MemberDecember 29, 2020 at 6:48 pm
I’m not sure whether this has been discussed before, or whether this is common knowledge, but I’ll just state my theory on what causes the relativistic mass effect. First off, I’ll state my fundamental assumptions:
- Forces expand at c with respect to their source.
- The mass of a particle actually remains constant.
From 1.) we can easily deduce that force fields are subject to the Doppler effect, specifically, the Doppler effect with a stationary emitter (since the force fields are emitted at constant relative speed within the reference frame of the emitting particle). If we imagine force fields as consisting of field lines (or rather, field spheres, since we’re in 3D space) that expand outwards at c, then it is easy to see that the Doppler effect will cause an observer moving towards the emitter particle to experience denser field lines than an observer moving away from the emitter particle. The number of field lines crossed determines the amount of force that is applied to a particle.
</div><div>Now for the interesting part: as a particle moves away from the emitter particle at exactly c, it experiences the force field waves as stationary, and no longer crosses any field lines per time, and thus no longer receives and force from the force field. As it speeds up beyond c, it actually experiences the field lines as if they were travelling in reverse, and thus, it also receives a reverse force compared to sub-c particles. The formula for the applied force should be</div><div>
where v is the relative velocity of the particle to the emitter, c is the speed at which the force field expands, and F’ is the amplitude of the force field at the moving particle’s location. For v=0, F=F’, and for v=(1+n)c, F=-nF’, and for v=-(1+n)c, F=-nF’. This would explain why particles cannot be accelerated faster than c using just a single force field without having to bend space and time and inventing magical mass. Thus, instead of relativistic mass, or some ether drag, we have a dampening effect within force fields that limits their effectiveness as something approaches c and slows things down that exceed c. This dampening effect can be circumvented as follows:
It should be possible to achieve FTL particle speeds in accelerators if the acceleration was not applied from behind the particle, but perpendicular to the particle’s movement direction. After all, if a particle moves only in x-direction, then it has a relative y-speed of 0, and thus, a force emitted in y-direction will be applied to the particle at full effectiveness without dampening its x-velocity (Fy=1-vy/c*Fy’). The x-velocity is not dampened because the particle does not pass any field lines from the inside to the outside in x-direction, and thus does not receive any reverse force.
MemberJanuary 15, 2021 at 2:23 am
Update: I previously didn’t consider that the forces themselves also seem to be acting with a speed of ±c respective to the emitter of the force, otherwise, it would not account for the speed limit of c on accelerators that pull particles instead of pushing them. So, the previous formula for the Doppler effect should only affect the field density, but not the effectiveness of the force applied to a particle.
A force moving with speed v acts upon an object moving with speed w with effectiveness e=(v-w)/v = (1-w/v). This means for faster velocities w>v, it has a negative effectiveness and slows down the object towards v. Thus, the new formula for the force applied to a particle should now be:
F = F'·f·e = F'·|1-v/c|·(1-v/c) = F'·±(1-v/c)²
where f=|1-v/c| is the perceived frequency/density of the field waves, and F’ is the original force/intensity of the force field, and e=(1-v/c) is the effectiveness of the applied force at the particle’s current relative velocity. The frequency is now an absolute value, since negative frequencies aren’t really possible, and the reversed force effect is now sufficiently described by the new term e.
The implication of this new formula would be that particles always are accelerated/decelerated towards ±c with respect to the force emitter.
Log in to reply.