# Why can't a particle rotate about its axis? [duplicate]

When treating an atom as a point partcle, it is usually assumed that it cannot rotate about its axis. Is it that anything considered as a point particle cannot rotate about its own axis? If so, why is that the case?

It is because of quantization of the angular momentum. If an atom rotates, that would mean a higher orbital angular momentum for the electrons. Usually, that is a state several electron-volts higher than the ground state. Not accessible at ordinary temperatures.

• I don't have much knowledge of quantum theory. What about a classical single particle? Can such a particle rotate about its axis?
– Toba
Sep 24 '20 at 8:51
• Doesn't quite answer the question I think. The question was about point particles. An atom which can get excited into a state with non-zero orbital angular momentum for the electrons is, for that very reason, not a point particle. Sep 24 '20 at 8:53
• @AndrewSteane The question was about atoms. The standard explanation for that is not that it cannot rotate because "an atom is a point particle".
– user137289
Sep 24 '20 at 8:55
• The question reads "Is it that anything considered as a point particle cannot rotate about its own axis? If so, why is that the case?" Sep 24 '20 at 8:56
• I have edited the question. I hope it is clearer now
– Toba
Sep 24 '20 at 8:58

The aim of this answer is to give a physical intuition without quantum mechanics (The answer by Pieter should be the correct rigorous answer):

A monoatomic gas contains per definition of single atom particles. That means these particles consist of the core of the Atom (with the whole mass, roughly point particle in comparison to the size of an atom) and the electronic orbitals (Contain a negligible amount of mass compared to the core).

I think the intuitive way of thinking is to say, by 'rotating' this negligible small mass (the electrons) you are nowhere near the energy scale of the displacement degrees of freedom (DOF) of the atom.

In comparison in a two atomic gas, only if the two atomic cores are rotating around the rotation axis the DOF is considered. We have the same case as for the one atomic gas if both cores lie on the rotation axis.

Hope this gives you some intuition behind this consideration in stat. phys. .

• Why can't the core of the atom rotate about its axis?
– Toba
Sep 24 '20 at 8:34
• It could, but intuitively an atomic core would have a small moment of inertia en.wikipedia.org/wiki/Rotational_energy en.wikipedia.org/wiki/Moment_of_inertia and therefore you could store far less energy in the rotation of a core Sep 24 '20 at 8:39
• The moment of inertia is dependent on the expansion of the object, therefore in point particles with roughly no expansion we have a negligible moment of inertia and therefore negligible rotational energy Sep 24 '20 at 9:54
• The moment of inertia was derived based on the assumption that the object contains many particles. How is the derivation modified for the case in which the object is a single particle?
– Toba
Sep 24 '20 at 13:59
• There is no. That is the whole point. For one point particle, you go in the integration of the density $r\rightarrow 0$ and the moment of inertia is zero, which means the rotational energy is zero. Sep 24 '20 at 14:01

If the particle is charged, then rotation would mean acceleration of charge. Rotation means constantly changing the direction of the velocity, ergo acceleration. An accelerating charge would radiate (emit photons), so this state would never be stable. The idea of spin came about because we observed stable charged particles that interacted with magnetic fields as if they were rotating, but weren't emitting any photons. So this classical interpretation led to a contradiction, and we knew it couldn't be the correct picture.

In general, small things in classical theories can rotate. But that's just not how particles actually behave.

A point has a location but zero size aka a radius/diameter/dimension of zero.

In order for something to rotate, some part of the thing must orbit around some other part of that thing. But if something has zero size, that means the entirety of the object exists in the same infinitesmal location in space. So how can multiple parts of an object orbit each other when all those parts are located at the same infinitesmal point in space?

Why can't a particle rotate about its axis? for elementary particles.

If you look at the table, almost all elementary particles are axiomatically assigned spin. Spin in classical mechanics is a rotation about an axis going through the center of mass of the particle. Thus , in a classical picture elementary particles do rotate.

BUT elementary particles are quantum mechanical entities, and their position in space is controlled by a probability distribution, which is predicted by the wavefunction of the solution for the boundary conditions of the specific problem, scattering or decay. So what does spin mean for elementary particles?

Spin had to be assigned in elementary particle interactions, particle by particle in the table, in order fot the law of angular momentum conservation to apply at the quantum mechanical frame.

