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PHY450H1S. Relativistic Electrodynamics Lecture 27 (Taught by Prof. Erich Poppitz). Radiation reaction force continued, and limits of classical electrodynamics.

Posted by peeterjoot on May 3, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Reading.

Covering chapter 8 section 65 material from the text [1].

FIXME: Covering pp. 198.1-200: (last topic): attempt to go to the next order (v/c)^3 – radiation damping, the limitations of classical electrodynamics, and the relevant time/length/energy scales.

Radiation reaction force.

We previously obtained the radiation reaction force by adding a “frictional” force to the harmonic oscillator system. Now its time to obtain this by continuing the expansion of the potentials to the next order in \mathbf{v}/c.

Recall that our potentials are

\begin{aligned}\phi(\mathbf{x}, t) &= \int d^3 \mathbf{x} \frac{\rho\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right)}{{\left\lvert{\mathbf{x} - \mathbf{x}}\right\rvert}} \\ \mathbf{A}(\mathbf{x}, t) &= \frac{1}{{c}}\int d^3 \mathbf{x} \frac{\mathbf{j}\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right)}{{\left\lvert{\mathbf{x} - \mathbf{x}}\right\rvert}}.\end{aligned} \hspace{\stretch{1}}(2.1)

We can expand in Taylor series about t. For the charge density this is

\begin{aligned}\begin{aligned}\rho&\left(\mathbf{x}', t - {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}/c\right) \\ &\approx \rho(\mathbf{x}', t) - \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \rho(\mathbf{x}', t) + \frac{1}{{2}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 \frac{\partial^2 {{}}}{\partial {{t}}^2} \rho(\mathbf{x}', t) - \frac{1}{{6}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3 \frac{\partial^3}{\partial t^3} \rho(\mathbf{x}', t) \end{aligned},\end{aligned} \hspace{\stretch{1}}(2.3)

so that our scalar potential to third order is

\begin{aligned}\phi(\mathbf{x}, t) &=\int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &+ \frac{1}{{2}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 \frac{\partial^2 {{}}}{\partial {{t}}^2} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{6}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3 \frac{\partial^3}{\partial t^3} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &=\int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- {\frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} }\\ &+ \frac{1}{{2}} \frac{\partial^2 {{}}}{\partial {{t}}^2} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}\left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^2 - \frac{1}{{6}} \frac{\partial^3}{\partial t^3} \int d^3 \mathbf{x} \frac{\rho(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \left(\frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c}\right)^3  \\ &= \phi^{(0)} + \phi^{(2)} + \phi^{(3)}\end{aligned}

Expanding the vector potential in Taylor series to second order we have

\begin{aligned}\mathbf{A}(\mathbf{x}, t) &=\frac{1}{{c}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{c}} \frac{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}{c} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}} \\ &=\frac{1}{{c}} \int d^3 \mathbf{x} \frac{\mathbf{j}(\mathbf{x}', t) }{{\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}}- \frac{1}{{c^2}} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) \\ &= \mathbf{A}^{(1)} + \mathbf{A}^{(2)} \end{aligned}

We’ve already considered the effects of the \mathbf{A}^{(1)} term, and now move on to \mathbf{A}^{(2)}. We will write \phi^{(3)} as a total derivative

\begin{aligned}\phi^{(3)} = \frac{1}{{c}} \frac{\partial {}}{\partial {t}} \left( - \frac{1}{{6 c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t){\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2\right)= \frac{1}{{c}} \frac{\partial {}}{\partial {t}} f^{(2)}(\mathbf{x}, t)\end{aligned} \hspace{\stretch{1}}(2.4)

and gauge transform it away as we did with \phi^{(2)} previously.

\begin{aligned}\phi^{(3)'} &= \phi^{(3)} - \frac{1}{{c}} \frac{\partial {f^{(2)}}}{\partial {t}} = 0 \\ \mathbf{A}^{(2)'} &= \mathbf{A}^{(2)} + \boldsymbol{\nabla} f^{(2)} \end{aligned} \hspace{\stretch{1}}(2.5)

\begin{aligned}\mathbf{A}^{(2)'} &= - \frac{1}{{c^2}} \frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) - \frac{1}{{6 c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t)\boldsymbol{\nabla}_{\mathbf{x}} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2 \\ \end{aligned}

Looking first at the first integral we can employ the trick of writing \mathbf{e}_\alpha = {\partial {\mathbf{x}'}}/{\partial {x^{\alpha'}}}, and then employ integration by parts

\begin{aligned}\int_V d^3 \mathbf{x} \mathbf{j}(\mathbf{x}', t) &=\int_V d^3 \mathbf{x} \mathbf{e}_\alpha j^\alpha (\mathbf{x}', t)  \\ &=\int_V d^3 \mathbf{x} \frac{\partial {\mathbf{x}'}}{\partial {x^{\alpha'}}}j^\alpha (\mathbf{x}', t)  \\ &=\int_V d^3 \mathbf{x} \frac{\partial {}}{\partial {x^{\alpha'}}} \left( \mathbf{x}' j^\alpha (\mathbf{x}', t) \right)-\int_V d^3 \mathbf{x}\mathbf{x}' \frac{\partial {}}{\partial {x^{\alpha'}}} j^\alpha (\mathbf{x}', t) \\ &=\int_{\partial V} d^2 \boldsymbol{\sigma} \cdot \left( \mathbf{x}' j^\alpha (\mathbf{x}', t) \right)-\int d^3 \mathbf{x} \mathbf{x}' -\frac{\partial {}}{\partial {t}} \rho(\mathbf{x}', t) \\ &=\frac{\partial {}}{\partial {t}} \int d^3 \mathbf{x} \mathbf{x}' \rho(\mathbf{x}', t) \\ \end{aligned}

