How Does Angular Momentum Relate to Magnetic Moments in Physics?

[bobie]

"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?
(http://upload.wikimedia.org/wikiped...etic_moment.svg/200px-Magnetic_moment.svg.png)
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

 

[Simon Bridge]

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.

 

[bobie]

The direction of L has a physical effect yes.

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?

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

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?

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

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

 

[Simon Bridge]

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 p*r indicate, if not the force we must apply to rotate the wheel?

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.

What is your education level?

4) if we consider an electron in 1H, the vector points down? so the north pole is down?
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?

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.

Does this experiment demonstrate angular or magnetic moment or both? does it prove that the spin is constantly flipping from one side to the other?

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.

 

[bobie]

It demonstrates both -
where you have a magnetic moment you have angular momentum,.

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?

 

[bobie]

...spins do not constantly "flip from one side to the other" so no experiment will demonstrate that.

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?

 

[Simon Bridge]

I have read also in this forum...

Please provide a link to where you have read this.

...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.

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.

 

[bobie]

Subsequent measurements will get the same value.

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)?

 

[Simon Bridge]

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?

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

how does this [Stern Gerlach] experiment prove the existence of an angular momentum L(e)?

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.

 

[bobie]

...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?
...You appear to be trying to understand quantum phenomena in terms of classical mechanics.

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.
Try to follow me:
As I said, QM says that there is no real spinning and that L_e 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 L_e (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 L_e and a$\mu$_e, when we apply a torqueand try to rotate the pseudovector(s), do the two values sum up or multiply?

Thanks, Simon, for your help

 

[Simon Bridge]

there seems to be no evidence of Le (or is there? )

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

I ask, if there were both a Le and aμe, when we apply a torqueand try to rotate the pseudovector(s), do the two values sum up or multiply?

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.

 

[bobie]

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

What is you education level so I can pitch my replies appropriately?

My physics level is A-level.

(http://journals.aps.org/prl/abstract...Lett.92.190801 ,) I couldn't access this article without a subscription, (I'll try google later), but
- how do you tell if a phenomenon is due to to L_e and not to L_o)?.

- 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?

 

[Simon Bridge]

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.

- 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?

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 classical physics do the momenta sum up?

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?

I couldn't access this article without a subscription, (I'll try google later),

... 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.

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

... 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.

 

[jtbell]

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

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.

 

[bobie]

In 2D yes. .

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?

 

[Simon Bridge]

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.

 

[bobie]

... 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.

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?

 

[Simon Bridge]

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.

And if we have 1H atoms here, instead of a Stern-Gerlach machine, what happens?

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.

 

[bobie]

...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".

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

 

[bobie]

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

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

 

[Simon Bridge]

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

This is not correct.

There is only one angular momentum direction - this can be resolved into three components just like any vector. QM takes account of all of the components. The focus is on the z-component because that is the one that matters. (i.e. we call the component that matters "the z-component".)

Te link I gave you shows the relationship between the components and the total angular momentum.

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

The wikipedia link was about a forbidden transition in hydrogen, not about beams of H atoms.
You asserted that electrons can only have one spin configuration wrt the proton in a hydrogen atom. This is not correct.

This means that it is possible for an H atom to have a net spin.
It would anyway though since it also has orbital angular momentum from the spatial distribution of the electron and not just spin angular momentum intrinsic to the electron and proton themselves.

If a particle has a net magnetic moment, then it will exhibit a stern gerlach effect - splitting into two beams in the apparatus. In one beam the total magnetic moment points "up" and in the other the total magnetic moment points "down".

This should answer your question.

I'll need to revisit my previous comments about particles in a well... hang on...

 

[Simon Bridge]

Remember how I was saying that there is no work in flipping a spin-state for a single particle confined to a potential well since there is no difference in energy?

In the hydrogen atom, the proton and the electron have a magnetic dipole - so there is a spin-sensituve component to the potential well produced by the proton. The effect is that the proton measures the electron spin in a manner similar to the stern-gerlach apparatus. The electron spin is aligned or anti-aligned with the proton spin.

