Is angular momentum of a quantum object, metaphorical?

[bland]

OK, I understand that 'spin' is just an unfortunate name some someone gave to a particular quantum property and that particles do not actually spin. I also accept that I may never be able to understand what spin is without also understanding the mathematics involved.

However, I am also told that 'spin' is an adequate analogy for the lay person because it gives rise to another quantised property of quantum objects called 'angular momentum'. This is what I would like to clear up or understand if possible. Is the 'angular momentum' of a particle also a metaphor for a particular mathematical property, or is a particle with angular momentum, able to exhibit a force that really is angular momentum.

I have read thread no. 886279 which discussed the divorcing of angular momentum from spin, but I'm hoping to get a clearer picture.

 

[PeterDonis]

is a particle with angular momentum, able to exhibit a force that really is angular momentum.

What do you mean by "a force that really is angular momentum"?

 

[bland]

What I meant was, and correct me if I'm wrong here but I am given to understand that Newton's f=ma equation can be applied to electrons as if they were little billiard balls if they move along a wire faster i.e. mass times acceleration then they have more force, that is they hit other atoms in the wire harder causing them to vibrate faster. So if an electron had angular momentum that momentum could manifest as a force if the electron was a billiard ball and the angular momentum was harnessed, stop me if I'm talking rubbish. That is what I meant about trying to figure out if the 'angular momentum' of a quantum object is real as opposed to the metaphorical spin of an electron. That is can the angular momentum of an electron be harnessed to do work, in the same way that a classical object with angular momentum could do work. And if so how would this angular momentum actually be used to do work. Or is angular momentum just a metaphorical term and it's not an intrinsic property that is in any way analogous to classical angular momentum, discounting of course that it is quantised.

 

[bhobba]

As just about any QM book explains if you apply the quantum rules to the equations of a classical spinning particle you get quantum spin - see for example page 142 of Dirac

Its not metaphysics - its in fact deeply related to rotational symmetry but that is an advanced area - however its vital for understanding. See two books - Landau - Mechanics and Chapter 3 Ballentine.

After you will have a deep appreciation of one of, if not the greatest discovery of modern physics, it's way way up there. Its deep, profound and well worth your while coming to grips with it.

After it will be all put together by Noether Beautiful theorem (it shocked Einstein - its that deep and profound):
https://www.amazon.com/dp/0801896940/?tag=pfamazon01-20

Philosophers rarely if ever talk about it, but physicists/mathematicians when exposed to it often (and correctly) sit in shocked awe.

Thanks
Bill

 

[vanhees71]

Yes, if you want to really understand physics, you cannot avoid group theory. Ironically Pauli called it "die Gruppenpest", when more and more mathematicians and physicists used group-theoretical methods in quantum theory (Weyl, van der Waerden, Wigner). Later he himself became a master in using these methods with profound findings (spin-statistiscs and CPT theorem).

In general you can say, that physics becomes the more difficult the less you are willing to learn the adequate math. I don't understand why, but often math is taken as some "annoying evil" in teaching physics. However, the contrary is true! It's great fun to learn math together with its application in physics!

In fact in high school we learned about the Schrödinger equation, and also about the momentum operator being realized as $\hat{p}_x=-\mathrm{i} \partial_x$, and I asked my physics teacher why this is the right operator describing momentum (it was derived a bit handwavy using the usual arguments a la Einstein and de Broglie). Since I've self-learnt a lot of math beyond the high school stuff, she gave me a theoretical-physics book (Weizel, "Theoretische Physik"; a somewhat older but great German textbook in two big volumes). There I learned about "symmetry". It took me a great effort to understand it, but then it really made "click" :-)).

