{z}-\mathfrak{E}_{z}\mathfrak{H}_{y}\right)=-\frac{i}{c}\mathfrak{S}_{x}\ \mathrm{etc.,}\\ S_{44} & =\frac{1}{2}\left(\mathfrak{E}^{2}+\mathfrak{H}^{2}\right)=w \end{aligned} $

Spacetime-thread as an image of a body.
The totality of the world lines of the points of a body is called a spacetime-thread by.

In parameter representation:

$x^{(\alpha)}=x^{(\alpha)}\left(\xi^{(1)}\xi^{(2)}\xi^{(3)}\xi^{(4)}\right),\quad(\alpha=1,2,3,4)$

where $$\xi^{(4)}$$ is a timelike parameter and the $$\xi^{(1)}\xi^{(2)}\xi^{(3)}$$ are varying within certain limits given by the spatial extension.

We consider a piece of a spacetime-thread, for instance between two values $$\xi_{1}^{(4)}$$ and $$\xi_{2}^{(4)}$$, and we form

$\begin{aligned}\int dS_{4}\underset{\beta}{\sum}F_{\alpha\beta}\mathrm{P}^{(\beta)} & =\int dS_{4}\sqrt{1-\mathfrak{w}^{2}/c^{2}}F_{\alpha}\\ & =\int dS_{4}\underset{\beta}{\sum}\frac{\partial}{\partial y^{(\beta)}}S_{\alpha\beta}=\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta} \end{aligned} \text{(by § 2)},$|undefined

the latter over the boundary of the spacetime-thread. By application to a rectilinear hyper-cylinder, whose generators are parallel to the $$x^{(4)}$$-axis, and whose base spaces are plane and parallel to the $$x^{(1)}x^{(2)}x^{(3)}$$–space, one obtains from that (at the passage to the limit at infinitesimal height of the cylinder) the known representation of the Newtonian total-force by means of the Maxwell stresses and by the electromagnetic momentum or the energy theorem.

Infinitely thin spacetime-threads as an image of a pointlike charge.
pointlike charges in the mathematical sense of this word do not exist; potentials and fields would indeed become infinitely large within them. The employment of pointlike sources in potential theory is only legitimate in the integral representation of the potential, or otherwise as approximation at distances which are not infinitely small. The same also holds in our case. Therefore we have to imagine the pointlike charge as physically very small. However, the individual points of this charge will exert forces upon each other. How can they be computed?

The reaction force of radiation.
Now, it cannot be done without making assumptions on the shape of the infinitely thin spacetime-thread. Since we do not want to make them, we have to confine ourselves with an approximation. So let the world line of the charge $$x$$ be given, for instance as a function $$x^{(\alpha)}(s)$$ of the arc. The pointlike charge cannot act on itself; because the distance 0 has to be excluded when the pointlike sources in the potential theory are used, and is excluded here as well. It also cannot act upon a later position of itself; because this would require superluminal speed. Outside of the world line $$x$$ there should by now charges though. If we therefore extend the integral

$\int dS_{4}\underset{\beta}{\sum}\frac{\partial}{\partial y^{(\beta)}}S_{\alpha\beta}$|undefined

over a piece of $$S_{4}$$ which does not contain the world line $$x$$, it will (being extended over a charge-free space) vanish. This is indeed a known property of the ponderomotive force of the electron theory of Lorentz. However, if the piece of $$S_{4}$$ contains a part of the world line $$x$$, then in accordance with the things previously said (and since the integrand would become infinite) it is to be excluded from the integration area by a small hypercylinder which encloses it, whose intersection may be taken as infinitely small. If one executes the integration in this way and then passes to the limit, by letting the hypercylinder contract itself upon the world line, then one finds zero again, as it has to be. Nevertheless, we will be able to use this integral for our purposes; because the charge is pointlike only when seen from infinitely large distances. Thus when we contract the considered piece of $$S_{4}$$ as follows: the boundary of the after-cone at point $$s=s_{1}$$ or $$s=s_{2}$$ of the wordline $$x$$ and if it is cut by a hypercylinder, whose intersection is assumed to be very large, then the field of a non-pointlike charge will not be computable upon the two after-cones by means of the  formulas, but it will be computable upon the mantle of the hypercylinder, as long as we assume its intersection to be very large. As it is now

$\int dS_{4}\underset{\beta}{\sum}\frac{\partial}{\partial y^{(\beta)}}S_{\alpha\beta}=\underset{\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}}+\underset{\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}}+\underset{\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}}$|undefined

so the third integral can be evaluated by means of the formulas, if the distance

$r=\text{spatial distance }\overleftarrow{xy}=iR^{(4)}$

is afterwards increased over all measures.

