How is angular momentum conserved when electron spirals towards the nucleus in Rutherford's model of the atom?

According to Rutherford, electrons orbit around nucleus just like planets. Since they are revolving, their motion is accelerated. When a charge particle (electron) accelerates it give off radiation and looses it energy. So the electron of Rutherford's model will continuously radiate- loose energy and spiral towards center but no external force is being exerted on the atom so the angular momentum should be conserved of the system. So when the electron initially starts to spiral inwards, its velocity will increase to compensate for the reduced radius but when finally it hits the nucleus it will have zero velocity. And angular momentum of the nucleus remains zero the whole time. So how will angular momentum remain conserved when it hits the nucleus?

• the leaving radiation takes angular momentum which has to be summed up. conservation laws are for closed systems. Commented Nov 30, 2017 at 14:03
• Your choice of username suggests that you should know the answer to your question. The electron never "hits the nucleus" because it eventually reaches the ground state where it ceases to radiate. Commented Nov 30, 2017 at 15:31

I think the Rutherford model should be completely consistent, being defined in terms of classical electrodynamics. That it doesn't correctly describe the atom is beside the point.

In this classical electrodynamical model there shouldn't be any difference between electrons around a nucleus and a light ball around a heavy ball (or other shapes you want to give them), as long as the masses and charges are scaled accordingly, and also the radius if you want to know exactly when the collision takes place. Note that if they were both points, the distance would get ever smaller but never 0 and there would be no collapse.

What would happen exactly depends on what happens when the electron hits the nucleus, something that is not part of the model. If the surface of the nucleus acts as an impenetrable and frictionless barrier, it would start sliding or bouncing around it. If on the other hand the two would somehow fuse into something new, the combined new object would be spinning.

EDIT As AnnaV and Emilio Pisanty remarked, the radiation will carry away part of the angular momentum, and it is not true that the electron-nucleus system would have conserved angular momentum. It would never lose all of it though.

• There does need to be angular momentum radiated away - the orbit cannot decay without it. Circular orbits have minimal energy for their angular momentum (and maximal angular momentum for their energy). If you want to decrease the energy of the orbit by radiating away, then you also need to get rid of angular momentum. This is also radiated away, and becomes part of the angular momentum content of the radiated fields. Commented May 15, 2019 at 8:45
• That is right, good point. I thought about it, but since it can never carry away all angular momentum, I assumed the qualitative description wouldn't change, but as you rightly remark, the orbits wouldn't change either. I'll edit, thanks! Commented May 15, 2019 at 8:51

For the model you're describing, the angular momentum of the system is conserved in the same way that energy is: it isn't.

More precisely, neither energy or angular momentum is conserved for the proton+electron system by itself, because it is not an isolated system, as it is in contact with the radiation field. However, once you include the energy and angular momentum content of the radiation, then you do get an isolated system, and both quantities are conserved.

It's important to emphasise that there absolutely needs to be angular momentum radiated away - the orbit cannot decay without it. Circular orbits have minimal energy for their angular momentum (and maximal angular momentum for their energy). If you want to decrease the energy of the orbit by radiating away, then you also need to get rid of angular momentum. This is also radiated away, and becomes part of the angular momentum content of the radiated fields.

At the endpoint of the spiral, you're strolling right out of the physics postulated by Rutherford, so what happens next is basically "in which way do you want to extend the model?" - and the answer will depend on the details of your choices when building that extension. But if you want a model where the nucleus has finite size and the electron merges into it, from a tightly knit high-velocity circular orbit, then the nucleus will need to start spinning, and that system will have a conserved total angular momentum within that interaction.

Finally, a word about the epistemology here. As has been pointed out, the Rutherford model is wrong. At best, it is a stepping stone to the Bohr model, which is less wrong (but still wrong); the dynamics of electrons in atoms are governed by quantum mechanics, period. The Rutherford model is encased within classical electrodynamics, and as such it does not describe reality. However, precisely because it is encased within classical electrodynamics, it is possible (it has to be possible) to provide a full analysis of the configuration within that classical electrodynamical framework, which explains what the model predicts will happen (even if it is inconsistent with reality, as in, say, an electron inspiral towards the nucleus) while fully respecting the internal rules of the formalism (including, in particular, the conservation of angular momentum).

• "At the endpoint of the spiral, you're strolling right out of the physics known to Rutherford, so what happens next is basically "which version of fictional physics do you think will happen?"." It is a 100% sensible question to ask when the "fictional physics" is "classical EM theory plus (relativistic?) Newtonian mechanics", which comprise, together, a well-defined mathematical system, even if one that becomes, as you say, "fictional" past a certain point because we are talking about just that - how a wrong (thus "fictional") model, which is built from those components, works . Commented May 15, 2019 at 8:42
• Even if the final answer that model gives is "singularity" or "undefined", that would still, imo, constitute an "answer" to this question and, moreover, I think the question is how that the AM is conserved through the inspiral, not what happens at the final point where the model mathematically kills itself. Commented May 15, 2019 at 8:43
• The Rutherford model says nothing about the structure or composition of the nucleus. The model simply has no answers to what will happen at the point of the collision, because it contains no information about the nucleus beyond its mass and charge. It is perfectly possible to extend the model, if you really want to, but the results will depend on the choices you make when extending the model. Commented May 15, 2019 at 8:50
• OK then, what happens if we just limit the scenario to "up to the point where the electron reaches whatever region Rutherford would have called a 'nucleus'" and stop right there? (Mathematically, just some finite $R > 0$ such that when the electron's $r = R$, we halt) And then ask, what happens to angular momentum during that part of the model? After all, he clearly knew of the finite size of the nucleus and it being much smaller than the whole atom, that's the whole point of his gold foil experiment that led him to posit the model to begin with. Commented May 15, 2019 at 8:53
• That's precisely what this answer explains. Commented May 15, 2019 at 8:55

the important point here, it seems to me, is that rutherford's model was incorrect- inasmuch as it predicted effects which were not observed, as you today and others back in the day pointed out. Because of this, there isn't any fundamental way to make sense of it, as you have discovered.

• care to explain the downvote? Commented Dec 3, 2017 at 7:56
• Actually the Rutherford model contains the genesis of the Bohr model. Rutherford is credited with discovering the nucleus, i.e. explaining the results of a scattering experiment by suggesting the positive charge is concentrated in a very small volume while the electrons are "around" it. I'm not sure what's wrong with this. Indeed AFAIK Rutherford was perfectly aware that his nuclear model was incompatible with the planetary model of an electron orbiting the nucleus, but he was not trying to explain the stability of atoms: he left that job to Bohr. Commented Dec 3, 2017 at 18:41
• in expressing his confusion, OP cites the unphysicality of the rutherford model. That rutherford may himself have been aware of this is not what I was responding to: My intent was simply to point out the fact that the OP managed to successfully apply the same line of reasoning as was used by others at the time to show that you couldn't make sense of the model (i.e., it predicts things that weren't observed, like radiative collapse of the electron's "classical orbit during inspiral). Commented Dec 3, 2017 at 18:51
• ...and I appreciate the opportunity you provided for me to explain myself BTW. Commented Dec 3, 2017 at 18:52
• Well... I was only commenting. I did find your use of "wrong" to be strong. You might want to edit your answer accordingly and maybe the downvote will eventually be reversed. Commented Dec 3, 2017 at 18:56