 

[bhobba]

I just ordered with great excitement the following:
https://www.amazon.com/dp/3319192000/?tag=pfamazon01-20

Thanks
Bill

 

[PeterDonis]

Newton's f=ma equation can be applied to electrons as if they were little billiard balls if they move along a wire faster

Only as an approximation that works under certain very restricted circumstances. (I'm not even sure electrons flowing in a wire is one of them--do you have any references?)

if an electron had angular momentum that momentum could manifest as a force if the electron was a billiard ball and the angular momentum was harnessed

I'm not aware of any scenarios where this is useful for single electrons. But the angular momentum of electrons certainly contributes to the angular momentum of larger objects that contain electrons.

can the angular momentum of an electron be harnessed to do work, in the same way that a classical object with angular momentum could do work

For any classical object with angular momentum, part of that angular momentum is coming from the angular momentum of the electrons inside it. In that sense, certainly the angular momentum of electrons can be harnessed to do work.

For single electrons, their intrinsic angular momentum can certainly be harnessed to do work on them--that's how the direction of their angular momentum is measured, by using the magnetic moment associated with their intrinsic angular momentum to deflect them (for example, look up the Stern-Gerlach experiment). But trying to view electrons as little billard balls has, as above, very limited usefulness.

 

[jtbell]

So if an electron had angular momentum that momentum could manifest as a force if the electron was a billiard ball and the angular momentum was harnessed,

There is experimental evidence that the intrinsic angular momentum ("spin") of electrons contributes to the total macroscopic angular momentum of a magnetized object. I first read about this in the Feynman Lectures on Physics many years ago:

http://www.feynmanlectures.info/docroot/II_37.html#Ch37-F3

See Fig. 37-3 and the paragraph preceding it, at the link above. This is known as the Einstein-de Haas effect. It's analogous to a common introductory physics lecture demonstration in which someone sits on an initially stationary but rotatable stool holding a spinning bicycle wheel with its axis aligned vertically. If the wheel spins clockwise as seen from above, and the person flips the wheel over so it is now spinning counterclockwise, the person and stool start to rotate clockwise so as to conserve the total angular momentum of the wheel + person + stool.

When Einstein and de Haas studied this in the mid 1910s, electron spin was unknown (along with the rest of quantum mechanics as we know it!), so they interpreted this phenomenon in terms of "bound currents" associated with magnetization (polarization of magnetic dipoles) in classical electrodynamics. Now we interpret it in terms of electron spin.

 

[secur]

]can the angular momentum of an electron be harnessed to do work, in the same way that a classical object with angular momentum could do work.

It seems to me the answer is "no". When a classical object with angular momentum is used to do work, it must lose some of its angular momentum, transferring it to the object upon which the work is done. But spin can't be lost, or reduced. Quantity of an electron's spin angular momentum is constant (of course the direction can change). Therefore it can't be transferred to anything else, therefore it can't do work. In the examples above (e.g. Stern-Gerlach) the work is being done on the electron by the externally generated magnetic field. The spin is the "handle" (you might say) by which the field "grabs" the electron. But the spin angular momentum itself is not doing any work.

Note - considering your original question - that doesn't mean it's "metaphorical"! It's not.

 

[bland]

I really appreciate the thought put into the replies. I am satisfied that my question has been adequately answered. I will not pretend to myself that I can learn the math involved, I've tried but I do my best. But while I struggle with the math, I try to make up for it by understanding the historical context and human stories concerning quantum physics and science in general. It's funny how many brave people chickened out, like Dirac not outright declaring that anti matter must exist, or more famously and curiously Einstein not trusting his general relatively to predict the expansion, or even the irrepressible Pauli, being hesitant about the neutrino.

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That's a great answer, thanks. I have listened many times to the Feynman Lectures where he goes into the Stern-Gerlach experiment, and he refers to devices as Stern-Gerlach 'apparatii', which is kinda cute.

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I have run into Noether's work many times in various lectures on symmetry and also the history of science, and even though I won't be able to understand the math, even I am astounded at the relationship between the conservation laws and symmetry. But I will have a look at one of the biographical books on her lower down on the Amazon page.

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Pauli called it "die Gruppenpest", I did actually lol at that. What about when someone maybe Born, said Pauli had to be at lectures by 10am, he said he didn't think he could stay up that late! That's a nice story you told and it makes me want to put more effort in.

