Question 6.5: Consider a relativistic charged particle of rest mass mo mov...

Solution of (a). To determine the trajectory of a particle of charge q moving in a magnetic field of constant strength B, we recall Newton’s law in (5.34) and consider the relation

\mathbf{F}(P, t)=\frac{d \mathbf{p}(P, t)}{d t}=\frac{d}{d t}[m(P) \mathbf{v}(P, t)]                              (5.34)

\frac{d}{d t}(\mathbf{p} \cdot \mathbf{B})=\frac{d \mathbf{p}}{d t} \cdot \mathbf{B}=\mathbf{F}_m \cdot \mathbf{B}=0,                  (6.17a)

wherein (6.16) is the total force on q. Therefore , the component of the  momentum in the direction of B is constant:

\mathbf{F}_m=q \mathbf{v} \times \mathbf{B}                     (6.16)

\mathbf{p} \cdot \mathbf{B}=m \mathbf{v} \cdot \mathbf{B}=C \text {, a constant. }                (6.17b)

Since the magnitudes of p and B are constant, (6.l7b) implies that the angle between the fixed axis of B and the tangent to the space curve along which q  moves is constant everywhere along the path. Consequently, as described in  Example 1.14, the path is a circular helix, a space curve of constant curvature ; therefore, \left|\mathbf{F}_m\right|=q v_0 B \sin \langle\mathbf{v}, \mathbf{B}\rangle   is constant. Conversely, it follows from (6.16) that if  \mathbf{F}_m  has a constant magnitude, \sin \langle\mathbf{v}, \mathbf{B}\rangle  is constant and hence the path is a circular helix .

The initial velocity \mathbf{v}_0  may be considered arbitrary. If the velocity is initially perpendicular to B, then, by (6.17b), \mathbf{p} \cdot \mathbf{B}=0  always, and the path is a circle in the plane perpendicular to B. If the initial velocity  \mathbf{v}_0  is parallel to B, the  constant force \mathbf{F}_m=\mathbf{0} ;  the motion is uniform and the path is a straight line along the axis of B. The circle and the line are degenerate kinds of helices . In summary, the trajectory of a charged particle which is given an arbitrary initial velocity in a constant magnetic field is a circular helix .

Solution of (b). The path of a charge q in a plane motion perpendicular to the constant vector B is a circle. To describe this circle, we apply Newton’s law in (6.16) to write  d \mathbf{p} / d t=d(q \mathbf{x} \times \mathbf{B}) / d t.  Integration yields  \mathbf{p}-\mathbf{q x} \times \mathbf{B}=\mathbf{A},  a constant vector. Let B = Bk and consider a plane motion perpendicular to B, so that \mathbf{x}=x \mathbf{i}+y \mathbf{j} . Then  \mathbf{p}=\left(A_1+q B y\right) \mathbf{i}+\left(A_2-q B x\right) \mathbf{j},  and with \mathbf{p} \cdot \mathbf{p}=|\mathbf{p}|^2=m^2 v_0^2,  a constant, this yields the equation of a circular orbit of radius  R \equiv m v_0 / q B:

\left(x-\frac{A_2}{q B}\right)^2  +  \left(y  +  \frac{A_1}{q B}\right)^2=R^2 .                 (6.l7c)

We thus find with (6.9) that a charged relativistic particle in a uniform magnetic field moves on a circular orbit with angular speed  \omega=v_0 / R=q B / m=\left(q B / m_0\right) \sqrt{1-\beta^2}.   This is known as the circular cyclotron frequency.

m=\gamma m_0=\frac{m_0}{\sqrt{1  –  \beta^2}} \quad \text { with } \quad \beta \equiv \frac{\dot{s}}{c} \text {. }                      (6.9)