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


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 E1 to E2 where E1-E2 would be the brehmstrahlung energy. That is the way interactions of neutrinos were treated when they were thought to be massless.

  • $\begingroup$ I'd thought about this, but rejected it for the following reason: if I constrain the particle to move in a circular path by a magnetic field (in a synchrotron for example) I could cause it to radiate away all of it's energy- leaving me with a) nothing (which seems to violate conservation of charge/angular momentum-if charge is half-integer spin) or b) a particle with zero energy but spin (if once again half-integer) and charge (which seems to not make sense, as it implies zero momentum and thus deceleration) I'd voice similar questions to @Vladimir. Is there a flaw in my reasoning? $\endgroup$ – user1567 Apr 3 '11 at 16:18
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    $\begingroup$ @jaskey13: at some point, you'd probably get some quantum mechanical ground state. Just like how electrons in a magnetic field (or the hydrogen atom) eventually have ground state energies, and cannot radiate a photon to decay to a lower energy. I don't know this for sure, and don't have time to run the calculation, but I would assume that a system like this would have a ground state energy that would prevent further decay--it's a bound quantum state, after all. $\endgroup$ – Jerry Schirmer Apr 3 '11 at 18:03
  • $\begingroup$ @Jerry Good point!! And that pretty much exhausts my questions. Though I still have a "gut" feeling that somehow these particles are precluded by the laws of physics- I've worked my reasoning to it's end with all the help from you guys. Thanks- it's truly been a learning experience! $\endgroup$ – user1567 Apr 4 '11 at 1:11

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.


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.

  • $\begingroup$ Normally we use Maxwell (wave) rather than "mechanical" equations to calculate the radiated power. In your case one cannot do without renormalizations. On the other hand, the energy-frequency relation follows from a free Dirac equation. $\endgroup$ – Vladimir Kalitvianski Apr 3 '11 at 18:27
  • $\begingroup$ You are two users: physics.stackexchange.com/users/1352/lawrence-b-crowell which has a lot of reputation and physics.stackexchange.com/users/2919/lawrence-b-crowell. I suspect it is the "unregistered" that does the mischief. $\endgroup$ – anna v Apr 3 '11 at 18:30
  • $\begingroup$ The point of the calculation is to demonstrate the existence of Bremsstrahlung. Of course in a most general setting we have the Dirac equation and the Maxwell equations coupled to each other. $\endgroup$ – Lawrence B. Crowell Apr 3 '11 at 20:27
  • $\begingroup$ @Anna: I resist giving out information on the web. I stopped doing financial transactions on the web some years ago and I avoid signing up with things like facebook. I scaled a lot of that stuff back after I decided I did not like what I saw coming. I might have to register here, since this is the second time this has happened. I avoid a lot of these things because I suspect we may be only a few technological steps away from becoming “THE BORG.” I want to delay my “assimilation” as long as possible. $\endgroup$ – Lawrence B. Crowell Apr 3 '11 at 20:27
  • $\begingroup$ Thank you for this illustration that the radiation can be mathematically formed within physical law. $\endgroup$ – user1567 Apr 4 '11 at 1:14

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