The assignment in the tables has been found to be consistent with all the up to now data. Thus the answer for elementary particles to the title is also: yes , elementary particles rotate in a mathematically convoluted probabilistic way.

In quantum mechanics and particle physics, spin is an intrinsic form of angular momentum carried by elementary particles, composite particles (hadrons), and atomic nuclei. extends to composite particles and also to atoms

So the argument above holds for atoms too. That is why one has nuclear magnetic resonance phenomenon.

All isotopes that contain an odd number of protons and/or neutrons (see Isotope) have an intrinsic nuclear magnetic moment and angular momentum, in other words a nonzero nuclear spin, while all nuclides with even numbers of both have a total spin of zero.

Is it that anything considered as a point particle cannot rotate about its own axis? If so, why is that the case?

Strictly, if one considers the center of mass of the particle as a point, and the particle has a volume about that point, there will be a possibility of acquiring angular momentum when, if it is a dust particle, it hits another dust particle; this will give a linear momentum and can (if not head on) also induce rotation about an axis that goes through the center of mass. But if, for simplicity of mathematics to be derived, it is really considered as a particle with no extension about the center of mass point, there is no way an impact can transfer angular momentum, by construction of the model.

• Really nice answer. Sep 25 '20 at 15:26

As requested by Toba, I would like to summarise my thoughts on the moment of inertia of a point particle and how it leads to a trivial consideration of the energetics of the rotation of a point particle.

This consideration is purely classical.

Moment of Inertia:

The moment of inertia is defined as follows: $$I=\int_V r_{\perp}^2\rho(\vec r)$$ The integral over a particle density $$\rho(\vec r)$$ multiplied with the square of the distance from the rotational axis $$r_\perp$$.

The moment of inertia is related to the rotational energy of the object described by the density rotation with an angular velocity $$\omega$$ as follows: $$E_{\text{rot}}=\frac{1}{2}I\omega^2$$

And also related to the net angular momentum of the system via: $$L=\frac{I}{\omega}$$

Moment of inertia of a point particle:

Now let's consider a point particle, that means we consider a particle density that is peaked at a single point $$\rho(\vec r)=m \delta(\vec r)$$ (That means it is described by a delta distribution, coordinate frame is chosen in the way that the particle is located in the origin). $$m$$ is the mass of the point particle. We can now use this definition of the moment of inertia to calculate it for the point particle. $$I_{\text{point particle}}=\int_V r_\perp^2\rho(\vec r)=m \int_V r_\perp^2 \delta(\vec r)=0$$ As the delta distribution gives a non-zero contribution only at $$\vec r = 0$$ where $$r_\perp=0$$.

Interpretation:

As we see the moment of inertia of a point particle is always zero, therefore, a point particle never has any intrinsic rotational energy or intrinsic angular momentum. So the internal rotational DOF of a point particle do not contribute in the statistical mechanical as they do not change the energy of the particles.

Angular velocity

It does also not make much sense to talk about the angular velocity of the particle because it because the angular velocity is defined by the angle the object rotates around the rotational axis in a certain time. If the particle is only present at the rotational axis the common definitions of the angular velocity fail:

$$\vec \omega = \frac{r\times v}{r^2}$$

As it describes the rotational between points which is no longer possible in a point particle.

Conclusion

• Rotational DOF are irrelevant for point particles as $$E_{\text{rot}}=0$$
• It does also not make much sense to talk of the rotation of a point particle because in order to define rotation you need two distinct points in the object.

What changes towards the two atomic gas

In the two-atomic gas (two particles of mass $$m$$, and distance $$2\vec d$$ perpendicular to the rotation axis, rotation axis between the particles) we have the following particle density \rho = m\delta(\vec r + \vec d) \delta(\vec r - \vec d) therefore the moment of inertia of this system is: $$I_{\text{2 atomic gas}}=\int_V r_\perp^2\rho(\vec r)=m \int_V r_\perp^2 \delta(\vec r + \vec d)\delta(\vec r - \vec d)=m|d|^2+m|d|^2 = 2 m|d|^2$$ So we have a non zero moment of inertia.

Also the angular velocity $$\vec \omega = \frac{r\times v}{r^2}$$ can be defined and calculated for the system. Therefore the rotation of those particles with respect to the rotational axis matters and leads to a change of the rotational energy of the gas constituent and has to be included in the thermodynamic considerations