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So, we all know the world famous equations $$E^2=\left(mc^2\right)^2+(pc)^2\tag{1}$$ $$E=hf={hc\over{\lambda}}{\rm\ (for\ light)}\tag{2}$$ $$p={h\over{\lambda}}\tag{3}$$

We can realise in eq.2 and 3. The $hc\over{\lambda}$ can be replaced by $pc$ And the $E=pc$ can be used in eq.1 to obtain $(mc^2)^2=0$ or the rest energy is $0$.

This can conclude that light if at rest would reach $0~\rm K$. This is a major contradiction in our non-utopian world. And thus light cannot be at rest. Is this reasoning correct?

Qmechanic
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user182947
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  • Einstein reputedly wondered in his youth what a light-wave would look like if you could run alongside it. Considered in light of Maxwell's equations this is a very interesting question with implication for what you ask. – dmckee --- ex-moderator kitten Jan 28 '18 at 01:53
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    Obligatory reminder that nothing about the physical world can be proven mathematically. A particular phenomenon can be demonstrated to be consistent (or not) with a particular model, but that's about it. – J. Murray Jan 28 '18 at 04:44

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Light cannot be at rest. Can this be proven mathematically?

Given Maxwell's equations based on four observational laws, yes.That the velocity of light should be c in vacuum is proven.

Then you go on:

$$E=hf={hc\over{\lambda}}{\rm\ (for\ light)}\tag{2}$$

That is wrong, this is not for light, it is for photons. Photons build up the electromagnetic field in a superposition of their wavefunctions.. The classical electromagnetic field, light, emerges from zillions of photons.

Photons are quantum mechanical particles, not bits of light. A building is made up of bricks. Bricks are not a building.

So the question you are proving morphs :

Photons cannot be at rest. Can this be proven mathematically?

You derive :

or the rest energy is 0.

Look again at your functions.There is the alternative or $m=0$

So you just reached an algebraic relationship.

This can conclude that light if at rest would reach $0K$ .

If the mass is zero, a photon cannot be at rest from the mathematics of Lorenz transformations. If it has a mass, it will have the energy of the mass which together with the uncertainty principle will not allow $0K$ .

So this is no proof.

That the photons have zero mass is an assumption that makes consistent macroscopic data and microscopic data.

Again I stress , do not confuse light with photons. The velocity of light comes out of the Maxwell equations in a clean proof.

I should also stress that physics tests the frontiers, and so there are people checking on the mass of the photon, example Photon and Graviton Mass Limits. So any proof needs feedback with experiment and observations.

anna v
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  • Correct me if I'm wrong, but you say that it's not possible to reach 0K. This is the basis of why i state that photons cannot be at rest. So you agree to the point i make? And could it be then right for photons(accepted not light)? – user182947 Jan 28 '18 at 09:56
  • The impossibility of a particle of mass m to be at rest absolutely depends on the heisenberg uncertainty principle, because due to special relativity m has energy ,https://physics.stackexchange.com/questions/56170/absolute-zero-and-heisenberg-uncertainty-principle . The argument is no good for photons, which have mass zero and it is the Lorenz transformations that impose c on mass zero particles. read the link, generally 0K cannot be reached by massive particles ( and by construction by 0 mass particles) – anna v Jan 28 '18 at 11:35
  • Ok; but the algebraic manipulations aren't wrong right? And if at step 0 i were to assume that photons do have mass; arriving at the final step would lead to the conclusion photons are massless. Does that somehow make sense? – user182947 Jan 28 '18 at 11:45
  • The algebra is ok for either mass 0 or c=0. If at some future time it is found that photons have a very very small mass, (as we found out about neutrinos) they can be at rest "classically"but as they are quantum mechanical entities the HUP enters, so they cannot form an ensemble with 0 kelvin. no massive particles can anyway. – anna v Jan 28 '18 at 14:06
  • please remember that temperature is thermodynamic, emerging from statistical mechanics . http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/temper.html – anna v Jan 28 '18 at 14:43