How Does Angular Momentum Relate to Magnetic Moments in Physics?

[Simon Bridge]

Good question - you have to look closer.
How would you double the length of the string?

Note that it takes some work to spin-up such a system in the first place.

 

[bobie]

Good question - you have to look closer.
How would you double the length of the string?
Note that it takes some work to spin-up such a system in the first place.

The work done is the Ke of the stone.

Once the stone is spinning, (if) there is no friction and the string is strong, it spins forever at v=5, r= 4.
the string is fastened to the centre, we just let it loose by 4 (or any other length) : it slips easily, no work done.
Now, if L is conserved (=20) , and r (4*2 =8) , p=v=2.5 is halfed ,Ke = 3.125, the stone lost 9.375 energy. right?
Who gained that energy? I cannot figure out.
But v cannot diminish, because if we keep loosing the string all the way , v stays the same.

 

[Simon Bridge]

Good example, let's use it
... but you need to divide the motion in the example into four stages, not two.

I. At the start, you have uniform circular motion at speed v1 of mass m at radius r1.

The force on the mass is purely radial: $$\vec F\!_{c1}=-\frac{mv_1^2}{r_1}\hat r$$
We can also say that $\vec L=mr_1v_1\hat k,\; K_1=\frac{1}{2}mv_1^2,\; p_1=mv_1$
... notice I have only given the magnitude of the linear momentum, it's direction is constantly changing under the unbalanced force $\vec F\!_{c1}$

Lets imagine mass m is on a perfectly flat ideal friction-less table and the string is also ideal and mass-less and passes through a small hole in the table and we can hold on to it from underneath or let it unreel and so on. The mass is in circular motion about the hole OK?

II. the string has been released - now there are no forces on the mass, it continues with the same energy and momentum and angular momentum as the instant that it was released (see post #25), a situation that continues until m reaches radius r₂ > r₁.

III. at the point the new radius is reached, the string delivers a specific impulse to the mass, changing it's momentum. A quick sketch will show you that the impulse is not delivered perpendicularly to the motion - there is actually a component opposite the direction of motion.

You should be able to do some maths here.
See what happened to the energy?

IV. the system again executes uniform circular motion (unlikely in real life) at the new radius.

But we don't really need to do all this - we know that total energy is conserved.
Since kinetic energy vanished, it must have been changed into something else.
The energy can be radiated away (sound or light - but maybe we are doing this classically in a vacuum), get contained internally (i.e. as vibrations or heat - but maybe it's a classical point mass) or transmitted away ... i.e. along the string into whatever is holding the string. Bingo!

Just as well or we'd have to give up an idealization.

 

[Simon Bridge]

Hopefully you can see the kind of careful thinking that is needed here - it is too easy to overlook something.
Now you should relax a bit and let what you've hopefully learned settle in for a bit before asking more questions any time soon.

Your original question has been answered.

Look at "classical mechanics" in these notes:
http://home.comcast.net/~szemengtan/
... I expect they are too advanced for you, but you should get an idea about what you have yet to learn about.

 

[bobie]

Hopefully you can see the kind of careful thinking that is needed here - it is too easy to overlook something.
Now you should relax a bit and let what you've hopefully learned settle in for a bit before asking more questions any time soon.

Thanks, Simon, for this great seminar.
I surely need time digest this stuff, but, just to wrap up angular momentum in 1H, can you tell me if L_p is intrinsic or real?, the proton should be really spinning. I read it has the same value of L_e =\hbar/2
So L_o=\hbar, L_e=L_o=\hbar/2. Correct?

If you are willing and have some time and patience yet to spare, I have some other questions about the magnetic moment
Thanks again for your great help

 

[Simon Bridge]

The proton, like any hadron, is understood in terms of it's constituent quarks - each of which have an intrinsic spin. Protons are not well modeled as classically rotating objects.

Protons are also "Fermions", they have half-integer spin just like electrons.
At your level you will have only had a glimpse of the standard model of particle physics ... which is a huge subject all by itself.

We have to be very careful about what we call "real" in quantum mechanics ... the usual definition is that it is real if you can measure it. Intrinsic spin is real angular momentum in the sense that we can measure it and it gives rise to rotations that we can measure.

