Translation:On the Theory of the Experiment of Trouton and Noble

by Max von Laue

On the Theory of the Experiment of Trouton and Noble;

by M. Laue.

I.

Michelson's interference experiment is on many occasions considered as the only support of the theory of relativity against the "absolute theory". However, this is by no means the case. Also the Trouton-Noble experiment^[1] decides in the same sense; and when it somewhat fell into oblivion, despite of the esteem which was expressed to it by H. A. Lorentz^[2] in 1904, then this is mainly due to the fact, that its theory lacks that kind of vividness which so much distinguishes the theory of the Michelson experiment. This deficiency shall be corrected by the following study.

The subject of the experiment is the question, whether a charged plate condenser experiences a torque due to its motion with Earth. Relativity and absolute theory accordingly are affirming this; of course, relativity theory does this only in the case of a non-co-moving reference system $K$; in the "rest system" $K^{0}$ there is no torque. With the absolute theory, we can now expect a rotation in consequence of the torque, since its dynamics is that of Newton; according to this theory, the condenser tries to locate itself, so that the direction of velocity is tangential to the plates. Yet it was explained by the author^[3] on many occasions, that according to relativity theory, this torque causes no change of motion because the material parts of the condenser (like all elastically stressed bodies) require a torque for uniform translatory motion; at this place, only the procedure shall be considered, by which that torque is calculated in both theories.

Previously this happened as follows^[4]: If $E^{0}$ denotes the electrostatic energy of the condenser field in the case of rest, ${\mathfrak {q}}$ the velocity, $\vartheta ^{0}$ the angle between the velocity and the normal of the plate related to system $K^{0}$ in which the condenser is at rest, then it can be shown that the electromagnetic momentum of the field has a component of magnitude $q\sin 2\vartheta ^{0}E^{0}/c^{2}$ perpendicular to the velocity. From that it follows by the theorem of the conservation of angular momentum, that this torque given by the vector product $[{\mathfrak {Eq}}]$ with respect to magnitude and direction, amounts to

$N={\frac {q^{2}}{c^{2}}}E^{0}\sin 2\vartheta ^{0}$.

Although this consideration is unconditionally conclusive, one still can ask: which are the force couples producing this torque, and where do they act? This shall be answered in the following. The difference of both ways of view is analogous to the theory of radiation pressure. There, the theorem of the conservation of momentum fully suffices. If the latter is calculated for the incident and the reflected ray, the (vectorial) difference directly gives the momentum added to the mirror.^[5] The way chosen here, however, corresponds to the other procedure by Abraham,^[6] which interprets the pressure as the result of Maxwell's stresses occurring at the mirror, or also to the procedure of Planck,^[7] who determines the electromagnetic field in the mirror and then calculates the force exerted upon its charges and currents.

In the following we presuppose, that the condenser plates are rectangles whose sides $a$ and $b$ are great compared to the plate distance $d$, and that furthermore the plate thickness is also small compared to $d$, and it can even considered as infinitely small.^[8] The space between the plates shall be free of matter, and also the supports which must be installed between the plates to maintain the distance between them, shall have the dielectric constant and permeability 1, so that they don't differ electromagnetically from empty space. Velocity ${\mathfrak {q}}$ shall be perpendicular to the sides of length $b$.

II.

First we calculate the forces exerted upon the resting condenser. Attraction is the by far strongest force-action exerted by the plates upon each other; it has (as far as the field in the interior can be considered as homogeneous) the strength ${\tfrac {1}{2}}\left.{\mathfrak {E}}^{0}\right.^{2}$ per unit area, when ${\mathfrak {E}}^{0}$ is the field strength. If we neglect the marginal effects, then these forces produce the resultants $\pm {\mathfrak {K}}_{1}^{0}$ of strength

(1)	$\left|{\mathfrak {K}}_{1}^{0}\right|={\frac {1}{2}}ab\left.{\mathfrak {E}}^{0}\right.^{2}={\frac {E^{0}}{d}}$

whose points of contact are the midpoints of the plates. Far less strong, but still useful for our purposes, are the traction-forces acting at the rim of the plates.

