# Rotating electron as a dipole is this right?

1. May 21, 2013

### mesa

An electron as shown by the Stern Gerlach experiment behaves like a dipole (albeit only in one of two states). I have been trying to figure out how this is so and drew up the following sketches. A few assumptions were made about electrons such as 'distribution' of charge assuming static equilibrium resulting in a shell of charge. Charge distribution could also be spread out evenly or by some gradient from the center however the general result would still be the same.

Figure one shows the F.o.R. with our shell rotating about the x axis in a clockwise direction when viewed from the positive x towards the negative. This rotation is the same for all three figures. Figure 2 is a cutaway about the y/z plane and a slice to it's right with direction of rotation shown again. So essentially the rotation of our charge distribution of some radius will act like a dipole just like a current loop with this case having a north pole facing the positive x direction and the south the negative x for all 'slices'.

The issue here is that an electrons dipole could point in any direction depending on its rotation and the magnitude of the Bf of each dipole would similarly be dependent on the speed of that rotation for whatever F.o.R. is used. As I understand it the Stern Gerlach experiments show otherwise as the electron can have only on of two possible directions of the dipole and of equal magnitude. Is this correct?

2. May 21, 2013

### Meir Achuz

The electron's magnetic dipole moment does not result from rotation like in your pictures.
It is a strictly relativistic quantum effect, following from the Dirac equation.

3. May 21, 2013

### mesa

I am sure you are correct, but we are in Classical Physics section not QM and these sketches are just a thought experiment about induced dipoles of electrons using a classical approach. As stated in the posting it does not work as compared to observation but that doesn't make it any less fun to tinker with.

Either way, thank you for the information.

4. May 21, 2013

### WannabeNewton

It looks like you are describing the dipole moment of a charged rotating spherical shell (whose cross sections behave like circular current loops) using classical EM hence not the dipole moment of an electron so what is the point of this? I'm having trouble seeing what your overall question is. The dipole moment of an electron is a result of its spin - it isn't rotating in the classical sense such as in your drawings.

5. May 21, 2013

### mesa

That is correct. I used a shell instead of an even distribution of charge because charge distributions in static equilibrium on a conductive solid sphere behave this way.

Use a classical mechanics/EM view point to 'build' an electron that makes sense and see if it has a dipole. No surprise it doesn't match what has been observed with the Stern Gerlach experiments but it is still good to play with. As I understand it QM didn't become common place until classical physics resulted in only dead ends. Why not follow the same path, it's interesting and will make my comprehension of classical EM stronger while giving a better appreciation for QM (I understand it gets really hairy).

As I understand it you are correct. The 'spin' is really represented by angular momentum but there is no actual rotation of a 'particle'.

6. May 21, 2013

### Staff: Mentor

7. May 21, 2013

### mesa

Without even getting into the math you can see how it doesn't match experimentation, but interesting none the less. Thanks for the link, although I will try to solve it on my own first it will be handy for a check.

8. May 22, 2013

### Meir Achuz

But there is correct Classical Physics and WRONG Classical Physics.
It is more 'fun' to learn physics and then tinker with it.

9. May 22, 2013

### mesa

I didn't realize I made such grievous errors. Since you believe you have a better understanding how would you tackle the problem from a classical approach?

10. May 22, 2013

### Meir Achuz

Some quantum physics has no reasonable 'classical approach'.
Trying to do so is what leads to most of the QM 'paradoxes'.
A better understanding has to wait for an understanding of QM.

11. May 22, 2013

### mesa

I understand that, you must have missed my post:

The point being a classical approach was done first, it failed which is why we have QM but this I believe is still a good exercise and should make the prospect of learning QM even more enticing.

So presuming we live in a world where QM does not yet exist where are my errors from a strictly classical approach? I thought the drawings were pretty straight forward but maybe not? Or perhaps you have something to add or believe it should be drawn out differently?

12. May 23, 2013

### Meir Achuz

We are still in a state of mutual confusion.
The project you state, the magnetic moment of a rotating charge distribution, is given as a homework problem in EM texts. It is more straightforward than your pictures imply. What are you adding to an old student exercise?

13. May 24, 2013

### mesa

That's alright, we are both sticking around trying to figure out what the point of the other person is. So as I now see it, you believe this topic is covered in our textbooks and they give an even simpler description of how a rotating electron will have a dipole. I wouldn't think that is possible so this is great, let's see it!

14. May 24, 2013

### mikeph

You're essentially describing a curiously-shaped solenoid, which can be solved to find a magnetic field according to electrostatics.

I don't personally see the link to QM, even in classical physics I don't know why an electron would be modelled as a spherical conductor. Spin is just one of those things that has no classical analogue.

I would happily be proven wrong, but I can't see a connection between this model and an electron of quantum physics.

15. May 24, 2013

### WannabeNewton

The term rotating electron has no classical meaning in the sense you are thinking of as has been stated, and as you yourself have recognized. However for a rotating uniformly charged spherical shell of radius $R$, calculating the magnetic dipole moment is a common textbook problem (e.g. Griffiths) and is quite easy to do. Reorienting the coordinate system so that the rotation axis coincides with the $z$-axis, you can break up the shell into infinitesimal circular current loops each of which will have magnetic moment $d\mathbf{m} = \mathbf{A}dI$ with $dI = (\sigma v)( R\sin\theta d\theta) = \sigma \omega R^{2}\sin\theta d\theta$ and $\mathbf{A} = \pi(R\sin\theta)^{2}\hat{z}$. This gives $\mathbf{m} = \pi R^{4} \omega \sigma \hat{z}\int _{0}^{\pi}\sin^{3}\theta d\theta = \frac{4}{3}\pi R^{4} \omega \sigma \hat{z}$.

The reason for why an electron has a magnetic dipole moment is given by QM; it is not something you would calculate using classical mechanics.

16. May 24, 2013

### mesa

We are in agreement, there is no 'link'. This exercise along with others are about taking a classical approach to quantum objects and treating them as reasonably as possible to particles using systems analogous to what is observed in the macro. We know that classical physics fails and QM is the only viable description for these systems, by exploring the limits of classical physics it will help improve my understanding of it while making the prospect of studying QM even more alluring.

Jtbell had posted a very similar solution, they are both quite clever.

This is understood. The Stern Gerlach experiment was the 'line in the sand' and validation for QM as opposed to a classical approach for the quantum. It is fascinating the behavior of electrons, and atoms in this device. Playing with classical mechanics in the quantum is fun but these exercises are to 'see' the limitations of classical physics just as the scientists of 85 years ago had.

17. May 24, 2013

### WannabeNewton

Sure, you are free to do whatever you wish with classical mechanics as long as you see the limitations, which you personally do, so I don't see a problem. Good luck and cheers, till next time!

18. May 24, 2013

### Spinnor

19. May 25, 2013

### mesa

"...a fatal flaw of this picture that the speed of rotation... was in excess of the speed of light.", that sounds like a pretty big limitation of classical mechanics :)

Apparently they used a rotating solid body as a representative charge distribution for the electron, which from a classical point of view makes sense. This is a very interesting result, must have been quite a shock to the scientific community at the time.

As for the remainder of the paper, it will take some time to acquire the skills for this new physics...