Does Spin Have Rotational Kinetic Energy?

referframe

Gold Member
Say we have an uncharged particle of spin ½ that is at rest.

Of course, there is an energy associated with the particle’s rest mass.

But is there an additional (rotational kinetic) energy associated with it’s spin?

As always, thanks in advance.

Related Quantum Physics News on Phys.org

Bill_K

If you're talking about a single particle, the rest mass is defined to be the total energy when the particle is at rest - there is no way to separately discuss contributions to this energy. Furthermore, "spin" does not represent a degree of freedom - there is no motion associated with it, and hence no kinetic energy.

Some compound particles on the other hand have genuine rotational degrees of freedom. A deformed (non-spherical) nucleus can rotate, and may therefore possess rotational bands: excited states with increasing angular momentum, and associated rotational kinetic energy.

dextercioby

Homework Helper
There's no energy assigned to the spin, because there's no real spinning and the spin angular momentum is not the consequence of motion.

referframe

Gold Member
There's no energy assigned to the spin, because there's no real spinning and the spin angular momentum is not the consequence of motion.
Yet if the spin ½ particle has a charge then the particle also has a magnetic moment, a concept that is independent of mass and therefore independent of angular momentum (½ h-bar).

So, in my opinion, to say that the particle is “not really spinning” is more a matter of semantics.

DrDu

Note that also a compound particle like a neutron or a hydrogen atom has a spin, namely the angular momentum in its rest frame. It is obviously possible to write down the angular momentum contribution to the kinetic energy of a hydrogen atom. With elementary particles the problem is that you cannot compare with other states of the particle which have different spin as spin really defines the particle.

Bill_K

So, in my opinion, to say that the particle is “not really spinning” is more a matter of semantics.
referframe, Back in the early days, the days of the "old" quantum theory, when people were still coming to terms with quantum mechanics and how much it differs from classical mechanics, they imagined elementary particles to be objects that had most of the properties of classical objects, differing only because they were very very small. Electrons went around the nucleus like planets in well-defined orbits. Sometimes they even imagined orbits that were elliptical! And you can find descriptions in older books of the "vector model" of the atom, in which spin and orbital angular momentum were represented by arrows. But since they didn't add up the way that vectors usually add, they were supposed to be tilted to each other at an angle, and wobble or precess about the z axis.

None of these ideas survived, Partly because they involved concepts that were inexplicable, partly because they involved things that were unmeasurable, even in principle. If an electron really spins, for example, what is its angular velocity? And is this angular velocity a universal constant? Or is it different for a muon?

In general, in quantum mechanics, a thing that is time-independent is described by a time-independent wavefunction, and does not in any sense "move". An electron does not orbit the nucleus. A particle in the ground state of a harmonic oscillator does not slosh back and forth. And elementary particles with spin do not rotate.

jetwaterluffy

v=ωr. An electron is a point particle, so r=0. This means v=0*ω, therefore v=0. KE=mv2=m02=0. Therefore, KE=0. So no, it doesn't.

phyzguy

v=ωr. An electron is a point particle, so r=0. This means v=0*ω, therefore v=0. KE=mv2=m02=0. Therefore, KE=0. So no, it doesn't.
By this logic, its moment of inertia would be zero, and hence its angular momentum would be zero also, which it isn't, so I don't think this is a valid argument.

jetwaterluffy

By this logic, its moment of inertia would be zero, and hence its angular momentum would be zero also, which it isn't, so I don't think this is a valid argument.
OK then, what is the hole in my argument that prevents that from happening? Why should there be angular momentum anyway? I think it would make more sense to me if its spin is the equivalent of mω.

phyzguy

OK then, what is the hole in my argument that prevents that from happening? Why should there be angular momentum anyway? I think it would make more sense to me if its spin is the equivalent of mω.
We don't have a working model for elementary particles. All we can say is that they have certain properties, i.e. mass, spin, charge, magnetic moment, ...

LostConjugate

OK then, what is the hole in my argument that prevents that from happening? Why should there be angular momentum anyway? I think it would make more sense to me if its spin is the equivalent of mω.
I don't think it can be said that r=0 just because the particle demonstrates properties of a point particle. The concept of a point particle is still theoretical. Even if it was found that space was discrete and could have "points" would those points not have a volume associated with them? (dx)(dy)(dz)

Naty1

The prior three posts seem ok.

An electron is REPRESENTED mathematically as an idealized point particle.

as Phrak posted recently: "particle" is a word used to distinguish one field from another having different quantum numbers.

bbbeard

In general, in quantum mechanics, a thing that is time-independent is described by a time-independent wavefunction, and does not in any sense "move". An electron does not orbit the nucleus. A particle in the ground state of a harmonic oscillator does not slosh back and forth. And elementary particles with spin do not rotate.
There is certainly a well-defined sense in which an electron orbits a nucleus -- at least, some electrons.