 

[bobie]

Intrinsic spin is real angular momentum in the sense that we can measure it ...
Protons are also "Fermions", they have half-integer spin just like electrons.
.

But the proton is really spinning on its internal axis, the electron makes it spin,right?
So, is the total angular momentum of 1H (\hbar+1/2+1/2) =2 \hbar =h/\pi?

 

[Simon Bridge]

No. You should be able to look that stuff up ;)
Note: electrons in an atom have orbital angular momentum as well as spin.

 

[bobie]

electrons in an atom have orbital angular momentum as well as spin.

So, is the total angular momentum of 1H (\hbar+1/2+1/2) =2 \hbar =h/\pi?

I have considered L_orbit = \hbar
According to what you said I considered a
vector at the nucleus : L_o = \hbar + L_p (you said) =1/2 \hbar =3/2 \hbar
Ant the vector of the spin at the electron L_e = 1/2 \hbar
So the total momentum of the ¹H atom should be 3/2+1/2 = 2 \hbar= h/\pi

I have read that the the spin being 1/2 implies the electron behaving like on a Moebius strip, is that true? why so?
if it where ,say, 32 couldn't it run on a Moebius strip?

 

[Simon Bridge]

I have read...

...where?!

...that the the spin being 1/2 implies the electron behaving like on a Moebius strip...

... in what way? "like" is such a vague term.
Sounds like a reference to Dirac's Plate Trick
http://en.wikipedia.org/wiki/Plate_trick

Also see:
http://scienceblogs.com/principles/2010/07/26/electron-spin-for-toddlers/
... since it is still bothering you.

Hydrogen:
http://en.wikipedia.org/wiki/Hydrogen_atom#Angular_momentum
... when you are thinking of the angular momentum of an atom, you have to consider what state it is in.
Usually you want to do this for the ground state.

https://www.physicsforums.com/showthread.php?t=69992

Beware of articles which go out of their way to make QM sound weird and mysterious.
That would include most pop-science and about 99% of the material found on the Discovery Channel.

 

[bobie]

Usually you want to do this for the ground state.
Beware of articles which go out of their way to make QM sound weird and mysterious.
.

Thanks for the precious links, Simon. I'm trying to do my homework.
What I meant is:
1/2 \hbar it's just a numerical value, it could be 2π \hbar or 32 or anything.
Why does this particular value determine the fact that it rotates twice before pointing in the same direction? or I misinterpreted the article? how do we notice that it rotates by 720°?

- I read that in ground state 1s there is no angular momentum, this being one of the main differences from Bohr's model, is this true?
- if so, as there is no pseudovector L(o) and no plane of an orbit xy, how can we detect the spin component on z? if there is no xy where is z?
- I do not think QM is weird, but it seems to have rules that cannot be verified and contradict the rules outside the atom. It's difficult to know where it is just theory.

Thanks.

 

[Simon Bridge]

The value of angular momentum is not arbitrary if that's what you mean.
It is determined empirically and there is a deeper mathematical model that makes sense of it.
Right now you are exploring the results of that model.

The value assigned to mean a particular symmetry is arbitrary though, and there are several schemes.
So you could be asking "why use that particular scheme?"

Consider:
It would be natural and intuitive, for many people, to classify rotational symmetry by the angle A you have to rotate through for the object to map on to itself. It's not nice to use this method in practice (give it a go) because you really want more symmetry to mean a bigger number and you want to avoid fractions if you can help it. But to see how that may work we have to be careful about what we mean when we talk about angles.

The angle is the pointy bit in a corner - the size of the angle would be the inverse of it's sharpness. The less pointy the corner, the bigger the angle.

The size of the angle is most conveniently defined on the unit circle as follows - the length of the circumference of the unit circle that lies inside the angle is called "the size of the angle". But that begs the question of the definition of the unit circle.

A unit circle may be defined with the radius, diameter, or circumference equal to 1. Pick one.
The choice is arbitrary - so long as you are consistent.

Having decided on how to measure angles, we can talk about rotational symmetry.

