# Angular and magnetic momentum

1. ### bobie

689
"As seen from the definition, the derived SI units of angular momentum are newton meter seconds (N·m·s or kg·m2/s) or joule seconds (J·s). Because of the cross product, L is a pseudovector perpendicular to both the radial vector r and the momentum vector p and it is assigned a sign by the right-hand rule."

I have a few simple questions:
- has the direction of L any meaning (as with magnetic moment), or just a convention?
- if we say that an electron L is $\hbar$/2, what does it mean? that if we want to rotate the plane of the orbit (by how many degrees?) we must apply a force of $\hbar$/2?, and is this true in either direction?

In the case of magnetic moment, the direction of $\mu$ here refers to a positive charge? is it the the North or South Pole?
and does it tell us that it takes a force of $\hbar$/4 to make the plane of the orbit rotate by 180° or what?

Thanks

Thanks

Last edited: Mar 31, 2014
2. ### Simon Bridge

15,474
The direction of L has a physical effect yes.
The right-hand screw rule comes from the cross product in 3D.

##L=\hbar/2##units is the angular momentum of the electron.
It is used in conservation of angular momentum calculations the same way as always.
But electrons do not "orbit" atoms so "rotating the plane of the orbit" does not mean anything.

Each electron in an atom contributes "orbital" angular momentum to the total angular momentum of the atom.

The direction of the magnetic moment is from south to north, as if there was a current loop.

To see an example of how quantum angular momentum works, see the Stern Gerlach experiment.

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3. ### bobie

689
Thanks, Simon,
1) What is that effect?
2) if we consider a bycicle wheel what is the difference if L points one way or the other?
3)What does the value L (p*r) indicate , if not the force we must apply to rotate the wheel?

4) if we consider an electron in 1H, the vector points down? so the north pole is down?
5) If there is both an orbit angular and magnetic moment, if we try to rotate the orbit, do L and $\mu$ sum up or multiply?

6) Does this experiment demonstrate angular (Le) or magnetic moment ( $\mu$e)or both or also of the orbital angular momentum (Lo) and magnetic moment ($\mu$o) ?
7) does it prove that the spin is constantly flipping from one side to the other?

Thanks a lot

Last edited: Mar 31, 2014
4. ### Simon Bridge

15,474
Quantum angular momentum works a little differently to the bike wheel if only because there is no axle on an atom to hang on to.
But you are right - the torques are usually provided by a magnetic field.

The choice of direction for the L pseudovector changes a sign in the equations. It amounts ot a change from a right-handed to a left handed coordinate system - so the choice is as arbitrary as the choice of coordinates and what you hope to achieve.

The vector points in a random direction unless there is something to tell the atom which way "up" and "down" are.
That is usually provided by a magnetic field - in which case the moment will align either along the field or opposite. There is no time during which the psuedovector can be thought of as "rotating" like you change the orientation of the bike wheel.

It demonstrates both - where you have a magnetic moment you have angular momentum, and spins do not constantly "flip from one side to the other" so no experiment will demonstrate that.

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5. ### bobie

689
Thanks, I ' ll discuss each problem separately. I 'll start with this which is the most difficult for me:

I do not understand that, how can a magnetic field prove the exixtence of an angular momentum? .Suppose there were none (Le), what would happen? what would change? How do we deduce the result is not due exclusively to $\mu$ e? As Le (and$\mu$e) is considered an intrinsic property and the electron is not really spinning, couldn't the result of the experiment be determined only by the magnetic moment?

Last edited: Mar 31, 2014
6. ### bobie

689
I have read also in this forum that if you run one of the two beams (say, spin up) of a Stern-Gerlach into another similar machine it will split , in its turn , into two beams up/down.
If this is true , doesn't it prove that at any instant spins are changing direction?

7. ### Simon Bridge

15,474

This only happens if the second apparatus is rotated wrt the first. i.e. the z-direction changes. Maybe you have misread, or mistaken the context. Maybe the author got it wrong. Without a proper reference I cannot see what is going on.

Whatever, there are plenty of resources on this experiment to put you right.
Even wikipedia relates the multiple experiments case correctly - well mostly.
http://en.wikipedia.org/wiki/Stern–Gerlach_experiment#Sequential_experiments

The measurement of a particular spin-value establishes the system in that spin eigenstate.
Subsequent measurements will get the same value.

Last edited: Mar 31, 2014
8. ### bobie

689
So, if we run H atoms into a machine and move one of the two beams into a box do we get magnetization at room temperature?
and also,
how does this experiment prove the existence of an angular momentum L(e)?

9. ### Simon Bridge

15,474
Good question.
Ideally, that would be the case, the magnetic moments would all be aligned.
This requires that the idealized H atoms interact only with the idealized "box" and nothing else.