The matrix of the stresses from the formulas.
$\begin{aligned}\left(\frac{4\pi}{e}\right)^{2}S_{\alpha\beta}= & -\frac{1}{(RV)^{4}}\left\{ R^{(\alpha)}\frac{dV^{(\beta)}}{ds}+R^{(\beta)}\frac{dV^{(\alpha)}}{ds}\right\} \\ & -\left\{ \frac{1}{(RV)^{4}}\left(\frac{dV}{ds}\right)^{2}+\frac{\left(1+R\frac{dV}{ds}\right)^{2}}{(RV)^{6}}\right\} R^{(\alpha)}R^{(\beta)}\\ & +\frac{1+R\frac{dV}{ds}}{(RV)^{5}}\left\{ R^{(\alpha)}V^{(\beta)}+R^{(\beta)}V^{(\alpha)}\right\} \\ & -\frac{1}{2}[\alpha\beta]\frac{1}{(RV)^{4}} \end{aligned} $|undefined

Introduction of a generalized system of coordinates.
If we anticipate the developments of § 7, then every curve in $$S_{4}$$ possesses a comoving tetrad, whose four directions in the respectively considered curve point are given by the tangent $$c_{1}^{(\alpha)}$$, the principal normal $$c_{2}^{(\alpha)}$$, the binormal $$c_{3}^{(\alpha)}$$ and the trinormal $$c_{4}^{(\alpha)}$$, which form a system of four mutually pairwise perpendicular unit vectors, changing their directions from curve point to curve point. In the case of a world line, the tangent must have a timelike direction, i.e. by the definition made in § 4: the quantities $$c_{1}^{(1)}c_{1}^{(2)}c_{1}^{(3)}$$ are purely imaginary, $$c_{1}^{(4)}$$ is real. Now, it is known that

$\frac{dx^{(\alpha)}}{ds}=c_{1}^{(\alpha)}=V^{(\alpha)},\ \frac{d^{2}x^{(\alpha)}}{ds^{2}}=\frac{dc_{1}^{(\alpha)}}{ds}=\frac{c_{2}^{(\alpha)}}{\mathrm{R}_{1}}=\frac{dV^{(\alpha)}}{ds},\quad\alpha=1,2,3$|undefined

where $$\mathrm{R}_{1}$$ is the radius of the first curvature, which is computed as follows:

$\sum_{\alpha=1}^{4}\left(\frac{d^{2}x^{(\alpha)}}{ds^{2}}\right)^{2}=\left(\frac{1}{\mathrm{R}_{1}}\right)^{2}=\left(\frac{dV}{ds}\right)^{2}$|undefined

As one can see, $$\mathrm{R}_{1}$$ is a real quantity and is related to the acceleration $$b$$ by

$b=\sqrt{\sum_{\alpha=1}^{4}\left(\frac{d^{2}x^{(\alpha)}}{d\tau^{2}}\right)^{2}}=-c^{2}\sqrt{\sum_{\alpha=1}^{4}\left(\frac{d^{2}x^{(\alpha)}}{d\tau^{2}}\right)^{2}}=-\frac{c^{2}}{\mathrm{R}_{1}}.$.|undefined

Now we want to organize the points $$y$$ of the integration area with respect to the points $$x$$ of the world line, which are connected to them as light-points, and we write

$y^{(\alpha)}=x^{(\alpha)}+ir\cdot c_{1}^{(\alpha)}+r\cos\vartheta\cdot c_{2}^{(\alpha)}+r\sin\vartheta\cos\varphi\cdot c_{3}^{(\alpha)}+r\sin\vartheta\sin\varphi\cdot c_{4}^{(\alpha)},\quad\alpha=1,2,3,4$