Let us for example consider the course of the force lines in the vicinity of the rim (Fig. 1). We take this figure from the known solution of Helmholtz^[9] of the electrostatic problem for the rim of a condenser, which indeed presupposes the plates as being infinitely great semi-planes, which we nevertheless can use without second thoughts due to our assumption concerning the ratio of $a$ and $b$ to $d$. We see upon it, that all force lines are ending perpendicularly upon the lines giving the intersection of the plates with the drawing plane.

Fig. 1.

Only the force line acting at the plate edges is an exception. It exerts a traction into the outside, namely it has a finite amount despite of the infinitely small thickness of the plates, because the surface density of the electric allocation is infinitely great there. To calculate this force, however, we must take a roundabout way due to the singularities arising at the rim. Instead of that single plate, we imagine any other contour line as a conductive limitation of the field, and calculate the force action experienced by it. The solution of our actual problem then occurs by a passage to the limit.

If we lay the $x$-axis in the symmetry line of Fig. 1, and if we choose the origin so that the plates form that part of lines $y=\pm {\tfrac {1}{2}}d$ reaching from $x=-\infty$ to $x=-d/2\pi$, and if we also denote the potential function by $\varphi$ and the potential difference of the plates by $2V$ (the line $y=+{\tfrac {1}{2}}d$ corresponds to the plate of potential $+V$), then the cited solution of Helmholtz is contained in the equation:

(2)	$x=iy={\frac {d}{2\pi }}\left\{{\frac {\pi }{V}}(\psi +i\varphi )+e^{{\frac {\pi }{V}}(\psi +i\varphi )}\right\}$

or in the equivalent relations:

(3)

$\left\{{\begin{array}{c}x={\frac {d}{2\pi }}\left\{{\frac {\pi }{V}}\psi +e^{{\frac {\pi }{V}}\psi }\cos \left({\frac {\pi }{V}}\varphi \right)\right\},\\\\y={\frac {d}{2\pi }}\left\{{\frac {\pi }{V}}\varphi +e^{{\frac {\pi }{V}}\psi }\sin \left({\frac {\pi }{V}}\varphi \right)\right\}.\end{array}}\right.$

The curves $\psi$=const. are the force lines. The force acting upon the line element $dl$ of a contour line $\varphi =\varphi _{0}\left(0<\varphi _{0}<V\right)$ imagined as conductive, amounts to

${\frac {1}{2}}\left.{\mathfrak {E}}^{0}\right.^{2}dl={\frac {1}{2}}\left({\frac {\partial \varphi }{\partial n}}\right)^{2}dl$

and has the direction $n$, when $n$ is the perpendicular of $dl$ directing to the outside. The $x$-component of this force is consequently:

${\frac {1}{2}}\left({\frac {\partial \varphi }{\partial n}}\right)^{2}\cos n\ x\ dl$

and the integral action upon the total contour line is

(4)	${\mathfrak {K}}_{x}^{0}={\frac {1}{2}}\int \left({\frac {\partial \varphi }{\partial n}}\right)^{2}\cos n\ x\ dl$

If we choose the integration direction by beginning in the interior of the condenser at an infinite distance from the rim, then $dl$ and $dn$ are mutually located as $x$ and $-y$. Consequently it is

${\frac {\partial \varphi }{\partial n}}=-{\frac {\partial \psi }{\partial l}}$;

since ${\tfrac {\partial \psi }{\partial n}}=0$, we can put ${\tfrac {\partial \psi }{\partial l}}dl=d\psi$, and in the equation

${\mathfrak {K}}_{x}^{0}=-{\frac {1}{2}}\int {\frac {\partial \varphi }{\partial n}}\cos n\ x\ d\psi$

we have to extend the integration from $-\infty$ to $+\infty$, since in the interior it is $\psi =-\infty$. On the other hand, since $\varphi$ only varies in direction $n$,

${\frac {\partial \varphi }{\partial n}}\cos n\ x={\frac {\partial \varphi }{\partial x}}$.

