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Photons travel at the speed of light. Is there a known explanation of this phenomenon, and if yes, what is it?

Edit: To be clearer, my question is why do photons travel at all. Why do they have a speed?

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Are you asking why photons travel at all? Or why they travel at the speed of light? –  DJBunk Aug 20 '12 at 19:28
    
I am sorry I was not clear. I meant why do they travel at all. –  BoD Aug 20 '12 at 22:13
    
Related: physics.stackexchange.com/q/16018/2451 –  Qmechanic Aug 20 '12 at 22:39
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5 Answers

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The electromagnetic wave travels at speed of light which is explained by Maxwell's equations. However, in many cases, electromagnetic wave can also exhibit the particle-like behaviour (wave-particle duality), e.g. Einstein's photoelectric effect, so physicists have to model it as particle, known as photons. So what we know as electromagnetic wave is also particle with common property, i.e. speed. Both theories are correct and we don't know exactly what "it" really is.

In summary, photon is just a model for explaining when electromagnetic wave behaves like particle. Asking why photon travels with speed of light is the same as asking why electromagnetic wave travels with speed of light.

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Kind of as an expansion on what drake said, this can be explained in several ways. For example:

In electromagnetism, we know that Maxwell's equations govern electromagnetic radiation. From Maxwell's equations you can derive the EM wave equation

$$\frac{\partial^2\vec{E}}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\vec{E}}{\partial t^2}$$

(and the same for $\vec{E}\to\vec{B}$) which has solutions corresponding to waves that travel at light speed. As the quanta of these waves, photons will also travel at light speed.

In special relativity, the energy of a particle is related to its mass via $E = \gamma mc^2$. Photons are massless, but they have finite energy. The only way both of these facts can be true without rendering $E = \gamma mc^2$ outright incorrect is if $\gamma$ is undefined, and since $\gamma = 1/\sqrt{1 - v^2/c^2}$, the only way to make $\gamma$ undefined is to have $v = c$.

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There is a kind of circular argument here. If you modified your wave equation changing $c$ by $c'$ or adding a mass term, then photons would still propagate at the speed of light. Another issue is why photons or light propagates at the invariant speed of Lorentz transformations. And one can answer the latter by means of an experimental fact or because of the Lorentz invariance of Maxwell equations (but in this case the question is why Maxwell equations must be Lorentz invariant or why Special Relativity holds in nature). –  drake Aug 20 '12 at 19:43
    
In the case one added a mass term, the equation would be Lorentz invariant. So the question would be why one has to use light or photons to synchronize clocks. –  drake Aug 20 '12 at 19:54
    
Well, photons travel at the speed of light by definition, like you said ;-) so there isn't really any non-circular argument to make, other than that this is how we measured it. I just thought I'd allude to some of the other parts of the circle. –  David Z Aug 20 '12 at 19:59
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Because photons are light quanta. So the claim 'photons travel at the speed of light' is the same as (or the quantum counterpart of) 'light travel at the speed of light'. It is almost by definition, that is, by definition and by the fact that photons are light quanta.

I am sorry I was not clear. I meant why do they travel at all. – BoD 24 mins ago

David has answered this question.

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Light travels as a wave and interacts as a particle so to be pedantic there is no way to answer the questuon.

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Light waves are an ensemble of an enormous number of photons/particles. Individual photons have also the particle wave duality, and the wave quantities ot the individual photons cohere when in large numbers to give the classical wave description. I have given a link in my answer of how this is done. Of course there is an answer. –  anna v Nov 21 '13 at 6:57
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As other answers have mentioned light in classical electricity and magnetism is described very well by the solutions of Maxwell's equations which combine electrostatics and magnetism and describe a traveling wave of energy propagating at a speed c.

This speed is not arbitrary but, as it comes out from the equations, depends on the electric and magnetic constants of materials or the vacuum:

speed of light

the magnetic permeability mu_0, and the electric permittivity epsilon_0 , which are well defined numbers in the set of units used and we get the speed of light in vacuum as 299,792,458 meters/second

Photons on the other hand are elementary particles and are necessary at the basic framework where quantum mechanics is the description of nature.

Photons appear in elementary and nuclear interactions, taking away energy and momentum and are seen in subsequent interactions giving up their energy and momentum.

gamma interactions

In this photo in a bubble chamber gammas are entering from the top and here one has interacted with an atomic electron scattering it away and generated an e+e- pair and a secondary photon, the whole first vertex conserves energy and momentum. We know the existence of the secondary photon because a second e+e- pair was by an interaction with an electron or a nucleus where the transfer of energy to the target was too small to appear in the photo.

So this is experimental evidence that photons travel.

As a particle at that level it is also a wave, and there exist consistent derivations of the formation of the classical wave from a large ensemble of photons.

For mathematical consistency, photons also travel in vacuum at speed c.

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How is it proof that photons travel? why can't a light wave travel that distance? –  Jitter Nov 21 '13 at 6:21
    
@Jitter In this photo time grows from top to bottom. The velocity of the charged particles can be found be the curvature they have in the magnetic field . The photon between the two vertices displays its particle properties, it has ans (x,y,z) defined track and interacts at a point.Light on the other hand is composed by zillions of photons of much lower energy than the one of the single photon between the two vertices. If the vertex were a light source radiating the energy taken by the single photon, it would radiate radially and no secondary vertex would be found because the individual –  anna v Nov 21 '13 at 6:53
    
because the individual photons comprising the light wave would be zillions . –  anna v Nov 21 '13 at 6:54
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