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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

3 Answers

up vote 1 down vote accepted

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 Zaslavsky Aug 20 '12 at 19:59

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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