# If time stops at the speed of light is a photon 'everywhere' at once? [duplicate]

I am not a physicist so excuse my question if it's paticularly stupid. As a particle gets closer to the speed of light time slows down as for that particle as compared to a reference from the surrounding environment. Therefore for a photon travelling at the speed of light time has stopped for the particle or in a sense does not even exist for the particle....correct?? So is it true that a particle travelling at the speed of light would be everywhere that it does exist at the same time from an outside reference? For example if it travels in a line from point A to B at the speed of light wouldn't the particle be at every point along that line between A and B at the same time?

I always read that photons exists as a probability wave but is it equally true to say that a photon exists everywhere that it will possibly exist at the same time? Presumably when a photon interacts with another particle (or is observed) it slows down and no longer travels at the speed of light i.e. no longer existing 'everywhere' at once. Would this explain the retrocausality of the delayed double split experiment? I think this would also explain duality. When unobserved the particle is at the speed of light and therefore exists everywhere that it could exist between A and B. When observed it no longer travels at the sppeed of light and therefore only exists at a point.

Where's the mistakes? Thanks for your thoughts.

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## marked as duplicate by Ben Crowell, dmckee♦May 25 '13 at 16:17

Possible duplicates: physics.stackexchange.com/q/27794/2451 , physics.stackexchange.com/q/54162/2451 and links therein. –  Qmechanic May 25 '13 at 13:10
voting to close as an exact duplicate of physics.stackexchange.com/q/27794/2451 –  Ben Crowell May 25 '13 at 15:25
$$\tau^2 = t^2 - x^2$$
The proper time of a light ray is $0$ since $t = x$, meaning that, in its own reference frame, it is not 'moving' through time.