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## PHY450H1S. Relativistic Electrodynamics Tutorial 3 (TA: Simon Freedman). Relativistic motion in constant uniform electric or magnetic fields.

Posted by peeterjoot on February 4, 2011

# Motion in an constant uniform Electric field.

Given

\begin{aligned}\mathbf{E} = E \hat{\mathbf{x}},\end{aligned} \hspace{\stretch{1}}(1.1)

We want to solve the problem

\begin{aligned}\mathbf{F} = \frac{d\mathbf{p}}{dt} =e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right) = e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.2)

Unlike second year classical physics, we will use relativistic momentum, so for only a constant electric field, our Lorentz force equation to solve becomes

\begin{aligned}\frac{d\mathbf{p}}{dt} = \frac{d (m \gamma \mathbf{v})}{dt} = e \mathbf{E}.\end{aligned} \hspace{\stretch{1}}(1.3)

In components this is

\begin{aligned}\dot{p}_x &= e E \\ \dot{p}_y &= \text{constant}\end{aligned} \hspace{\stretch{1}}(1.4)

Integrating the $x$ component we have

\begin{aligned}e E t + p_x(0)=\frac{m \dot{x}}{\sqrt{1 - (\dot{x}^2 + \dot{y}^2)/c^2}} \end{aligned} \hspace{\stretch{1}}(1.6)

If we let $p_x(0) = 0$, square and rearrange a bit we have

\begin{aligned}\frac{m^2}{(e E t)^2} \dot{x}^2 = 1 - \frac{\dot{x}^2 + \dot{y}^2}{c^2}\end{aligned} \hspace{\stretch{1}}(1.7)

For

\begin{aligned}\dot{x}^2 = \frac{c^2 - \dot{y}^2}{1 + (\frac{mc}{eEt})^2}.\end{aligned} \hspace{\stretch{1}}(1.8)

Now for the $y$ components, with $p_y(0) = p_0$, our equation to solve is

\begin{aligned}\frac{m \dot{y}}{\sqrt{1 - (\dot{x}^2 + \dot{y}^2)/c^2}} = p_0.\end{aligned} \hspace{\stretch{1}}(1.9)

Squaring this one we have

\begin{aligned}\frac{c^2 m^2}{p_0^2} \dot{y}^2 = c^2 - \dot{x}^2 - \dot{y}^2 ,\end{aligned} \hspace{\stretch{1}}(1.10)

and

\begin{aligned}\dot{y}^2 = \frac{ c^2 - \dot{x}^2}{1 + \frac{m^2 c^2}{p_0^2}}\end{aligned} \hspace{\stretch{1}}(1.11)

Observe that our energy is

\begin{aligned}\mathcal{E}^2 = p^2 c^2 + m^2 c^4,\end{aligned} \hspace{\stretch{1}}(1.12)

and for $t=0$

\begin{aligned}\mathcal{E}_0^2 = p_0^2 c^2 + m^2 c^4.\end{aligned} \hspace{\stretch{1}}(1.13)

We can then write

\begin{aligned}\dot{y}^2 = \frac{ c^2 p_0^2 (c^2 - \dot{x}^2)}{ \mathcal{E}_0^2 }.\end{aligned} \hspace{\stretch{1}}(1.14)

Some messy substitution, using 1.8, yields

\begin{aligned}\boxed{\begin{aligned}\dot{x} &= \frac{c^2 e E t}{\sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }} \\ \dot{y} &= \frac{c^2 p_0 }{\sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }}\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.15)

Solving for $x$ we have

\begin{aligned}x(t) = c^2 e E \int \frac{dt' t'}{\sqrt{ \mathcal{E}_0^2 + (e c E t')^2 }}\end{aligned} \hspace{\stretch{1}}(1.16)

Can solve with hyperbolic substitution or

\begin{aligned}x(t) = c^2 e E \int \frac{dt' t'}{\sqrt{ \mathcal{E}_0^2 + (e c E t')^2 }}\end{aligned} \hspace{\stretch{1}}(1.17)

\begin{aligned}d(u^2) = 2 u du \implies u du = \frac{1}{{2}} d(u^2)\end{aligned} \hspace{\stretch{1}}(1.18)

\begin{aligned}x(t) = \frac{c^2 e E}{2 \mathcal{E}_0} \int \frac{d (u^2)}{\sqrt{ 1 + \left(\frac{e c E}{\mathcal{E}_0}\right)^2 u^2 }}\end{aligned} \hspace{\stretch{1}}(1.19)

