# How does an electron move around the nucleus according to classical mechanics?

If an electron moves around a nucleus in an elliptical path, is the moment of inertia of the electron with respect to the nucleus a constant w.r.t time?

I think that both the electron and nucleus must be in motion so as to conserve linear momentum. If I assume that both move in elliptical paths around the COM, then the abovementioned moment of inertia can be a constant only if they move in their respective elliptical trajectories while maintaining constant separation between them. Is such a motion possible? If yes, then how are the ellipses related to each other?

• classically, the electron would radiate and spiral into the nucleus Commented Aug 2 at 21:16
• What if it didn't radiate? Or if instead of electron and nucleus, we consider 2 masses, would they sill collide?
– Tom
Commented Aug 2 at 21:39
• I have heard that the ratio of moment of inertia and angular momentum stays constant for isolated systems. If that is true then moment of inertia must be conserved because angular momentum is conserved for isolated systems. Does this mean that the separation between the two masses stay constant?
– Tom
Commented Aug 2 at 21:42
• Forget radiation, forget QM. That's clearly totally unrelated to what the OP is asking. The OP is just asking how two masses centrally attracted by a $1/r^2$ potential would move. It's exactly the two-body gravity problem. It's a question about classical Newtonian physics, not E&M, not QM. Commented Aug 3 at 4:49
• @Jagerber48 Then he should ask how a 1kg copper sphere carrying a charge of 1e-6C orbits a 1000kg copper sphere carrying a charge of 14e-6C. THAT would be a classical physics problem and then you would be correct. An electron does NOT move like these hypothetical copper spheres would. Commented Aug 3 at 5:36

So a hypothetical question in which the hypothetical breaks a law of physics can always be problematic, since you need to decide how to break the law of physics, and that choice cannot be justified by physics---nevertheless, ppl will attack your choice based on it violating physics, which it has to do.

tl;dr: any comments about ignoring a law of physics will ignored with equal vigor.

Your supposition is correct, for an atom with a nuclear mass $$M$$, one can posit:

$$\frac 1 2 \le \frac{m_e} M \le 294$$

which starts at the exotic atom positronium, and tops out at Oganesson-294.

That means the nucleus is not stationary.

Labeling the two particles' positions $$\vec r_N$$, $$\vec r_e$$, we have a hamiltonian (in the Coulomb approximation...and likewise for the position):

$$H(\vec r_N, \vec r_e) = \frac{p^2_N}{2M} + \frac{p^2_e}{2m_2} + V(|\vec r_N - \vec r_e|)$$

which is not fun.

We note that the position dependence can be expressed in term of the vector:

$$\vec r \equiv \vec r_e - \vec r_N$$

moreover, it only depends on the magnitude:

$$r= |\vec r|$$

(which means we have spherical symmetry).

Step 1 is to defined the center of mass affine position:

$$\vec R = \frac{M\vec r_N + m_e \vec r_e} {M+m_e}$$

with conjugate momentum $$\vec P$$.

Step 2 is define the reduced mass, $$\mu$$, associated with the coordinate vector $$\vec r$$:

$$\frac 1 {\mu} = \frac 1 M + \frac 1 {m_e}$$

The Hamiltonian is then:

$$H = \frac{P^2}{2(M+m_e)} + \frac{p^2}{2\mu} + V(r)$$

which contains a free-particle term:

$$H_{CoM} \equiv \frac{P^2}{2(M+m_e)}$$

and a pure Coulomb term:

$$H \equiv \frac{p^2}{2\mu} + V(r)$$

where:

$$V(r) = \frac 1 {4\pi\epsilon_0} \frac{Ze^2} r$$

At this point, we ignore the center-of-mass motion, and solve for the orbit (classical) or hydrogen-like atom (quantum).

As you point out in the post: yes, the nucleus undergoes 'reciprocal' motion. This means it classically is also in an elliptical orbit about the barycenter (aka: CoM), and in quantum mechanics, it is also in an orbital with a wave function $$\Psi_{nlm}(\vec r_N)$$ that has the same functional form as the atomic orbitals $$\psi^{(e)}(\vec r_e)$$.

Nevertheless, super-imposing the two motions is messy, so we call the atomic orbital

$$\psi_{nlm}(\vec r)$$

and compute it with mass $$\mu$$.

The original question: "how are the two ellipses related" is that they are related by dilation symmetry, where the scale factor is a ratio of the two masses.

In the case of positronium:

$$\frac{m_e}{M} = \frac{m_e}{m_e}=1$$

and they are identical, while for the largest atom:

$$\frac{m_e}{M} \approx \frac{m_{AMU}/1823}{294m_{AMU}} \approx \frac{1}{536,000}$$

so the size of the ellipse (or orbital) of the nucleus is:

$$a_N \approx 0.1\,{\rm fm} = \frac{R_p}{12}$$

where:

$$R_p = 0.88\,{\rm fm}$$

Regarding the hypothetic part: a classical orbit radiates EM waves because charges are accelerated. Since they electron moves at:

$$v \approx \alpha c$$

$$a_0 =\lambda_C/\alpha$$

where $$\alpha \approx 1/137$$ is the fine structure constant, and

$$\lambda_C = \frac h {m_ec} \approx 4\times 10^{-13} \,{\rm m}$$

is the Compton wavelength of the electron, we can ball park the acceleration:

$$a = \frac{v^2} r = \frac{\alpha^2 c^2}{\lambda_C/\alpha}$$

$$a = \frac{m_e}{h}\big(\alpha c\big)^3 \approx 10^{22}\, g$$

Wow. That is big, and that is for $$Z=1$$. I'll leave it as an exercise to estimate the result for $$Z=118$$.

Quantum mechanics of course precludes this state from radiating.

(As an aside: as $$Z\alpha \rightarrow 1$$ this estimate runs into relativity problems as $$v\rightarrow c$$, and the naive radius gets down to the Compton wavelength)