Thus it becomes

(5)	${\mathfrak {K}}_{x}^{0}=-{\frac {1}{2}}\int \limits _{-\infty }^{+\infty }{\frac {\partial \varphi }{\partial n}}d\psi$

Now it follows from (2) or (3) by means of simple calculations:

${\frac {\partial \varphi }{\partial x}}=-{\frac {2V}{d}}{\frac {e^{{\frac {\pi }{V}}\psi }\sin \left({\frac {\pi }{V}}\varphi \right)}{1+2e^{{\frac {\pi }{V}}\psi }\cos \left({\frac {\pi }{V}}\varphi \right)+e^{{\frac {2\pi }{V}}\psi }}}$

If we introduce as integration variable

$u=e^{{\frac {\pi }{V}}\psi }$

then we find that we have to integrate along line $\varphi =\varphi _{0}$:

${\begin{array}{ll}{\mathfrak {K}}_{x}^{0}&={\frac {V^{2}}{\pi d}}\sin \left({\frac {\pi }{V}}\varphi _{0}\right)\int \limits _{0}^{+\infty }{\frac {du}{1+2u\cos \left({\frac {\pi }{V}}\varphi _{0}\right)+u^{2}}}\\\\&={\frac {V^{2}}{\pi d}}\left[\operatorname {arctg} {\frac {\cos \left({\frac {\pi }{V}}\varphi _{0}\right)+u}{\sin \left({\frac {\pi }{V}}\varphi _{0}\right)}}\right]_{0}^{\infty }={\frac {V\varphi _{0}}{d}}\end{array}}$

and in the limiting case $\varphi _{0}=V$, in which the contour line coincides with the condenser plate:

(6)	${\mathfrak {K}}_{x}={\frac {V^{2}}{d}}$

Here it was calculated at first, as if the condenser had the extension 1 perpendicular to the drawing plane. However, upon any of the plates of length $b$, a force $\pm {\mathfrak {K}}_{2}^{0}$ dragging into the outside is acting at their center, amounting to:

(7)	$\left|{\mathfrak {K}}_{x}^{0}\right|={\frac {bV^{2}}{d}}={\frac {1}{2}}{\frac {E^{0}}{a}}$

because the electrostatic energy can be set in the previous approximation

$E^{0}={\frac {1}{2}}abd\cdot \left({\frac {2V}{d}}\right)^{2}=2{\frac {ab}{d}}V^{2}$

The corresponding holds for the four edges of length $a$; though the forces acting on it come not into consideration here, since they themselves as well as the connecting line of their contact points are perpendicular to velocity ${\mathfrak {q}}$ by the previous presuppositions.

III.

The result of section II is illustrated in Fig. 2, representing a cross-section (perpendicular to the edges $b$) through the center of the condenser. $K^{0}$ is chosen as reference system. Forces $\pm {\mathfrak {K}}_{1}^{0}$ are acting at the plate-centers $M$ and $M'$, forces $\pm {\mathfrak {K}}_{2}^{0}$ are acting at the centers $A,B,A',B'$ of the edges of length $b$; their amounts are given by (1) and (7). The lengths of the distances representing these forces, are not drawn in accordance with (1) and (7). Nevertheless this and all following figures correctly represent the directions of the forces. It is obvious that no torque is exerted.

The cross-axis $x^{0},y^{0},z^{0}$ is chosen so that the $z^{0}$-axis is parallel to the edges $b$ and thus to the drawing plane; the $x^{0}$-axis shall form the angle $\vartheta ^{0}$ with the normal of the plane. The following relation thus holds for points $A,B,A',B'$:

(8)	$y_{M}^{0}-y_{M'}^{0}=d\sin \vartheta ^{0},\ y_{A}^{0}-y_{B}^{0}=y_{A'}^{0}-y_{B'}^{0}=a\cos \vartheta ^{0}$

Furthermore it is:

(9)	${\mathfrak {K}}_{1x}^{0}=-{\frac {E^{0}}{d}}\cos \vartheta ^{0},\ {\mathfrak {K}}_{2x}^{0}=-{\frac {1}{2}}{\frac {E^{0}}{d}}\sin \vartheta ^{0}$

Now we transform into a system $K$, with respect to which $K^{0}$ has the velocity $q$ in the $x$-direction. The Lorentz transformation reads:

(10)	$x^{0}={\frac {x-qt}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}}},\ y^{0}=y,\ z^{0}=z,\ t^{0}={\frac {t-{\frac {q}{c^{2}}}x}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}}}$

The transformation formulas for the forces read^[10]:

(11)	${\mathfrak {K}}_{x}={\mathfrak {K}}_{x}^{0},\ {\mathfrak {K}}_{y}={\mathfrak {K}}_{y}^{0}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}},\ {\mathfrak {K}}_{z}={\mathfrak {K}}_{z}^{0}{\sqrt {1-{\frac {q^{2}}{c^{2}}}}}$

Fig. 3. illustrates the deformation of the condenser due to Lorentz contraction, and the direction of forces by (11). (In the figures, it is chosen:

$\vartheta ^{0}={\frac {\pi }{4}},\ {\sqrt {1-{\frac {q^{2}}{c^{2}}}}}={\frac {1}{2}}$, i.e., $q=0,866c$;

Fig. 2, 3, 4 and 5 are drawn in the same measure.) It can be seen without further ado, that a torque about the $z$-axis is exerted in the sense given in the figure:

Fig. 2.