Then we have in any case

$\sum_{\alpha=1}^{4}\left(x^{(\alpha)}-y^{(\alpha)}\right)^{2}=\sum_{\alpha=1}^{4}\left(-ir\cdot c_{1}^{(\alpha)}-r\cos\vartheta\cdot c_{2}^{(\alpha)}-r\sin\vartheta\cos\varphi\cdot c_{3}^{(\alpha)}-r\sin\vartheta\sin\varphi\cdot c_{4}^{(\alpha)}\right)=0$

since it is indeed

$\sum_{\alpha=1}^{4}c_{i}^{(\alpha)}c_{k}^{(\alpha)}=[ik],$,

by assumption.

The $$y$$ are to be seen as functions of four parameters:

$r,\vartheta,\varphi,s$;

the latter is the timelike coordinate. The integration area is limited:


 * 1) by the after-cone $$s=s_{1}$$,


 * 1) by the mantle of the (spherical) hypercylinder

$r=\text{const}=a$


 * 1) by the after-cone $$s=s_{2}$$.

Only the integral over 2) can be evaluated by us, if we let $$a$$ go towards $$\infty$$ after finished evaluation.

Evaluation of the integral over the mantle $$r=a$$.
With the aid of the formulas (see § 7, (19)) one finds the value of

$\frac{dy^{(\alpha)}}{ds}=c_{1}^{(\alpha)}\left(1-\frac{r\cos\vartheta}{\mathrm{R}_{1}}\right)+c_{2}^{(\alpha)}\left(\frac{ir}{\mathrm{R}_{1}}-\frac{r\sin\vartheta\cos\varphi}{\mathrm{R}_{2}}\right)+c_{3}^{(\alpha)}\left(\frac{r\cos\vartheta}{\mathrm{R}_{2}}-\frac{r\sin\vartheta\sin\varphi}{R_{3}}\right)+c_{4}^{(\alpha)}\frac{r\sin\vartheta\cos\varphi}{\mathrm{R}_{3}}$.|undefined

Furthermore

$\begin{matrix}\frac{dy^{(\alpha)}}{dr}= & c_{1}^{(\alpha)}\cdot i & +c_{2}^{(\alpha)}\cos\vartheta & +c_{3}^{(\alpha)}\sin\vartheta\cos\varphi & +c_{4}^{(\alpha)}\sin\vartheta\cos\varphi\\ \frac{dy^{(\alpha)}}{d\vartheta}= & \ast & -c_{2}^{(\alpha)}r\sin\vartheta & +c_{3}^{(\alpha)}r\cos\vartheta\cos\varphi & +c_{4}^{(\alpha)}r\cos\vartheta\sin\varphi\\ \frac{dy^{(\alpha)}}{d\varphi}= & \ast & \ast & -r\sin\vartheta\sin\varphi & +c_{4}^{(\alpha)}r\cos\vartheta\cos\varphi \end{matrix}$|undefined

If it is now assumed that the determinant is

$\left

then it follows from it

$\frac{\partial\left(y^{(1)}y^{(2)}y^{(3)}y^{(4)}\right)}{\partial\left(r\vartheta\varphi s\right)}=r^{2}\sin\vartheta>0$

For the mantle $$r=a$$ we have, if $$N$$ is the normal going to the exterior, i.e. $$(N,\ y-x)>0$$:

