Is there Bremsstrahlung radiation for a charged massless particle? This is a follow up question from: Massless charged particles
Since by definition such a particle would interact with photons- resulting in some change in momentum- would the particle emit Bremsstrahlung during this interaction? If it does, it seems that there is a problem as the particle cannot decelerate, yet the Bremsstrahlung would necessarily carry away some of the energy. Or perhaps the fact the particle is massless precludes such emissions. Is this a real problem? What would the physical effects of Bremsstrahlung radiation be for this particle?
Note: for some reference- this question came out of a discussion in the comments after my answer to   Massless charged particles
 A: From an experimental point of view, we know one mass less particle, the photon.
We cannot describe the photon relativistically by $E=mc^2$. Its energy is $E=h× \nu$ , When it interacts and loses energy, it is the frequency that changes.
Thus I would expect, if a massless charged particle could exist on shell, a corresponding energy definition would give it a change in its wave "frequency".
It is simpler to say that it would loose energy and go from $E_1$ to $E_2$ where $E_1-E_2$ would be the brehmstrahlung energy. That is the way interactions of neutrinos were treated when they were thought to be massless.
A: It is not so hypothetical situation if you consider very relativistic electrons, for example. The velocity is always $c$ but the energy-momentum changes. Quantum mechanically it means a change in the De Broglie wave frequency. So a less energetic relativistic electron is like a less energetic photon - it has lower frequency. Classically it correspond to changes of energy-momenta that are determined not only with particle velocity.
A: The Dirac current obeys a continuity equation
$$
\partial_\mu({\bar\psi}\gamma^\mu\psi)~=~0,
$$
which for electromgnetism we generalize $\partial_\mu~\rightarrow~\partial_\mu~+~ieA_\mu$ in a covariant manner.  The $J^0~=~{\bar\psi}\gamma^0\psi$ $\psi^\dagger\psi~=~\rho$, which is the density. The spatial part is
$$
{\vec j}~=~\psi^\dagger{\vec\alpha}\psi
$$
The application of the momentum operator ${\hat p}~=~-i\hbar\nabla~-~ie{\vec A}$ gives a divergence of the 3-current 
$$
{\hat p}\cdot{\vec j}~=~i\hbar\big((\nabla\psi^\dagger)\cdot{\vec\alpha}\psi~+~\psi^\dagger{\vec\alpha}\cdot\nabla\psi\big)~-~ie{\vec A}\cdot\psi^\dagger{\vec\alpha}\psi,
$$
where the spatial component $\alpha^i~=~\gamma^0\gamma^i$.  This continuity equation tells us this equals $\partial\rho/\partial t$ and the quantum current divergence through a volume is equal to the rate the quantum density changes in that volume.  
To consider the Bremsstrahlung question we need to address the force ${\vec F}~=~d{\vec p}/dt$ with respect to the current flow.  The time derivative is evaluated with the Dirac Hamiltonian
$$
H~=~{\vec\alpha}(\cdot{\hat p}~-~ie{\vec A}),
$$
where the absence of mass should be noted.  The time derivative is evaluated according to the commutator $[{\hat p},~H]$ and the calculation result in
$$
\frac{d}{dt}{\hat p}\cdot{\vec j}~=~\langle F\rangle~=~e\langle{\vec E}~+~{\hat\alpha}\times\nabla\times{\vec A}\rangle.
$$
The vector ${\vec\alpha}$ plays the role of a velocity of the particle, where the matrices are units in space with unit magnitude or $v~=~c$ in the Lorentz equation $({\vec v}/c)\times{\vec B}$.
So we have now that a massless particle which carries a charge will obey electrodynamics.  Then what about the Bremsstrahlung?  We can now use this to compute the energy of work of this force $W~=~\int\langle F\rangle\cdot d{\vec r}$  The power is the time derivative of this 
$$
P~=~dW/dt~=~\int\langle F\rangle\cdot d{\vec v}
$$ 
We then consider the “c velocity” given by $\vec\alpha$ as a rotation given by the $SO(3)$ matrix elements $M$ that ${\vec\alpha}’~=~M{\vec\alpha}M^T$, where the time derivative on the velocity is due to a rotational $dM/dt$ and $dM^T/dt$.  So the velocity is a particle at the speed of light moving in a circular path --- ignoring the electric field.  This will then result in an acceleration $\dot a$ from the time derivatives of these matrices.  At this point I am largely sketching this out but it is clear that a form of the Abraham-Lorentz force and Bremsstrahlung can be derived.
Rats, I see this has slit me from my previous identity here --- am back to 1.