Fig. 3.

We calculate its magnitude by the formula:

$N=\sum \left(x{\mathfrak {K}}_{y}-y{\mathfrak {K}}_{x}\right)$

Now, the torque in the rest system is:

(11a)

$N^{0}=\sum \left(x^{0}{\mathfrak {K}}_{y}^{0}-y^{0}{\mathfrak {K}}_{x}^{0}\right)=0$

It is to be noticed for the evaluation of the corresponding sum for system $K$, that oppositely acting forces are always occurring pairwise, so that it is not about the absolute value of $x$ and $y$, but only about the differences $x_{a}-x_{b}$, $y_{a}-y_{b}$, which are to be formed at equal $t$, and consequently by (10) they have to be set to (Lorentz contraction):

$\left(x_{a}^{0}-x_{b}^{0}\right){\sqrt {1-{\frac {q^{2}}{c^{2}}}}},\ y_{a}^{0}-y_{b}^{0}$

Thus by (11) and (11a)

(12)	$\left\{{\begin{array}{ll}N=\sum \left(x{\mathfrak {K}}_{y}-y{\mathfrak {K}}_{x}\right)&=\sum \left(x^{0}{\mathfrak {K}}_{y}^{0}\left(1-{\frac {q^{2}}{c^{2}}}\right)-y^{0}{\mathfrak {K}}_{x}^{0}\right)\\\\&=-{\frac {q^{2}}{c^{2}}}\sum y^{0}{\mathfrak {K}}_{x}^{0}\end{array}}\right.$

Regardting the force couple $\pm {\mathfrak {K}}_{1}$, the torque [[according to (8) and (9)] amounts to:

$-{\frac {q^{2}}{c^{2}}}\left(y_{M}^{0}-y_{M'}^{0}\right){\mathfrak {K}}_{1x}^{0}={\frac {1}{2}}{\frac {q^{2}}{c^{2}}}E^{0}\sin 2\vartheta ^{0}$

yet for any of the two force couples $\pm {\mathfrak {K}}_{2}$

$-{\frac {q^{2}}{c^{2}}}\left(y_{A}^{0}-y_{B}^{0}\right){\mathfrak {K}}_{2x}^{0}={\frac {1}{4}}{\frac {q^{2}}{c^{2}}}E^{0}\sin 2\vartheta ^{0}$

In the whole we thus find a torque

(13)	$N={\frac {q^{2}}{c^{2}}}E^{0}\sin 2\vartheta ^{0}$

in agreement with the previous way of calculation.

The forces (mentioned at the end of section II) acting upon the center of the sides of length $a$, are parallel to $z^{0}$ and remain parallel to $z$ at the transformation, as well as the connecting line of their contact points. Thus they produce no torque.

Since it is presupposed that $a\gg d$, then it is $\left|{\mathfrak {K}}_{2}^{0}\right|\ll \left|{\mathfrak {K}}_{1}^{0}\right|$ by (1) and (7). When despite of this the two force couples $\pm {\mathfrak {K}}_{2}$ contribute to half of the torque $N$, then this is [according to (8)] due to the fact, that they act at a lever-arm proportional to $a$, while forces $\pm {\mathfrak {K}}_{1}$ only act at a lever-arm proportional to $d$.

IV.

The status of our consideration with respect to the absolute theory is as follows. As to the transformation of the forces arising at this place, the absolute theory agrees with the relativity theory since these forces are of electromagnetic origin. But the Lorentz transformation is unknown to it, so that the condenser has the same force according to this theory, as in Fig. 2. Nevertheless, the previous way of conclusion can be transferred; because also the absolute theory is familiar with a procedure of projection, by which the problem of the electromagnetic field of uniformly moving carriers of charge can be reduced to an electrostatic problem.

