Isn't the standard proof of a classical electron's orbit invalid?

Question

There is an excellent answer to the question of classical electrons falling into nuclei here. But the very first step is to state that $ma=m\frac{v^2}{r}$, Newton's law for a centripetal force/acceleration. I think that isn't valid. There's two arguments (that are fundamentally similar):

The post states as a known fact that $P=\frac{2}{3} \frac{e^2a^2}{4\pi\epsilon_0c^3}$, but a power implies a force in the tangential direction, i.e. even if it is travelling in a circle initially, it won't be for long, thus making the centripetal acceleration expression ($a=\frac{v^2}{r}$) invalid.

The other argument is that the situation is not electrostatic and thus the electric field may be applying a tangential force. Once again, $a=\frac{v^2}{r}$ isn't valid. I should point out that I don't know any advanced E&M, just enough to know this is a thing.

You could alternatively accept the most obvious and glaring fact of them all. If the electron is spiraling in, it is not going in a circle because spirals aren't circles.

I can respect that the proof is still probably within an order of magnitude. But still, I feel like a conspiracy theorist because I haven't been able to find anything else on this even though to me it seems obvious that the proof isn't valid.

To be more clear of what I asking for: is the proof valid; if the proof is invalid, how would one validate it; if it isn't invalid, why can $a=\frac{v^2}{r}$ be used while its spiraling in.

Edit What I'm getting from the comments is that it is indeed an approximation to use $a=\frac{v^2}{r}$; the orbit is nearly circular which is good enough as a heuristic analysis. I guess this is all I wanted, a confirmation that this is technically not exact.

Edit 2 There is a phenomenal answer by G. Smith that explains that although the circular orbit is only an approximation, it is a very good one. It answers the question well, I am marking it as the answer. Thanks!

Additionally, for an interesting discussion of the history of classical theories of the atom see Ponder Stibbons' answer.

Answer 1

No, it is not invalid. The circular approximation is a very good one. I’ll explain why.

Using the usual approximation that the classical spiral orbit of the electron can be considered a circle with slowly-shrinking radius, let's calculate how much the radius shrinks, as a fraction of the radius, per orbit.

We start by using Newton's second law,

$$F=ma,$$

where the acceleration for a circular orbit is

$$a=\frac{v^2}{r}$$

and the electrostatic force is

$$F=\frac{e^2}{4\pi\epsilon_0r^2}.$$

So we have

$$m\frac{v^2}{r}=\frac{e^2}{4\pi\epsilon_0r^2}$$

as the relationship between $v$ and $r$. We can use this to calculate everything in terms of just $r$.

To start, we find that

$$v=\sqrt{\frac{e^2}{4\pi\epsilon_0 m r}}$$

and

$$a=\frac{e^2}{4\pi\epsilon_0 m r^2}.$$

The power radiated by the accelerating electron is given by Larmor's formula,

$$P=\frac23\frac{e^2 a^2}{4\pi\epsilon_9 c^3}=\frac23\frac{e^6}{(4\pi\epsilon_0)^3 m^2 c^3 r^4}.$$

The period of the orbit is

$$T=\frac{2\pi r}{v}=2\pi\frac{(4\pi\epsilon_0)^{1/2}m^{1/2}r^{3/2}}{e}$$

so the energy lost in one revolution is

$$\Delta E=-PT=-\frac{4\pi}{3}\frac{e^5}{(4\pi\epsilon_0)^{5/2}m^{3/2}c^3r^{5/2}}.$$

The total energy (kinetic energy plus electrostatic potential energy) of the orbit is

$$E=\frac12mv^2-\frac{e^2}{4\pi\epsilon_0r}=-\frac12\frac{e^2}{(4\pi\epsilon_0)r}$$

so the fractional change in energy per orbit is

$$\frac{\Delta E}{E}=\frac{8\pi}{3}\frac{e^3}{(4\pi\epsilon_0)^{3/2}m^{3/2}c^3r^{3/2}}.$$

(Note: The total energy of the electron is negative. It gets more and more negative as energy is radiated away.)

Since the radius and the energy are inversely related, we have

$$\frac{\Delta r}{r}=-\frac{\Delta E}{E}=-\frac{8\pi}{3}\frac{e^3}{(4\pi\epsilon_0)^{3/2}m^{3/2}c^3r^{3/2}}$$

for the fractional change in the radius per revolution.

Putting in numbers, and assuming an initial radius equal to the Bohr radius for the ground state of hydrogen, we get

$$\frac{\Delta r}{r}=-3.3\times10^{-6}$$

indicating that the actual spiral trajectory would start out as a very tight spiral. In other words, the circular approximation is excellent at the beginning.

(By the way, if we use the expression from quantum mechanics for the Bohr radius, we find that this initial fractional change in the radius is just $-8\pi\alpha^3/3$ where $\alpha$ is the famous dimensionless fine-structure constant! The value of $\alpha$ is quite small, about $1/137$, and its cube is very small.)

What about near the end?

As the electron spirals toward the proton, the spiral gets increasingly less tight, because $\Delta r/r$ increases as $1/r^{3/2}$. For example, by the time the electron has spiraled 99% of the way toward the proton, we have

$$\frac{\Delta r}{r}=-3.3\times10^{-3}.$$

So it is still quite tight, but not as tight as at the beginning.

By the time the circular approximation becomes poor, the electron is practically at the proton. (Also remember: The proton isn't a point; it has a nonzero radius.)

The lesson of this calculation is that the circular approximation is very good. The calculation demonstrates a common technique in physics: Make an approximation, then use the calculated result to demonstrate that the approximation is justified.

Answer 2

In the core orthodox literature it is taken granted that the instability of the classical electron orbit has been established - essentially because of the principle that an accelerating charge radiates and so must loose energy, regardless of the details of the trajectory. This argument does not really need the details of the radiation computation to make its qualitative point.

However, a fuller reading of the theory of the Abraham-Lorentz force and some of the work under Stochastic electrodynamics suggests that it is possible, even within this context, that absorption of radiation from a background thermal bath could offset the loss due to radiation from acceleration - resulting in a stochastically stable orbit - albeit not an exact circle but something closer to a random walk.

Also, there is the idea that an electron does not actually radiate due to acceleration but due to jerk - again see the Abraham-Lorentz force theory. Keep in mind that an electron in a circular orbit does not have a constant acceleration, so it is consistent with the principle of the spiral paths in a particle detector that the radiation comes from the jerk rather than the acceleration.

Another interesting point is that a loop of current does not radiate. So, if the electron was like a drop of fluid and spread out around the nucleous in a loop, then it would not have to radiate, and so could be stable, depending on the details that one assumes for the non electromagnetic interactions within the substance of the electron. This is all tied up with the unsolved problem of the self-force of the electron.

While the classical computation of the rate of the spiral illustrates an idea about the stability of the classical orbit, there is at least some circularity in the argument in that it assumes a lot about the details of the electron and its trajectory.

At its core is a rigorous enough demonstration that a point-particle electron which is given to be travelling in a circular orbit in the absence of background radiation will radiate energy and hence would require a source of power to continue in that orbit. Assuming no source for that power, such an electron orbit would not persist. For this conclusion, it is not required that one assumes that the result is a spiral - the rigorous core is that the circular orbit will not persist.