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Archive for February 4th, 2011

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

Posted by peeterjoot on February 4, 2011

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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.

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