Fig. 4.

Fig. 5.

For that, we only have to start with a resting condenser, having a form (with equal total charge) which emerges from the one drawn in Fig. 2 by an elongation in the $x$-direction by the factor $1/{\sqrt {1-q^{2}/c^{2}}}$ (Fig. 4). When we (exactly as before) pass from rest system $K^{0}$ to $K$ (Fig. 5), then this elongation is exactly compensated by the contraction of all dimensions parallel to $x$ by the factor ${\sqrt {1-q^{2}/c^{2}}}$. Thus also Abraham was forced, as he tried to study the field of a rigid spherical electron in motion, to assume a static field of an elongated ellipsoid of rotation.^[11]

Now, as to the forces acting upon the condenser of Fig. 4, they are of course of different amount than the one calculated with respect to Fig. 2. For example, the forces in $A$ and $B'$ as well as $A'$ and $B$ are still mutually equal, but the force in $A$ is stronger than the one in $A'$. The positive torque exerted by these four forces around the $z$-axis is compensated by the fact, that the forces perpendicular to the plates are not lying in one line as in Fig. 2 any more. Although their contact points (as shown by a simple consideration) are lying nearer at $A$ and $B$ than at $A'$ and $B$, they must produce a negative torque. Because the sum of all torques is of course zero as in any static field. Fig. 5 now emerges from Fig. 4 by exactly the same construction as Fig. 3 from Fig. 2. It represents the forces on the moving condenser in accordance with the absolute theory. One sees without further ado, that a torque is exerted in Fig. 5 having the same sense as in Fig. 3. Its amount, however, could not be calculated so easily in general. Yet if one considers, that the velocity $q$ is small against $c$ in the Trouton-Noble experiment, it can be seen that by the rotation leading from Fig. 2 to Fig. 4, the points of the condenser suffer displacements that are small of second order, i.e. they are proportional to $q^{2}/c^{2}$. Consequently, the forces drawn in Fig. 4 can be represented with respect to increasing powers of $q^{2}/c^{2}$ by series expansion, whose first terms independent of $q^{2}$ are represented by the previously calculated forces ${\mathfrak {K}}_{1}^{0}$ and ${\mathfrak {K}}_{2}^{0}$. This causes that the torque exerted upon the condenser in Fig. 5, differs from the one previously given only by terms of fourth and higher order. Because in expression (12)

$N=-{\frac {q^{2}}{c^{2}}}\sum y^{0}{\mathfrak {K}}_{x}^{0}$

$y^{0}$ as well as ${\mathfrak {K}}_{x}^{0}$ only differ by terms of second order from the values in (8) and (9). In the same way, the electrostatic energy $E^{0}$ can be represented in the form of such a series. Therefore (except terms of fourth and higher order) relation (13) remains

$N={\frac {q^{2}}{c^{2}}}E^{0}\sin 2\vartheta ^{0}$

V.

Until now we have presupposed, that the plates of the condenser are completely embedded in a dielectric of dielectric constant 1. This presupposition was not satisfied in the experiment. We want to discuss the modifications consequently to be made with respect to the previous considerations.

a) Let us assume that a medium of dielectric constant $\epsilon$ fills (besides the space between the plates) also the entire surrounding of the condenser to such an extent, as its field has a noticeable intensity. Then all ponderomotive forces are of $\epsilon$-times the amount as above, if it is presupposed that the stress difference between the plates still amounts $2V$; for example by (7) it is now:

(14)	$\left|{\mathfrak {K}}_{2}^{0}\right|={\frac {\epsilon bV}{d}}$

Thus the same holds by (12) for torque $N$; but on the other hand also for the electrostatic energy $E^{0}$. Thus relation (13) between $N$ and $E^{0}$ is maintained, for the absolute theory as well as for the relativity theory.

b) The following presupposition fits better to the actual circumstances than assumption a), namely that the dielectric only fills the interior space, and that its boundary to the outside is formed by a cylindric surface formed by electric force lines, thus equation $\psi$ = const. in the mode of expression of II. As to the appearance of the field, nothing will be changed by that. Because both in the dielectric as well as in the exterior space, the conditions of the electrostatic field are satisfied by equation (2) and (3); at the limit, however, not only the tangential components of the field strength are steady (since they are tangential themselves), but also the normal component of the electric displacement, since it is zero on both sides; thus also the limiting conditions are satisfied.