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I've just heard it mentioned so many times as an explanation of why a wire glows when it resists electrons flowing through it. I have found the lectures on classical mechanics of Steven Pollock useful. But I feel that I have a better understanding of the angular momentum that I sought after reading this thread.

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That's a fantastic resource you have linked to and I see that I can view all the other pages too, thanks.

 

[strangerep]

[...] Pauli called it "die Gruppenpest", [...]

Wasn't that Schrodinger?

 

[Jimster41]

It seems to me the answer is "no". When a classical object with angular momentum is used to do work, it must lose some of its angular momentum, transferring it to the object upon which the work is done. But spin can't be lost, or reduced. Quantity of an electron's spin angular momentum is constant (of course the direction can change). Therefore it can't be transferred to anything else, therefore it can't do work. In the examples above (e.g. Stern-Gerlach) the work is being done on the electron by the externally generated magnetic field. The spin is the "handle" (you might say) by which the field "grabs" the electron. But the spin angular momentum itself is not doing any work.

Note - considering your original question - that doesn't mean it's "metaphorical"! It's not.

this bent my head. I would appreciate some clarification - In collision experiments don't particles with some spin get busted up into particles with different (but constrained) collections of spin?

When you bust an particle up into smaller/other particles is spin perfectly conserved? Does some portion always have to go to entropy? Is that portion always constrained to be of a minimum size (a valid group member)?

 

[PeterDonis]

When you bust an particle up into smaller/other particles is spin perfectly conserved?

Total angular momentum is perfectly conserved. In the kind of experiments you are talking about, the only angular momentum present is the intrinsic spins of the particles, so yes, the sum of all the incoming spins is equal to the sum of all the outgoing spins. I believe there are other scenarios where intrinsic spin can be exchanged with orbital angular momentum, so intrinsic spin by itself is not conserved (but total angular momentum, the sum of intrinsic spin and orbital angular momentum, is).

 

[secur]

Spin is always conserved in collisions, decays, or anything else. Yes, "particles with some spin get busted up into particles with different (but constrained) collections of spin" but the total must always be equal. No, none is lost to entropy.

Dirac says, in "Principles of Quantum Mechanics" 4th edition sec. 70 p 266, "The spin angular momentum does not give rise to any potential energy". Therefore, I figure, it can't do work. See my previous post for more explanation.

[EDIT] Didn't see above post until now. I didn't know there are scenarios where intrinsic spin is not conserved - that would make seem to make a difference - wouldn't that allow spin to do work? Maybe it depends on the (somewhat vague) definition of the word "work". Usually in QM we talk only about different forms of "energy".

 

[PeterDonis]

I didn't know there are scenarios where intrinsic spin is not conserved

I might have been wrong about that; I thought I remembered reading about such situations a while back, but I haven't been able to find any references.

 

[Jimster41]

Some dumb questions:
If intrinsic spin is always perfectly conserved where does entropy come in?

Assuming the spin_in = spin_out + angular_momentum_out case:
So the curving paths of the particles resulting from the collision could account for some portion of incoming purely intrinsic spin?

If we collide two particles exactly the same way twice would the results be the same? I can imagine it is effectively impossible but Is there a theoretical prohibition against being able to do this?

 

[secur]

If intrinsic spin is always perfectly conserved where does entropy come in?

They're different things. Forget about spin for a moment, consider energy in general. Energy is conserved but entropy increases (or, can stay constant; and, this is only true statistically, and of closed systems). Entropy is a measure of "disorder" (there are many other ways to look at it) and as such not explicitly related to energy - or spin. It is however related to the "disorder" of the spins of a bunch of particles or atomic systems.

Assuming the spin_in = spin_out + angular_momentum_out case: So the curving paths of the particles resulting from the collision could account for some portion of incoming purely intrinsic spin?

I think not. The rotational energy, or angular momentum, of the curve of the particle, comes from the magnetic field. AFAIK.

If we collide two particles exactly the same way twice would the results be the same? I can imagine it is effectively impossible but Is there a theoretical prohibition against being able to do this?

It's about the same chance as finding two identical snowflakes. However - there is a theoretical prohibition against being able to know for certain that the two outcomes were the same! (Uncertainty principle).