Thus the spin-states have a different amount of energy, so it does take work to flip the spins over.
electron spin-down has the lower energy so to go from down to up costs energy and to go from up to down gains energy. We could do this, for instance, by shooting a photon at it.

But satisfying conservation of energy and angular momentum is not good enough - the symmetry has to work too... so the lowest energy transition is forbidden. This just means you have to use a higher energy transition to flip the spin over.

For the particle in a box example, it does not have a magnetic moment - so it is uncharged ... so how it gets confined to a box in the first place is something of a mystery. Whatever - the symmetry condition prevents zero-work spin flips ... to flip the spin it has to change energy levels. If it starts out in the lowest energy level, then that means we have to do work - but it my start out in a higher energy level. In which case the spin-flip does work on whatever triggers it ...

Isn't this fun?

 

[Simon Bridge]

I have a couple of examples for you:

Classically:
Say you have a classical ideal rigid pendulum and you bolt a classically idea flywheel to it - the wheel is spinning and we bolt it in place so that it's angular momentum vector always points along the length of the pendulum.
What does that do the the motion of the pendulum?

Quantum:
Same pendulum, but instead of a flywheel we have a box of particles prepared in the same spin state - the total angular momentum points in the same direction as the flywheel previous.
Now what?

 

[bobie]

Thanks Simon,
but in order to fully understand your explanation I need to clarify a couple of concepts in classical theory:

Suppose we have a stone (m=1) tied to a string (r= 4) flying at v = 5, it has momentum p=5.
If the other end of the string is tiet to a pivot, it has L = 20, and L points along the pivot and not tangentially, right?, but
does m retain p=5? , if it hits another (tangent) stone (m₂=1) , m₂ gets p = 5, right?
when they say that L is conserved
"
The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of mass of her body closer to the axis she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase.
"
what is conserved here is not the vector along the pivot, but tangential (m)v , same speed occurring on a shorter circumference makes angular velocity ω increase, that seems all,
where is the role of L , the vector along the axis of rotation.?

 

[Simon Bridge]

Thanks Simon,
but in order to fully understand your explanation I need to clarify a couple of concepts in classical theory:

Suppose we have a stone (m=1) tied to a string (r= 4) flying at v = 5, it has momentum p=5.

That is not correct - you have written <a vector> = <a scalar>.

If the other end of the string is tiet to a pivot, it has L = 20, and L points along the pivot and not tangentially, right?,

correct.

but
does m retain p=5? ,

don't forget that the direction component is constantly changing.

if it hits another (tangent) stone (m₂=1) , m₂ gets p = 5, right?

Only if stone #1 is completely stopped by the collision.

when they say that L is conserved
"
The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of mass of her body closer to the axis she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase.
"
what is conserved here is not the vector along the pivot, but tangential (m)v , same speed occurring on a shorter circumference makes angular velocity ω increase, that seems all,
where is the role of L , the vector along the axis of rotation.?

It's all the same - if the rotation stops then the angular momentum from stone #1 is transferred to stone #2 at the instant of the collision. But stone #2 is not acted on by an unbalanced force - so it flies off in a straight line re Newton's 1st law.

What happens to rxp? Hang on for some maths!

Initially traveling in a loop with linear momentum magnitude p and radius R:

$\vec L_i = Rp\hat k$

During this motion: $\vec r = (R\cos\omega t, R\sin\omega t, 0)$ and $\vec p = (-p\sin\omega t, p\cos\omega t, 0)$

At the instant t=nT (after n complete revolutions) $\vec r = (R,0,0),\; \vec p = p\hat \jmath$ it is let go and flies off at a tangent to the circle - traveling in a straight line at constant velocity v=p/m .

$\vec r = (R,pt\!/\!m,0),\; \vec p = (0,p,0)$

Notice that now the position vector r is now a function of time?
We can work out the angular momentum as a function of time from it's definition.

$$\vec L_f = \vec r\times \vec p
= \left| \begin{matrix} \hat \imath & \hat \jmath & \hat k\\
R & \frac{pt}{m} & 0 \\
0 & p & 0
\end{matrix}\right| = Rp\hat k = L_i$$... i.e. L is conserved.
The determinant in the middle is the 3D definition of the cross product - something you may not have encountered yet.

