# Energy conserved… or not? Confused!

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?

[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?

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The equation of motion you have written does not include the radiation reaction force. Basically, when the charge starts radiating $F^a_{\ b}$ no longer has the simple form you've given. You really need to solve Maxwell + EOM together to do it properly. Or you could take a shortcut and use the Abraham-Lorentz force, which is not completely kosher, but includes the radiation reaction. –  Michael Brown May 16 '13 at 1:14
Ah right, thanks, @MichaelBrown! –  Diana Stevens May 16 '13 at 1:29
Unfortunately, I still can't quite do the question... –  Diana Stevens May 16 '13 at 1:31
What Mike said below: it's actually really hard to do this problem properly, but you don't need to know the details of the particle motion to get a good approximation of the energy radiated in a small time interval (compared to the time it takes for all the energy to be radiated away). –  Michael Brown May 16 '13 at 2:56
Thanks again, @MichaelBrown ! :) I still need a bit of help, I'm afraid :-/ please c.f. my comment under Mike's answer... –  Diana Stevens May 16 '13 at 9:46

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.

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Thanks, Mike! :) I think I have now figured 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 $$a=\frac{v^2}{r}$$ The I can substitute it into Larmor's formula. Is that right? But is it not relativistic? I always get confused when doing relativistic cases :( –  Diana Stevens May 16 '13 at 9:44
Do I simply throw in a $\gamma$ in front of $v$? –  Diana Stevens May 16 '13 at 12:12
The problem is that Larmor's formula is only derived for the non-rel. case. You can actually modify it in this particular case, but unless you've seen the Liénard–Wiechert formula I linked to, you can't derive the correct expression -- which is synchrotron radiation, as xaxa pointed to. But even then, there are fancier things you can do. I guess what I'm trying to say is that you can make this problem arbitrarily hard, but I'll bet that your teacher just wanted you to solve the easy problem that you've shown us you can do -- unless synchrotron radiation has already come up in your class. –  Mike May 16 '13 at 13:49
BTW, welcome to physics.SE! This was a really nicely formulated question. And latex skills are always appreciated. :) –  Mike May 16 '13 at 13:52
Also, thanks for your patience! –  Diana Stevens May 16 '13 at 14:12
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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!

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Thanks, xaxa :) I should probably have a look at the mentioned textbook! –  Diana Stevens May 16 '13 at 14:11