Question

Three vector potential for uniform magnetic field in x direction lagrangian

Answer 1

In three dimensions with cartesian Coordinates, this can be written as d dt. (. ∇vL. ) − ∇L = 0. (3). Here, ∇v ... The magnetic field B can be derived from a vector potential A: B = ∇ × A. (7). If we plug this into Eq charged particles in the presence of electric and magnetic fields known as functions of position and time. Thus, the .... But, since ∇ · B = 0, we can define a magnetic vector potential A by B = ∇ × A and write (1.5.2)

Answer 2

The vector potential arising from the nuclear spin IK of nucleus K at position RK is normally given as

(200)AK(ri)=(

μ0

4π

)

ˆ

M

K×riK

r

3

iK

;   riK=ri−RK

where MK = ΓKIK is the corresponding magnetic dipole moment and ΓK is the magnetogyric ratio of nucleus K. To understand a bit the origin of this expression and inherent approximations we may assume that the current density generated by the nuclear spin has the form

(201)jk = ∇ × {ψ

†

K

 

ˆ

M

KψK}

in analogy with the spin contribution (131) to the non-relativistic current density of the electron. In the above expression ψK is the (unknown) nuclear wave function and

ˆ

M

K is the magnetic dipole moment operator. The current density is purely transversal and so we may use (96) to find the corresponding vector potential in Coulomb gauge. The general solution in the static case is

Ak(ri)=(

μ0

4π

)∫

jk(R)

|ri−R|

dτR=(

μ0

4π

)∫

{ψ

†

k

ˆ

M

kψk}×(ri−R)

|ri−R|3

dτR

where the last term has been obtained through integration by parts. If we assume a point-like magnetic dipole the magnetization density becomes