# Why do electrons have to fall on the nucleus in the Rutherford atomic model?

As I read on Wikipedia, the Rutherford atomic model is not correct according to classical electrodynamics, as it states that electron must radiate electromagnetic waves, lose energy and fall onto the nucleus.

I don't understand this explanation.

It is clear to me that with given acceleration directed to nucleus and proper speed, electron can move around the nucleus.

I don't understand explanation about energy, but I understand that there must be some force directed to nucleus. Also this force must not be constant because if it is, a larger speed could keep electron moving around the nucleus.

So what is that force? Why does this explanation on Wikipedia and on other resources operate with energy, not with force?

Well, I dont see a problem in any of those answers here, but, since you want in force terms... lets go.

The Lorentz force is: $$\mathbf F = q(\mathbf E + \mathbf v\times\mathbf B)$$

Lets assume the nice and simple atom of hydrogen. A single electron is classically orbiting it. Lets say there is no magnetic field. Only electric. The electric field is a central field, meaning it is pointing only radially, meaning it will result in an orbit. And more: Its a kepler orbit (same of the planets).

$$\mathbf F = q\mathbf E$$

But then, the electron when accelerated irradiates electromagnetic energy. Conservation of energy must apply, such that the irradiation takes away the energy of the electron. The electron loses then its energy. Energy is proportional to the momentum (kinetic energy). Thus, electron loses momentum. Changing in momentum is force. If we take Larmor Formula and make this process, we will arrive at Abraham-Lorentz force.

Now the complete force of this is: $$\mathbf F = \frac{d\mathbf p}{dt} = m\frac{d^2\mathbf r}{dt} = q\mathbf E(\mathbf r) + \frac{\mu_0 q^2}{6\pi c}\frac{d^3\mathbf r}{dt^3}$$

Note that, for a circular orbit in xy-plane: $$\mathbf r = r(\cos\omega t, \sin\omega t, 0)$$, and thus: $$\omega^2\mathbf r = -\frac{d^2\mathbf r}{dt^2} \quad\Longrightarrow\quad \omega^2\frac{d\mathbf r}{dt} = -\frac{d^3\mathbf r}{dt^3} \quad\Longrightarrow\quad \mathbf F = q\mathbf E - \frac{\mu_0 q^2}{6\pi c}\omega^2\mathbf v$$

Meaning, the third order derivative has a relationship with the speed. And not only that: Has a minus sign over there, indicating a drag force: A force always opposite to the velocity, and thus will tend to stop the motion. So, an electron orbiting a proton with no magnetic field present, will drag because this force, spiral in, and collapse into the proton.

• Yes, this is what I was looking for - explanation in terms of force, thank you. And now I realized what made and still makes me confused. I thought that Lorentz force is the only one that affects charged particles in electromagnetic field. You see, I thought, that only this force, no more. And now it turns out to be more foreces. Commented Oct 13, 2015 at 17:25

Excellent question, which is rarely addressed in introductions to quantum mechanics. Maxwell's equations clearly show that the electron in a classical Rutherford atom would radiate EM fields with a total power given by the Larmor formula. But it does not immediately follow that this radiation would cause the electron to spiral in toward the nucleus.

For that, you need to separately postulate a radiation reaction force (often, but not always, assumed to take the form of the Abraham-Lorentz-Dirac force) that supplements the usual Lorentz force. There is no universally agreed-upon way to do this, or even a consensus among physicists as to whether it's necessary to do at all. There are different versions of the theory of classical EM that handle the radiation reaction in different ways, each of which has (somewhat subjective) pros and cons. There is no single "correct" way to do it, because classical EM is just an approximate theory that can't fully capture the (ultimately quantum) nature of physical reality at tiny length scales.

(It's often claimed that conservation of energy logically requires that the Lorentz force law be supplemented with a radiation reaction force for internal consistency, but this isn't actually true. Since the self-energy of a point particle is formally infinite, it's actually mathematically self-consistent - although perhaps subjectively distasteful - for a charged particle to radiate forever without ever changing its trajectory, because strictly speaking, it has a bottomless reserve of potential energy to borrow from.)

So it's actually an oversimplification to say that the radation emitted by a classical Rutherford atom's electron necessarily causes it to fall inward towards the nucleus. Whether or not that's true depends on your choice of postulated radiation reaction mechanism, for which there is no single canonical form.

• There can be no discussion about how a classical rotating dipole radiates energy and how much drag force this creates on the two charged bodies. Both quantities are experimentally measurable. You may not like the necessary numerical math if there is no exact solution (which I suspect), but that this problem does have a classical answer is trivial. Commented Apr 25, 2023 at 20:20
• @FlatterMann I disagree. Yes, it's straightforward to calculate the electromagnetic radiation from a classical rotating dipole if you specify the dipole's internal structure. But the drag force is definitely not straightforward to calculate; it depends on your assumed modification to the Lorentz force law. Nor is it experimentally measurable, because there are no classical dipoles in nature; the interactions that hold the charges together are quantum and make an important contribution to the back-reaction. This problem does not have a unique classical solution. Commented Apr 26, 2023 at 13:55
• A dipole doesn't have any internal structure that we care about. The charges are concentrated in a "small enough" volume of space (small compared to the size of the separation) at all times, which makes their actual distribution irrelevant up to a small error term. That the solution to Maxwell's equations is hard to calculate is fine. The only thing that matters is that it produces a radiation term. We can try to hide behind the electromagnetic self-energy problem if we like, of course, but it's not relevant to the question: a rotating classical dipole radiates and it loses angular momentum. Commented Apr 26, 2023 at 14:09
• @FlatterMann I would define a dipole to be any charge or current configuration that produces a field with a nonzero dipole moment, and what you are describing as an ideal dipole. But that's just semantics and not relevant to the main point. I disagree that the solution to Maxwell's equations is particularly hard to calculate; there are known formulas that work very well. A rotating classical dipole certainly radiates, but whether or not it loses angular momentum depends on if and how we choose the generalize the Lorentz force law. It does not necessarily lose angular momentum. Commented Apr 27, 2023 at 0:25
• @FlatterMann Maxwell's equations tell you which EM fields the rotating dipole produces, but they do not tell you how those EM fields back-react on the dipole. That is determined by an (a priori independent) choice of dynamical equation of motion for the electric sources. (E.g. the Lorentz force law, or the Abraham-Lorentz force, etc.) Commented Apr 27, 2023 at 0:27

why is it not possible that the electric force between electron and nucleus be simply centripetal force and electron maintain constant speed in its orbital? this way there is no acceleration, no radiation of EM wave.

• Because two charges in orbit around each other form a rotating electric dipole. We can calculate and measure how much energy and angular momentum that dipole radiates in form of electromagnetic waves. Energy and angular momentum conservation do the rest. In quantum mechanics the ground state is stabilized by the fact that the electromagnetic field can't absorb energy without also absorbing angular momentum and in the ground state an atom has no more angular momentum to give. So basically what they are telling you in QM 101 about atoms is (a little) wrong, but you aren't ready, yet, to do QED. Commented Apr 25, 2023 at 20:25
• this is really the partial answer I am seeking for, although my EM knowledge never touched the radiation pattern of dipole rotation. But now I see where the changing electric field comes from for EM radiation. Thanks a lot. Commented Apr 25, 2023 at 21:52