If you consider electrons in states with L>0, then those states with ml≠0 have non-zero orbital angular momentum. You can evaluate the "orbital velocity" of the electron wave function by computing the expectation values of the probability density and the http://en.wikipedia.org/wiki/Probability_current" [Broken] for the given energy eigenstate |L,ml>. ISTR that if you then compute the orbital velocity at the expected value of radial position (e.g. the Bohr radius for the ground state of the hydrogen atom), you get a value that accords with the semi-classical model that pictures the electron as a tiny BB of mass m, orbiting in a circular orbit of radius rBohr (or appropriate multiple), with angular momentum L. For the |1,1> state I think you find that the orbital velocity is alpha*c, where alpha is the fine-structure constant 1/137.036....

However, the semi-classical model breaks down when you try to apply it to intrinsic spin and a BB-shaped electron with the http://en.wikipedia.org/wiki/Classical_electron_radius" [Broken]. When you go down to that tiny scale, where renormalization effects become significant, the computed surface velocity is well in excess of the speed of light....

Last edited by a moderator:

Jeff Byram

Folks,

Please note this is not a homework question! This is true curiosity.
I have two questions regarding rotational kinetic energy. I know that rotational kinetic energy is defined as:

KE = ½ I Ω^2

Where KE is the rotational kinetic energy [in units of: joules; kg*m^2/s^2], I is the Moment of inertial [in units of: kg* m^2], and Ω is angular velocity [in units of: radians/sec^2 or 1/s^2] so that the units check.

If (Electron or Nuclear) Spin is defined as J, with values equal to a half integer * h/2π, where h is Plank’s constant [in units of: Joule *seconds or kg*m^2/s],

#1, Is the rotational kinetic energy associated with Spin: KE = ½ J Ω ? (so the units check).

Since V = Ω R, where V is velocity, and R is the radius (or the Electron or Nucleus), so Ω = V/R, and since the fastest possible speed is the speed of light, c,

#2, Is the greatest possible rotational kinetic energy associated with Spin: KE = ½ J c/R ? (so the units check).

Sincerely,
Jeff Byram

phyzguy

There is no detailed model of elementary particles where we can say: X% of the mass of the electron is associated with the kinetic energy of spin, Y% is associated with the energy of the electromagnetic field, etc. It is not clear (at least to me) whether or not such a model is even possible. So no one knows how to answer your question. If you plug numbers into what you suggested, and use the classical electron radius, you would calculate:
$$KE = \frac{\hbar c}{4 r_e} = \frac{\hbar c m_e c^2}{4 \alpha} = \frac{137}{4}m_e c^2$$

which is about 30 times the mass of the electron.

qsa

Say we have an uncharged particle of spin ½ that is at rest.

Of course, there is an energy associated with the particle’s rest mass.

But is there an additional (rotational kinetic) energy associated with it’s spin?

As always, thanks in advance.
there are off beat theories ( that tries to calculate mass) that conjecture that spin is responsible for part of that mass. my own theory which relies on computer simulation (which I hope you will like that since you are also an IT person) also indicate that. I will PM you some of my findings, since the mentors will be very annoyed if I say anything about that, to say the least.

yoron

Isn't the speed of an electron 'spin' faster than light :) when classically seen? I thought that was established, and discarded as a reason a long time ago? And isn't it so that a atoms mass can 'fluctuate' intrinsically too?

If 'energy' can be translated into mass I would rather go with that than with 'speeds'. But then you also need to come up with a reason why it fluctuate. 'Virtual photons' maybe.

ytuab

There is no detailed model of elementary particles where we can say: X% of the mass of the electron is associated with the kinetic energy of spin, Y% is associated with the energy of the electromagnetic field, etc. It is not clear (at least to me) whether or not such a model is even possible. So no one knows how to answer your question. If you plug numbers into what you suggested, and use the classical electron radius, you would calculate:
$$KE = \frac{\hbar c}{4 r_e} = \frac{\hbar c m_e c^2}{4 \alpha} = \frac{137}{4}m_e c^2$$

which is about 30 times the mass of the electron.

You mean that only the kinetic energy (KE) is more than 30 times the "big" rest mass energy (= mc^2) ?
According to Virial theorem, the potential energy needs to be minus 2 times the kinetic energy (V = - 2 x KE).
So the potential energy becomes minus 60 times the rest mass energy.
This means the total energy (V+KE+rest mass energy) is about minus 30 times rest mass energy (- 30 x mc^2 ).

But the Dirac equation for hydrogen atom, (which predicts fine structure) shows the energy level ( V+KE+rest mass energy) is

$$E_{n,j} = \frac{m_0 c^2}{\sqrt{1+\frac{\alpha^2}{(n-(j+1/2) + \sqrt{(j+1/2)^2 - \alpha^2})^2}}} \qquad (\alpha = 1/137)$$

For example, the energy of the 1S1/2 state (j = 1/2, n = 1) is

$$E_{1,1/2} = \frac{1}{1.0000266} m_0 c^2 \approx m_0 c^2$$

This result of Dirac equation is completely different.
So it is impossible that the we treat spin as "spinning", I think. Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving