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Lorentz Force Trajectory.

Posted by peeterjoot on September 10, 2011

Solving the Lorentz force equation in the non-relativistic limit.

The problem.

[1] treats the solution of the Lorentz force equation in covariant form. Let’s try this for non-relativistic motion for constant fields, but without making the usual assumptions about perpendicular electric and magnetic fields, or alignment of the axis. Our equation to solve is

\begin{aligned}\frac{d}{dt} \left( \gamma m \mathbf{v} \right) = q \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right),\end{aligned} \hspace{\stretch{1}}(1.1)

so in the non-relativistic limit we want to solve the matrix equation

\begin{aligned}\mathbf{v}' &= \frac{q}{m} \mathbf{E} + \frac{q}{ m c } \Omega \mathbf{v} \\ \Omega &=\begin{bmatrix}0 & B_3 & -B_2 \\ -B_3 & 0 & B_1 \\ B_2 & -B_1 & 0 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.2)

First attempt.

This is very much like the plain old LDE

\begin{aligned}x' = a + b x,\end{aligned} \hspace{\stretch{1}}(1.4)

which we can solve using integrating factors

\begin{aligned}x' - b x &= a \\ e^{b t} \left( x e^{-b t} \right)' &=\end{aligned}

This we can rearrange and integrate to find a solution to the non-homogeneous problem

\begin{aligned}x e^{-b t} = \int a e^{-b t}.\end{aligned} \hspace{\stretch{1}}(1.5)

This solution to the non-homogeneous equation is thus

\begin{aligned}x - x_0 = e^{b t} \int_{\tau = 0}^t a e^{-b \tau} = \frac{a}{b} \left(e^{bt} - 1 \right).\end{aligned} \hspace{\stretch{1}}(1.6)

Because this already incorporates the homogeneous solution $x = C e^{b t}$, this is also the general form of the solution.

Can we do something similar for the matrix equation of 1.2? It is tempting to try, rearranging in the same way like so

\begin{aligned}e^{ \frac{q}{m c}\Omega t} \left( e^{-\frac{q}{m c} \Omega t} \mathbf{v} \right)' = \frac{q}{m} \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.7)

Our matrix exponentials are perfectly well formed, but we will run into trouble attempting this. We can get as far as the integral above before running into trouble

\begin{aligned}\mathbf{v} - \mathbf{v}_0 = \frac{q}{m}e^{ \frac{q }{m c} \Omega t}\left( \int_{\tau = 0}^te^{ -\frac{q }{m c} \Omega \tau}\right) \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.8)

Only when $\text{Det}{\Omega} \ne 0$ do we have

\begin{aligned}\int e^{ -\frac{q }{m c} \Omega \tau} =- \frac{m c}{q} \Omega^{-1} e^{ -\frac{q }{m c} \Omega \tau},\end{aligned} \hspace{\stretch{1}}(1.9)

but in our case, this determinant is zero, due to the antisymmetry that is built into our magnetic field tensor. It appears that we need a different strategy.

Second attempt.

It’s natural to attempt to pull out our spectral theorem toolbox. We find three independent eigenvalues for our matrix $\Omega$ (one of which is naturally zero due to the singular nature of the matrix).

These eigenvalues are

\begin{aligned}\lambda_1 &= 0 \\ \lambda_2 &= i{\left\lvert{\mathbf{B}}\right\rvert} \\ \lambda_3 &= -i{\left\lvert{\mathbf{B}}\right\rvert}.\end{aligned} \hspace{\stretch{1}}(1.10)

