Page:IgnatowskiBemerkung.djvu/5

 element at its motion, transformed to rest, conserves its form. This is 's differential equation. If we keep this condition, then superluminal velocity is given. For the time being we don't know as to how the body will actually move since we can make no statement about the interior forces of the body. 's differential equations is only one condition which must be satisfied by the motion of rigid bodies.


 * I want to add only some words to the remark concerning rigid bodies. Within such a body, every action is indeed propagating momentarily for a co-moving observer, and thus with superluminal velocity for a moving observer. It it were possible to show, that it contradicts the relativity principle by "telegraphing into the past" in this way, then the concept of the rigid body must surely be abandoned. The differential expressions by whose disappearance I have defined rigidity would still be useful, since they are a measure of the deformation of the volume element, so they should provide the foundation for an elasticity theory in accordance with the relativity principle. As to how the "propagation with superluminal velocity" mentioned by the lecturer arises, can be made quite illustrative in 's representation. I imagine a rod of length $$l$$ (in the following the speaker is drawing at the table) and lay it parallel to the $$y$$-axis; perpendicular to the $$xy$$-plane I include time as the third coordinate in order to describe the successive locations of the rod. When the rod is at rest, it will obviously represented by primes lying one above the other in the $$t$$-direction, and which can be combined to a plane belt parallel to the $$yt$$-plane. If one now gives the rod a motion in the direction of the $$x$$-axis, then this results in bending the belt towards the $$x$$-direction. If the rod should be at rest again at the end, then the belt will eventually be parallel to the $$yt$$-plane again. In order to show now that the process is propagating with superluminal speed, I take a new coordinate system in motion, that is, I introduce oblique parallel coordinates into our figure in accordance with the Lorentz transformation, at which the $$t'$$-axis is inclined against the $$t$$-axis and the $$x'y'$$-plane against the $$xy$$-plane. In this system, simultaneous events are represented by planes which are lying obliquely to the old $$xy$$-plane. If one intersects the bent sheet previously constructed with such an oblique plane, then the intersection is evidently not straight, but has a bending that is moving to the right when one moves the intersecting plane towards above, that is, it is moving to the right with superluminal velocity. The other example mentioned by, concerns the motion denoted by me as hyperbolic motion (the following is again illustrated by a drawing). It will be represented in the $$xt$$-plane by a bundle of hyperbolas having line $$x=\pm ct$$ as their asymptote. It satisfies my differential equations of rigidity and one can easily see, that the points having the same velocity are lying on the line passing through the origin. However, these lines are oblique and their inclination is smaller than that of the line $$x=\pm ct$$. The place where the velocity $$q$$ is present, thus propagates with superluminal velocity through the body.


 * I have wanted to add some words about this matter. arrives at the following formula (Annalen der Physik 23, 381, 1907),

$T=l\frac{1-Wv/V^{2}}{W-v}$|undefined

and says, that when $$W$$ is larger than $$V$$ ( denoted by $$V$$ which was denoted by $$c$$ in this lecture), one can always choose $$v$$ so that $$T$$ is smaller than zero, thus $$T$$ becomes negative. That is, one could telegraph into the past. However, if one sets $$W$$ equal to $$V^{2}/v$$ ($$c^{2}/q$$ in this lecture), then the numerator becomes equal to zero here. That is, the velocity becomes infinite for the coordinate system in which the rod is located. Because this velocity $$(c^{2}/q)$$ is nothing other than the velocity necessary to catch up the time in the moving coordinate system (in which the rod is located) as seen from our viewpoint (in the resting system). Because if we had a clock moving with this velocity as seen from our viewpoint, then it moves with infinitely large velocity for the coordinate system (in which the rod is located) according to the above; this boils down to the circumstance, that everywhere in the system it is simultaneously present. Thus there is an arbitrary amount of synchronous clocks, from which it follows that this velocity is actually the velocity to catch up the time in the moving system. Thus the value $$c^{2}/q$$ is the limiting velocity, at which the numerator of the previous formula becomes zero, thus $$T$$ equal to $$O$$, but not negative. A larger velocity is just not imaginable.