In electromagnetism we often have a perpendicular constant magnetic field causing a charge to move in a circle. My question is, how do we formally solve this differential equation which involves a cross product?
$\frac{\mathrm{d} \mathbf{v}}{\mathrm{d}t} = \frac{qB_0}{m}\mathbf{v}\times{{\hat{\mathbf{k}}}}$
So I think I have made progress by splitting into 3 equations. Using that cartesian unit vectors are linearly independent and don't change with time. $\frac{\mathrm{d} v_x}{\mathrm{d}t} = \frac{qB_0}{m}v_y$ and $\frac{\mathrm{d} v_y}{\mathrm{d}t} = -\frac{qB_0}{m}v_x$ and $\frac{\mathrm{d} v_z}{\mathrm{d}t} = 0$
$v_z = v_z(0)$, which is trivial. But what about the others? Also I can see that the derivative is a linear transformation of the vector $v_x,v_y$ and have found the matrix. What now?
Best Answer
Given that
$\dot v_x = \dfrac{dv_x}{dt} = \dfrac{qB_0}{m}v_y, \tag 1$
and
$\dot v_y = \dfrac{dv_y}{dt} = -\dfrac{qB_0}{m}v_x, \tag 2$
since $q$, $B_0$, and $m$ are constant, we may differentiate each equation with respect to $t$ and so obtain
$\ddot v_x = \dfrac{d^2v_x}{dt^2} = \dfrac{qB_0}{m} \dot v_y, \tag 3$
$\ddot v_y = \dfrac{d^2v_y}{dt^2} = -\dfrac{qB_0}{m} \dot v_x, \tag 4$
the we substitute (2) into (3) and find
$\ddot v_x = \dfrac{d^2v_x}{dt^2} = -\dfrac{q^2B_0^2}{m^2} v_x; \tag 5$
likewise, we may substitute (1) into (4):
$\ddot v_y = \dfrac{d^2v_y}{dt^2} = -\dfrac{q^2B_0^2}{m^2} v_y. \tag 6$
Each of the equations (5)-(6) in fact describes a simple harmonic motion of the form
$\ddot w + k^2 w = 0, \tag 7$
where
$k = \dfrac{qB_0}{m}, \tag 8$
and as such they may be readily solved for $v_x$ and/or $v_y$, giving rise to solutions of the general paradigm
$w(t) = A \cos kt + B \sin kt. \tag 9$
It should be observed that the initial conditions for (5)-(6) at $t = t_0$ may be completely determined by specifying the components $v_x(t_0)$, $v_y(t_0)$, since then the derivatives $\dot v_x(t_0)$ and $\dot v_y(t_0)$ are had via (1)-(2); thus in principle specifying
$\vec v(t_0) = (v_x(t_0), v_y(t_0), v_z(t_0))^T \tag{10}$
is sufficient to determine the evolution of $\vec v(t)$ for all time. This should come as no surpise, since (1)-(2) is in fact a first order system in the two variables $v_x(t)$, $v_y(t)$; we only need the derivatives $\dot v_x(t_0)$, $\dot v_y(t_0)$ when we seek to exploit the decoupled form (5)-(6). Furthermore, (1) and (2) enforce the appropriate phase relations on the components of $\vec v(t)$ to ensure that it in fact follows a circular motion; indeed we have
$\dfrac{d}{dt}(v_x^2 + v_y^2) = v_x \dot v_x + v_y \dot v_y = \dfrac{qB_0}{m} (v_x v_y - v_y v_x) = 0, \tag{11}$
which integrates to
$v_x^2(t) + v_y^2(t) = v_x^2(t_0) + v_y^2(t_0), \; \text{ a constant}; \tag{12}$
this shows that $(v_x(t), v_y(t))$ is constrained to lie in a circle of radius $\sqrt{v_x^2(t_0) + v_y^2(t_0)}$, a fact which is of course related to the conservation of kinetic energy.
