Translation:Concerning Relativistic Statics

Concerning Relativistic Statics.

by.

While the dynamics of relativity theory was substantially extended by the foundational papers of and , the field of statics – which became important also practically by the  experiment – experienced no attention from theoretical side up to now. Only quite recently, the relativistic statics was discussed by ; the view developed in the present report is essentially different from 's one, and can be considered as its supplement. The static conditions of equilibrium for an observer moving arbitrarily relative to the considered system, are formally derived in §§ 2 and 3, whereas § 1 contains a physical explanation of these relations.

§ 1. The broken lever of Lewis and Tolman.
We start on the basis of a thought experiment given by and. A broken lever is considered, which is affected by two forces. Let arms and forces be equal from the standpoint of the co-moving observer.

$a'=b',\ f_{x'}=f_{y'}$

Thus there is equilibrium. Let the lever now move with velocity $$q$$ in direction $$x$$. The authors take the standpoint of the resting observer and apply the law of the lever. Since arm $$OA$$ has now (due to Lorentz contraction) the length

$a=a'\cdot\sqrt{1-\beta^{2}},\ \beta=\frac{q}{c}$,|undefined

and since the equilibrium (kinematically seen as the state of rest) must exist independently from the reference system, it can be concluded that for the resting observer, the ratio of forces is expressed in this way:

$\frac{f_{x}}{f_{y}}=\sqrt{1-\beta^{2}}$|undefined

On the other hand, from the transformation equations of relativistic mechanics it follows:

This paradox was already discussed by, who showed that this difficulty can be solved when one assumes a reaction effect of the energy current arising in the lever. We want to consider these relations from another standpoint and first pose the following question: Does this case contain something unexpected, or is mechanics allowing us to predict, that the static moments of the affecting forces are not relevant for equilibrium? We will see, that under the considered circumstances, the application of the law of the lever is unjustified, and that the reason for this lies in the tensor character of mass.

For this purpose, we first want to discuss a somewhat simpler case. A massless rod is rotatable mounted at point $$O$$, and a mass point of rest mass $$m_0$$ at which force $$f$$ is acting, is mounted at point $$M$$. It shall be for the co-moving observer:

$\frac{f_{x'}}{f_{y'}}=\operatorname{tg}\ \alpha'$|undefined

The total force is acting in the direction of the connecting line $$OM$$, thus there is equilibrium. For the resting observer it is

thus

$\frac{f_{x}}{f_{y}}=\frac{\operatorname{tg}\ \alpha}{1-\beta^{2}}$|undefined

The total force is not lying any more in the direction of connecting line $$OM$$.

However, if we consider the accelerations $$j_{x},\ j_{y}$$ exerted by the forces $$f_{x},\ f_{y}$$ upon mass point $$M$$, then due to the tensor character of mass

$\begin{array}{c} m_{\mathrm{long}.}=m_{0}\left(1-\beta^{2}\right)^{-3/2},\ m_{\mathrm{tr}.}=m_{0}\left(1-\beta^{2}\right)^{-1/2},\\ \\j_{x}=\frac{f_{x}}{m_{\mathrm{long}.}}=\frac{f_{x}}{m_{0}}\left(1-\beta^{2}\right)^{3/2},\\ \\j_{y}=\frac{f_{y}}{m_{\mathrm{tr}.}}=\frac{f_{y}}{m_{0}}\left(1-\beta^{2}\right)^{1/2},\end{array}$|undefined

therefore

$\frac{j_{x}}{j_{y}}=\operatorname{tg}\ \alpha.$|undefined

The (virtual) total acceleration lies in the direction of connecting line $$OM$$. We know that no rotation is actually taking place, and it is surely difficult to see why the rod should be rotating, since point $$M$$ tends to move into the direction $$OM$$ (according to the acceleration just determined).

