# Why don't photons split up into multiple lower energy versions of themselves?

A photon is an elementary particle. As much elementary and as much particle as the electron .

A single elementary particle has a fixed mass and cannot emit another particle without violating energy conservation, because its mass is fixed. In the center of mass of a massive elementary particle, electron, there is no energy for an emission , for a radiating electron in a field the energy is supplied by the field.

If a zero mass elementary particle like the photon could split into two, suddenly an invariant mass will appear and the before the split has zero invariant mass, after the split a measurable invariant mass. This means both momentum and energy conservation **are** violated, as the invariant mass is the measure of the four vector, before and after the split. A photon can also interact with a field in higher order diagrams , but cannot split in the sense you envisage.

Edit after discussion in comments:

Assume a photon could decay into two photons.

These photons will have four vectors. There are two situations: their three momenta are parallel in the laboratory to the original photon, or there exists an angle of the three momenta with the original photon and also between them. In the latter case the two decay photons define a center of mass ( similar to a pi0 at rest). In this system the two momenta add up to zero, but there will be energy giving an invariant mass to the system, which violates energy conservation as the original photon had 0 invariant mass, i.e. cannot supply this energy. The original photon in the center of mass of the decayed photons will still be moving with velocity c, and so have a momentum different than zero, thus momentum conservation is also violated.

In the case of two collinear photons in the lab , their invariant mass will be zero at the limit of the angle between them being exactly 0, otherwise the above argument holds. If it is exactly 0 no center of mass can be defined because a zero mass system moves with the velocity of light.

So the question becomes: why a photon of frequency nu does not turn into two exactly collinear in the lab photons of lower frequency. Experimentally this has not been observed so if it can happen it is a very very low probability process. In the comments Lubos Motl gives this statement :"For photons, this amplitude is 0 due to the Abelian gauge symmetry and other symmetries." I am still looking for a link on this.

In the next answer the collinear case is excluded by special relativity,

Mathematically, the reason is that the Lorentz group is non-compact, which means that the parameter gamma can take any value from [1, infinity) but not infinity itself which would correspond to a coordinate frame moving at lightspeed with all massive particles having infinite kinetic energy.

After the hypothetical split, 2 photons with the same energy would be propagating at an angle ok with momentum conservation. Then there would be a rest frame where the angle is 180 degrees. Now if you stay in this restframe and go back in time before the split, your single photon would be at rest. However, that is not possible: According to relativity, speed of light is constant for all frames. Thus, there can be not split of a single photon into two in vacuum (i.e. without momentum transfer during split). Mathematically, the reason is that the Lorentz group is non-compact, which means that the parameter gamma can take any value from [1, infinity) but not infinity itself which would correspond to a coordinate frame moving at lightspeed with all massive particles having infinite kinetic energy.

Photons come with chirality, so you should consider angular momentum conservation as well. For $1\gamma \to 2\gamma$ scattering, this will not be possible. (I'm assuming production of collinear photons only; it's obvious when two are not collinear, energy and momentum conservation will be violated)