Does this vector equation have a solution

Question

I am working with a modified version of the Lorentz force equation wherein I am trying to find generalized E and B field components that produce a specific particle path. I have replaced E with $$\vec{E}=\frac{\vec{v} ^{2}}{2\vec{r}}$$ so my only unknowns are the x,y,z components of the B field.

I want to solve the following equation now for Bx, By, and Bz where a,v, r (and E) are known. $$ \vec{a} = \vec{E}+\vec{v}\times \vec{B} $$

I am not sure if the following matrix can be used to solve this equation, but, if it is, it doesn't seem to have a solution. (due to the 0 determinant). I figured that there would be a solution because there are three unknowns (Bx, By, Bz) and three equations, but I am trouble proving this.

$$ \begin{pmatrix} 0& -v_{z} &v_{y} \\ v_{z}&0 &-v_{x} \\ -v_{y}& v_{x}& 0 \end{pmatrix}\begin{pmatrix} B_{x}\\ B_{y}\\ B_{z} \end{pmatrix} =\begin{pmatrix} a_{x}-E_{x}\\a_{y}-E_{y} \\ a_{z}-E_{z} \end{pmatrix} $$

So mainly I want to know if the matrix I have set up represents the equation above it and if a solution for the B components can be achieved.

Answer 1

The cross product $\vec{v}\times$ is indeed not invertible, due to the fact that it has a non-trivial nullspace (all vectors parallel to $\vec{v}$).

There are thus two possibilities:

• $\vec{a}-\vec{E}$ is perpendicular to $\vec{v}$: a solution exists where the components of $\vec{B}$ perpendicular to $\vec{v}$ are uniquely specified but the component of $\vec{B}$ parallel to $\vec{v}$ is arbitrary.
• $\vec{a}-\vec{E}$ is not perpendicular to $\vec{v}$: your equation is a contradiction, no solution exists.