The above presents the classical approach to this problem from classical mechanics and electromagnetism; there is, however, a somewhat slicker solution available, based on the identification of the vector cross product $\vec A \times \vec B$ with a certain matrix-vector multiplication: with
$\vec A = (A_x, A_y, A_z)^T$ and so forth we may define the matrix
$\vec A_\times = \begin{bmatrix} 0 & -A_z & A_y \\ A_z & 0 & -A_x \\ -A_y & A_x & 0 \end{bmatrix}; \tag{13}$
then
$\vec A \times \vec B = \begin{pmatrix} A_y B_z - A_z B_y \\ A_z B_x - A_x B_z \\ A_x B_y - A_y B_x \end{pmatrix} = \begin{bmatrix} 0 & -A_z & A_y \\ A_z & 0 & -A_x \\ -A_y & A_x & 0 \end{bmatrix} \begin{pmatrix} B_x \\ B_y \\ B_z \end{pmatrix} = \vec A_\times \vec B; \tag{14}$
then the equation
$\dfrac{d \vec v}{dt} = \dfrac{qB_0}{m} \vec v \times \vec k = \dfrac{qB_0}{m} (-\vec k \times \vec v) \tag{15}$
becomes
$\dfrac{d \vec v}{dt} = \dfrac{qB_0}{m}\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \vec v; \tag{16}$
from this form we may immediately write the solution as
$\vec v(t) = \exp \left ( \dfrac{qB_0}{m}\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} (t - t_0) \right ) \vec v(t_0); \tag{17}$
it is easy to evaluate the matrix exponential here if we write the argument in block form
$\dfrac{qB_0}{m}\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} (t - t_0) \equiv \dfrac{qB_0}{m}\begin{bmatrix} J & 0 \\ 0 & 0 \end{bmatrix} (t - t_0), \tag{18}$
where
$J = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}; \; J^2 = -I; \tag{19}$
then
$\exp \left ( \dfrac{qB_0}{m}\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} (t - t_0) \right ) \equiv \exp \left (\dfrac{qB_0}{m}\begin{bmatrix} J & 0 \\ 0 & 0 \end{bmatrix} (t - t_0) \right )$ $= \begin{bmatrix} \exp \left (\dfrac{qB_0}{m}J(t - t_0) \right ) & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \cos \left (\dfrac{qB_0}{m}(t - t_0) \right )I + \sin \left (\dfrac{qB_0}{m}(t - t_0) \right )J & 0 \\ 0 & 0 \end{bmatrix}; \tag{20}$
since the bottom row of the coefficient matrix consists entirely of $0$s, we see via (17) that
$\dot v_z = 0; \tag{21}$
the block on the upper left yields
$\cos \left (\dfrac{qB_0}{m}(t - t_0) \right )I + \sin \left (\dfrac{qB_0}{m}(t - t_0) \right )J = \begin{bmatrix} \cos \left ((\dfrac{qB_0}{m}(t - t_0) \right ) & \sin \left (\dfrac{qB_0}{m}(t - t_0) \right ) \\ -\sin \left (\dfrac{qB_0}{m}(t - t_0) \right ) & \cos \left ((\dfrac{qB_0}{m}(t - t_0) \right ) \end{bmatrix}; \tag{22}$
it follows therefore from (17) that
$\begin{pmatrix} v_x(t) \\ v_y(t) \end{pmatrix} = \begin{bmatrix} \cos \left ((\dfrac{qB_0}{m}(t - t_0) \right ) & \sin \left (\dfrac{qB_0}{m}(t - t_0) \right ) \\ -\sin \left (\dfrac{qB_0}{m}(t - t_0) \right ) & \cos \left ((\dfrac{qB_0}{m}(t - t_0) \right ) \end{bmatrix}\begin{pmatrix} v_x(t_0) \\ v_y(t_0) \end{pmatrix}, \tag{23}$
which gives the complete solution to (1)-(2), as the reader may readily verify by direct differentiation of (23).
Nota Bene: It might be helpful to add a few comments on the establisment of the equation
$\exp \left (\dfrac{qB_0}{m}J(t - t_0) \right ) = \cos \left (\dfrac{qB_0}{m}(t - t_0) \right )I + \sin \left (\dfrac{qB_0}{m}(t - t_0) \right )J, \tag{24}$
which occurs in (20); the key idea here is that, since $J^2 = -I$, expressions of the form $\exp (\alpha J)$ behave in a manner strictly parallel to $e^{i\alpha}$ for scalars $\alpha$, where $i$ is the usual "imaginary" root of $-1$, $i^2 = -1$. When one develops the power series for $\exp(\alpha J)$, (24) emerges in a straighforward way, just as does
$e^{i \alpha} = \cos \alpha + i \sin \alpha \tag{25}$
from the corresponding exponential power series for (complex) scalars. End of Note.