While reasoning that why a particle can not be accelerated to light speed $c$, it is argued that the mass/momentum approaches infinity as speed approaches $c$. I think it is per GR.
I am sure this also fits into mathematics, otherwise people would not be making this argument.
I may be wrong, and please feel free to correct me if you think so. But I do not think that is the case - i.e. mass/momentum does not approach infinite.
My simple argument is - if the mass/momentum of a moving particle approaches infinite and such a particle moving at speeds close to $c$, then it would be almost impossible to stop that particle. In other words, it should be equally difficult/impossible to slow it down.
We all know that though it is not possible to accelerate the particle further, but it is no big deal to slow it down. Slowing down an infinite mass/momentum would not be that easy. Infinite mass reasoning must apply both ways - in speeding up as well as in slowing down. Has it been experimentally shown that it also applies to slowing down at limits close to $c$?
Therefore, I can argue that mass/momentum does not approach infinite, it is the forces that are rendered ineffective at such speeds because the force itself propagates at $c$ and can not accelerate anything as fast as itself, or faster. Force is rendered ineffective only in direction of motion (acceleration), not in opposite direction (slowing down).
Analogy how force may become ineffective - In a way, we can not accelerate a car that is already going at 300 miles/hr by pushing with our hands, because humans can not move their hand as fast. But we can accelerate a car going at 5 miles an hour. As the speed gets closer and closer to that of force $c$, the force can not push it any more. Same way as we can not move our hand faster than 300 miles/hr and can not accelerate that car by pushing on it. But slowing down would be effective, dangerous and fatal though.
Please correct if I am missing something, instead of blank down voting.
Considering formula given by John Rennie in his answer -
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The momentum of an object of mass $m$ moving at velocity $v$ is:
$$ p = \gamma m v = \frac{mv}{\sqrt{1 - \frac{v^2}{c^2}}} $$
which goes to infinity as $v \to c$. In the limit of $v \ll c$ the Lorentz factor $\gamma \approx 1$ and we recover the Newtonian approximation.
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Same math can be applied to effectiveness of the force. Only thing is that v is the velocity (only positive) component in the direction of the force. So, for slowing down, it will be 0, or $\gamma \approx 1$
The effective force $F1$ when particle is moving at velocity $v$ and a force $F$ is applied:
$$ F = \gamma F1 = \frac{F1}{\sqrt{1 - \frac{v^2}{c^2}}} $$
This way, the math does not change either.
So at limits close to $c$, the force must be fully effective in slowing down and pretty much ineffective in accelerating.
I am proposing below experiment to prove/disprove the concept. If someone is aware of such an experiment being done, please share the results.
Make a particle accelerate at ~highest speed that the accelerator can achieve.
Once this ~speed is achieved, continue to apply the force for another 1 minute. The particle should gain negligible speed during this 1 minute, but should gain a lot of momentum (per momentum formula)
Now stop the accelerating force and start an equal slowing force. I.e. reverse the force.
Per the current (infinite mass/momentum) explanation, 1 minute of slowing should reduce the speed by negligible – same speed that was gained during last 1 minute of acceleration. Because force is rate of change of momentum and same force in both directions should cause same change of momentum/speed during same amount of time.
But per my explanation, a lot more slowing down will take place during the 1 minute because gamma becomes zero for slowing down.
I think evidence and results of such experiment being done, can answer this question definitively. But equivalent other answers would help too - like evidence of the 7 Tev energy of protons being physically measured rather than just being calculated via the momentum formula.