Energy conserved... or not? Confused!

Question

I am confused. Could someone kindly explain what's going on in this question?

A particle of mass $m$ and charge $e$ moves in the $x,y-$ plane. There is a constant magnetic field $B$ that points in the $z$ direction. Initially the particle has energy $\gamma m$ in the rest frame of the observer. After a time elapse of $\Delta t$, the observed energy is $\gamma' m$ (again in the rest frame of the observer). How are $\gamma,\gamma',t$ related?

Some thoughts:

Since we have an accelerated charge, energy is radiated. Since the question does not specify that the motion is non-relativistic, I shall use the relativistic equations.

Larmor's formula:

$$P=\frac{\mu_0 q^2}{6\pi m^2}\frac{dp^a}{d\tau}\frac{dp_a}{d\tau}$$

To get $\frac{dp^a}{d\tau}$ I need the equation of motion: $$\frac{dp^a}{d\tau}=eF^a{}_b\frac{dx^b}{d\tau}=\frac{e}{m}F^a{}_bp^b$$ where $$F^a{}_b=\begin{pmatrix}0&0&0&0\\ 0 &0& B&0\\0 &-B& 0&0\\0&0&0&0\end{pmatrix}$$ This gives 4 coupled ODEs. Actually the first 2 are trivial such that $$p_0=p_{0, init}\\p_4=p_{4,init} $$ and $$\frac{dp_1}{d\tau}=\frac{qB}{m}p_2\\ \frac{dp_2}{d\tau}=-\frac{qB}{m}p_1$$

But then doesn't $$p_0=p_{0, init}=\gamma m$$ say that the energy is conserved? It is also not physically surprising if it were since the magnetic field induces a force perpendicular to the velocity of the particle, therefore does no work on it.

What's gone wrong?

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[Update 1] With Mike's and Michael's help, I have proceeded to figured out how to do it for the non-relativistic case (but I'm not very sure how to extend it to the relativistic case :-/ ) -- if I'm not mistaken, the NR case should be something of the following: $$F=e \vec v\times \vec B$$ Assuming circular motion and the magnetic field being perpendicular to the plane of motion, $$r=\frac{mv_{init}}{eB}$$ Therefore acceleration is $$a=\frac{v^2}{r}$$ Then I can substitute it into Larmor's formula and so on. How do I generalize it to the relativistic case?

[Update 2] Do I simply add a $\gamma$ factor in front of the $v$'s?

Answer 1

The question you have formulated is not an easy one to answer (correctly). But the question you've formulated isn't quite the question that I see. The good news is that the text of the question you've posted implies a much simpler question; it's just asking for the energy change.

You can probably assume that the acceleration is dominated by the circular motion induced by the magnetic field. You know how to calculate that circular motion given the field strength $B$ and the velocity (derived from $\gamma$). And if you're happy with the Larmor formula (rather than the more general Liénard–Wiechert formula), you can easily get the radiated power. Multiply that by $\Delta t$, and you get the change in energy. (Integration is probably overkill, though you'll be a better judge of that than I.)

You do seem to understand the physics behind what's happening, and you have a good idea what formulas are needed and how to use them. But if you need more clarifications, just ask.

Answer 2

Are you sure you really need to take radiation in this question? =)

In case you're sure, have a look at synchrotron radiation (relativistic case) and cyclotron radiation (non-relativistic case).

In short, radiation intensity in relativistic case (assuming radiation is not much intense and the speed changes slowly) is (in CGS units, $c = 1$): $$ I = \frac{2q^4 H^2 v^2}{3m^2(1-v^2)} = \frac{2q^4 H^2}{3m^4} (E^2-m^2) $$ where $E$ is total energy, $E=\gamma m$. So energy loss is $$ \frac{dE}{dt} = -I = -\frac{2q^4 H^2 }{3m^4} (E^2-m^2) $$

Solving this you get $$ \gamma = E/m = \coth (\frac{2q^4 H^2}{3m^3} t + const) $$ where $const$ comes from initial conditions.

I'm using Landau & Lifschitz Course of theoretical physics, vol.2, §74.

Hope this helps!

Answer 3

When in relativistic frames energy is no longer the conserved quantity, only the interval (Usually denoted 'S') is. Energy seems to be a time phenomenon, like gravity, so when dealing with "conservations" relativistic effects must be taken into account.