The corresponding orthonormal eigenvectors are found to be

\begin{aligned}\mathbf{u}_1 &= \frac{\mathbf{B}}{{\left\lvert{\mathbf{B}}\right\rvert}} \\ \mathbf{u}_2 &=\frac{1}{{{\left\lvert{\mathbf{B}}\right\rvert} \sqrt{ 2(B_1^2 + B_3^2) } }}\begin{bmatrix}i B_1 B_2 - B_3 {\left\lvert{\mathbf{B}}\right\rvert} \\ -i(B_1^2 + B_3^2) \\ i B_2 B_3 + B_1 {\left\lvert{\mathbf{B}}\right\rvert} \\ \end{bmatrix} \\ \mathbf{u}_3 &=\frac{1}{{{\left\lvert{\mathbf{B}}\right\rvert} \sqrt{ 2(B_1^2 + B_3^2) } }}\begin{bmatrix}-i B_1 B_2 - B_3 {\left\lvert{\mathbf{B}}\right\rvert} \\ i(B_1^2 + B_3^2) \\ -i B_2 B_3 + B_1 {\left\lvert{\mathbf{B}}\right\rvert} \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.13)

The last pair of eigenvectors are computed with the assumption that not both of $B_1$ and $B_3$ are zero. This allows for the spectral decomposition

\begin{aligned}U &=\begin{bmatrix}\mathbf{u}_1 & \mathbf{u}_2 & \mathbf{u}_3\end{bmatrix} \\ D &= \begin{bmatrix}0 & 0 & 0 \\ 0 & i {\left\lvert{\mathbf{B}}\right\rvert} & 0 \\ 0 & 0 & -i {\left\lvert{\mathbf{B}}\right\rvert}\end{bmatrix} \\ \Omega &= U D U^{*}\end{aligned} \hspace{\stretch{1}}(1.16)

We can use this to decouple our equation

\begin{aligned}\mathbf{v}' = \frac{q}{m} \mathbf{E} + \frac{q}{ m c } U D U^{*} \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.19)

\begin{aligned}\mathbf{w} &= U^{*} \mathbf{v} \\ \mathbf{F} &= U^{*} \mathbf{E} \\ \mathbf{w}' &= \frac{q}{m} \mathbf{F} + \frac{q}{ m c } D \mathbf{w}.\end{aligned} \hspace{\stretch{1}}(1.20)

Written out explicitly, this is a set of three independent equations

\begin{aligned}w_1' &= \frac{q}{m} F_1 \\ w_2' &= \frac{q}{m} F_2 + \frac{q i {\left\lvert{\mathbf{B}}\right\rvert}}{ m c } w_2 \\ w_3' &= \frac{q}{m} F_3 - \frac{q i {\left\lvert{\mathbf{B}}\right\rvert}}{ m c } w_3\end{aligned} \hspace{\stretch{1}}(1.23)

Utilizing 1.6 our solution is

\begin{aligned}w_1 - w_1(0) &= \frac{q}{m} F_1 t \\ w_2 - w_2(0) &= - \frac{i c F_2 }{{\left\lvert{\mathbf{B}}\right\rvert}} \left( e^{ \frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ m c } } - 1 \right) \\ w_3 - w_3(0) &= \frac{i c F_3 }{{\left\lvert{\mathbf{B}}\right\rvert}} \left( e^{ -\frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ m c } } - 1 \right)\end{aligned} \hspace{\stretch{1}}(1.26)

Reinserting matrix form we have

\begin{aligned}\mathbf{w} - \mathbf{w}(0) =\begin{bmatrix}\frac{q}{m} \mathbf{e}_1^\text{T} U^{*} \mathbf{E} t \\ \frac{2 c \mathbf{e}_2^\text{T} U^{*} \mathbf{E} }{{\left\lvert{\mathbf{B}}\right\rvert}}e^{ \frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } } \sin \left( \frac{q {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } \right) \\ \frac{2 c \mathbf{e}_3^\text{T} U^{*} \mathbf{E} }{{\left\lvert{\mathbf{B}}\right\rvert}}e^{ -\frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } } \sin \left( \frac{q {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } \right) \\ \end{bmatrix}\end{aligned} \hspace{\stretch{1}}(1.29)

with

\begin{aligned}f_1 &= \frac{q}{m} t \\ f_2 &= \frac{2 c }{{\left\lvert{\mathbf{B}}\right\rvert}} e^{ \frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } } \sin \left( \frac{q {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } \right) \\ f_3 &= \frac{2 c }{{\left\lvert{\mathbf{B}}\right\rvert}} e^{ -\frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } } \sin \left( \frac{q {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } \right)\end{aligned} \hspace{\stretch{1}}(1.30)