In general, if the object must be rotated through angle A to map onto itself, then we can define a rotational symmetry value is:
$S=1/A$ (circumference = 1)
$S=\pi/A$ (diameter = 1, unused)
$S=2\pi/A$ (radius =1: this is "radians")
$S=360/A$ (using degrees: 1deg = 1/360th of the unit circumference.)

... this is a useful way to define a symmetry number, because it gets bigger the more symmetry you have and, for all common experience, it will always be bigger than one (no fractions). Notice that S, defined this way, will usually be an integer? (The S value ends up being the same as the number of times the object maps onto itself when you rotate it once. i.e. a pentagon would have S=5.)

In a purely abstract way, though, you can ask yourself what happens for non-integer values of S. i.e. what is S=8/3? That's bigger than 1 and not an integer. What is S were irrational?

But that's just pure maths - there was no reason, before QM experiments, to suppose that non-integer symmetries would have any meaning in Nature.

Physics uses Maths as a language for describing Nature.
The formal language involves going through this process of making definitions and proposing theorems.
Maths is a powerful language capable of describing things that do not exist in Nature too.
We use the principles of empirical inquiry to check.

 

[bobie]

It is determined empirically .

I found a nice link that explains in detail how \mu_e is determined as 1.001156 \mu_B(=h/4π) in a Penning trap:
http://gabrielse.physics.harvard.edu/gabrielse/overviews/ElectronMagneticMoment/ElectronMagneticMoment.html .
Can you give me a link or explain how L_e is concretely determined?
How can you detect and determine mechanical angular momentum at distance, without manipulating a body?

 

[Simon Bridge]

That is pretty much the definitive experiment: how much more "concrete" do you need?

 

[bobie]

That is pretty much the definitive experiment: how much more "concrete" do you need?

Do you mean that spin angular momentum S (L_e) and electron magnetic moment \mu_e are the same thing? if they are,
L_e is = 1* h/2*2π (= \hbar/2), whereas \mu_e is 1.001156 * h/2*2 π, which value is right?

 

[Simon Bridge]

We've been over this before.

 

[bobie]

We've been over this before.

I'm lost, in which post(s)?
Can we determine concretely the magnetic moment of a bigger object?
Suppose we have a coil where current is flowing, we have a formula to find the value of B and \mu, but is it possible to verify the real value of \mu? what apparatus do we need? a magnetic field of any strength would do or we need a critical value? any direction will do? do we need more than a measurement?
Thanks again for your patience, Simon

 

[Simon Bridge]

If you have a charge with an angular momentum you must also have a magnetic moment.
They go together.

The other demonstration for angular momentum is a torque experiment - where a spin-polarized beam is incident on a target, the target starts to rotate.

You seem to be going in circles now.
I think you need to have more time to go through these materials.
Intrinsic angular momentum is an established property in physics.

 

[bobie]

You seem to be going in circles now..

Probably I give this impression to you, but I need a couple of notions, before I can read further, I cannot make a synthesis of what I learned so far.Of course I know that a rotating body has L and it has charge it must have a \mu, too
I'd appreciate if you could answer my previous question (is this the usual method? :http://www.serviciencia.es/not-apli/NAS01-i.pdf)

Suppose we have a coil where current is flowing,... the real value of \mu?

What I do not understand is the following:
suppose we make that coil spin around the axis of \mu, then that coil will have a magnetic moment \mu = k * \mu_B and a mechanichal angular momentum L (=mvr) = j * \hbar, just like an electron in a ¹H atom, am I right so far?

Now, if we measure \mu when the coil is rotating, I assume we get a different value of \mu, surely greater, since the applied field must win the resistance to torque also offered by L.
- is that right? naively I assume that, now, the value of \mu must be k+j, or (for some obscure reason) k*j ?

I'd appreciate very much if you could tell me if anything is wrong there. That would save me a lot of more stupid or circular questions. If my assumptions are right, it will be clear to you the fact that I do not understand why:
- if intrinsic angular momentum of the e on the z-axis is \hbar
- and \mu is 1.001156\hbar
when we put a ¹H atom in a Stern-Gerlach machine or a single electron in a Penning trap we measure in the first case only L and in the second only \mu (ì 1.1156 L)
Why not always both, and why angular momentum in Stern-Gerlach?
Do you follow me, Simon? probably I should start a new thread on this.

Thanks for your patience!