# Can a photon move at another speed that's not the speed of light?

I was reading an article about the new collider photon-photon, and the writer says "the scientists accelerate photons in a very high speed".

It's non sense to me, because as far I know a photon only moves on the speed of light.

So my question is: Can a photon move at another speed that's not the speed of light?

EDIT:

This is the article about photon-photon collider:

http://www.theverge.com/2014/5/18/5724658/photon-collider-could-turn-light-into-matter

I found the snippet about accelerating photons at a very high speed in this Portuguese article:

http://meiobit.com/287429/colison-foton-foton-proposta-transformar-luz-em-materia/

• Photons only move with the speed of light. Can you give the link to the article please? May 19, 2014 at 19:30
• probably protons not photons May 19, 2014 at 19:34
• Photons, not protons May 19, 2014 at 20:05
• I put the link of portuguese article in the question May 19, 2014 at 20:12
• I don't speak Portuguese well, but the sentence "a colisão de dois fótons a uma velocidade altíssima resultaria na criação de um elétron e um pósitron" seems to phrase the collision as being "high-speed"; it probably was supposed to be "high-energy", but the writer just mixed the words. May 19, 2014 at 20:30

Yes, normally $c$ is for speed of light in vacuum at which light travels at its maximum speed. However if I were to put, say, atmosphere or diamond in from of a light beam, the light would travel slower than $c$, therefore I conclude yes its possible to travel slower than the TRUE speed of light. Next, to elaborate, in 1905, Einstein's miracle year, he proposed his mass-energy equivalence theory which is $E=mc^2$. This must mean, with some simple rearranging, we understand that mass can be converted in energy and vice-versa. That in mind, a photon is an packet of energy in basic definition. That in mind, we can convert the photons into mass using the formula rearranged: $m=E/c^2$ which means we can even make light travel even at 0 indirectly.