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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.

  1. Make a particle accelerate at ~highest speed that the accelerator can achieve.

  2. 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)

  3. 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.

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  • $\begingroup$ In addition to John's answer, this blog article is very helpful, imo: profmattstrassler.com/articles-and-posts/… $\endgroup$ – user140606 Jan 20 '17 at 16:49
  • $\begingroup$ Large and infinity are different things. And the reasoning applies both ways: the amount of momentum that you need to transfer to speed up the particle is the same you need to stop it. $\endgroup$ – user126422 Jan 20 '17 at 17:28
  • $\begingroup$ "Only thing is that v is the velocity component in the direction of the force. So, for slowing down, it will be 0, or $\gamma\approx 1$" No. You use $-v$, which gives you the same gamma because velocity enters squared. Seriously. See my comment under John's answer about energy recovery linacs. We really do know what we are talking about with the stuff. $\endgroup$ – dmckee Jan 22 '17 at 20:02
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    $\begingroup$ @kpv If a force is applied opposite to a velocity, their dot product (i.e. the component of the force in the direction of the velocity) is certainly not zero. Its magnitude is the same as when the vectors are parallel, but its sign is different. Is that the source of the confusion here? $\endgroup$ – Mark Mitchison Jan 22 '17 at 20:37
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    $\begingroup$ @kpv It "breaks Newtonian mechanics" (actually it breaks Math 101) because you are defining antiparallel vectors to be orthogonal. $\endgroup$ – Mark Mitchison Jan 22 '17 at 20:52
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The question is founded on an incorrect assumption.

The math absolutely is symmetric between acceleration and deceleration (because velocity enters in to the Lorentz factor squared), and we have machines that take advantage of this fact.

Energy recovery linacs work in exactly the manner linacs usually work, only the field timing is maintained 180 degrees out of phase from the acceleration mode. This means that instead of the particle gaining energy at the expense of the field, the field gains energy at the expense of the particle. The forces are the same as in the accelerating case only opposed to the direction of motion, and the particle exhibits the same magnitude of coordinate acceleration (i.e. very little because it is highly relativistic) in the lab frame only slowing rather than speeding up.

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The mass does not approach infinity. The mass, more precisely the invariant mass, is a constant. What happens is the the momentum approaches infinity.

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.

Force is the rate of change of momentum, so if you apply a constant force the momentum will increase linearly with time. However for any finite force applied for a finite time the momentum can never reach infinity so the velocity can never reach $c$.

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  • $\begingroup$ So, if we keep applying force for a long time, and say speed does not change, or changes just a little to remain below c, where does the energy imparted by force goes to? Does it keep accumulating with the particle? Secondly, after the speed stabilizes at close to c, keep applying the force for one more hour. Suppose speed changes negligible during this one hour. Now reverse the same force to slow it down. Will speed decrease by same negligible amount during one hour? What does math says and what does experiments say if there have been any? $\endgroup$ – kpv Jan 20 '17 at 17:00
  • $\begingroup$ @kpv the energy is given by $E^2 = p^2c^2 + m^2c^4$, so as $p \to \infty$ also $E \to \infty$ $\endgroup$ – John Rennie Jan 20 '17 at 17:02
  • $\begingroup$ Please see if you can answer second part of my previous comments. $\endgroup$ – kpv Jan 20 '17 at 17:07
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    $\begingroup$ The LHC accelerates particles very, very close to $c$ every day (though I don't think it slows them down again) so the process I've described is verified all the time. $\endgroup$ – John Rennie Jan 20 '17 at 17:39
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    $\begingroup$ @kpv The field effect laser installation at Jefferson Lab is an example of a facility that slows a highly relativistic beam using the same mechanism they initially used for speeding it up. They even recover some of the energy input that way. See en.wikipedia.org/wiki/Energy_recovery_linac . $\endgroup$ – dmckee Jan 22 '17 at 20:00
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According to my understanding, you can't calculate these separately, you always have to hold the complete picture in mind, thus not only the momentum ("mass") approach infinity, but also Length Contraction. and Time dilation happens.

This means that as momentum, is $p=mv$ and $v=l/t$,

  1. Length contraction;
    So as this Length $l$ approaches zero, then any time would cause the velocity be zero too.

  2. Time Dilation;
    or this $t$ approaches infinity, and again any lenght would cause the velocity to be zero.

But if velocity is zero, then also $p=mv$ should be zero, no matter what is the mass! So something must be wrong here! (not to forget that all this stuff is still "agreeing with the experiments")

The Mass is wrong. There is no mass at all. And thus the "momentum" is in troubles with photon. Photon has momentum, but no mass. There is no mathematically solid explanations for this.

What happens, in the speed of light, or in photon emission, is that the volume of a particle goes to zero. This is easy to understand with the length contraction.

$$ L = L_0\sqrt{1 - \frac{v^2}{c^2}} $$

No matter what is the area, if height is zero, volume is zero. When volume approaches zero, the density approaches infinity, if we hold the idea of some "invariant mass". And this is what is observed in the experiments.

But mass is a pseudo thing, If you just think the Newton's 3rd law for while, you might realize that; total mass is zero, as every "mass" has equal negative "mass".

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Einstein's mass-velocity relationship, as given above by John Rennie, is a simple mathematical formula that comes from the special theory of relativity. It does not take into account the matter-wave duality and energy-temperature relationship, as given by the Maxwell-Boltzman formula.

$E = m(v) \cdot c^2$; $E= hf$ and $E= KT$ are three fundamental relationships. As the energy of a particle is increased, the associated matter-wave frequency and temperature also increase. Both of these factors increase the energy radiated by the mass to the surrounding. If these two properties of matter will not be there, it will not be possible to convert matter into energy. Einstein's mass-velocity formula, also does not take into account the fact that all matter contain an equal amount of positive and negative charges. These charges produce rotating Electro-Magnetic waves. Increasing the velocity of a particle produces EM waves in the form of spirals -- like the grooves in a cork-screw...These also contribute to the radiation of Energy ,in the form of EM waves to the surrounding.

So , as the velocity of a particle increases ,the Energy radiate by it to the surrounding also increases.
This problem , has not been studied by the Scientists in detail ...So no Mathematical formula relating Particle Velocity to the Energy Radiated by it , is available at present .
This is precisely what , has prevented scientists and Engineers from generating Fusion Energy , with the help of Tokamak.s .

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    $\begingroup$ I promise you can answer this question without thermodynamics. $\endgroup$ – Jahan Claes Jan 20 '17 at 19:22
  • $\begingroup$ When you are talking about Physical objects taking Infinite value , because of application of Mathematics , opinions will vary ..Mathematicians use " Infinite" freely , but when you consider applied Mathematics , concept of Infinite , gives rise to a lot of problems. Sometimes it is not possible to experimentally verify these mathematically derived conclusions. eg.-an Ideal Voltage Source has Infinite Charge and Infinite Capacitance --unrealisable ,but an integral part of Electrical Engg.Theory . $\endgroup$ – b.sahu Jan 20 '17 at 19:58
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    $\begingroup$ Sahu. Recommend you take a physics 1 course, you'll learn something and eventually see how wrong you are. As the energy of a particle is increased, in vacuum (there's no consideration of air or anything else needed to answer this basic physics question) the temperature does not change. Also your first equation is wrong. Best to start from scratch. $\endgroup$ – Bob Bee Jan 21 '17 at 6:29

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