Also the law of the lever is proven in ordinary mechanics on the basis of proportionality of force and acceleration, thus it can impossibly be applied in the case considered at the beginning. Indeed, we can show by application of a mechanical principle, which doesn't use the force concept (for example ' principle of least constraint), that equilibrium must exist also in the case of the broken lever, even though the forces doesn't satisfy the law of the lever.

Let us assume, that the lever of Fig. 1. is rotated in the very small time $$\tau$$ about the rest angle $$\varphi'$$ under the influence of these forces, then ' principle requires

$\sum md^{2}=\mathrm{Minimum}$,

where $$d$$ is the distance between the end-location after time $$\tau$$, which is actually occupied by a mass point, and the end-location which it would occupy when it were free. We denote by $$m_{x'},\ m_{y'}$$ the rest mass, at which forces $$f_{x}$$ or $$f_{y}$$ are directly acting. If mass point $$m_{x}$$ were free, then it should move in the small time $$\tau$$ by the distance

$j_{x}\cdot\frac{\tau^{2}}{2}=\frac{f_{x}}{m_{x}}\cdot\frac{\tau^{2}}{2}=\frac{f_{x'}}{m_{x'}}\left(1-\beta^{2}\right)^{3/2}\cdot\frac{\tau^{2}}{2}$|undefined

(according to (1) and (2)) in the $$x$$-direction. However, the lever is actually rotating about the rest angle $$\varphi'$$, so that the components of motion of point $$m_{x}$$ are:

$\alpha'\sin\varphi'\sqrt{1-\beta^{2}}$ and $a'(1-\cos\varphi')$.|undefined

The corresponding quantities for mass point $$m_{y}$$ are

$\frac{f_{y'}}{m_{y'}}\cdot\left(1-\beta^{2}\right)\cdot\frac{\tau^{2}}{2},\ a'\sin\varphi',\ a'(1-\cos\varphi')\cdot\sqrt{1-\beta^{2}}$|undefined

These two mass points thus give the amount for expression $$\sum md^{2}$$:

$\begin{array}{l} \frac{m_{x'}}{\left(1-\beta^{2}\right)^{3/2}}\left\{ \left[\frac{f_{x'}}{m_{x}}\cdot\left(1-\beta^{2}\right)\frac{\tau^{2}}{2}-a'\sin\varphi'\right]^{2}\cdot\left(1-\beta^{2}\right)+a'^{2}\cdot(1-\cos\varphi')^{2}\right\} \\ \\\quad+\frac{m_{y'}}{\left(1-\beta^{2}\right)^{1/2}}\cdot\left\{ \left[\frac{f_{y'}}{m_{y}}\cdot\frac{\tau^{2}}{2}\left(1-\beta^{2}\right)+a'\sin\varphi'\right]^{2}+a'^{2}(1-\cos\varphi')^{2}\left(1-\beta^{2}\right)\right\} \end{array}$|undefined

In case the remainder of the lever is massless, then this expression must become a minimum, thus its derivative with respect to $$\varphi'$$ must vanish.

$\begin{array}{l} m_{x'}\left\{ -\left[\frac{f_{x'}}{m_{x}}\cdot\frac{\tau^{2}}{2}\left(1-\beta^{2}\right)-a'\sin\varphi'\right]\cdot\cos\varphi'+\frac{a'(1-\cos\varphi')\sin\varphi'}{1-\beta^{2}}\right\} \\ \\\quad+m_{y'}\left\{ \left[\frac{f_{y'}}{m_{y}}\cdot\frac{\tau^{2}}{2}\cdot\left(1-\beta^{2}\right)+a'\sin\varphi'\right]\cdot\cos\varphi'+a'(1-\cos\varphi')\sin\varphi'\left(1-\beta^{2}\right)\right\} =0.\end{array}$|undefined

By definition it is $$f_{x'}=f_{y'}$$, therefore $$\varphi'=0$$ satisfies this equation. Yet this means, that the lever is not rotating, what was to be proven.

