Consider a particle (point charge) with charge $q$ and mass $m$ that crosses into a uniformly charged sphere (with charge $Q$ and radius $R$). The trajectory of the particle is a diameter of the sphere, and when it crosses the surface of the sphere it has a velocity $\mathbf{v_0}$. Assuming the particle only interacts electromagnetically with the sphere, and considering Larmor's radiation formula, what is the motion of the particle?

First I have to find the electric field $\mathbf{E}$ and the potential $\varphi$ inside the sphere. This can be done easily with Gauss' law, and I obtain:

$$\mathbf{E}=\frac{Q}{4\pi \epsilon_0 R^3}r\mathbf{\hat{r}} $$

Let $k=Q/4\pi\epsilon_0 R^3$, so $\mathbf{E}=kr\mathbf{\hat{r}}$. To find the potential, I consider spherical symmetry so $\mathbf{E}=-\frac{\mathrm{d}\varphi}{\mathrm{d}r}\mathbf{\hat{r}} $, therefore:

$$\varphi = -\frac{kr^2}{2}$$

The energy of the particle is:

$$\mathcal{E}=\frac{1}{2}m\dot{r}^2-\frac{qkr^2}{2} $$

And Larmor's formula is:

$$P=\frac{q^2\ddot{r}^2}{6\pi\epsilon_0 c^3}=-\frac{\mathrm{d}\mathcal{E}}{\mathrm{d}t}$$

So we have:

$$\frac{q^2\ddot{r}^2}{6\pi\epsilon_0 c^3}=-m\dot{r}\ddot{r}-qkr\dot{r}$$

Now, according to Newton's third law, and considering that the particle only moves radially:


Substituting $\ddot{r}$, I get (setting $L=1/6\pi\epsilon_0 c^3$):


(Oh, the $L$ has nothing to do with angular momentum, I just called it $L$ for Larmor (as in the big constant in Larmor's formula). Anyway! I get:


This equation has the solution:


This makes sense, I guess? Until I consider the sign of the constant inside the exponential. If $q$ and $Q$ have opposite signs, I would assume that the particle falls, but it doesn't. Also I find it peculiar that this solution doesn't depend on the initial velocity $\mathbf{v_0}$. So the trajectory always decays and never actually crosses the center, regardless of the initial speed? I know there's something I'm missing. What's wrong here?

EDIT: I think the problem with the particle shooting away instead of falling in has to do with me setting the Larmor radiation $P=-\frac{\mathrm{d}\mathcal{E}}{\mathrm{d}t}$. However, from the formula, $P\geq 0$, and radiation makes a particle lose energy, hence why I set it as the negative of the derivative of energy. There must be something else.


Okay, so Floris noted that that Newton step:


Was not accurate, because it assumes that the acceleration of the particle is only due to the electric field, while clearly the loss of energy through radiation must also change the kinetic energy of the particle, thereby causing some kind of acceleration.

So, disregard Newton and acquire another differential equation from $P=-\frac{\mathrm{d}\mathcal{E}}{\mathrm{d}t}$:

$$q^2 L \ddot{r}^2 + m\dot{r}\ddot{r} -qkr\dot{r}=0$$

Or rather,

$$\ddot{r}^2 + \frac{m}{q^2L}\dot{r}\ddot{r} -\frac{k}{qL}r\dot{r}=0$$

This is a quadratic on $\ddot{r}$, so I uh...

$$\ddot{r}=-\frac{m}{2q^2L}\dot{r} \pm \sqrt{\frac{m^2}{4q^4L^2}\dot{r}^2+\frac{k}{qL}r\dot{r}} $$

This is assuming, of course, that the thing inside the square root is non-negative. So, now this became a differential equations problem. That equation is non-linear, and it doesn't allow the usual linearization method since the critical points are $\dot{r}=0$, and at that point the derivative $\frac{\mathrm{d}\ddot{r}}{\mathrm{d}\dot{r}}$ doesn't exist. How does one solve that equation? Or perhaps not solve but just gain insight on the system?

  • $\begingroup$ How do you figure the "according to Newton's third law" part - the radial acceleration is not simply a function of position because of the radiative losses no? $\endgroup$ – Floris Mar 12 '15 at 23:44
  • $\begingroup$ @Floris True, that is probably the problem. Newton's third law should read something like ma=qE+F, where F is an effective force due to radiation. Maybe don't even consider Newton's law at all and just try to solve for P=-dE/dt. $\endgroup$ – squinterodlr Mar 13 '15 at 2:50
  • $\begingroup$ Yes - just drop that Newton step in the middle; otherwise you seem to be on the right track. $\endgroup$ – Floris Mar 13 '15 at 3:02
  • $\begingroup$ Just a minor note, I think you missed the vacuum permittivity in the electric field. $\endgroup$ – Ivan Lerner Mar 13 '15 at 5:45
  • $\begingroup$ @IvanLerner yep. I'll fix it. $\endgroup$ – squinterodlr Mar 13 '15 at 9:33

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