You can try it in terms of L=r.p.sinθ remembering that r and θ change with time.

 

[bobie]

"The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of mass of her body closer to the axis she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase."
If we want to refer this to the rotating stone , we should consider not a collision but a shortening of the string

It's all the same - if the rotation stops then the angular momentum from stone #1 is transferred to stone #2 at the instant of the collision. But stone #2 is not acted on by an unbalanced force - so it flies off in a straight line re Newton's 1st law..

"Suppose we have a stone (m=1) tied to a string (r= 4) flying at v = 5, it has momentum p=5.
If the other end of the string is tiet to a pivot, it has L = 20, and L points along the pivot and not tangentially, right?, ..."
if we shorten the string by a half (r=2) we have conservation of mass (=1) of KE (12.5) , of speed v (=5) and consequently of (mv) p (=5), angular velocity doubles and, although there is no torque applied, angular momentum is not conserved but becomes (m=1*v=5*r=2) 10.
Isn't it so?

 

[Simon Bridge]

if we shorten the string by a half (r=2) we have conservation of mass (=1) of KE, of speed v (=5) and consequently of (mv) p (=5), angular velocity doubles and, although there is no torque applied, angular momentum is not conserved but becomes (m=1*v=5*r=2) 10.
Isn't this so?

... therefore - this is not what happens.

For the case that a mass twirling on a spring experiences a shortening string, with no applied torque, then the speed v, and thus the magnitude of the linear momentum p, increases. Linear momentum is not conserved in this system because it is being acted on by an unbalanced force: the centripetal force. (In fact - to reel in the string like that, the centripetal force must increase...)

You can readily observe this by tying a string to a pole (mass on tother end) and swinging it so that the string wraps around the pole. Observe how the speed increases as the string gets shorter.

If speed stays the same then there must be a torque from somewhere.

Notice, my own example with the math above was the same thing going the other way - in that case the "string" length r was being increased in such a clever way that the linear momentum was a constant (equivalent to cutting the string) so there was no more circular motion.

You could do it the other way around too and get the same result.

 

[bobie]

... therefore - this is not what happens.
For the case that a mass twirling on a spring experiences a shortening string, with no applied torque, then the speed v, and thus the magnitude of the linear momentum p, increases..

So , the angular velocity in this case is the quadruple (v*2/.5)? pulling the string is a torque? it gives KE to the stone (+ 37.5, (v= 5*2=10))? But we are exerting a force perpendicularly to the tangent, so there should be no net work done. It is really amazing, Simon, thanks!

 

[Simon Bridge]

So , the angular velocity in this case is the quadruple (v*2/.5)?

... work it out - what conserves angular momentum?

$L_{before}=mrv\\ L_{after}=m(r/2)v_2\\ L_{after}=L_{before}\implies v_2=2v\\ \omega_1 = v/r,; \omega_2=4v/r = 4\omega_1$
... so yeah: the angular velocity quadruples if the radius halves.

You can also see this from $L=I\omega = mr^2\omega$ ... for a stone on a string radius r.
You halve the radius you quarter the moment of inertia and you quadruple the angular velocity.

pulling the string is a torque?

Normally (in normal circular motion) there is no torque pulling on the string: pulling the string is where the centripetal force comes from - but to reel the string in what has to happen?

it gives KE to the stone (+ 37.5, (v= 5*2=10))? But we are exerting a force perpendicularly to the tangent, so there should be no net work done. It is really amazing, Simon, thanks!

Work is also the change in energy. That energy still had to come from someplace.
You have to look more closely at how reeling in the string increases that energy.
Another way to shorten the radius is to put a peg in the way - the string wraps around the peg: this changes the position of the pivot.

 

[bobie]

Work is also the change in energy. That energy still had to come from someplace.
You have to look more closely at how reeling in the string increases that energy.
.

But if we just do the reverse, if we let loose the string by 2, v should decrease by a half (=2.5), right?, where does that Ke (-9.375) go?

 

[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