Rothwell Bronrowan physbron@t-online.de This paper begins with an example chosen to indicate the difficulties associated with determining which of several plausible, alternative frames of reference - if any - is the correct frame of reference in the given setup. These findings are then used to propose a new, universal frame of reference. The example envisages a long table in a railway carriage with a smooth, magnetic surface on (or above) which a puck-shaped object can float. It is supposed that all friction between the "puck" and the table surface is thereby eliminated. The puck rebounds from the table edges, again with no significant loss of energy. The train is initially stationary. The puck is pushed from one side of the table to the opposite side, where it rebounds and returns again to the starting position. Trajectory of puck ( + ) Starting point Fig. 1 (Motion of puck on a table in a stationary railway carriage) The puck is then held still while the train is accelerated to a speed of 100 km/h. Then the procedure is repeated and the puck follows the same trajectory as when the train was stationary (Fig. 1). The behaviour of the puck would initially seem to suggest, since there is no apparent difference in its trajectory for these two attempts despite different train speeds, that the correct frame of reference for the motion of the puck must be the table - or the railway carriage, both of these representing the same frame of reference. 21.11.12 1 (7)
But what other alternatives are there? Two obvious alternatives suggest themselves: (1) the planet earth; (2) the stationary ether, or "empty space" as its non-ether equivalent. Seen in the context of the earth (1), the above two experimental phases can be readily described. In the first phase - in which the train is stationary (Fig. 1) - there would be no noticeable difference in the trajectory of the puck, whether in the frame of the earth or in the frame of the carriage, since both frames are essentially identical. In the second phase, however, the puck would no longer describe a straight-line path in the earth frame on the outbound journey and the same path on the return journey - as in the frame of the carriage - but would now describe longer paths equivalent to two sides of a triangle: Motion of train Trajectory of puck Motion of train Starting point Fig. 2 (Motion of puck on a table in a moving railway carriage, seen in the frame of the earth, blue indicating the motion of the table/train) Seen in the context of the frame of reference of the stationary ether (or of empty space), however, the paths described in these two phases would both be equivalent to the two sides of a triangle. These sides would differ from each other, though not by very much (Fig's 3 & 4, below). Both of them would be much longer, however, than in the second phase of the earth frame, since they also take the motion of the earth into account. Fig's 3 & 4 (Path described by the puck for each phase in the frame of the ether/space, exaggerated) 21.11.12 2 (7)
So now we have three optional frames to choose from, though the first option (the carriage frame) does seem to suggest itself as the likeliest candidate. But suppose someone who favours the earth frame of reference complains that the puck only appears to be in the frame of reference of the carriage because it was held tight while the carriage was accelerating and thereby "took on" the speed of the carriage before being pushed over the table again. Repeating the experiment to account for this, the puck, which is again pushed over the table while the train is stationary thereby tracing its original trajectory in both of these frames, is now allowed to continue its trajectory while the train is accelerated to a speed of 100 km/h. And this time it changes its trajectory in the frame of the carriage (though a bit wobbly at first) to the same triangle as that already described in the frame of the earth (Fig. 2), though now in the opposite direction: Trajectory of puck Motion of train Starting point Fig. 5 (Table in a moving railway carriage, seen in the frame of the carriage) This now seems like an argument in favour of the frame of the earth. Those favouring the frame of the carriage may nevertheless claim that it is normal that things in this frame be accelerated along with the carriage - otherwise everything would fall out of it! - and that this does not disqualify their claim. In fact, they might even insist that things are no different in the frame of the earth, since initially the puck, which was stationary in the frame of the stationary carriage, was only stationary in the frame of the moving earth because it had already "taken on" the earth's motion, or velocity. At this point those preferring the ether/space frame of reference might well join the argument, claiming that, in a moving frame such as that of the earth, the true trajectory described by the moving puck is never anything but a triangle-type trajectory (Fig's 3 & 4), and that their frame is the only one that always describes such a trajectory. So what now? Can we resolve the dispute with a clear decision in favour of one of these options? 21.11.12 3 (7)
