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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

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3 Answers 3

up vote 6 down vote accepted

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.

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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? –  user1567 Apr 3 '11 at 16:18
@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. –  Jerry Schirmer Apr 3 '11 at 18:03
@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! –  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.

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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.

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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. –  Vladimir Kalitvianski Apr 3 '11 at 18:27
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. –  anna v Apr 3 '11 at 18:30
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. –  Lawrence B. Crowell Apr 3 '11 at 20:27
@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. –  Lawrence B. Crowell Apr 3 '11 at 20:27
Thank you for this illustration that the radiation can be mathematically formed within physical law. –  user1567 Apr 4 '11 at 1:14

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