Translation:On the spacetime lines of a Minkowski world/Paragraph 4

Spacetime lines.
We now turn to the exclusive consideration of pointlike charges. It is known, in which way to each of them in 's $$S_{4}$$ a curve (having a tangent which is timelike throughout) is related, its spacetime line or world line. Their “orthogonal” projection on an arbitrarily chosen $$S_{3}$$, whose normal has to be timelike though, will be seen in every reference system as the trajectory of the charge in the sense of ian Mechanics, whose $$S_{3}x^{(4)}=0$$ coincides with the chosen $$S_{3}$$. It is clear that this trajectory, not the original wordline, depends on the reference system. For instance, a timelike straight line can be related to a uniformly moving material point; its projection can be a straight line or a point; the latter is the case in the “rest system”, where the $$x^{(4)}$$-axis is directed parallel to the world line. Furthermore, a material point of plane world line is connected to a plane or straight trajectory depending on the choice of reference system, the latter being the case when the $$x^{(4)}$$-axis is directed parallel to the plane of the world line; in this case, however, the rectilinear trajectory is traversed with non-uniform velocity. The similar is true for world lines which can be placed in a plane $$S_{3}$$. The most general case is of course the three-fold curved wordline.

The direction cosines of the tangent of a spacetime line.
At first, in the non-Euclidean $$M_{4}\ (x\ y\ z\ u=ct)$$, for an arbitrary reference system as the representation of a Newtonian trajectory it is given

$x=x(t),\ y=y(t),\ z=z(t)$

thus in terms of the arc-law valid in this $$M_{4}$$

$d\sigma^{2}=dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}$

and the assumed timelike direction of the connected world line, i.e. the curve

$x=x(t),\ y=y(t),\ z=z(t),\ u=ct$,

and for the arc of this curve

$dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}=\left(\mathfrak{v}_{x}^{2}+\mathfrak{v}_{y}^{2}+\mathfrak{v}_{z}^{2}-c^{2}\right)dt^{2}=\left(\mathfrak{v}^{2}-c^{2}\right)dt^{2}<0$

It is known that this is a consequence of the postulate of the theory of relativity, that no speed $$|\mathfrak{v}|$$ can surpass the speed of light $$c$$. If we now pass from $$M_{4}\ (x\ y\ z\ ct)$$ to the Euclidean space $$S_{4}\ \left(x^{(1)}x^{(2)}x^{(3)}x^{(4)}\right)$$ by means of the mapping

$x^{(4)}=iu=ict$ and $x^{(1)}=x,\ x^{(2)}=y,\ x^{(3)}=z$

then every timelike direction for which we now have

$ds^{2}=\left(dx^{(1)}\right)^{2}+\left(dx^{(2)}\right)^{2}+\left(dx^{(3)}\right)^{2}+\left(dx^{(4)}\right)^{2}<0$

becomes an imaginary one. The unit vector of the respective direction will have (as in $$M_{4}$$) the magnitude $$\sqrt{-1}$$, because every vector is indeed equal to the unit vector of its direction multiplied by a positive number. However, in order to still formulate direction cosines even for such vectors, we will divide all components of the unit vector by $$i=\sqrt{-1}$$, at which occasion we will always be aware of the fact, that this is an arbitrary convention defined because of formal reasons, because a vector of timelike direction with magnitude $$+1$$ cannot exist.

Our world line is now

$x^{(1)}=x^{(1)}(t),\ x^{(2)}=x^{(2)}(t),\ x^{(3)}=x^{(3)}(t),\ x^{(4)}=ict,$

$\frac{dx^{(1)}}{dt}=\mathfrak{v}_{x},\ \frac{dx^{(2)}}{dt}=\mathfrak{v}_{y},\ \frac{dx^{(3)}}{dt}=\mathfrak{v}_{z},\ \frac{dx^{(4)}}{dt}=ic.$|undefined

Consequently

$ds^{2}=-dt^{2}\left(c^{2}-\mathfrak{v}^{2}\right)$

thus

$ds=ic\ dt\sqrt{1-\mathfrak{v}^{2}/c^{2}}=ic\ d\tau,$|undefined

in which one recognizes 's proper time multiplied by $$ic$$, which here appears to be connected in the most simple way with the (imaginary) Euclidean arc of the world line. For the previously defined direction cosines of the tangent we now have