IRL: Thermal interactions, scattering off the sides of the real physical box etc. would probably randomize the magnetic moments.

But you can show the spin angular momentum that way ... if you caught the beam/particles in "a box", and the process randomized the magnetic moments, then some angular momentum would get transferred to the box. The box would start to turn. That good enough for you?
http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.92.190801

Technically it demonstrates the quantization of angular momentum.
The experiment would not work without angular momentum ... can you can think of a way to get a magnetic moment out of a distribution of charge without rotations, without angular momentum as well?

But you may be thinking of the macroscopic case where you can have a macroscopic magnetic dipole without the magnet turning? This is where things get quantum on you :)

Where is all this coming from?
What is you education level so I can pitch my replies appropriately?

You appear to be trying to understand quantum phenomena in terms of classical mechanics.
This cannot be done - if it could, we wouldn't need QM. The quantum terms do not have a 1-1 correspondence to the classical ones. They are like them, but not exactly like.

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10. ### bobie

689
I am just a student trying to get an insight into physics.
Actually, I thought I was trying to do the reverse, to carry QM to its strictest logical consequences.
As I said, QM says that there is no real spinning and that Le and$\mu$e are intrinsic properties. That is plausible for $\mu$e as the electron could be a natural dipole with a South and North pole and ,consequently, a magnetic moment. An intrinsic angular momentum is a bit more hard to justify in principle, right?

So, as there seems to be no evidence of Le (or is there? ), if no experiment can isolate the angular momentum S from $\mu$e,
- why can't we attribute the response of the electron only to a $\mu$, solving in this way also the sterile dispute of e- spinning faster than C?
- Then, I ask, if there were both a Le and a$\mu$e, when we apply a torqueand try to rotate the pseudovector(s), do the two values sum up or multiply?

Last edited: Apr 1, 2014
11. ### Simon Bridge

15,474
There is very definite evidence for the intrinsic angular momentum of an electron. The magnetic moment is itself evidence of it.

You way you are asking the question that only makes sense in classical mechanics.
We can ask how much work it takes to change an angular momentum state.

That is the difference in energy from one state to another.

But what I think you are interested in is whether angular momentum and magnetic moment are separate phenomenon or not. The answer is that they are not separate.
You can have a spin without a magnetic moment but not the other way around.

You are right - the angular momentum is an intrinsic quality of a fundamental particle.
There is no classical rotation happening. There is no classical analog.

OTOH: angular momentum is happening, as is shown in conservation of angular momentum experiments such as the link provided. It is easy to find such experiments in Google Scholar.

What the QM is showing us is that our concept of angular momentum applies even when there are no rotations.
But intrinsic angular momentum is a quantum concept with it's won rules.

What is you education level so I can pitch my replies appropriately?
I don't think I can properly help you further without this information.

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12. ### bobie

689
My physics level is A-level.

- how do you tell if a phenomenon is due to to Le and not to Lo)?.

- can you tell me, besides any evidence, why, in principle, an intrinsic magnetic moment that implies no actual rotation must imply an intrinsic angular moment, too?

- In classical physics do the momenta sum up?

Last edited: Apr 1, 2014
13. ### Simon Bridge

15,474
A-Level - good: that explains some of the talking past each other.
You've done the bike wheel thing, and you have some calculus and probability theory.

The gyromagnetic effect relates magnetic moment to angular momentum.

eg. the dipole moment for a single charge moving in a loop is $$\vec\mu = q(\vec r\times\vec v)=\frac{q}{m}\vec L_q$$ ... i.e. the definition of the dipole moment includes the definition of angular momentum. That's why its called a "moment".

Similarly in QM: for electron with spin angular momentum ##\vec S##, the dipole is: $$\vec\mu_s = -\frac{e}{2m}g_e\vec S$$

We think of the particles carrying angular momentum even though it doesn't have any rotation, much the same way as we think of light carrying linear momentum even though it doesn't have any mass.

In 2D yes. In 3D, rotational motion gets a bit more complicated, even for a classically rigid body.
http://ocw.mit.edu/courses/aeronaut...fall-2009/lecture-notes/MIT16_07F09_Lec26.pdf
http://people.rit.edu/vwlsps/IntermediateMechanics2/Ch9v5.pdf

Note: classically - if the bike wheel also carried a charge (say you'd hooked it to a van-der-graaf generator) and you spun it up, would it take more work to turn the wheel?

... but you could read the abstract and the journal name?
You can check the reputation of the journal re peer review?
Physical Review Letters is something of a gold standard for publishing physics research.

... context.

http://en.wikipedia.org/wiki/Angular_momentum_operator
... wikipedia summarizing how angular momentum is handled in quantum mechanics.
Gives you something to aim for: it handles angular momentum as a "generator of rotations".

http://en.wikipedia.org/wiki/Magnetic_moment#Magnetic_moment_and_angular_momentum
... relationship between magnetic moment and angular momentum.