Nothing changes at this result, when we consider a ponderable lever. The contribution of all other masses of the lever, except those of points $$m_{x},\ m_{y}$$ for function $$\sum md^{2}$$ which we want to call $$\sum\limits _{R}md^{2}$$, namely corresponds to the same function for a lever, upon which no forces act at all. Thus, $$\sum\limits _{R}md^{2}$$ is also made to a minimum by $$\varphi'=0$$.

Thus we arrived at the result, that equilibrium must exist, even though the sum of torques doesn't vanish.

§ 2. The general equilibrium condition in relativity theory.
We considered it as interesting, to allude to those connections and to emphasize, that mechanics fully accounts for this, yet we cannot conceal to ourselves, that this mode of consideration is practically highly inconvenient, since the force concept loses its ordinary sense with which we are familiar – in both dynamics and statics. It is therefore of great advantage, to introduce a supplementary force, which is so to be arranged, that the total force constructed by the parallelogram law, coincides with the direction of acceleration again. This is for example performed by a compensating force attached at the $$x$$-direction

$f_{x}^{k}=-\beta^{2}f_{x}$,

by which the $$x$$-component is weakened in the ratio $$1-\beta^{2}$$, which generates the proportionality between force components and acceleration components according to equation (3). At the same time, the tensor-mass is removed from the world; under consideration of force $$f_{x}^{k}$$, the mass is now $$m_{0}/\sqrt{1-\beta^{2}}$$ in all directions.

We now want to transfer our considerations into 's four-dimensional world, namely in terms of the modification with an imaginary time component $$l=ict$$ introduced by. If we look at the four-force $$\mathfrak{K}$$ (see note at p. 780) instead of the ian force $$f$$, then the compensating force to be attached is:

However, since force $$\mathfrak{K}$$ is always perpendicular upon the world line, and if $$x$$ shall be the direction of motion (thus $$dy=dz=0$$)

Therefore

Therefore we can apply ordinary statics, when we assume a ponderomotive action (into the $$x$$-direction) of the time-component of the force, whose magnitude is $$i\beta\mathfrak{K}_{l}$$.

This view can conveniently inserted into 's world, when we see the space- and time components of $$\mathfrak{K}$$ as equally valid in their ponderomotive action.

Indeed, we have already mentioned in the beginning, that equilibrium must exist in all reference systems, when it exists for a co-moving observer. Yet for the latter, the equilibrium condition is equal to the disappearance of the torque around any possible rotation axis, or in other words, the disappearance of the torque in the plane $$E$$ perpendicular to the considered axis. This plane is corresponding at any moment to a definite plane in the world; the resting observer has to project the lever arms and forces (including time components) which are measured in this system, to the plane just mentioned: The criterion of equilibrium is equal to the disappearance of the torque obtained in this way. Thus we state the following rule:

There is equilibrium around an axis defined in an arbitrary reference system, when the sum of the torques of all forces in the plane (perpendicular to that axis and to the world-line direction of their affecting points) is disappearing in the world.

Since the world-line direction coincides with the time axis of the co-moving system, then the plane to be defined is identical with the previously mentioned plane $$E$$. We want to apply the rule just obtained upon the broken lever of and. From the standpoint of the resting observer, the lever arms are: $$a$$ in the $$x$$-direction and $$b$$ in the $$y$$-direction (for the sake of generality, we assume them as being different). Force $$\mathfrak{K}_{y}$$ acts upon the first one; forces $$\mathfrak{K}_{x}$$ and $$\mathfrak{K}_{l}$$ upon $$b$$. The $$x'y'$$-plane of the co-moving system is rotated towards plane $$xy$$ by the angle $$\varphi=\operatorname{arctg}\ i\beta$$ around the $$yz$$-plane. The projections of the lever arms upon this plane, thus read $$a\cos\varphi$$, $$b$$, and those of the forces $$\mathfrak{K}_{y}$$, $$\mathfrak{K}_{x}\cos\varphi$$, $$\mathfrak{\ K}_{l}\sin\varphi$$. The equilibrium condition is therefore

or

$\begin{array}{rr} b\left(\mathfrak{K}_{x}+\mathfrak{K}_{l}\operatorname{tg}\varphi\right)= & a\mathfrak{K}_{y},\\ b\left(\mathfrak{K}_{x}+i\beta\mathfrak{K}_{l}\right)= & a\mathfrak{K}_{y}\end{array}$

and according to (4')

$b\left(1-\beta^{2}\right)\mathfrak{K}_{x}=a\mathfrak{K}_{y}$

in exact agreement with our previous assumption (3) about the addition of a compensating force.