{{MathForm2|(6)|$$\left.\begin{aligned}V^{(1)} & =\frac{dx^{(1)}}{ds}=\frac{1}{ic}\frac{dx^{(1)}}{d\tau}=\frac{\mathfrak{v}_{x}}{ic}\frac{1}{\sqrt{1-\mathfrak{v}^{2}/c^{2}}},\\ V^{(2)} & =\frac{\mathfrak{v}_{y}}{ic}\frac{1}{\sqrt{1-\mathfrak{v}^{2}/c^{2}}},\\ V^{(3)} & =\frac{\mathfrak{v}_{z}}{ic}\frac{1}{\sqrt{1-\mathfrak{v}^{2}/c^{2}}},\quad V^{(4)}=\frac{1}{\sqrt{1-\mathfrak{v}^{2}/c^{2}}} \end{aligned} \right\} $$}}

which we will also denote as the components of the velocity vector $$V$$. Therefore, we are now completely within Euclidean geometry.

The formulas of .
Let now the world line of a pointlike charge be given, whose coordinates shall always be denoted by $$x$$; we look for the potentials and fields in an arbitrary reference point, which is not allowed to be taken infinitely close to the respective location $$x$$ of the light-point, and whose coordinates shall be denoted by $$y$$. If the radius vector from the reference-point to the light-point is given by

$R^{(\alpha)}=x^{(\alpha)}-y^{(\alpha)},\quad\alpha=1,2,3,4$

where $$R$$ is a minimal vector, then it must be

$R^{2}=0$,

so it is known that for the potentials in $$y$$ the formulas of hold:

where $$RV$$ means the scalar product

$RV=R^{(1)}V^{(1)}+R^{(2)}V^{(2)}+R^{(3)}V^{(3)}+R^{(4)}V^{(4)}$

from the minimal vector $$R$$ and the velocity vector $$V$$ of the world line of the charge at the light-point $$x$$.

These formulas can be derived in the shortest way as follows: The static case, i.e. a straight world line parallel to the $$x^{(4)}$$-axis, is only a specialization of the coordinate system for the potentials $$\Phi_{\alpha}$$ of a pointlike charge $$x$$ in $$y$$, as one can see e.g. from the solution of, because

$\mathrm{P}^{(1)}=\rho\frac{\mathfrak{v}_{x}}{c}=i\rho\sqrt{1-\mathfrak{v}^{2}/c^{2}}\cdot\frac{\mathfrak{v}_{x}}{ic}\frac{1}{\sqrt{1-\mathfrak{v}^{2}/c^{2}}}=i\rho_{0}V^{(1)}$ etc.|undefined

thus

$\mathrm{P}^{(\alpha)}=i\rho_{0}V^{(\alpha)}$

where $$\rho_{0}$$ means the “rest”-density, and the assumption of the static case is equivalent with a rotation of the coordinate system in the plane $$\left[Vx^{(4)}\right]$$ (Lorentz transformation), so that

$\bar{V}^{(1)}=\bar{V}^{(2)}=\bar{V}^{(3)}=0,\ \bar{V}^{(4)}\equiv1$.

However, since it is indeed

$\bar{R}^{(4)}=\bar{x}^{(\alpha)}-\bar{y}^{(\alpha)}=ic\left(t_{L}-t_{A}\right)=-ir$,

the solution is

$\bar{\Phi}_{1}=\bar{\Phi}_{2}=\bar{\Phi}_{3}=0,$

$\bar{\Phi}_{4}=\frac{ei}{4\pi r}=\frac{e}{4\pi R^{(4)}}=\frac{e\bar{V}^{(4)}}{4\pi(\overline{RV})}$.|undefined

Thus we have for the general case of a single world line

$\Phi_{\alpha}=\frac{e}{4\pi}\frac{V^{(\alpha)}}{RV}$|undefined

From that, the fields in $$y$$ have to be computed by means of the formulas

$F_{\alpha\beta}=\frac{\partial\Phi_{\beta}}{\partial y^{(\alpha)}}-\frac{\partial\Phi_{\alpha}}{\partial y^{(\beta)}}$.|undefined

The differentiation has already been carried out by albeit in a somewhat complicated way. A displacement of the reference point $$y$$ namely must cause a displacement of the light-point $$x$$, since both have to remain connected by a minimal vector. In other words, the $$x$$ are therefore functions of the $$y$$. The general and most simple formulas for this dependency, which naturally have a great importance in the theory of a pointlike charge, have not been given by. We will see, that his method looses all of its inconvenience with these formulas.

