Does electron moving in a circular motion inside a magnetic field experience acceleration? [closed]

Question

I always thought that object turning means acceleration but what if an electron is caught inside a magnetic field that simply changes it path into a circular one so no electric field is allowed. The electron would be doing circle at the same speed so it seems contradictory, how can I resolve this problem?

Answer 1

$\vec{v}(t)=V \left( \vec{e}_y \cos \omega t - \vec{e}_x \sin \omega t \right)$ with $|\vec{v}(t)| = V= \rm const.$, but $\dot{\vec v}(t)=-V \omega \left(\vec{e}_y \sin \omega t +\vec{e}_x \cos \omega t \right) \ne \vec{0}$ solves your puzzle. Velocity (the vector $\vec{v}$) $\ne$ speed (its magnitude $|\vec{v}|$).

Answer 2

In a constant magnetic field, velocity and canonical momentum are different entities:

$$H = \frac{1}{2m} (p - e A)^2 = \frac{m}{2}v^2$$

by $v = \nabla_p H$

The vector potential can be e.g. in the circular gauge $$\vec B=(0,0,B): \ A = \frac{1}{2} \ B \times x = \frac{B}{2}\ (- y, x, 0)$$

eg.in Mathematica code

    Curl[1/2 Cross[{0, 0, B}, {x, y, z}], {x, y, z}] =={0,0,B}

or more simple

$$A = B\ (0, x, 0),$$ the linear gauge.

In first case the 2d- reduced Hamiltonian is

$$ H = \frac{1}{2m}\left( (p_x - \frac{1}{2}\ e\ B\ y)^2+ (p_y +\frac{1}{2}\ e \ B x)^2\right)$$

yielding the momentum change equation

$$\dot p =-\nabla_{x,y} H = \frac{e B}{m} \ \left (v_y, \ - v_x \right)$$ Replacing $$\dot p =\frac{d}{dt} ( m v + e A(x))$$ yields, in cartesian coordinates,

$$\frac{d}{dt} \left( m \ (v_x, v_y) + \frac{1}{2} \ e \ \ B (-x,y) \right) = \frac{e B}{m} (v_y,-v_x)$$ reducing to

$$(\ddot x, \ddot y ) = \omega (-\dot y, \dot x)$$

In the coordinate-velocity picture, the acceleration seems to be pure gauge, like the Coriolis force, linear in and orthogonal to the velocity, kinetic energy $H$ is a constant.

In quantum theory the situation is even more open to interpretation. Using the second form

$$H= \frac{1}{2m} (p_x )^2 + (p_y+ m \omega x)^2 $$ we observe that $p_y$ is a constant of motion, that can be fixed by a plane wave $e^{i k y}$ and the effective Hamiltonian is $$H= \frac{1}{2m}p_x^2 + \frac{m\omega^2}{2} (x-x_0)^2$$ that is, a linar oscillator in x-direction and a plane wave in y-direction, the center of the oscillator determined by the wave vector $k_y$.

This picture is used in solid state physics and makes the spectacular appearance in the quantum Hall effect, when current in direction $p_y$ inevitably is coupled rigidly to a transversal electric field $E_x$ by a Lorentz boost and the magnetic field is so strong that all electrons are in the ground state of the y-oscillator, but must occupy different states of the plane wave and the midpoint coordinate of the oscillator, following the Pauli principle.