I am not a physicist so please excuse me if this is a dumb question. As far as I understand Relativity, as observer (in this case a photon) travels at the speed of light, time stops. So how can the photon have wave properties as it propagates through space?
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$\begingroup$ I'm not sure you can make the observer a photon. For a photon is nothing other than light, and light is precisely what does travel at the speed of light, i.e. without making time stop. $\endgroup$– David ReishiCommented May 12, 2016 at 5:32
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3$\begingroup$ Possible duplicates: physics.stackexchange.com/q/241769/2451 , physics.stackexchange.com/q/29082/2451 , physics.stackexchange.com/q/16018/2451 and links therein. $\endgroup$– Qmechanic ♦Commented May 12, 2016 at 7:24
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2 Answers
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You misunderstand special relativity. For objects that are moving at large speeds, the time runs more slowly for the object compared to the observer who measured the speed. To observe the motion of the object, you don't have to go to its coordinate frame and observe from there. You observe from the outside, that's how you measured the speed in the first place (to use your clock and your ruler to get the time and the distance).
The same is true for light. You, as the observer, measure the distances, the times, the electric and magnetic fields, write down the Maxwell equations in your coordinates and measure the time with your clock. In this case, you actually cannot go into the observation frame of the photon, because there is no Lorentz transform that gets you there. So there is no issue... true, if you take the mathematical limit of the time dilation, you will get the result that for the photon, the time doesn't move forward. But you're not the photon. In your coordinate frame, you don't even need coordinate transforms from special relativity to describe light. You have the Maxwell equations, you know how EM wave behaves, you have the wave equation that tells you how fast the wave is going, and that's it.
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Let me explain light.
Classically, light is electromagnetic radiation.
There exists a field permeating all of spacetime called the electromagnetic field. Charges create curvature in this field. When charges accelerate, waves are created in this field. These waves are what we perceive as light.
A little more specifically, let us examine Maxwell's equations in free space.
Faraday's law, \begin{equation} \nabla \times {\bf E} = -\frac{\partial {\bf B}}{\partial t} \end{equation} tells us that a time varying magnetic field induces a circulating electric field.
Whereas the Ampere-Maxwell law, \begin{equation} \nabla \times {\bf B} = \frac{1}{c^2}\frac{\partial {\bf E}}{\partial t} \end{equation} tells us that a time varying electric field induces a circulating magnetic field.
The symmetry between electric and magnetic fields in these two equations implies that a wave can propagate freely.
Combining these two equations, we find,
\begin{equation} \frac{1}{c^2}\frac{\partial^2 {\bf E}}{\partial t^2} - \nabla^2 {\bf E} = 0 \end{equation}
\begin{equation} \frac{1}{c^2}\frac{\partial^2 {\bf B}}{\partial t^2} - \nabla^2 {\bf B} = 0 \end{equation}
which give us the equations of an electromagnetic wave propagating in free space at speed c.
Quantum mechanically, the photon is the quantum of the electromagnetic field. Electromagnetic interactions between charged particles are effectively described by the exchange of 'virtual' photons. The photon has no mass, so there is no problem with it traveling at c. In the appropriate classical limit, this reduces to the description given above.