Formulas for the differential quotient of the coordinates of the light-point $$x$$ with respect to the coordinates of the reference-point $$y$$.
We set

$dx^{(\alpha)}=\underset{\beta}{\sum}\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}dy^{(\beta)},\quad\alpha=1,2,3,4$|undefined

In whatever way the $$dy$$ may be constituted, the light-point cannot be displaced except along its given world line. The $$dx^{(\alpha)}$$ thus must always be proportional to the $$V^{(\alpha)}$$ or

$\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}=\rho_{\beta}V^{(\alpha)},\quad(\alpha,\beta)=1,2,3,4$.|undefined

Furthermore, the new location $$x+dx$$ shall again be effective for $$y+dy$$, which shall also apply to the $$dy$$. Therefore we can differentiate the relation

$R^{2}=\sum\left(x-y^{2}\right)\equiv0$

with respect to $$y$$; it follows

$0=-\left(x^{(\beta)}-y^{(\beta)}\right)+\underset{\alpha}{\sum}\left(x^{(\alpha)}-y^{(\alpha)}\right)\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}=0,\quad\beta=1,2,3,4$|undefined

or

$0=-R^{(\beta)}+\rho_{\beta}\underset{\alpha}{\sum}R^{(\alpha)}V^{(\alpha)},\quad\beta=1,2,3,4$

by which we have found:

The denominator $$RV$$ can never vanish; a timelike vector and a minimal vector can never be mutually perpendicular; because all vectors perpendicular to a timelike vector are spacelike. Therefore it follows

$\frac{\partial}{\partial y^{(\beta)}}=\left(\frac{\partial}{\partial y^{(\beta)}}\right)_{\mathrm{const}x^{(1)}x^{(2)}x^{(3)}x^{(4)}}+\frac{R^{(\beta)}}{RV}\underset{\alpha}{\sum}V^{(\alpha)}\frac{\partial}{\partial y^{(\alpha)}}$|undefined

or, since it is indeed $$V=\frac{dx}{ds}$$,

The formulas.
It follows that

$\begin{aligned}\frac{\partial V^{(\alpha)}}{\partial y^{(\beta)}} & =\frac{dV^{(\alpha)}}{ds}\cdot\frac{R^{(\beta)}}{RV},\\ \frac{\partial R^{(\alpha)}}{\partial y^{(\beta)}} & =-[\alpha\beta]+\frac{R^{(\beta)}}{RV}V^{(\alpha)},\\ \frac{\partial(RV)}{\partial y^{(\beta)}} & =\left(R\frac{dV}{ds}\right)\frac{R^{(\beta)}}{RV}-V^{(\beta)}+\frac{R^{(\beta)}}{RV} \end{aligned} $|undefined

Thus from the formulas (7) follow the  formulas:

here it is known that

$[RV]^{(\alpha\beta)}=R^{(\alpha)}V^{(\beta)}-R^{(\beta)}V^{(\alpha)}$ etc.

Confirmation of the relation $$\mathrm{Div}_{y}\Phi=0$$ and $$\square_{y}\Phi=0$$.
The happens also by the aid of the formulas, of which one can be convinced by simple calculation.

The principle as a special case of formulas (9).
If we write our formulas by avoiding the imaginary [notation], and if we set

$\begin{aligned}R^{(1)} & =x^{(1)}-y^{(1)}=\mathfrak{r}_{\overrightarrow{AL}}^{(1)}=x_{L}-x_{A},\\ R^{(2)} & =y_{L}-y_{A}=\mathfrak{r}^{(2)},\\ R^{(3)} & =z_{L}-z_{A}=\mathfrak{r}^{(3)},\\ R^{(4)} & =-ir \end{aligned} $|undefined

it follows:

where

$\frac{d}{dt_{L}}=\sqrt{1-\mathfrak{v}^{2}/c^{2}}\frac{d}{d\tau_{L}}$|undefined

is the known hydrodynamic operator for the time differentiation at one and the same individual particle; $$\frac{d}{d\tau_{L}}$$ taken with respect to proper time gives the velocity of the processes in a system where $$L$$ is momentarily at rest; thus for a process which does not depend on the conditions at the reference-point, i.e. when $$\left(\frac{\partial}{\partial t_{A}}\right)=0$$, the velocity is to be multiplied by

$\frac{\sqrt{1-\mathfrak{v}^{2}/c^{2}}}{1+\frac{(\mathfrak{rv})}{cr}}=\frac{\sqrt{1-\mathfrak{v}^{2}/c^{2}}}{1+\frac{|\mathfrak{v}|}{c}\cos(\mathfrak{rv})}$,|undefined

if it is observed from another system in which $$L$$ momentarily has the velocity $$\mathfrak{v}$$ (fourth of the previous formulas).