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### Staff: Mentor

For an experiment involving electron spins in a magnetic material, see the Feynman Lectures:

http://www.feynmanlectures.caltech.edu/II_37.html#Ch37-F3

in particular Fig. 37-3 and the paragraph that precedes it.

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15. ### bobie

689
Thanks, Simon, for your excellent explanations.
Now, we feed 1H atoms in a stern-Gerlach machine, $\mu$ vectors points in random directions.
Suppose this (http://upload.wikimedia.org/wikiped...etic_moment.svg/200px-Magnetic_moment.svg.png) is the vector of the proton $\mu$p spin-up,
we know (http://en.wikipedia.org/wiki/Hydrogen_line#Cause) that the vectors of $\mu$e can be only antiparallel to $\mu$p spin-down.
When we apply a magnetic field, what happens? $\mu$e cannot smiply flip over , because the parallel position is forbidden, so the whole atom must rotate? wouldn't this affect the energy involved , shouldn't it be different from $\hbar$? And the magnetic moment of the orbit ($\mu$o) is not involved?

Last edited: Apr 2, 2014
16. ### Simon Bridge

15,474
You are still trying to apply classical ideas to a quantum system.

Treat it in general first - you have a bunch of particles with non-zero spin and magnetic moment.

Classically, before the particles enter, we don't know what their spin state is. But the particles know what their spin state is - each particle's spin is randomly orientated wrt the apparatus and each other.

... the apparatus exerts a torque on the dipoles so they turn to align to the field one way or the other.
There is also a translational force on the dipole that depends on it's orientation. (You can look up the physics for how a dipole interacts with an external magnetic field.)

Since they are moving fast, some particles exit the apparatus before they have had time to turn all the way.
So the particles should exit at different angles in a fan-shape which depends on the initial distribution of their spins. i.e. A particle that entered with spin at 45deg to the apparatus should deflect less than one that happens to enter completely aligned.

Do the experiment and that is not what happens!

In practice, two well-defined beams emerge from the apparatus.
One at the angle predicted for a 0deg spin and the other for a 180deg spin.
These are called "spin up" and "spin down".

This shows us the quantum nature of the spins.

The QM description goes like this:
before entering the apparatus, the particle spins are "uncertain" - not "randomly aligned".
that is - each individual particle does not know what it's spin is.

The apparatus "measures" the spin state.
The measurement "discovers" a spin state up or down - no rotation required.

How the state goes from "uncertain" to definite QM does not say.

I've been trying to work out if there is a situation where flipping the spin requires work.
A single particle in the ground state in a potential well may have spin-up or spin-down.
Both orientations have the same energy level - so flipping the spin over has a net zero change in energy.
Thus no work required.

If there are many particles, things can change because the spins may interact with each other ... I don't think that would help you, and I'm not sure this talk of states and energy levels makes sense at A-level.

Note: you have misread the wikipedia article about the possible spin states in the H atom.

Last edited: Apr 2, 2014
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17. ### bobie

689
Just to mark the difference with macroscopic world,
suppose we have tiny magnets inside a solenoid, they are randomly oriented. If we switch on the current, do they align in the same way or spin up-down?
And if we have 1H atoms here, instead of a Stern-Gerlach machine, what happens?

Last edited: Apr 2, 2014
18. ### Simon Bridge

15,474
If you put magnets in a solenoid, switch it on, then the magnets will experience a torque as if they were a current loop in the uniform field. You've seen it in an electric motor. The torque depends on the orientation of the magnet ... so it is likely that they would end up executing rotational SHM.

Lets see: ##\vec \tau = \vec\mu\times\vec B = I\vec \alpha##

This gives: ##I\ddot\theta =\mu B \sin\theta \approx \mu B \theta## (small angle approx) - that's SHM all right.

If they are in a damping medium, they'll end up aligned to the magnetic field.
There are two stable orientations: can you see what they are?

Note: a pair of classical magnets close to each other in the solenoid will try to align so opposite poles are close to each other. This is the low energy configuration. But there is a configuration where they are both exactly aligned. In the classical world this is unstable.

A box of H atoms instead of S-G magnet?
H is very light so there would simply be multiple scattering of the beam particles and the H atoms.
The thing about the magnet in the SG experiment is that it is big and heavy and the dipoles it is composed of are not easily knocked out of alignment by other dipoles passing through. When you do a scientific experiment you have to control variables like that.

Last edited: Apr 2, 2014
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19. ### bobie

689
There are 3 magnetic momenta in 1H, why does QM take into account only one, i.e. $\mu$e?

20. ### bobie

689
Does this mean that in one of the two beams $\mu$e is antiparallel to the magnetic moment of the proton?