In this case, to forces ${\mathfrak {K}}_{2}^{0}$ in $A$ and $A'$ also a force ${\mathfrak {K}}_{3}^{0}$ is added, which also stresses the material parts of the condenser tangential to the plates. It acts upon the boundary of the dielectric. If the boundary of the dielectric lies in the vicinity of the rim of the plate, then the sum $2\left|{\mathfrak {K}}_{2}^{0}\right|+\left|{\mathfrak {K}}_{3}^{0}\right|$ plays now the same role for the calculation of torque $N$, as it was earlier played by the two forces ${\mathfrak {K}}_{2}$. Namely, a pressure $(\epsilon /2)\left.{\mathfrak {E}}^{0}\right.^{2}$ perpendicular to the force lines is present in the dielectric, and outside of it there is a pressure ${\tfrac {1}{2}}\left.{\mathfrak {E}}^{0}\right.^{2}$. From the difference of these pressures, the force (parallel to the plates due to reasons of symmetry) is given:

(15)	$\left|{\mathfrak {K}}_{3}^{0}\right|={\frac {1}{2}}b(\epsilon -1)\int \left.{\mathfrak {E}}^{0}\right.^{2}\cos n\ x\ d\ l$

where the integration over the whole force line $\psi$ is to be extended from the plate upon potential $-V$, to the one of potential $+V$. $n$ is the outermost normal of the limiting surface. Thus (see Fig. 1) $n$ and $dl$ are located to each other as $x$ to the $y$-direction, so that

${\frac {\partial \varphi }{\partial l}}={\frac {\partial \psi }{\partial n}}$

However, since $\left.{\mathfrak {E}}^{0}\right.^{2}=(\partial \varphi /\partial l)^{2}$ and due to the invariability of $\varphi$ in the $n$-direction and $\psi$ in the $l$-direction, it can be set:

${\frac {\partial \varphi }{\partial l}}dl=d\varphi ,\ {\frac {\partial \psi }{\partial n}}\cos nx={\frac {\partial \psi }{\partial x}}$

thus it follows from (15):

(16)	$\left|{\mathfrak {K}}_{3}^{0}\right|={\frac {\epsilon -1}{2}}b\int \limits _{-V}^{+V}{\frac {\partial \psi }{\partial x}}d\varphi$

However, by (2) or (3) it is:

${\frac {\partial \psi }{\partial x}}={\frac {2V}{d}}{\frac {1+e^{{\frac {\pi }{V}}\psi }\cos {\frac {\pi }{V}}\varphi }{1+e^{2{\frac {\pi }{V}}\psi }+e^{{\frac {\pi }{V}}\varphi }\cos {\frac {\pi }{V}}\varphi }}$

If we set for abbreviation

${\frac {\pi }{V}}\varphi =u,\ e^{{\frac {\pi }{V}}\psi }=s$

hence

$\left|{\mathfrak {K}}_{3}^{0}\right|=(\epsilon -1){\frac {bV^{2}}{\pi d}}\int \limits _{-\pi }^{+\pi }{\frac {1+s\cos u}{1+s^{2}+2s\cos u}}du$

Furthermore it is known^[12], that

$\int \limits _{-\pi }^{+\pi }{\frac {1+s\cos u}{1+s^{2}+2s\cos u}}du={\begin{array}{ccc}2\pi &\mathrm {if} &s^{2}<1,\\\pi &''&s^{2}=1,\\0&''&s^{2}>1;\end{array}}$

thus

(17)	$\left|{\mathfrak {K}}_{3}^{0}\right|={\begin{array}{ccc}2(\epsilon -1){\frac {bV^{2}}{d}}&\mathrm {if} &\psi <0,\\(\epsilon -1){\frac {bV^{2}}{a}}&''&\psi =0,\\0&''&\psi >0;\end{array}}$

The force lines, for which $\psi <0$, are now located between the plates, while the ones with positive $\psi$ start at the outside of the plates. The limit is formed by line $\psi =0$ starting from the edge of the plate. Thus, if the boundary of the dielectric lies in the interior of the condenser ($\psi <0$), then the rims of the plates are in empty space, so that by (7)

$\left|{\mathfrak {K}}_{2}^{0}\right|={\frac {bV^{2}}{d}}$

and by (17)