You may have noticed that the differences between these various trajectories can also be associated with other factors not yet explicitly mentioned, namely the times taken and the forces involved. In the case of the stationary carriage, for example, the frames of the carriage and of the earth seem to be identical. In the second phase, however, they differ. This difference is one we can arguably use to help determine which of the two frames remains essentially unchanged vis-à-vis the trajectory of the puck. We have three situations to consider in this connection: 1) that of the stationary carriage (Fig. 1); 2) that of the moving carriage in which the puck has taken on the speed of the carriage (Fig. 2 [earth's frame] - though arguably also as in Fig. 1 [carriage frame], since the carriage is really only stationary in the earth's frame); 3) that of the moving carriage in which the puck has not taken on the speed of the carriage (Fig. 5). For situations 2 and 3, the force required in situation 1 to transfer the puck to the opposite side of the table and back in a given time serves as a common comparative value for the different frames. So let's keep the force constant and measure the change in times. In situation 2, if the time taken for the puck to cross the table and return under the same force as applied in situation 1 is identical to that of situation 1, then the carriage is the correct frame. If the time taken is longer, then the earth is the correct frame. As already indicated, though, if the puck has been held steady during the period of acceleration of the train and has thereby taken on the speed of the train, this could compensate for a part of any additional force required for the outbound and return journeys. So in this case we cannot say for certain that, if the time taken is identical to that of situation 1, this indicates that the carriage is indeed the correct frame of reference. The second point, however, should hold, i.e. if the time required for the outbound and return journeys is longer, this suggests that the correct frame of reference is that of the earth. Why? Because the reason for the extra time taken is that the true trajectory is not that indicated in Fig. 1, but is that shown in Fig. 2. Here the distances actually travelled, as indicated by the lengths of the arrows, are longer, so that a greater force is required if the puck is to complete the outbound and return journeys in the same time as in situation 1. This in turn means that if we keep the force unaltered, more time will be required to travel the longer distance, thus indicating that the correct frame is that of the earth. In situation 3, which is indicated in Fig. 5, it seems apparent - even in the frame of the carriage! - that the correct frame of reference is that of the earth. The reason is that the 21.11.12 4 (7)
puck, at each stage of its outbound and return journeys, remains above the same point on the earth, under the moving train. This is why the arrows point backwards. However, the question as to whether more time is now required for the longer journey or whether the puck travels the greater distance in the same time - implying a greater applied force - must still be answered. At first it seems fairly clear that if the force applied remains unchanged and the distance covered is greater, the time taken must be longer. If one considers a situation in which a pen is mechanically drawn from one side of a piece of paper to the other under the same applied force, however, one can now see that by pulling the paper in one direction one can indeed increase the length of the trajectory marked without altering the force applied and without requiring more time for the supposedly longer journey. This last example makes it fairly clear that it is the moving train that "creates" the longer journey. In the frame of the earth the force applied and the time taken for the journey should therefore both remain unchanged! So in other words - and in contrast to what was argued above! - since the longer journey is due solely to the moving train, the time taken for the longer journey will not increase in the frame of the earth, nor will the force applied. On the other hand, the argument made above against the carriage frame and for the earth's frame, namely that the puck had already taken on the velocity of the carriage before phase 2 of the experiments was undertaken, can also be made against the earth's frame - since the puck had taken on the velocity of the earth before the experiments began - in favour of the ether/space frame of reference. So can we determine the speed of the earth in the ether/space frame? In this connection a zero velocity would seem to be required, since otherwise the velocity of the earth would only be measurable against some other object or feature, which would arguably also have to be taken into account. So let's now take a closer look at force as such and resulting or corresponding speeds, to see whether such an approach can help us resolve this issue, or at least provide an alternative. Force can be defined as the energy required to increase the velocity of a massive body at a certain rate of increase, which we call acceleration, i.e. F = ma, where F is the force (or energy) applied, m is the mass of the body and a is the resulting acceleration, or rate of increase in velocity. This equation, however, is clearly flawed, since the rate of increase in velocity is not constant for a constant F and m. 21.11.12 5 (7)