$\begin{aligned}N_{1}dS_{3} & =+\frac{\partial\left(y^{(2)}y^{(3)}y^{(4)}\right)}{\partial\left(\vartheta\varphi s\right)}d\vartheta\ d\varphi\ ds=a^{2}\sin\vartheta\sin\vartheta\cos\varphi\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)d\vartheta\ d\varphi\ ds,\\ N_{2}dS_{3} & =-\frac{\partial\left(y^{(1)}y^{(3)}y^{(4)}\right)}{\partial\left(\vartheta\varphi s\right)}d\vartheta\ d\varphi\ ds=a^{2}\sin\vartheta\sin\vartheta\sin\varphi\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)d\vartheta\ d\varphi\ ds,\\ N_{3}dS_{3} & =+\frac{\partial\left(y^{(1)}y^{(2)}y^{(4)}\right)}{\partial\left(\vartheta\varphi s\right)}d\vartheta\ d\varphi\ ds=a^{2}\sin\vartheta\cos\vartheta\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)d\vartheta\ d\varphi\ ds,\\ N_{4}dS_{3} & =-\frac{\partial\left(y^{(1)}y^{(2)}y^{(3)}\right)}{\partial\left(\vartheta\varphi s\right)}d\vartheta\ d\varphi\ ds=-ia^{3}\sin\vartheta\frac{\cos\vartheta}{\mathrm{R}_{1}}, \end{aligned} $|undefined

in which it was set for an instant:

$c_{3}\Vert x^{(1)},\ c_{4}\Vert x^{(2)},\ c_{2}\Vert x^{(3)},\ c_{1}\Vert x^{(4)}$

Generally it is therefore:

$\begin{aligned}N_{\alpha}dS_{3} & =a\sin\vartheta\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)\left\{ a\sin\vartheta\cos\varphi\cdot c_{3}^{(\alpha)}+a\sin\vartheta\sin\varphi\cdot c_{4}^{(\alpha)}+a\cos\vartheta\cdot c_{2}^{(\alpha)}+iac_{1}^{(\alpha)}\right\} d\vartheta\ d\varphi\ ds\\ & \ -ia^{2}\sin\theta\cdot c_{1}^{(\alpha)}d\vartheta\ d\varphi\ ds,\quad(\alpha=1,2,3,4) \end{aligned} $|undefined

One notices that it is indeed:

$\begin{aligned}\sum_{\alpha=1}^{4}\left(N_{\alpha},\ x^{(\alpha)}-y^{(\alpha)}\right)dS_{3} & =a\sin\vartheta\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)\sum\left(x^{(\alpha)}-y^{(\alpha)}\right)^{2}d\vartheta\ d\varphi\ ds+a^{3}\sin\vartheta\ d\vartheta\ d\varphi\ ds\sum\left(c_{1}^{(\alpha)}\right)^{2}\\ & =a^{3}\sin\vartheta d\vartheta\ d\varphi\ ds>0 \end{aligned} $|undefined

$$N$$ is thus the exterior normal. Now one has also

$N_{\alpha}dS_{3}=-a\sin\vartheta\ d\vartheta\ d\varphi\ ds\cdot\left(1+R\frac{dV}{ds}\right)R^{(\alpha)}+a\sin\vartheta\ d\vartheta\ d\varphi\ ds\cdot\left(R^{(\alpha)}V\right)V^{(\alpha)}$.

Thus it follows

$\begin{aligned}\left(\frac{4\pi}{e}\right)^{2}dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta} & =-a\sin\vartheta\ d\vartheta\ d\varphi\ ds\left(\frac{4\pi}{e}\right)^{2}\underset{\beta}{\sum}S_{\alpha\beta}\left\{ \left(1+R\frac{dV}{ds}\right)R^{(\beta)}-\left(RV\right)V^{(\alpha)}\right\} \\ & =-a\sin\vartheta\ d\vartheta\ d\varphi\ ds\left\{ -\frac{1}{2}\frac{1+R\frac{dV}{ds}}{(RV)^{4}}R^{(\alpha)}+\frac{1}{(RV)^{2}}\frac{dV^{(\alpha)}}{ds}-\right.\\ & \quad\left.-\left[\frac{1}{(RV)^{2}}\left(\frac{dV}{ds}\right)^{2}+\frac{\left(1+R\frac{dV}{ds}\right)\left(R\frac{dV}{ds}\right)}{(RV)^{4}}\right]R^{(\alpha)}+\frac{\frac{1}{2}+\left(R\frac{dV}{ds}\right)}{(RV)^{3}}V^{(\alpha)}\right\} \end{aligned} $|undefined