Now we have something of the form

\begin{aligned}\int \frac{d v}{\sqrt{1 + a v}} = \frac{2}{a} \sqrt{1 + a v},\end{aligned} \hspace{\stretch{1}}(1.20)

so our final solution for $x(t)$ is

\begin{aligned}x(t) = \frac{1}{{e E}} \sqrt{ \mathcal{E}_0^2 + (e c E t)^2 }\end{aligned} \hspace{\stretch{1}}(1.21)

or

\begin{aligned}x^2 - c^2 t^2 = \frac{\mathcal{E}_0^2}{ e^2 E^2 } = a^{-2}.\end{aligned} \hspace{\stretch{1}}(1.22)

Now for $y(t)$ we have

\begin{aligned}y(t) = c^2 p_0 \int \frac{dt}{ \sqrt{\mathcal{E}_0^2 + (e c E t)^2 }}\end{aligned} \hspace{\stretch{1}}(1.23)

\begin{aligned}t = \frac{\mathcal{E}_0}{ e c E} \sinh(u) \end{aligned} \hspace{\stretch{1}}(1.24)

\begin{aligned}dt = \frac{\mathcal{E}_0}{ e c E} \cosh(u) du\end{aligned} \hspace{\stretch{1}}(1.25)

\begin{aligned}y(t) &= \frac{c^2 p_0}{\mathcal{E}_0} \int \frac{dt}{\sqrt{1 + (\frac{e c E}{\mathcal{E}_0})^2 t^2 }} \\ &= \frac{c^2 p_0}{\mathcal{E}_0} \frac{\mathcal{E}_0}{ e c E} \int \frac{ du \cosh u }{\sqrt{1 + \sinh^2 u }} \\ &= \frac{c p_0}{ e E} u \end{aligned}

A final bit of substitution, including a sort of odd seeming parametrization of $x$ in terms of $y$ in terms of $t$, we have

\begin{aligned}\boxed{\begin{aligned}y(t) &= \frac{c p_0}{ e E} \sinh^{-1} \left( \frac{e c E t}{\mathcal{E}_0} \right) \\ x(y) &= \frac{\mathcal{E}_0}{c E \cosh \left( \frac{y e E }{ c p_0} \right) }\end{aligned}}\end{aligned} \hspace{\stretch{1}}(1.26)

## Checks

FIXME: check the checks.

\begin{aligned}v \rightarrow c, t \rightarrow \infty\end{aligned} \hspace{\stretch{1}}(1.27)

\begin{aligned}v << c, t \rightarrow 0\end{aligned} \hspace{\stretch{1}}(1.28)

\begin{aligned}m v_x &= e E t + ... \\ x &\sim t^2 \end{aligned}

\begin{aligned}m v_y = p_0 \rightarrow y \sim t\end{aligned} \hspace{\stretch{1}}(1.29)

\begin{aligned}x(y) \sim y^2\end{aligned} \hspace{\stretch{1}}(1.30)

(a parabola)

## An alternate way.

There’s also a tricky way (as in the text), with

\begin{aligned}\mathbf{p} &= m \gamma \mathbf{v} \\ \mathcal{E} &= \gamma m c^2 \end{aligned} \hspace{\stretch{1}}(1.31)

We can solve this for $\mathbf{p}$

\begin{aligned}m \gamma &= \frac{\mathbf{p} \cdot \mathbf{v}}{\mathbf{v}^2} = \frac{\mathcal{E}}{c^2} \\ \mathbf{p} \times \mathbf{v} &= 0\end{aligned}

With the cross product zero, $\mathbf{p}$ has only a component in the direction of $\mathbf{v}$, and we can invert to yield

\begin{aligned}\mathbf{p} = \frac{\mathcal{E} \mathbf{v}}{c^2}.\end{aligned} \hspace{\stretch{1}}(1.33)

This implies

\begin{aligned}\dot{x} = \frac{c^2 p_x}{\mathcal{E}},\end{aligned} \hspace{\stretch{1}}(1.34)

and one can work from there as well.