Since the time component $$\mathfrak{K}_{l}$$ is proportional to the performance of work, then this mode of consideration is practically in agreement with the more physical view of , according to which any energy current is always accompanied by momentum. From the things said it follows, that it is about a definition which serves to maintain the parallelogram- and lever laws, that cannot hold without a similar extension of the force definition.

§ 3. The six-vector of torque.
A simple and elegant analytical form can be given to the equilibrium condition of the previous paragraph, when one extends the concept of static moment in a suitable way in terms of the four-dimensional world of ; this view I owe to Prof..

A six-vector or spacetime vector of second kind, can represented first as a plane section of certain magnitude and length in space of four dimensions. This geometric image is, however, too special in so far as it only has five independent specification parts. However, if one imagines (in the perpendicular plane to be considered ) a second plane section of certain magnitude, then the epitome of two such plane sections being mutually perpendicular, forms the general six vector. In the following, we only have to do with the special six vector, consisting of a single plane section; the vector formed by a plane section (in the normal plane) of same magnitude, we call (following ) its "dual" six-vector.

The torque (in the world) is to be defined as four-dimensional vector product

$$\mathfrak{K}$$ is the vector force known to us, and $$\mathfrak{R}$$ shall be denote the arm of the force measured from an origin being fixed in the world.

$$\mathfrak{M}$$ is therefore a parallelogram constructed by the four vectors $$\mathfrak{R}$$ and $$\mathfrak{K}$$ in the world, whose magnitude, location and rotation sense are determined, thus it is a special six-vector. The components of torque $$\mathfrak{M}$$ are obviously projections of the parallelogram upon the six coordinate planes

{{MathForm1|(7)|$$\left\{ \begin{array}{lll} \mathfrak{M}_{xy}=\mathfrak{R}_{x}\mathfrak{K}_{y}-\mathfrak{R}_{y}\mathfrak{K}_{x}, & & \mathfrak{M}_{xl}=\mathfrak{R}_{x}\mathfrak{K}_{l}-\mathfrak{R}_{l}\mathfrak{K}_{x},\\ \mathfrak{M}_{yz}=\mathfrak{R}_{y}\mathfrak{K}_{z}-\mathfrak{R}_{z}\mathfrak{K}_{y}, & & \mathfrak{M}_{yl}=\mathfrak{R}_{y}\mathfrak{K}_{l}-\mathfrak{R}_{l}\mathfrak{K}_{y},\\ \mathfrak{M}_{zx}=\mathfrak{R}_{z}\mathfrak{K}_{x}-\mathfrak{R}_{x}\mathfrak{K}_{z}, & & \mathfrak{M}_{zl}=\mathfrak{R}_{z}\mathfrak{K}_{l}-\mathfrak{R}_{l}\mathfrak{K}_{z}.\end{array}\right.$$}}

We assert now, that the equilibrium condition for the co-moving (primed) system, consists in this, that vector $$\sum\mathfrak{M}'$$ (the geometric sum of the four-dimensional torques of all forces) vanishes in all of its components. Indeed, $$(x',y',z',t')$$ are the components of $$\mathfrak{R}'$$, where $$l'$$ is constant for all moments of the sum, since they shall all be measured at the same time. Therefore we can pull out $$l'$$ before the sign of the sum. If we further consider that $$\mathfrak{K}_{l'}$$ vanishes (according to equation (4') of the previous paragraph), then we can write the condition just given:

{{MathForm1|(8)|$$\left\{ \begin{array}{c} \sum\mathfrak{M}_{x'y'}=\sum\left(x'\mathfrak{K}_{y'}-y'\mathfrak{K}_{x'}\right)=0,\\ \\\sum\mathfrak{M}_{y'z'}=\sum\left(y'\mathfrak{K}_{z'}-z'\mathfrak{K}_{y'}\right)=0,\\ \\\sum\mathfrak{M}_{z'x'}=\sum\left(z'\mathfrak{K}_{x'}-x'\mathfrak{K}_{z'}\right)=0,\\ \\\sum\mathfrak{M}_{x'l'}=l'\sum\mathfrak{K}_{x'}=0,\ \sum\mathfrak{M}_{y'l'}=l'\sum\mathfrak{K}_{y'}=0,\\ \\\sum\mathfrak{M}_{z'l'}=l'\sum\mathfrak{K}_{z'}=0;\end{array}\right.$$}}

when $$\mathfrak{r}'(x',y',z')$$ is the three-dimensional arm in space $$(x',y',z')$$, and $$f'=\mathfrak{K}'\cdot\sqrt{1-\beta^{2}}$$ the ian force, this reduces to

We see, that the two equilibrium conditions (9), which are heterogeneous in ordinary mechanics, can be uniformly combined by introduction of the six-vector $$\mathfrak{M}$$ into condition (8).

If we direct our attention to torque $$\sum\mathfrak{M}$$, as it is representing itself to the non-co-moving observer, then we must notice again, that all $$\mathfrak{R}_{l}$$ are equal, since the location of all affecting points are simultaneously determined. A calculation which is quite analogous to the one just executed, shows us that $$\mathfrak{R}_{l}$$ always occurs with the factor $$\sum\mathfrak{K}_{x}$$ (or $$\sum\mathfrak{K}_{y},\ \sum\mathfrak{K}_{z}$$), which identically vanish. Thus the value of $$\sum\mathfrak{M}$$ is completely independent from $$\mathfrak{R}_{l}$$, so that we can give any value to $$\sum\mathfrak{M}$$. We use this and set $$\mathfrak{R}_{l}=\mathfrak{R}_{l'}=0$$ for the following consideration, which should simplify them not insignificantly.

Under the assumption, that the transformation equations for both four vectors $$\mathfrak{R}$$ and $$\mathfrak{K}$$ are as follows:

the transformation equation of the special six vector $$\mathfrak{M}$$ is given in the following form

{{MathForm1|(11)|$$\left\{ \begin{array}{lll} \mathfrak{M}_{i'k'} & =\mathfrak{M}_{xy}\cdot\left|{\cos(i'x)\cos(i'y)\atop \cos(k'x)\cos(k'y)}\right| & +\mathfrak{M}_{yz}\cdot\left|{\cos(i'y)\cos(i'z)\atop \cos(k'y)\cos(k'z)}\right|\\ \\ & +\mathfrak{M}_{zx}\cdot\left|{\cos(i'z)\cos(i'x)\atop \cos(k'z)\cos(k'x)}\right| & +\mathfrak{M}_{xl}\cdot\left|{\cos(i'x)\cos(i'l)\atop \cos(k'x)\cos(k'l)}\right|\\ \\ & +\mathfrak{M}_{yl}\cdot\left|{\cos(i'y)\cos(i'l)\atop \cos(k'y)\cos(k'l)}\right| & +\mathfrak{M}_{xl}\cdot\left|{\cos(i'z)\cos(i'l)\atop \cos(k'z)\cos(k'l)}\right|\end{array}\right.$$}}

when $$i'$$ and $$k'$$ are two coordinate axes of system $$(x', y', z', l')$$.

We now want to test the presuppositions of equations (10) using Fig. 3. As to force $$\mathfrak{K}$$, the ordinary geometric projection rule of (5b) is actually correct in all cases. It is different for distance $$\mathfrak{K}_{x}$$ or $$\mathfrak{R}_{y}$$, $$\mathfrak{R}_{z}$$. Distances are always measured synchronously, i.e., the locations of the initial- and endpoints are determined simultaneously; therefore distance $$a'$$ in a co-moving system,



goes over into distance $$a$$ in the stationary one, which is cut from a line parallel to the $$x$$-axis by world lines $$W_1$$ and $$W_2$$ of the endpoints. Distances don't have a $$l$$-component at all. From that we see, that $$\mathfrak{R}$$ is transformed into $$\mathfrak{R}'$$ as follows

$\mathfrak{R}_{x'}=\mathfrak{R}_{x}\cos\varphi,\ \mathfrak{R}_{y'}=\mathfrak{R}_{y},\ \mathfrak{R}_{z'}=\mathfrak{R}_{z}$

Since it should be $$\mathfrak{R}_{l}\equiv0$$, the relations are in agreement with equation (10a), yet equation (10a) is not usable for the determination of time component $$\mathfrak{R}_{l}$$. Thus one sees, 1. that equation (10a) is only valid for the space components of $$\mathfrak{R}'$$, though not for $$\mathfrak{R}_{l'}$$, 2. that this relation is not reversible, i.e., one may not permute the primed quantities with the unprimed ones. From that consideration we conclude: The transformation equation (11) is only applicable to the pure space components of the six-vector $$\mathfrak{M}'\left(\mathfrak{M}_{x'y'},\ \mathfrak{M}_{y'z'},\ \mathfrak{M}_{z'x'}\right)$$ and is not reversible, i.e., it cannot serve to express the components of vector $$\mathfrak{M}$$ by those of vector $$\mathfrak{M}'$$. There (as well as in the following) we understand under $$\mathfrak{M}$$ the total torque, thus the sum of torques of all individual forces. We see, that $$\mathfrak{M}$$ is not a six-vector like the six-vector of electrodynamics. The latter is a structure fixed in the world, which is projected into the relevant coordinate system. However, torque $$\mathfrak{M}$$ (also considered in the world), is different for any reference system, and the transformation equation shall transform the $$\mathfrak{M}$$ of one system into the $$\mathfrak{M}$$ of the other one. It is clear without further ado, that this passage can have the sense of a geometric projection in one direction only.

If we consider, that the value of the direction cosine is as follows

$\begin{array}{l} \cos(yy')=\cos(zz')=1,\\ \cos(xx')=\cos(ll')=\cos\varphi,\\ \cos(x'l)=-\cos(l'x)=\sin\varphi,\\ \cos(x'y)=\mathrm{etc}.=0\end{array}$

then starting from formula (11), we arrive at the following transformation equations

Now, the question concerning the equilibrium conditions in any reference system, can be solved in the simplest way. First we consider the equilibrium of a body rotatable around one fixed point. As shown above, the conditiond for that are

$\mathfrak{M}_{x'y'}=\mathfrak{M}_{y'z'}=\mathfrak{M}_{z'l'}=0$

For the non-co-moving system, they become according to (12)

$\begin{array}{c} \mathfrak{M}_{xy}\cos\varphi-\mathfrak{M}_{yl}\sin\varphi,\ \mathfrak{M}_{yz}=0,\\ \mathfrak{M}_{zx}\cos\varphi+\mathfrak{M}_{zl}\sin\varphi=0.\end{array}$

According to (7) $$\left(\mathfrak{R}_{l}=0\right)$$, the first of these conditions (for the case of the broken lever considered in the previous paragraph) is identical to equation (10) at p. 785.

If we go over to the case of equilibrium of a body with two fixed points, which is thus rotatable around an axis $$P'$$, which shall enclose the angle $$\lambda',\mu',\nu'$$ with the coordinate axes $$x',y',z$$, then the equilibrium condition in the co-moving system is the disappearance of the torque around the axis

In the passage to the resting system, these angles become $$\lambda,\mu,\nu$$, where the cosines $$\cos\lambda',\ \cos\mu',\ \cos\nu'$$ are proportional to the quantities $$\cos\lambda\cos\varphi,\ \cos\mu,\ \cos\nu$$ respectively. Using (12), condition (13) therefore goes over into

$\begin{array}{ll} \left[\mathfrak{M}_{yz}\cos\lambda\right. & \left.+\mathfrak{M}_{zx}\cos\mu+\mathfrak{M}_{xy}\cos\nu\right]\\ & +\left[\mathfrak{M}_{zl}\cos\mu-\mathfrak{M}_{yl}\cos\nu\right]\cdot\operatorname{tg}\ \varphi=0.\end{array}$

The interpretation of this expression is near at hand. The first bracketed expression is the static moment of the space components of forces $$\mathfrak{K}$$ (calculated by the rules of ordinary mechanics), or also the projection of $$\mathfrak{M}$$ upon space $$x, y, z$$ ; the second term, however, contains the moment of the time components. According to (7) $$\left(\mathfrak{R}_{l}=0\right)$$, the latter expression can also be written

$\sum\left(\mathfrak{R}_{z}\cos\mu-\mathfrak{R}_{y}\cos\nu\right)\mathfrak{K}_{l}\cdot\operatorname{tg}\ \varphi$

The perpendicular from the affecting point of the considered force $$\mathfrak{K}_{l}$$ upon axis $$P$$ (in space $$x, y, z$$) is not within the brackets, but the projections of it upon the $$xy$$-plane. From that we see, that for the three-dimensional observer under consideration of forces $$\mathfrak{K}_{l}$$, the latter must be seen as affecting in the $$x$$-direction and as multiplied with $$\operatorname{tg}\ \varphi$$, in agreement with rule (4) stated in the previous paragraph.

§ 4. Appendix: Momentum and mass transformation.
At this occasion, I want to discuss with some words the other thought experiment of and, which shall serve to derive the transformation formula of mass from the momentum theorem. The validity of the results found at that occasion, has been recently criticized by, which was unjustified as we will see. However, the execution of the thought experiment was not entirely correct, yet this doesn't affect the results.

The following brilliant arrangement was stated. Let observer $$B$$ be in motion with constant velocity $$q$$ relative to observer $$A$$. $$A$$ throws an elastic ball towards system $$B$$ perpendicular to velocity $$q$$; mass and velocity of the ball (as measured from system $$A$$) shall be $$m_{A}=m,\ v_{A}=v$$. At the same time, $$B$$ throws a ball towards $$A$$ perpendicularly to $$q$$, whose mass $$m_{B}$$ and velocity $$v_{B}$$, as measured from system $$B$$, are exactly the same: $$m$$ and $$v$$. The experiment is now so executed that the balls are colliding, where the collision diameter is exactly perpendicular to $$q$$, so that the momentum components are uninfluenced by the collision towards $$q$$. Now, and  argued in the following way. From perfect symmetry it follows, that $$A$$ and $$B$$ (each one at its ball) measure the same velocity change $$v-v'$$ before and after the collision. However, for observer $$A$$, the velocity change of ball $$B$$ is $$(v-v')\cdot\sqrt{1-\beta^{2}}$$ according to 's transformation equation, and since (according to the momentum theorem) the masses shall be inversely proportional to the velocity changes, it is concluded that

when $$m$$ is the mass corresponding to $$q$$, and $$m_{0}$$ is the rest mass.

has rigorously tested the reasoning of and. It didn't miss his attention, that the balls by definition also have velocities $$v, v'$$ perpendicular to $$q$$, and that the influence of these velocities upon the masses remained unconsidered. This influence only then vanishes, when $$v, v'$$ are very small relative to $$c$$. comes to the conclusion, that formula (1) is in contradiction with the momentum theorem, 1. when the velocity $$v, v'$$ are not small, 2. when (opposite to the assumption of and ) the collision occurs so that the collision diameter lies in the direction of $$q$$.

