In spin orbit interaction, a magnetic field (internal ,I guess)comes into picture. So how this magnetic field is produced and by who - electron or nucleus?

In my book, it says electric field is produced by nucleus . Due to this electric field, moving electron experiences magnetic field. I am not able to understand this? We have studied till now moving charge produce electric field.

I don't know about special relativity but I read magnetic field is combination of electrostatics and special relativity. I don't know if it can answer my question or not?

Help me to understand.


2 Answers 2


So let me suppose that I measure no magnetic field, but I measure an electric field in the $x$-direction, and I measure you cruising past me at speed $v$ in the $y$-direction.

As you measure your fields, you will see two discrepancies with my measurements:

  1. First, your electric field is a little stronger. It's hard to notice this because it does some interesting things to not violate conservation of energy; in particular the same strength increase will also be experienced by you as a time decrease in the amount of time that the electric field is acting upon you... also because the strengthening at low velocities is by the factor $1 + (v/c)^2/2,$ so even if you're going 100mph (= ~160 kph) you will only notice this if you can measure forces to accuracies of one part in 45 trillion (on the short scale, so a trillion is a million million, no "milliards").

  2. Second, you will measure a magnetic field in the $z$ direction, and it will be approximately equal (barring the same sorts of high-accuracy equipment measured above) to the $v$ I measured times the $E$ I measured, divided by the speed of light squared. This is very weak for basically the same reason; $c$ is a big number and therefore $c^2$ is a big number. If you were traveling 100 mph past a $E$-field that was nearly strong enough to cause sparks through air (roughly 3 million volts per meter) then you would still only measure something like a billionth of a tesla. For comparison the magnetic field of the earth is about 30 millionths of a tesla, and there's a lab out there which specializes in high magnetic fields that can go all the way up to 100 tesla, with most other applications somewhere in the middle.

In other words, the "electric field" and "magnetic field" are not observer-independent as separate entities; if two observers are moving relative to each other then if one person says "the magnetic field did this, it's a magnetic effect," someone else might say "no, that's an electrical effect" and they might both be right in their respective coordinates. For this reason those two vector fields need to be combined into an "electromagnetic field tensor" which is observer-independent according to the coordinate transformations of special relativity.

Now if the electron is orbiting around the nucleus in the $\theta$-direction the $r$-directed electric field will appear a little bit like a $\phi$-directed magnetic field to the electron and can tend to interact with its spin. So that's where the spin-orbit couplings come from. If you froze the electron in a moment in time the proton would be moving relative to it and would therefore have a magnetic field circling around its direction of motion; this magnetic field is precisely the one that we're talking about the spin interacting with.

  • $\begingroup$ Thanks for helping. In my book, formula for magnetic field is written as (E*v/c) and not c squared. $\endgroup$
    – Ka Sikh
    Commented May 18, 2017 at 18:16
  • $\begingroup$ What would be the case if the moving particle is charged like electron? It has its own electric field. $\endgroup$
    – Ka Sikh
    Commented May 18, 2017 at 18:25
  • $\begingroup$ (1) This is one of those things which depends on how your units measure charge; what I wrote is true for the SI units where the Lorentz force is $F=q(E+v\times B)$; in normalized units $E$ and $B$ have the same dimension with the Lorentz force taking the form $F=q(E+\frac vc\times B)$ in which case you'll see $B=v\times E/c$ without the extra factor of $c$ in the denominator. (2) In principle electrons have a similar spin-orbit effect on the nuclear spin, but since this happens in the nucleus you'd have to observe that fine-structure in something like MRI experiments or so. $\endgroup$
    – CR Drost
    Commented May 18, 2017 at 19:19

A static charge produces a purely electric field. A moving charge produces a magnetic field as well.

Concerning Special Relativity: If you apply a Lorentz transformation to the field strength tensor of a static charge distribution (which will only have non-zero electric field components), you will see that in a moving frame also magnetic field components become non-zero, so a magnetic field appears if you move with respect to the charge distribution.

  • $\begingroup$ Could you explain in simple way and concerning my problem of spin-orbit interaction. $\endgroup$
    – Ka Sikh
    Commented May 17, 2017 at 20:34
  • $\begingroup$ Here is the calculation I was referring to: physicspages.com/2013/03/26/… You see, if in the equations (10)-(15) the components of $\vec B$ are zero, the components of $\vec B'$ are non-zero, if $\vec v$ is non-zero. A purely electric field leads to a magnetic field in a moving frame. This is exactly what happens in the atom: The nucleus generates a static electric field in the lab frame, but since the electron is moving w.r.t. the nucleus it also feels a magnetic field in its moving frame. $\endgroup$
    – Photon
    Commented May 18, 2017 at 6:52

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