 

[PeterDonis]

Assuming the spin_in = spin_out + angular_momentum_out case:
So the curving paths of the particles resulting from the collision could account for some portion of incoming purely intrinsic spin?

Not in a collision experiment. What can happen, though, is an exchange of angular momentum between the orbital angular momentum of the particles and the angular momentum stored in magnetic fields, if there are such fields present and they are deflecting the particles. So to properly account for angular momentum conservation you have to include the angular momentum stored in the fields.

 

[Jilang]

Not in a collision experiment. What can happen, though, is an exchange of angular momentum between the orbital angular momentum of the particles and the angular momentum stored in magnetic fields, if there are such fields present and they are deflecting the particles. So to properly account for angular momentum conservation you have to include the angular momentum stored in the fields.

Do magnetic fields store angular momentum then? I haven't come across this before.

 

[PeterDonis]

Do magnetic fields store angular momentum then?

Yes, they have to. Otherwise an electric motor or generator would violate angular momentum conservation.

 

[jtbell]

Do magnetic fields store angular momentum then?

Yes, electromagnetic fields carry both linear and angular momentum. See here for example:

http://www.physicspages.com/2014/06/17/angular-momentum-in-electromagnetic-fields/

 

[vanhees71]

One should emphasize that the split of total angular momentum $\vec{J}$ into orbital ($\vec{L}$) and spin ($\vec{S}$) angular momentum is possible in a unique way only in non-relativistic physics. In relativistic QFT this is not possible in the general case, and only total angular momentum is a well-defined Poincare covariant and gauge invariant property. If one talks about orbital and spin angular momentum (as, e.g., in the context of the structure of the proton and other hadrons) one has to carefully check which observables are meant by this and how they are precisely defined (e.g., certain generalized parton-distribution functions).

 

[jartsa]

That is can the angular momentum of an electron be harnessed to do work, in the same way that a classical object with angular momentum could do work. And if so how would this angular momentum actually be used to do work.

That's easy:

Feed electrons into a electron-destroyer-device. The device gets the angular momentum of the electrons.

Feed electrons into a rotating electron-destroyer-device. The device gets the angular momentum of the electrons, and loses or gains rotational energy, because rotational energy is proportional to angular momentum.

How to build the device: A spinning tank made of anti-matter, with a hatch, which closes after an electron goes in. The hatch is located on a pole, the electron goes in slowly, then annihilates at the opposite pole. By pole I mean a point that does not move.

The tank can be made of ordinary matter just as well, if the directions of spins of electrons inside the tank do not stay unchanged for long times, and if there's no preferred direction of spin.

Let's say the heat energy stays at the pole in the annihilation case. And the electrons stay at the pole in the other case. Now rotational inertia does not change.

 

[Jilang]

Yes, electromagnetic fields carry both linear and angular momentum. See here for example:

http://www.physicspages.com/2014/06/17/angular-momentum-in-electromagnetic-fields/

I can see the E X B term. I was asking about a magnetic field though. Can there be a B without an E?

 

[Mentz114]

I can see the E X B term. I was asking about a magnetic field though. Can there be a B without an E?

It depends on the derivatives of the 4-potential $A^\mu$. The field tensor is $F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$.

The time derivatives ($\mu=0$) are electric field components, the spatial terms are magnetic field components.

 

[jtbell]

Can there be a B without an E?

If the B field is static (not time-varying), yes. Such a field would carry neither linear nor angular momentum. Likewise for a static E field with B = 0.

 

[Jimster41]

Just trying to improve my cartoon here:

So any particle can have orbital angular momentum. But a particle with spin has intrinsic angular momentum. Intrinsic angular momentum is quantized. It's unit is that of force. But the way that intrinsic angular momentum interacts with the world is through the particle's magnetic dipole moment.

 

[PeterDonis]

It's unit is that of force.

No. The units of angular momentum are momentum times distance. The units of force are momentum divided by time.

 

[Jimster41]

Sorry, right. Newton meter second is unit of power.