This can be clearly seen in the case of a Formula 1 racing car, which requires more energy (in the form of fuel) to accelerate from 250 to 260 km/h than it requires to accelerate from 50 to 60 km/h. Einstein, who was aware of this need for more energy to maintain the same acceleration at higher speeds, sought to remedy the issue (a = F/m) by increasing the mass at higher speeds. Apart from the maths, however, he had no justification whatsoever for doing so. A more intuitive alternative is F net = ma, where F can be seen as a constant in terms of the energy expended, but not in terms of the net or effective energy (F net ) generating the acceleration, i.e. F - F net > 0. Seen is this context, a = F net /m, i.e. the acceleration decreases as the net force decreases, the mass - and the applied or expended force (F) - remaining constant. The explanation of this phenomenon is that, at higher speeds, more of the energy expended is required to maintain the speed already achieved - or to reach the speed already achieved, in a mechanical context, by the object to be pushed or pulled - leaving less and less energy (or force) for still-further acceleration. Think of this in terms of two vehicles, a trailer and a second vehicle pushing or pulling it. Before the second vehicle can further accelerate the trailer it must itself first reach the speed of the trailer. And since the energy or force expended in doing so (F) is finite, sooner or later it will all be required for this purpose (i.e. F - F net = F F net = 0 a = 0). A similar situation can also be envisaged in the context of a charged particle in an electromagnetic field. The particle can be expected to accelerate more quickly at the start (when the field is initially generated) than subsequently. And at some speed - its maximum or terminal speed within the force field - its acceleration will drop to zero. So by applying a known force to a massive body, the reaction of the massive body will give an indication of the net force applied such that this can be used to determine the velocity of the massive body. But things are not quite that simple. The direction of the applied force is important here. Coming back to the Formula 1 example, in which friction plays the major role in transferring forces, acceleration begins at a maximum rate, this rate diminishing until top speed is reached. Braking, by contrast, exhibits a deceleration (= negative acceleration, or rate of decrease in the velocity of the racing car) that is greater at all phases (e.g. 60-50 km/h) than the equivalent rate of acceleration (50-60 km/h). 21.11.12 6 (7)
The same can be expected to apply to a charged particle accelerating or decelerating in a force field. If we imagine the particle travelling at almost the speed of light (c), for example, any accelerating force must be applied from "behind" it and will have a minimal effect, since F net will be close to 0. The same force applied from in front of it, by contrast, will have a maximum, brakingtype effect. For a particle not within a force field, of course, this will not be seen in terms of braking, but also in terms of acceleration. So how can we determine whether we are applying our force from "ahead" or from "behind"? This should be easy to determine by using at least two experimental setups, one with a force field operating in one direction, the other with a force field operating in the opposite direction. If the reactions of particles of equal mass and equal velocity then differ, the direction with the greatest acceleration shows the "braking" reaction, i.e. the particle is "moving" in the direction opposite the direction of the applied force. If, on the other hand, there is no difference between the reaction of the two particles, then a) the forces have been applied from the sides, in which case rotate the setup through 90, or b) the particles are travelling at no great speed, i.e. v ~ 0. But even if this approach does allow us to determine the motion of small particles, can it help us determine the motion of planets? Large massive bodies, after all, are not responsive - in terms of changes in their orbits or trajectories - even to the largest forces available to mankind. The answer is yes, theoretically at any rate. If, for example, we would like to know the direction in which a planet or a solar system is moving and with what (average) velocity, all we need to do is to send a spacecraft in the same direction of travel with either a similar speed or a speed that can be calculated - and then corrected - to match that of the planet or system in question. Then a known force or force field can be applied in different directions to identical masses within the spacecraft in order to identify differences in their reactions, which can then be used, as indicated above, to determine their common initial speed in space, and hence that of the targeted planet or solar system. So this approach does arguably identify a universal frame of reference against which to measure the absolute motion of massive bodies: the universal frame of net force. ** 21.11.12 7 (7)