# An Example Concerning the Dynamics of the Theory of Relativity

 An Example Concerning the Dynamics of the Theory of Relativity  (1911)  by Max von Laue, translated by Wikisource
 In German: "Ein Beispiel zur Dynamik der Relativitätstheorie." Verhandlungen der Deutschen Physikalischen Gesellschaft 13: 513–518. Source

An Example Concerning the Dynamics of the Theory of Relativity

by M. Laue.

Fig. 1

In a paper of Lewis and Tolman[1] concerning the dynamics of relativity theory, the following consideration can be found, which should serve for the derivation of the transformation formulas of force: In the valid reference system $K^{0}$, an angle-lever $ABC$ (Fig. 1) with two equally long arms – being mutually perpendicular $\left(AB=BC=l^{0}\right)$ – it at rest. It is rotatable in $B$ around an axis perpendicular to its plane. $A$ and $C$ are affected by two forces $\mathfrak{K}_{1}^{0}$ and $\mathfrak{K}_{2}^{0}$ of equal magnitude; $\mathfrak{K}_{1}^{0}$ is parallel to $\overrightarrow{BC}$, and $\mathfrak{K}_{2}^{0}$ to $\overrightarrow{BA}$. The torques of these forces are $\pm\left|\mathfrak{K}_{1}^{0}\right|l^{0}$ and they compensate each other.

Let us consider this state from another reference system $K$, relative to which $K^{0}$ has the velocity $\mathfrak{q}$ and the direction $\overrightarrow{BC}$. Within it, the arm $AB$ has the length

 $l_{1}=l^{0}\,$ 1)

while, due to Lorentz contraction, $BC$ has just the length

 $l_{2}=l^{0}\frac{\sqrt{c^{2}-q^{2}}}{c}$ 2)

The torque of both forces in the sense given in the figure, is:

 $\left|\mathfrak{R}\right|=\left|\mathfrak{K}_{1}\right|l_{1}-\left|\mathfrak{K}_{2}\right|l_{2}=l^{0}\left\{ \left|\mathfrak{K}_{1}\right|-\left|\mathfrak{K}_{2}\right|\frac{\sqrt{c^{2}-q^{2}}}{c}\right\}$ 3)

From that, Lewis and Tolman conclude $\left|\mathfrak{R}\right|=0$ (since no rotation of the lever arises in $K$ as well), so that:

$\frac{\left|\mathfrak{K}_{1}\right|}{\left|\mathfrak{K}_{2}\right|}=\frac{\sqrt{c^{2}-q^{2}}}{c}$

[ 514 ] Though this conclusion is not correct[2]. It's known[3] that for the transformation of an arbitrary force $\mathfrak{K}$ in the passage from the primed system $K'$ to $K$, when $K'$ has the velocity $\mathfrak{v}$ relative to $K$ and $x$ and $x'$ are parallel to it, and when $y$ is chosen parallel to $y'$, and $z$ is parallel to $z'$:

$\begin{array}{c} \mathfrak{K}_{x}=\frac{\mathfrak{K}'_{x}+\frac{v}{c^{2}}(\mathfrak{q}'\mathfrak{K}')}{1+\frac{v\mathfrak{q}'_{x}}{c^{2}}},\ \mathfrak{K}_{y}=\mathfrak{K}'_{y}\frac{\sqrt{1-\beta^{2}}}{1+\frac{v\mathfrak{q}'_{x}}{c^{2}}},\ \mathfrak{K}_{z}=\mathfrak{K}'_{z}\frac{\sqrt{1-\beta^{2}}}{1+\frac{v\mathfrak{q}'_{x}}{c^{2}}}.\\ \\\left(\beta=\frac{v}{c}\right)\end{array}$

If $K'$ is replaced by $K^{0}$, then $\mathfrak{q}'=0,\ \mathfrak{v}=\mathfrak{q}$, thus

 $\mathfrak{K}_{x}=\mathfrak{K}_{x}^{0},\ \mathfrak{K}_{y}=\mathfrak{K}_{y}^{0}\frac{\sqrt{c^{2}-q^{2}}}{c},\ \mathfrak{K}_{z}=\mathfrak{K}_{z}^{0}\frac{\sqrt{c^{2}-q^{2}}}{c}$ 4)

However, in the present case $\mathfrak{K}_{1}$ is parallel to $\mathfrak{q}$, and $\mathfrak{K}_{2}$ is perpendicular to it, thus it follows from 4)

 $\mathfrak{K}_{1}=\mathfrak{K}_{1}^{0},\ \mathfrak{K}_{2}=\mathfrak{K}_{2}^{0}\frac{\sqrt{c^{2}-q^{2}}}{c},\ \frac{\left|\mathfrak{K}_{1}\right|}{\left|\mathfrak{K}_{2}\right|}=\frac{c}{\sqrt{c^{2}-q^{2}}}$ 5)