We have

\begin{aligned}\mathbf{v} - \mathbf{v}(0) = U\begin{bmatrix}f_1 \mathbf{e}_1^\text{T} U^{*} \mathbf{E} \\ f_2 \mathbf{e}_2^\text{T} U^{*} \mathbf{E} \\ f_3 \mathbf{e}_3^\text{T} U^{*} \mathbf{E} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.33)

Observe that the dot products embedded here can be nicely expressed in terms of the eigenvectors since

\begin{aligned}U^{*} \mathbf{E}= \begin{bmatrix}\mathbf{u}_1^{*} \cdot \mathbf{E} \\ \mathbf{u}_2^{*} \cdot \mathbf{E} \\ \mathbf{u}_3^{*} \cdot \mathbf{E}\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.34)

Our solution is thus a weighted sum of projections of the electric field vector $\mathbf{E}$ onto the eigenvectors formed strictly from the magnetic field tensor

\begin{aligned}\mathbf{v} - \mathbf{v}(0) = \sum f_i \mathbf{u}_i ( \mathbf{u}_i^{*} \cdot \mathbf{E} ).\end{aligned} \hspace{\stretch{1}}(1.35)

Recalling that $\mathbf{u}_i = \mathbf{B}/{\left\lvert{\mathbf{B}}\right\rvert}$, the unit vector that lies in the direction of the magnetic field, we have

\begin{aligned}\mathbf{v} - \mathbf{v}(0) = \frac{q t}{m} \hat{\mathbf{B}} (\hat{\mathbf{B}} \cdot \mathbf{E})+ \sum_{i=2}^3 f_i \mathbf{u}_i ( \mathbf{u}_i^{*} \cdot \mathbf{E} ).\end{aligned} \hspace{\stretch{1}}(1.36)

Also observe that this is a manifestly real valued solution since remaining eigenvectors are conjugate pairs $\mathbf{u}_2 = \mathbf{u}_3^{*}$ as are the differential solutions $f_2 = f_3^{*}$. This leaves us with

\begin{aligned}\mathbf{v} - \mathbf{v}(0) = \frac{q t}{m} \hat{\mathbf{B}} (\hat{\mathbf{B}} \cdot \mathbf{E})+ \frac{4 c }{{\left\lvert{\mathbf{B}}\right\rvert}} \sin \left( \frac{q {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } \right) \text{Real} \left(e^{ \frac{q i {\left\lvert{\mathbf{B}}\right\rvert} t}{ 2 m c } } \mathbf{u}_2 ( \mathbf{u}_2^{*} \cdot \mathbf{E} )\right).\end{aligned} \hspace{\stretch{1}}(1.37)

It is natural to express $\mathbf{u}_2$ in terms of the direction cosines $b_i$ of the magnetic field vector $\mathbf{B} = {\left\lvert{\mathbf{B}}\right\rvert}(b_1, b_2, b_3)$

\begin{aligned}\mathbf{u}_2 =\frac{1}{{\sqrt{ 2(b_1^2 + b_3^2) }}}\begin{bmatrix}i b_1 b_2 - b_3 \\ -i(b_1^2 + b_3^2) \\ i b_2 b_3 + b_1 \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(1.38)

Is there a way to express this beastie in a coordinate free fashion? I experimented with this a bit looking for something that would provide some additional geometrical meaning but did not find it. Further play with that is probably something for another day.

What we do see is that our velocity has two main components, one of which is linearly increasing in proportion to the colinearity of the magnetic and electric fields. The other component is oscillatory. With a better geometrical description of that eigenvector we could perhaps understand the mechanics a bit better.

References

[1] C. Doran and A.N. Lasenby. Geometric algebra for physicists. Cambridge University Press New York, Cambridge, UK, 1st edition, 2003.