On the other hand, if we rewrite the formulas

$dx^{(\alpha)}=\underset{\beta}{\sum}\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}dy^{(\beta)}=V^{(\alpha)}\frac{(R\ dy)}{Rv},\quad\alpha=1,2,3,4$,|undefined

and if

$dy^{(1)}=\frac{\mathfrak{w}_{x}}{ic}dy^{(4)},\ dy^{(2)}=\frac{\mathfrak{w}_{y}}{ic}dy^{(4)},\ dy^{(3)}=\frac{\mathfrak{w}_{z}}{ic}dy^{(4)}$,|undefined

it follows

{{MathForm2|(12)|$$\left.\begin{aligned}dx_{L} & =\mathfrak{v}_{x}\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}dt_{A},\\ dy_{L} & =\mathfrak{v}_{y}\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}dt_{A},\\ dz_{L} & =\mathfrak{v}_{z}\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}dt_{A},\\ dt_{L} & =\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}dt_{A}, \end{aligned} \right\} $$}}

for the co-displacement (connected to the motion of the reference-point) of the light-point along its given world line. It follows

$dx_{L}^{2}+dy_{L}^{2}+dz_{L}^{2}-c^{2}dt_{L}^{2}=-c^{2}d\tau_{L}^{2}=\left(\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}\right)^{2}dt_{A}^{2}\left(\mathfrak{v}^{2}-c^{2}\right)$

or

$d\tau_{L}=dt_{A}\sqrt{1-\mathfrak{v}^{2}/c^{2}}\frac{\mathfrak{rw}+cr}{\mathfrak{rv}+cr}$|undefined

which contains the Doppler principle for a moving reference-point.

Remarks concerning the formulas.
We notice the occurrence of only first $$\left(V=\frac{dx}{ds}\right)$$ and second derivatives $$\left(\frac{dV}{ds}=\frac{d^{2}x}{ds^{2}}\right)$$ of the coordinates $$x$$ of the light-point, thus in order to compute the fields, the curve in $$x$$ can be replaced by the circle osculating to second order (circle of curvature or curvature hyperbola in $$M_{4}\ (x\ y\ z\ ct)$$, ).

For the two “invariants” of the vector $$F_{\alpha\beta}$$ we have

and

$\frac{4\pi}{e}\underset{(\alpha_{1}\alpha_{2})}{\sum}F_{\alpha_{1}\alpha_{2}}F_{\alpha_{2}\alpha_{4}}\equiv0,\quad\alpha=1,2,3,4$,|undefined

thus

$\mathfrak{HE}=0$;

the vector is singular, and representable by the area piece

$\frac{4\pi}{e}F_{\alpha\beta}=\left[R,\ \frac{1}{(RV)^{2}}\frac{dV}{ds}-\frac{1+R\frac{dV}{ds}}{(RV)^{3}}V\right]=[R,\ T]\frac{1}{(RV)^{3}}$,|undefined

where $$T$$ is the vector important to us,

It lies at the osculating plane $$\left[V\frac{dV}{ds}\right]$$ and can be written:

$T^{(\alpha)}=(RV)\frac{dV^{(\alpha)}}{ds}-\left(R\frac{dV}{ds}\right)V^{(\alpha)}-V^{(\alpha)}$,|undefined

of which the first part, being $$V\frac{dV}{ds}=0$$, is perpendicular to

$(RV)V^{(\alpha)}+\frac{\left(R\frac{dV}{ds}\right)\frac{dV^{(\alpha)}}{ds}}{\left(\frac{dV}{ds}\right)^{2}}$|undefined

However, this is the component of the minimal radius vector $$R$$ falling into the osculating plane, to which the vector $$(RV)\frac{dV}{ds}-\left(R\frac{dV}{ds}\right)V$$ is perpendicular; it is of course also perpendicular to $$R$$ itself. For large $$r$$ this becomes important, because the second part $$V^{(\alpha)}$$ within $$T^{(\alpha)}$$ can be neglected with respect to $$(RV)\frac{dV^{(\alpha)}}{ds}-\left(R\frac{dV}{ds}\right)V^{(\alpha)}$$, and if follows