According to 3), the torque $\mathfrak{R}$ is thus by no means zero, but

 $\left|\mathfrak{R}\right|=l^{0}\left|\mathfrak{K}_{1}^{0}\right|\frac{q^{2}}{c^{2}}$ 6)

This fact is an example for the theorem, that an elastically stressed body – as the angle-lever at this place – requires a torque for a purely translatory motion, because its angular momentum $\mathfrak{L}$ changes in the course of this, and exactly as in Newton's dynamics, a torque

 $\mathfrak{R}=\frac{d\mathfrak{L}}{dt}$ 7)

is necessary for this[4]. Here, we can simply calculate the increase of $\mathfrak{L}$, and in this way confirm relation 7) by 6). There, the starting point is the theorem of the inertia of energy [ 515 ] in its most general form stated by Planck[5], which relates the momentum density $(\mathfrak{g})$ with the energy current $(\mathfrak{S})$ by

 $\mathfrak{g}=\frac{\mathfrak{S}}{c^{2}}$ 8)

At point $A$, force $\mathfrak{K}_{1}$ performs the work $\mathfrak{q}\left|\mathfrak{K}_{1}\right|$. An energy current of this magnitude (i.e. current density $\times$ cross-section) thus enters here into the lever, propagates in equal magnitude to pivot $B$, and there it crosses into the axis. This is because the axis exerts a force $-\left(\mathfrak{K}_{1}+\mathfrak{K}_{2}\right)$ upon the lever, and thus performs the work $-q\left|\mathfrak{K}_{1}\right|$. However, force $\mathfrak{K}_{2}$ performs no work, so that except the mentioned current, only the convective entrainment of the energy resting in the lever (heat, chemical energy, etc.) arises, which everywhere represents a current parallel to $\mathfrak{g}$. This comes not into consideration for our purposes (see below), and we therefore omit it.

To find the total momentum $\mathfrak{G}$ of the lever, we have to form the integral over its volume

$\int\mathfrak{g}dV=\frac{1}{c^{2}}\int\mathfrak{S}dV$

Arm $BC$ comes not into consideration here, and for $AB$ we already have executed the integration of the current density over the cross-section by calculating the current magnitude. Thus

$\left|\int\mathfrak{S}dV\right|=l_{1}q\left|\mathfrak{K}_{1}\right|$

and by 1) and 5)

 $\left|\mathfrak{G}\right|=\frac{q}{c^{2}}l_{1}\left|\mathfrak{K}_{1}\right|=\frac{q}{c^{2}}l^{0}\left|\mathfrak{K}_{1}^{0}\right|$ 9)

The direction of $\mathfrak{G}$ is that of $\overrightarrow{AB}$, thus it is perpendicular to velocity $\mathfrak{q}$. In this fact – which is excluded in Newton's dynamics by the definition of momentum $(\mathfrak{G}=m\mathfrak{q})$ – lies the explanation of the increase of angular momentum in translatory motion.[6]

[ 516 ] Namely, the angular momentum is (exactly as in Newton's dynamics) in relation to momentum density $\mathfrak{g}$ by

$\mathfrak{L}=\int[\mathfrak{rg}]dV$

Fig. 2

where $\mathfrak{r}$ is the line segment of the radius vector, drawn from any space point to $dV$. If point $O$ located in the extension of $BC$ (Fig. 2) is chosen here (Fig. 2), then it is evidently

$\left|\mathfrak{L}\right|=\left|\mathfrak{G}\right|\cdot OB$

namely the direction of $\mathfrak{L}$ is directed upwards, perpendicular to the plane of the lever. In time $dt$, however, the angle-lever moves from its starting position into the end position (see the dotted line), thus $OB$ grows from $qdt$, and $\left|\mathfrak{L}\right|$ increases per unit time according to 9)

 $\frac{d\left|\mathfrak{L}\right|}{dt}=q\left|\mathfrak{G}\right|=\frac{q^{2}}{c^{2}}l^{0}\left|\mathfrak{K}_{1}^{0}\right|$ 10)

Regarding the absolute values of $\left|\mathfrak{L}\right|$ and $\left|\mathfrak{R}\right|$, the agreement between 6) and 10) confirms equation 7). But also in terms of the direction, vectors $\mathfrak{R}$ and $\tfrac{d\mathfrak{L}}{dt}$ are in agreement, because both of them are directed upwards perpendicular to the drawing plane. At the same time one can see, that a component of $\mathfrak{G}$ (parallel to $\mathfrak{q}$) doesn't changes value 10), so that the omission of the convection current of energy is irrelevant. – In general, the increase of angular momentum is

$\frac{d\mathfrak{L}}{dt}=\left|\mathfrak{qG}\right|$[7]

by which the fact can be confirmed again, that the component of $\mathfrak{G}$ (parallel to $\mathfrak{q}$) doesn't matter.