# Motion in an constant uniform Magnetic field.

## Work by the magnetic field

Note that the magnetic field does no work

\begin{aligned}\mathbf{F} = \frac{e}{c} \mathbf{v} \times \mathbf{B}\end{aligned} \hspace{\stretch{1}}(2.35)

\begin{aligned}dW &= \mathbf{F} \cdot d\mathbf{l} \\ &=\frac{e}{c} (\mathbf{v} \times \mathbf{B}) \cdot d\mathbf{l} \\ &=\frac{e}{c} (\mathbf{v} \times \mathbf{B}) \cdot \mathbf{v} dt \\ &= 0\end{aligned}

Because $\mathbf{v}$ and $\mathbf{v} \times \mathbf{B}$ are necessarily perpendicular we are reminded that the magnetic field does no work (even in this relativistic sense).

## Initial energy of the particle

Because no work is done, the particle’s energy is only the initial time value

\begin{aligned}\mathcal{E} = .... + e A^0\end{aligned} \hspace{\stretch{1}}(2.36)

Simon asked if we’d calculated this (i.e. the Hamiltonian in class). We’d calculated the conservation for time invariance, the Hamiltonian (and called it $E$). We’d also calculated the Hamiltonian for the free particle

\begin{aligned}\mathcal{E} = \mathbf{p}^2 c^2 + (m c^2)^2.\end{aligned} \hspace{\stretch{1}}(2.37)

We had not done this calculation for the Lorentz force Lagrangian, so lets do it now. Recall that this Lagrangian was

\begin{aligned}\mathcal{L} = - m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} - e \phi + \frac{e}{c} \mathbf{v} \cdot \mathbf{A},\end{aligned} \hspace{\stretch{1}}(2.38)

with generalized momentum of

\begin{aligned}\frac{\partial {\mathcal{L}}}{\partial {\mathbf{v}}} = \frac{m \mathbf{v}}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \frac{e}{c} \mathbf{A}.\end{aligned} \hspace{\stretch{1}}(2.39)

Our Hamiltonian is thus

\begin{aligned}\mathcal{E} = \frac{m \mathbf{v}^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} + \frac{e}{c} \mathbf{A} \cdot \mathbf{v}+ m c^2 \sqrt{1 - \frac{\mathbf{v}^2}{c^2}} + e \phi - \frac{e}{c} \mathbf{v} \cdot \mathbf{A},\end{aligned}

which gives us

\begin{aligned}\mathcal{E} = e \phi + \frac{m c^2}{\sqrt{1 - \frac{\mathbf{v}^2}{c^2}}} \end{aligned}

So we see that our “energy”, defined as a quantity that is conserved, as a result of the symmetry of time translation invariance, has a component due to the electric field (but not the vector potential field $\mathbf{A}$), plus the free particle “energy”.

Is this right? With $\mathbf{A}$ and $\phi$ being functions of space and time, perhaps we need to be more careful with this argument. Perhaps this actually only applies to a statics case where $\mathbf{A}$ and $\phi$ are constant.

Since it was hinted to us that the energy component of the Lorentz force equation was proportional to $F^{0j} u_j$, and we can peek ahead to find that $F^{ij} = \partial^i A^j - \partial^j A^i$, let’s compare to that

\begin{aligned}e F^{0 j} u_j&=e (\partial^0 A^j - \partial^j A^0) u_j \\ &=e (\partial^0 A^\alpha - \partial^\alpha A^0) u_\alpha \\ &=e \left( \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} + \partial_\alpha A^0 \right) \frac{1}{{c}} \frac{dx_\alpha}{d\tau} \\ &=-e \left( \frac{1}{{c}} \frac{\partial {A^\alpha}}{\partial {t}} + \frac{\partial {\phi}}{\partial {x^\alpha}} \right) \frac{1}{{c}} \frac{dx^\alpha}{dt} \gamma,\end{aligned}

which is

\begin{aligned}e F^{0 j} u_j = e \left(\mathbf{E} \cdot \frac{\mathbf{v}}{c}\right) \gamma.\end{aligned} \hspace{\stretch{1}}(2.40)

So if we have

\begin{aligned}\frac{d\mathbf{p}}{dt} = e \left( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B} \right)\end{aligned} \hspace{\stretch{1}}(2.41)

I’d guess that we have

\begin{aligned}\frac{d(\mathcal{E}/c)}{d\tau} \propto e F^{0 j} u_j,\end{aligned} \hspace{\stretch{1}}(2.42)

which is, using 2.40

\begin{aligned}\frac{d(\mathcal{E}/c)}{dt} \propto e \left(\mathbf{E} \cdot \frac{\mathbf{v}}{c}\right) \end{aligned} \hspace{\stretch{1}}(2.43)

