Electron Spin Magnetic Dipole Moment

[@Bl4ckoctober]

Thank you! All the best from Romania! You should never stop!

[@AKLECTURES]

Bogdan Ghiorghiu Thank you! I appreciate that :)

[@rizqaminmahmudkhan4159]

Loved the video especially the manner of your speaking which is strangely soothing and helped me understand the concept better

[@jerryiuliano871]

A=electron magnetic moment. B=muon magnetic moment. C=tau magnetic moment. (erg/A/B/C/pi)+pi=e^pi

[@4pharaoh]

The Electron Spin as you have described it doesn’t seem to indicate an intrinsic property of the electron, but instead is presented as a product of the orientation of the electron as it just happens to enter the Stern- Gerlach device. It would seem, based on your explanation and the diagrams you drew, that the electrons could have just as likely re-orientated themselves so that it’s north magnetic point was oriented to the south pole of the S-G machine and thus all of the electrons would be pulled in the direction of the stronger of the magnetic fields. Now I know this is incorrect. So what am I missing?

[@admiral_xanadu]

I see what you mean. From my understanding, since the electrons were initially prepared at ground state, this means that the angular momentum number of all the electrons is zero. So from what we know, they have no angular momentum, and thus no magnetic dipole moment which could account for the splitting, when an external magnetic field is present. So what the scientists would expect to see, was a single trajectory of the electrons, without any form of splitting. However, there is a splitting because the electrons do have a magnetic dipole moment due to the intrinsic spin property which CANNOT be removed even if the electron is at ground state (although it can be prepared in one of the two states). This spin creates the same effect as quantized angular momentum, but establishing a magnetic dipole moment on the electron, however it should not be thought of as the electron rotating about a certain axis since electrons are considered to be a point mass. Just remember that spin is an intrinsic quantity, just like mass or charge. Hope that helps.

[@mdmobashshir432]

Thanks man. You are helping a lot of kids like us who can't AFFORD high quality universities. Your every video is very useful. I'm referring you to every student I meet. One request, I'd like a video that explains how detailed mathematical analysis of viscosity and thermal conductivity is carried out. I have read the proofs a lot of time, but it just didn't connect with me.

[@inox1ck]

You can do the classical prediction which won't necessarily show a continuous distribution. I' m sure a numerical computation would show two gaussian distributions rather than a continuous distribution. The simulation would depend on the magnets strength, the magnetic field strength, the velocity of magnets, shooting angle. In a strong field the magnets immediately orient towards north. Spinning magnets might act differently though.

[@岡安一壽]

I think that an electron has shift in coordinate (1/2π)msλ in r-direction. [ms: spin quantum number, λ: wave length] Accordingly, the angular momentum is as follows: {r+(1/2π)msλ}p=rp+(1/2π)msλp=rp+ms(h/2π)=L+S. [De Broglie equation: p=h/λ] And the magnetic moment is as follows: (area)･(electric current)＝{πr^2+2πr･(1/2π)msλ}･(q/m)p/(2πr)=(q/2m)pr+2(q/2m)ms(h/2π)= (q/2m)L+2(q/2m)S. Thus, we can know that the g-factor for spin angular momentum is equal to 2. I'm sorry that I'm not good at English.

[@Jarrod_C]

Ok so you are missing an explanation on why the plus 1/2 and negative 1/2 have those particular orientations, why arent they total opposites like 180 degrees opposites. Why are they mearly a reflection of each other. You should have explained why those particular orientiations or are they arbitraly chosen.

[@luccapfitzer3226]

Jarrod C the ms quantum numbers (+ or - 1/2) refer to the z-component of the Spin vector. The total length of the spin vector is given by |s|=(s(s+1))^1/2, which is due to s=1/2 (for electrons or fermions in General) the Square root of 3/4 (in Units of h bar). So the vector of the Spin can‘t be parallel to the z axis. Thats a general property of angular momentum vectors.

[@luccapfitzer3226]

So basically you can only know the z component, which means you can draw them any way you like, you only have to give them a z component of negative or positive 1/2

[@Nmag8217]

I like the video, little bit shouty though

[@adosar7261]

Why magnetic field must be inhomogenous?

[@pong301275cmu]

to make the total force on both magnetic poles not equal Zero. If the magnetic field is uniform, only torque affects on the tiny magnetic dipole moment. Then, the travel direction of atom will be on straight line which result in band spectra, from different orientation of magnetic dipole moments.

[@4pharaoh]

Question on Spin: Assume I have an electron subjected to a very strong magnetic field. For its magnetic dipole moment (μ) orientation to MAGNET (M) orientation I expect would align (μ)-south to (M)-NORTH, and (μ)-north to (M)-SOUTH, due to this electron’s PARTICULAR spin (let's say Spin UP) My question is: With a Spin DOWN electron is the μ to M orientation remain unchanged (μ)-south to (M)-NORTH, and (μ)-north to (M)-SOUTH) but the direction of spin reversed? If so, why do you (and so many others) draw it the spin Up and Sin Down the same way, just pivoted? surely Nature can have that electron spinning both ways, or is something else happening?

[@abhisekhkumar4948]

you could have been a rapper.. attitude is there.. nice video, thanks..

[@marxregis]

Good one sir!

[@peacekeeper9687]

Thank you

[@kiruikalistus5071]

Just wow

[@abzalzhumaseiit580]

Thanks, veru useful. Your explaining is very clear for me.

[@abzalzhumaseiit580]

+Abzal Zhumaseiit very

[@AKLECTURES]

+Abzal Zhumaseiit you're welcome!

[@bharatkarmur6887]

I don't understand your speaking style