However, an even more rigorous test shows, that despite of some incorrectnesses of the argumentation, the result of and  is without doubt correct: the Ansatz

where $$w$$ is to be understood as the total velocity, satisfies the momentum theorem for any velocity and collision direction. Any of both balls namely has before and after the collision, generally spoken, two different velocities in the direction perpendicular to $$q$$, thus it has two different masses before and after the collision ($$m_{A}$$ and $$m_{A'},$$ and $$m_{B}$$ and $$m_{B'}$$). By application of the momentum theorem in the direction perpendicular to $$q$$, thus one only arrives at the equation

when applied to ball $$B$$ in the $$q$$-direction it gives

Equation (2) is satisfied for arbitrary great values of velocities $$v, v'$$ by Ansatz (1'). This can be seen from the special form, which is assumed by the velocities for this case. If a body is moved (in reference system $$B$$ moving with velocity $$q$$ relative to reference system $$A$$) perpendicular to $$q$$ with velocity $$v$$ (measured in $$B$$), then the total velocity $$w$$ as measured from $$A$$, is connected with $$q$$ and $$v$$ as follows

Under consideration of this formula, the four masses occurring in equation (2) are

{{MathForm1|(4)|$$\left\{ \begin{array}{c} m_{A}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}},\ m'_{A}=\frac{m_{0}}{\sqrt{1-\frac{v'^{2}}{c^{2}}}},\\ \\m_{B}=\frac{m_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}\cdot\frac{1}{\sqrt{1-\beta^{2}}},\ m'_{B}=\frac{m_{0}}{\sqrt{1-\frac{v'^{2}}{c^{2}}}}\cdot\frac{1}{\sqrt{1-\beta^{2}}}.\end{array}\right.$$}}

Inserted into equation (2'), this gives $$v'=v$$, by which equation (2) is satisfied.

In order to solve the case of the collision directed to relative velocity $$q$$, we want to generalize the thought experiment in so far, by assuming that both balls $$A$$ and $$B$$ also have velocity components in the $$q$$-direction, which (any of them is measured by the corresponding observer) are equal and opposite. If we denote these velocity components with $$u_{A},\ u_{B}$$ (both measured from the standpoint of $$A$$), then it is according to 's addition theorem

$u_{B}=\frac{u_{A}+q}{1+\frac{u_{A}q}{c^{2}}}$|undefined

After the collision, the new velocities are $$u'_{A}$$ and $$u'_{B}$$; however, if $$u_{B}$$ and $$q$$ were of same direction, then $$u'_{B}$$ and $$q$$ are now oppositely directed, thus $$u'_{A}$$ is equal to the sum of $$u'_{B}$$ and $$q$$

$u'_{A}=\frac{u'_{B}+q}{1+\frac{u'_{B}q}{c^{2}}}$|undefined

The momentum theorem gives the condition

$m_{A}\cdot u_{A}+m_{B}\cdot u_{B}=m'_{A}\cdot u'_{A}+m'_{B}\cdot u'_{B}$

In the limiting case considered by, when the components $$u_{A}, v, v'$$ are so small, that one can neglect them when compared with $$q$$, it must considered that ball $$B$$ has the small velocity $$\left(u'_{B}\right)$$ after the collision. If one develops all four masses with respect to those small magnitudes, and if one neglects the latter with respect to $$q$$, then one arrives at the identity

$q\cdot m_{q}\equiv q\cdot m_{q}$

Thus no contradiction with (1') exists. That is arriving at a deviating result, can be explained by the fact, that he has overlooked the difference of masses before and after the collision, and has calculated (as  and ) with two masses instead of four.

We believe, that the pretty thought experiment of and  has even gained in terms of didactic value by the consideration made about it at this place. Namely, it provides a welcomed example for the application of 's addition theorem (3), which looks somewhat strange.

I want to express my gratitude to Prof. for multiple support.

(Received August 12, 1911.)