Can the left hand side be integrated to yield $e \phi$? Yes, but only in the statics case when ${\partial {\mathbf{A}}}/{\partial {t}} = 0$, and $\phi(\mathbf{x},t) = \phi(\mathbf{x})$ for which we have

\begin{aligned}\mathcal{E} &\propto e \int \mathbf{E} \cdot \mathbf{v} dt \\ &= -e \int (\boldsymbol{\nabla} \phi) \cdot \frac{\mathbf{x}}{dt} dt \\ &= -e \int \frac{\partial {\phi}}{\partial {x^\alpha}} \frac{\partial {x^\alpha}}{\partial {t}} dt&= -e \phi\end{aligned}

FIXME: My suspicion is that the result 2.43, is generally true, but that we have dropped terms from the Hamiltonian calculation that need to be retained when $\phi$ and $\mathbf{A}$ are functions of time.

## Expressing the field and the force equation.

We will align our field with the $z$ axis, and write

\begin{aligned}\mathbf{B} = H \hat{\mathbf{z}},\end{aligned} \hspace{\stretch{1}}(2.44)

or, in components

\begin{aligned}\delta_{\alpha 3} H = H_\alpha.\end{aligned} \hspace{\stretch{1}}(2.45)

Because the energy is only due to the initial value, we write

\begin{aligned}\mathcal{E}(t) = \mathcal{E}_0\end{aligned} \hspace{\stretch{1}}(2.46)

\begin{aligned}\mathbf{p} = \mathcal{E} \frac{\mathbf{v}}{c^2} = \mathcal{E}_0 \frac{\mathbf{v}}{c^2}\end{aligned} \hspace{\stretch{1}}(2.47)

implies

\begin{aligned}\mathbf{v} = \mathbf{p} \frac{c^2}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.48)

\begin{aligned}\dot{\mathbf{v}} = \dot{\mathbf{p}} \frac{c^2}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.49)

\begin{aligned}\dot{v}_\alpha = \frac{e c}{\mathcal{E}_0} \epsilon_{\alpha \beta \gamma} v_\beta H_\gamma\end{aligned} \hspace{\stretch{1}}(2.50)

write

\begin{aligned}\omega = \frac{e c H}{\mathcal{E}_0}\end{aligned} \hspace{\stretch{1}}(2.51)

Evaluating the delta

\begin{aligned}\dot{v}_\alpha = \omega \epsilon_{\alpha \beta 3} v_\beta \end{aligned} \hspace{\stretch{1}}(2.52)

\begin{aligned}\dot{v}_1 &= \omega \epsilon_{1 \beta 3} v_\beta = \omega v_2 \\ \dot{v}_2 &= \omega \epsilon_{2 \beta 3} v_\beta = - \omega v_1 \\ \dot{v}_3 &= \omega \epsilon_{3 \beta 3} v_\beta = 0\end{aligned} \hspace{\stretch{1}}(2.53)

Looks like circular motion, so it’s natural to use complex variables. With

\begin{aligned}z = v_1 + i v_2 \end{aligned} \hspace{\stretch{1}}(2.56)

Using this we have

\begin{aligned}\frac{d}{dt} ( v_1 + i v_2 ) &= \omega v_2 - i \omega v_1 \\ &= -i \omega ( v_1 + i v_2 ).\end{aligned}

which comes out nicely

\begin{aligned}\frac{dz}{dt} = -i \omega z\end{aligned} \hspace{\stretch{1}}(2.57)

for

\begin{aligned}z = V_0 e^{-i \omega z t + i \alpha}\end{aligned} \hspace{\stretch{1}}(2.58)

Real and imaginary parts

\begin{aligned}v_1(t) &= V_0 \cos( \omega z t + \alpha) \\ v_2(t) &= -V_0 \sin( \omega z t + \alpha)\end{aligned} \hspace{\stretch{1}}(2.59)

Integrating

\begin{aligned}x_1(t) &= x_1(0) + V_0 \sin( \omega z t + \alpha) \\ x_2(t) &= x_2(0) + V_0 \cos( \omega z t + \alpha)\end{aligned} \hspace{\stretch{1}}(2.61)

Which is a helix.
FIXME: PICTURE.