# Energy carried by photon not conserved?

In an imaginary frame of reference traveling with a photon, the length of the path traveled is 0. If the length of the path is 0, isn't it similar to say that the photon is either at the source or at the destination and nowhere in between?

• There is no co-moving reference frame for photons (see my answer to this question for more details), so we can't really make the assertion you make in the first sentence. – Kyle Kanos Jun 26 '14 at 20:43

No, it doesn't make sense. Who says the length of the path is 0? If you are talking about the spacetime interval $ds^2$ being zero for lightlike paths, this has no interpretation of this kind. Your whole...story rests on wrong assumptions:

There are no frames of reference travelling with a photon

This is because the speed of light appears the same in all reference frames, and a comoving frame of reference would by definition be one where the speed of the photon is 0 at any spacetime point, which contradicts the constancy of the speed of light.

Photons do not plan anything

Seriously, no elementary particle is smart enough to plan anything.

Photons do not warp

Photons do what photons do, they travel at the speed of light. They do not warp. They do not slow down (in empty space, anyways).

Energy conservation does not hold at relativistic scales

At relativistic scales, only the scalar product of the four-momentum with itself, which is, in usual non-relativistic notation, $E^2 + \vec{p}^2$, is conserved, not $E$ and $\vec{p}$ seperately.

The spacetime interval of photon worldlines is zero. In this regard your descriptions contain plenty of contradictions. As ACuriousMind is pointing out, the idea of a proper time of photons is generally rejected as meaningless.