Can a particle radiate in vacuum without external forces applied? I was reading about possible and forbidden processes involving photons. First of all, the annihilation process of a pair electron and positron cannot result into a single photon because it would be kinematically forbidden.
On the other hand, I was wondering if a single electron could radiate a photon in vacuum. The process would then be
$$e^- \to e^-+\gamma$$
But I'm making the calculations and finding that this process would also be kinematically forbidden, am I right?
Is there a process where a free particle in vacuum can radiate a photon without any external forces applied? 
 A: Go to the center of mass of the electron.
$$e^- \to e^-+\gamma$$
The only available energy to generate a photon from a mass at rest is the energy of the mass itself. As the electron is not composite, the value of its mass cannot change, and therefor, from energy momentum conservation this is a kinematically forbidden reaction
De-excitations can and do happen, in composite bodies, and two body decays are common. 
A: No, except for the case of the particle being unstable. But I suspect you're thinking of stable particles such as the electron in asking your question.
To radiate, the electron has to receive a photon, real or virtual, and very soon after radiate another photon.  The prime examples of this are synchrotron radiation, where any charged particle with a lifespan of at least a microsecond or so is flung around in a circle, and as a result loses some energy due to radiating photons.  But there's clearly a "external force" - a magnetic field. Another example is bremsstrahlung, where a charged particle slams into matter of any kind having electromagnetic structure.  Again, it's an external force, but explainable within QED by multiple photons real or virtual interacting with the subject of interest.
If a charged massive particle has some energy and momentum (E,p) then what different (E',p') could it have?   The change in momentum will always exceed the change in energy. Using units where c=1,
$$E^2=p^2+m^2$$
Thus, $ E>p$
Differentiate to obtain
$$2E dE= 2p dp$$
and so
$$\frac{dE}{dp} = \frac{p}{E} < 1$$
A photon can only change a particle's momentum and energy by the same amount, $dE=dp$.   There's no such thing as a massless charged particle.  Massive charged particle cannot interact with individual photons, except in a Heisenberg sort of way that really requires multiple photon interactions.