thus we have the integral

$\begin{aligned} & \int_{s_{1}}^{s_{2}}ds\int_{0}^{\pi}d\theta\int_{0}^{2\pi}d\varphi\cdot a\sin\vartheta\left\{ ic_{1}^{(\alpha)}\left[\frac{\cos\vartheta}{\mathrm{R}_{1}}\frac{\left(\frac{1}{2}-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)}{a^{2}}+\frac{1}{\mathrm{R}_{1}^{2}a}\right]\right.\\ & \quad+c_{2}^{(\alpha)}\left[\frac{\cos\vartheta}{\mathrm{R}_{1}^{2}a}-\frac{\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)\left(\frac{1}{2}-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)}{a^{3}}\cos\vartheta+\frac{1}{\mathrm{R}_{1}a^{2}}\right]\\ & \quad+c_{3}^{(\alpha)}\left[\frac{\cos\vartheta}{a^{2}\mathrm{R}_{1}^{2}}-\frac{\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)\left(\frac{1}{2}-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)}{a^{4}}\right]a\sin\vartheta\cos\varphi\\ & \quad\left.+c_{4}^{(\alpha)}\left[\frac{\cos\vartheta}{a^{2}\mathrm{R}_{1}^{2}}-\frac{\left(1-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)\left(\frac{1}{2}-\frac{a\cos\vartheta}{\mathrm{R}_{1}}\right)}{a^{4}}\right]a\sin\vartheta\sin\varphi\right\} \end{aligned} $|undefined

Thus it follows in addition

$\left(\frac{4\pi}{e}\right)^{2}dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}=\int_{s_{1}}^{s_{2}}ds\cdot c_{1}^{(\alpha)}\left[4\pi\cdot\frac{2}{3}\cdot i\cdot\frac{1}{\mathrm{R}_{1}^{2}}\right]+\int_{s_{1}}^{s_{2}}ds\cdot c_{2}^{(\alpha)}\left[6\pi\cdot\frac{1}{\mathrm{R}_{1}}\cdot\frac{1}{a}\right]$|undefined

and at the limit $$a=\infty$$

$\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}=\frac{e^{2}}{4\pi}\cdot\frac{2}{3}\int_{s_{1}}^{s_{2}}\frac{iV^{(\alpha)}}{\mathrm{R}_{1}^{2}}ds,\quad(\alpha=1,2,3,4)$|undefined

On the other hand, it was

$\int dS_{3}\underset{\beta}{\sum}S_{\alpha\beta}N_{\beta}=\int dS_{4}\sqrt{1-\mathfrak{w}^{2}/c^{2}}F_{\alpha},\quad(\alpha=1,2,3,4)$|undefined

Now, since $$F_{\alpha}$$ is vanishing everywhere except along the infinitely thin spacetime-thread of the charge, one can write the integral on the right also as follows:

$\begin{aligned}\int dS_{4}\sqrt{1-\mathfrak{w}^{2}/c^{2}}F_{\alpha} & =\int dx^{(1)}dx^{(2)}dx^{(3)}dx^{(4)}\sqrt{1-\mathfrak{v}^{2}/c^{2}}F_{\alpha},\\ & =\int_{s_{1}}^{s_{2}}ds\int dx^{(1)}dx^{(2)}dx^{(3)}F_{\alpha}=\int_{s_{1}}^{s_{2}}ds\ K_{\alpha}, \end{aligned} $|undefined

by introducing a vector of total force $$K_{\alpha}$$, thus

$\int_{s_{1}}^{s_{2}}K_{\alpha}ds=\frac{e^{2}}{6\pi}\int_{s_{1}}^{s_{2}}ds\frac{iV^{(\alpha)}}{R_{1}^{2}},\quad\alpha=1,2,3,4$|undefined

It is known that this result has already been derived by in a different way. Now, guided by dimensional considerations, there is still a supplement; to the previous expression we can add

$\frac{e^{2}}{6\pi}\int_{s_{1}}^{s_{2}}ds\frac{d^{2}V}{ds^{2}}i=i\frac{e^{2}}{6\pi}\left|\frac{dV}{ds}\right|_{s_{1}}^{s_{2}}$,|undefined

if

$\left(\frac{dV}{ds}\right)_{s=s_{1}}\equiv\left(\frac{dV}{ds}\right)_{s=s_{2}}\equiv0$,|undefined

i.e. if the motion in the location $$s_{1}$$ or $$s_{2}$$ is not accelerated.