Eventually, we imagine that the lever is in a shell, at which an axis of rotation is mounted, and from which – for example by strained springs – forces $\mathfrak{K}_{1}$ and $\mathfrak{K}_{2}$ are exerted. The energy current considered above, is continued in the axis after it left the lever at $B$, then it enters [ 517 ] the shell through the mounting locations of the axes, from there it enters the spring which affects the lever at $A$, and thus it is closed. If one forms the integral $\int\mathfrak{S}dV$ over the system completed in this way, one finds the value zero; for the same reason, at a stationary electric current, by which the integral of the current density vanishes over the entire current-area.[8] The total momentum stemming from this current, is also zero according to 8); the totality thus has no momentum component perpendicular to the velocity. However, the shell itself (including the axes and the springs) surely has such a component; it must be oppositely equal to the momentum calculated in 9). From that it follows, that the shall needs a torque for its translatory motion, which is opposite to the one given in 6). This is exerted by the lever by means of the springs and the axes upon the shell, as well as the shall conversely exerts the previous torque upon the lever.

The whole thing is an analogy to the moving condenser in the Trouton-Noble experiment[9]. We can compare its field with an angle-lever, its material parts with the shell. Neither the electromagnetic momentum of one part, nor the mechanical momentum of the other part, is parallel to the velocity, therefore both need a torque for their translatory motion. However, since momentum as a whole is parallel to velocity, both momentum components (perpendicular to the velocity) are oppositely equal to one another, and the same holds thus for the torque. The totality doesn't require any torque, it rather represents (as in the model consisting of angle-lever and shell) a "completely static system"[10], [ 518 ] which obeys the dynamics of the mass-point for stationary and quasi-stationary translatory motions.

That the momentum of the angle-lever has not the direction of its velocity, is also given by another consequence, which appears to be paradox in Newton's dynamics. Namely, if it is accelerated in a quasi-stationary way in the direction of its velocity (longitudinal), and if we denote by $\dot{q}$ the absolute amount of acceleration, then its momentum component perpendicular to $q$, increases according to 9) by

$\frac{d\left|\mathfrak{G}\right|}{dt}=\frac{\dot{q}}{c^{2}}l^{0}\left|\mathfrak{K}_{1}^{0}\right|$

According to the momentum theorem

$\mathfrak{K}=\frac{d\mathfrak{G}}{dt}$,

a component is involved, which is perpendicular (transverse) to the velocity of the accelerating force $\mathfrak{K}$; namely it doesn't vanish, when we pass to the limiting case $q=0$. Thus also in that limit, Newton's mechanics doesn't hold for the lever. Of course, a longitudinal force component for the increase of the longitudinal momentum component is additionally needed for this. The corresponding self-evidently holds for the shell alone. On the other hand, at longitudinal acceleration of the whole system, the momentum increase is purely longitudinal, so that it behaves as a mass point at quasi-stationary acceleration. Also this is a characteristic example for the behavior of all stressed bodies.

Munich, Institute for theoretical physics, July 1911.

1. Gilbert N. Lewis and Richard C. Tolman, Phil. Mag. (6) 18, 510, 1909.
2. I was alluded to the fact that the result of the mentioned authors is wrong, by Professor Sommerfeld.
3. See for example, M. Planck, Ann. d. Phys. (4) 26, 1, 1908, equation 21 or M. Laue, Das Relativitätsprinzip. Braunschweig, Friedr. Vieweg & Sohn, 1911. Equation 87a.
4. See M. Laue, l. c. § 27c, also Ann. d. Phys. (4) 35, 524, 1911.
5. M. Planck, Phys. ZS. 9, 828, 1909; Verh. d. D. Phys. Ges. 10, 728, 1908.
6. Only in hydrodynamics, at the motion of a solid body in a fluid, the corresponding case arises that the momentum of all moving masses is not parallel to the velocity of the solid body.
7. See M. Laue, l.c.
8. Namely, it is $\mathrm{div}\mathfrak{S}=0$; if this equation is integrated over a space, limited by an even and resting cross-section of the system, and also by any area outside of the system, then one finds $\int\mathfrak{S}_{n}d\sigma=0$, and this integral is extended over the entire boundary. However, at the part located outside of the system, it is $\mathfrak{S}_{n}=0$, thus also the cross-section alone $\int\mathfrak{S}_{n}d\sigma=0$. Now we sum up this relation over all cross-sections perpendicular to direction $n$, and thus find $\int\mathfrak{S}_{n}dV=0$, where $n$ denotes an arbitrary direction.
9. H. A. Lorentz, Proc. Amsterdam, p. 803; see also M. Laue, l. c. § 17 c.
10. M. Laue, l. c. § 27d.
This is a translation and has a separate copyright status from the original text. The license for the translation applies to this edition only.
Original:
 This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1960, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 50 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works.
Translation:
 This work is released under the Creative Commons Attribution-ShareAlike 3.0 Unported license, which allows free use, distribution, and creation of derivatives, so long as the license is unchanged and clearly noted, and the original author is attributed.