$2\left|{\mathfrak {K}}_{2}^{0}\right|+\left|{\mathfrak {K}}_{3}^{0}\right|=2\epsilon {\frac {bV^{2}}{d}}$

On the other hand, if one of the force lines $\psi >0$ bounds the dielectric, then the edges are completely embedded in the dielectric, so that by (14)

$\left|{\mathfrak {K}}_{2}^{0}\right|=\epsilon {\frac {bV^{2}}{d}}$

and by (17)

$2\left|{\mathfrak {K}}_{2}^{0}\right|+\left|{\mathfrak {K}}_{3}^{0}\right|=2\epsilon {\frac {bV^{2}}{d}}$

The simple consideration, showing in consequence of II that $2\left|{\mathfrak {K}}_{2}^{0}\right|+\left|{\mathfrak {K}}_{3}^{0}\right|$ has this value also in the case $\psi =0$, we don't want to carry out. The forces stressing the condenser tangentially to the plates, are thus in summa always the same as under assumption a), and since the corresponding holds for the electrostatic energy $E^{0}$ (within the approximation employed here, which neglects all marginal actions in the calculation of $E^{0}$), then also here the relation (13) between $N$ and $E^{0}$ remains as under assumption a).

That the absolute theory only adds to it terms of fourth and higher order, can be shown as it was done under IV.

The boundary of the dielectric now actually hasn't the form of a force line. If we would extend our investigation to this case, we would be faced with great mathematical difficulties, because one cannot use the solution of Helmholtz (section III) for the potential function at the rim any more. Our result can only qualitatively be transferred, so that forces are acting at the rim and the boundary of the dielectric (besides the attraction of the plates) which are tangentially stressing the plates in the state of rest, so that the totality of these forces produces the torque in the case of motion. In addition, our investigations show such a far reaching independence of the results from the special presuppositions, that our goal – the illustration of Lorentz's theory of the Trouton-Noble experiment – should be achieved.

Munich, Institute for Theoretical Physics, March 1912.

(Received March 9, 1912).

---

1. Fr. P. Trouton and H. R. Noble, Proc. Roy. Soc. 72. p. 132. 1903. The largest observable rotation amounted to 5% of the one calculated by the absolute theory. Unfortunately, the experiments only lasted over the period of 10 days, so that a proponent of the absolute theory can still object, that during this time the velocity of Earth against the Sun is canceled by the unknown motion of the Sun against the aether (see E. Budde, Physik. Zeitschr. 12. p. 978.
2. H. A. Lorentz, Proc. Amsterdam 1904, p. 809.
3. M. Laue, Das Relativitätsprinzip. Braunschweig. p. 168, also Verh. d. Deutsch. Physik. Ges. 13. p. 513. 1911 and Ann. d. Phys. 35. p. 524. 1911. See also P. Epstein, Ann. d. Phys. 36. p. 779. 1911.
4. H. A. Lorentz, l.c. M. Laue. Relativitätsprinzip. p. 99.
5. M. Abraham, Ann. d. Phys. (4) 14. p. 236. 1904. § 3. Theorie der Elektrizität II. p. 329 ff. 1908. See also M. Laue, Relativitätsprinzip, § 23.
6. M. Abraham, Ann. d. Phys. 14. p. 236. 1904. § 5.
7. M.Planck, Theorie der Wärmestrahlung. Leipzig 1906. § 53 to 58. W. Meissner, Dissert. Berlin 1907.
8. From the last condition we can free us by taking two contour lines $\pm \varphi =\mathrm {const} <V$ as cross-section of the plates in section II. Equation (1) and (7) remain unchanged. In the same way, one can arbitrarily vary the form of the condenser plates without essential changing the form, as long as one only assumes that their dimensions are great against $d$.
9. H. v. Helmholtz, Berliner Ber. 1868. p. 215. Ges. Abh. I. p. 146. See also Vorlesungen über theoretische Physik IV. p. 236. 1907.
10. M. Laue, Das Relativitätsprinzip, Gleichung (87a). One has to put ${\mathfrak {q}}'=0,\ v=q$ within it.
11. M. Abraham, Ann. d. Phys. 10. 5. 105. 1903.
12. See e.g. J. A. Serret, Differential- und Integralrechnung 2. p. 344. Leipzig 1899.