We remark, that the first solution

$K_{\alpha}=\frac{e^{2}}{6\pi}i\frac{V^{(\alpha)}}{\mathrm{R}_{1}^{2}}$|undefined

gives $$K_{\alpha}$$ the appearance of a timelike vector; furthermore, the orthogonality condition by which $$K\perp V$$, is not satisfied in this form, so that also $$K_{4}$$ is not representing the mere mechanical power of force $$K$$. Since we have no reason to allow non-mechanical energy at a single almost pointlike charge, we rather have to assume the second form

$K=i\frac{e^{2}}{6\pi}\left(\frac{d^{2}V}{ds^{2}}+\frac{V}{\mathrm{R}_{1}^{2}}\right)$,|undefined

since this one satisfies the orthogonality condition. Because it follows from $$(V)^{2}\equiv1$$ by two-fold differentiation with respect to $$s$$:

$\left(V\frac{dV}{ds}\right)=0$,

$\left(V\frac{d^{2}V}{ds^{2}}\right)+\left(\frac{dV}{ds}\right)^{2}=\left(V,\ \frac{d^{2}V}{ds^{2}}+\left(\frac{dV}{ds^{2}}\right)^{2}V\right)=0$.|undefined

As the reaction force of radiation we then have:

$-K_{1}=-\mathfrak{K}_{x},\ -K_{2}=-\mathfrak{K}_{y},\ -K_{3}=-\mathfrak{K}_{z}$

and as their purely mechanical power

$-K_{4}=-\frac{i}{c}(\mathfrak{Kv})$

The result stated above is obviously the only part of the ponderomotive forces exerted by the electron onto itself, which can be computed without assumptions about its shape. Because on the remaining boundaries, the cones $$s=s_{1}$$ or $$s=s_{2}$$, the electron cannot be seen as pointlike, because $$r$$ is decreasing to 0. Since assumptions on the shape should not be made, we have to refrain from further treating the dynamical problem of the forces exerted by the electron onto itself, and by that we have to refrain from treating its motions in terms of dynamics. We will rather consider them as given and then compute the respective potentials and fields at a sufficiently distant reference-point, from which the electron can be seen as pointlike.

The reaction force at Born's hyperbolic motion.
This motion is characterized by circles (equally sided hyperbolas in $$M_{4}\ :\ x\ y\ z\ t$$) as world lines (see § 6). Therefore it follows for the three radii of the curvatures

$\frac{1}{\mathrm{R}_{1}}=\mathrm{const},\quad\frac{1}{\mathrm{R}_{2}}=\frac{1}{\mathrm{R}_{3}}=0$|undefined

and for

$\frac{d^{2}V}{ds^{2}}+\frac{V}{\mathrm{R}_{1}^{2}}=\frac{d^{2}c_{1}}{ds^{2}}+\frac{1}{\mathrm{R}_{1}^{2}}c_{1}$|undefined

and from the formulas § 7 (19):

$\frac{dc_{1}}{ds}=\frac{c_{2}}{\mathrm{R}_{1}},\quad\frac{d^{2}c_{1}}{ds^{2}}=\left(-\frac{c_{1}}{\mathrm{R}_{1}}+\frac{c_{3}}{\mathrm{R}_{2}}\right)\frac{1}{\mathrm{R}_{1}}-\frac{c_{2}}{\mathrm{R}_{1}^{2}}\frac{d\mathrm{R}_{1}}{ds}$,|undefined

$\frac{d^{2}c_{1}}{ds^{2}}+\frac{c_{1}}{\mathrm{R}_{1}^{2}}=\frac{c_{3}}{R_{1}R_{2}}-\frac{c_{2}}{\mathrm{R}_{1}^{2}}\frac{d\mathrm{R}_{1}}{ds}\equiv0$.|undefined

The reaction force is identical with zero for hyperbolic motion, and only for it, if uniform motion is seen as its special case $$\left(\frac{1}{\mathrm{R}_{1}}\equiv0\right)$$.