Mass, Rest Frames & Neutrinos: Explained

[Lapidus]

So every particle with some mass, even if the mass is very, very close to zero has a rest frame. A neutrino, say, could sit right next to me. But a photon, a massless particle, of course, can't, it has to zip by with light velocity.

But when I would reduce the mass of a particle slowly towards zero, how is this sudden and abrupt change of behaviour, that there is no rest frame anymore for zero mass, explained?
(Which equation shows that best?)

Or does it require more and more energy to keep a ever lighter particle at rest? But then all frames a equal by Lorentz transformation, and there is always a rest frame for a particle which does not move with c...

confused!

thanks

 

[tiny-tim]

Hi Lapidus!

… But when I would reduce the mass of a particle slowly towards zero, how …

But we can't reduce (or increase) the mass of a particle …

the mass (ie rest-mass) of each particle is fixed. ​

 

[Dale]

So every particle with some mass, even if the mass is very, very close to zero has a rest frame. A neutrino, say, could sit right next to me. But a photon, a massless particle, of course, can't, it has to zip by with light velocity.

But when I would reduce the mass of a particle slowly towards zero, how is this sudden and abrupt change of behaviour, that there is no rest frame anymore for zero mass, explained?
(Which equation shows that best?)

in addition to tiny tim's comments, don't forget that light has energy. For any fixed E the v goes smoothly to c as m goes to 0.

 

[Lapidus]

in addition to tiny tim's comments, don't forget that light has energy. For any fixed E the

Right, v goes smoothly to c as m goes to 0, so that the denominator and numerator of the expression for the relativistic energy become both zero.

But isn't it that for very small masses quantum physics has to come into play?

Does not quantum physics, with its Broglie wavelength or even Comton wavelength for particles, blur the concept of localization and thus the concept of a rest frame?

Does quantum physics (relativistic quantum physics) give a more smooth transition to the massless case with no rest frame?

 

[Dale]

Right, v goes smoothly to c as m goes to 0, so that the denominator and numerator of the expression for the relativistic energy become both zero.

No, the relativistic energy is well-defined even for a massless particle. You do not get division by 0.

Does quantum physics (relativistic quantum physics) give a more smooth transition to the massless case with no rest frame?

For quantum physics I think the only answer is tiny tim's answer that mass of quantum particles does not vary so your situation is simply not possible quantum mechanically. For non-quantum relativistic physics the transition is smooth already, as I mentioned above.

 

[Bill_K]

To answer your original question, you can imagine a series of particles with smaller and smaller mass. As the mass becomes small, it becomes more and more difficult to keep the particle at rest. The slightest bit of energy will cause it to go scooting off at near light speed.