$\frac{4\pi}{e}\sum_{\alpha=1}^{3}F_{4\alpha}R^{(\alpha)}=\sum\frac{\left[R,\ (RV)\frac{dV}{ds}-\left(R\frac{dV}{ds}\right)V\right]^{(4\alpha)}}{(RV)^{3}}R^{(\alpha)}\equiv0$|undefined

or

$(\mathfrak{Er})=0$

for large $$r$$.

As one easily confirms, in general for any $$r$$ it applies:

$\underset{\beta}{\sum}F_{\alpha\beta}^{\ast}R^{(\beta)}=0,\quad\alpha=1,2,3,4$,

and from the fourth of these equations:

$\mathfrak{H}=\frac{1}{r}[\mathfrak{Er}]$

for any $$\mathfrak{r}=\overrightarrow{AL}$$.

In the wave zone for large $$r$$, i.e. when

$\frac{4\pi}{e}F_{\alpha\beta}=\left[R,\ (RV)\frac{dV}{ds}-\left(R\frac{dV}{ds}\right)V\right]^{(\alpha\beta)}\frac{1}{(RV)^{3}}$,|undefined

where the fields are of order $$\frac{1}{r}$$ (the neglected term $$-\frac{1}{(RV)^{3}}[RV]$$ is of order $$\frac{1}{r^{2}}$$), we have the known relations: the $$\mathfrak{E\ h\ r}$$ form a sequence of three mutually perpendicular directions; $$\mathfrak{E}$$ has also a radial component in closer vicinity. The magnitude of $$\mathfrak{E}$$ is in general larger as that of $$\mathfrak{H}$$, only in the wave zone they become equal.

formulas in generalized coordinates.
Now, let $$\bar{x}^{(\alpha)}$$ and $$\bar{y}^{(\alpha)}$$ be the generalized coordinates of the light-point and the reference-point, respectively. Then the method of differential invariants gives for the formulas:

where $$RV$$ are of course to be treated as invariants, thus

$RV=\underset{\alpha}{\sum}R^{(\alpha)}V^{(\alpha)}=\underset{\alpha,\beta}{\sum}\bar{c}_{\alpha\beta}\bar{R}^{(\alpha)}\bar{V}^{(\beta)}=(\overline{RV})$

In exactly the same way it can be found:

where again

$\left[\bar{R}\frac{d\bar{V}}{ds}\right]^{(\gamma\delta)}=\bar{R}^{(\gamma)}\frac{d\bar{V}^{(\delta)}}{ds}-\bar{R}^{(\delta)}\frac{d\bar{V}^{(\gamma)}}{ds}$|undefined

and where one could again introduce the reciprocal systems $$\bar{R}_{\alpha}$$, $$\frac{d\bar{V}_{\alpha}}{ds}$$, $$\bar{V}_{\alpha}$$. The differences of this representation compared to the method of vectorial splitting discussed in § 1, have to be remembered here.

The differentiation formulas $$\frac{\partial\bar{x}^{(\alpha)}}{\partial\bar{y}^{(\beta)}}$$ in generalized coordinates.
We had

$\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}=\frac{V^{(\alpha)}R^{(\beta)}}{RV}$.|undefined

Now it is

$\frac{\partial x^{(\alpha)}}{\partial y^{(\beta)}}=\underset{\gamma,\delta}{\sum}\frac{\partial x^{(\alpha)}}{\partial\bar{x}^{(\gamma)}}\frac{\partial\bar{x}^{(\gamma)}}{\partial\bar{y}^{(\delta)}}\frac{\partial\bar{y}^{(\delta)}}{\partial y^{(\beta)}}=\frac{V^{(\alpha)}R^{(\beta)}}{RV}$,|undefined

thus

where $$\bar{V}^{\left(\bar{x}^{(\gamma)}\right)}$$ is a component taken with respect to the generalized axes in $$x$$, and $$\bar{R}^{\left(\bar{y}^{(\varepsilon)}\right)}$$ is taken with respect to the axes in $$y$$. It follows

Retrospectively, it is easy to confirm using the differential formulas derived, that the formulas (15) and (16) satisfy the differential equations of § 3.