For the second integral, we have

\begin{aligned}\boldsymbol{\nabla}_{\mathbf{x}} {\left\lvert{\mathbf{x} - \mathbf{x}'}\right\rvert}^2 &= \mathbf{e}_\alpha \partial_\alpha (x^\beta - x^{\beta'})(x^\beta - x^{\beta'}) \\ &=2 \mathbf{e}_\alpha \delta_{\alpha \beta}(x^\beta - x^{\beta'}) \\ &= 2 (\mathbf{x} - \mathbf{x}'),\end{aligned}

so our gauge transformed vector potential term is reduced to

\begin{aligned}\mathbf{A}^{(2)'} &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t) \left(\mathbf{x}' + \frac{1}{{3}}(\mathbf{x} - \mathbf{x}') \right) \\ &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \rho(\mathbf{x}', t) \left(\frac{1}{{3}} \mathbf{x} + \frac{2}{3}\mathbf{x}' \right) \\ \end{aligned}

Now we wish to employ a discrete representation of the charge density

\begin{aligned}\rho(\mathbf{x}', t) = \sum_{b=1}^N q_b \delta^3(\mathbf{x}' - \mathbf{x}_b(t))\end{aligned} \hspace{\stretch{1}}(2.7)

So that the second order vector potential becomes

\begin{aligned}\mathbf{A}^{(2)'} &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \int d^3 \mathbf{x} \left(\frac{1}{{3}} \mathbf{x} + \frac{2}{3}\mathbf{x}' \right) \sum_{b=1}^N q_b \delta^3(\mathbf{x}' - \mathbf{x}_b(t)) \\ &= -\frac{1}{{c^2}} \frac{\partial^2}{\partial t^2} \sum_{b=1}^N q_b \left( {\frac{1}{{3}} \mathbf{x}} + \frac{2}{3}\mathbf{x}_b(t) \right) \\ &=-\frac{2}{3 c^2} \sum_{b=1}^N q_b \dot{d}{\mathbf{x}}_b(t) \\ &=-\frac{2}{3 c^2} \frac{d^2}{dt^2}\left( \sum_{b=1}^N q_b \mathbf{x}_b(t) \right).\end{aligned}

We end up with a dipole moment

\begin{aligned}\mathbf{d}(t) = \sum_{b=1}^N q_b \mathbf{x}_b(t) \end{aligned} \hspace{\stretch{1}}(2.8)

so we can write

\begin{aligned}\mathbf{A}^{(2)'} = -\frac{2}{3 c^2} \dot{d}{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.9)

Observe that there is no magnetic field due to this contribution since there is no explicit spatial dependence

\begin{aligned}\boldsymbol{\nabla} \times \mathbf{A}^{(2)'} = 0\end{aligned} \hspace{\stretch{1}}(2.10)

we have also gauge transformed away the scalar potential contribution so have only the time derivative contribution to the electric field

\begin{aligned}\mathbf{E} = -\frac{1}{{c}} \frac{\partial {\mathbf{A}}}{\partial {t}} - {\boldsymbol{\nabla} \phi} = \frac{2}{3 c^2} \dddot{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.11)

To O((v/c)^3) there is a homogeneous electric field felt by all particles, hence every particle feels a “friction” force

\begin{aligned}\mathbf{f}_{\text{rad}} = q \mathbf{E} = \frac{2 q}{3 c^3} \dddot{\mathbf{d}}(t).\end{aligned} \hspace{\stretch{1}}(2.12)

Moral: \mathbf{f}_{\text{rad}} arises in third order term O((v/c)^3) expansion and thus shouldn’t be given a weight as important as the two other terms. i.e. It’s consequences are less.

Example: our dipole system

\begin{aligned}m \dot{d}{z} &= - m \omega^2 a + \frac{2 e^2}{3 c^3} \dddot{z} \\ &= - m \omega^2 a + \frac{2 m}{3 c} \frac{e^2}{m c^2} \dddot{z} \\ &= - m \omega^2 a + \frac{2 m}{3} \frac{r_e}{c} \dddot{z} \\ \end{aligned}

Here r_e \sim 10^{-13} \text{cm} is the classical radius of the electron. For periodic motion

\begin{aligned}z &\sim e^{i \omega t} z_0 \\ \dot{d}{z} &\sim \omega^2 z_0 \\ \dddot{z} &\sim \omega^3 z_0.\end{aligned}

The ratio of the last term to the inertial term is

\begin{aligned}\sim \frac{ \omega^3 m (r_e/c) z_0 }{ m \omega^2 z_0 } \sim \omega \frac{r_e}{c} \ll 1,\end{aligned} \hspace{\stretch{1}}(2.13)

so

\begin{aligned}\omega &\ll \frac{c}{r_e}  \\ &\sim \frac{1}{{\tau_e}}  \\ &\sim \frac{ 10^{10} \text{cm}/\text{s}}{10^{-13} \text{cm}}  \\ &\sim 10^{23} \text{Hz} \\ \end{aligned}

So long as \omega \ll 10^{23} \text{Hz}, this approximation is valid.

Limits of classical electrodynamics.

What sort of energy is this? At these frequencies QM effects come in

\begin{aligned}\hbar \sim 10^{-33} \text{J} \cdot \text{s} \sim 10^{-15} \text{eV} \cdot \text{s}\end{aligned} \hspace{\stretch{1}}(3.14)

\begin{aligned}\hbar \omega_{max} \sim 10^{-15} \text{eV} \cdot \text{s} \times 10^{23} \frac{1}{{\text{s}}} \sim 10^8 \text{eV} \sim 100 \text{MeV}\end{aligned} \hspace{\stretch{1}}(3.15)

whereas the rest energy of the electron is

\begin{aligned}m_e c^2 \sim \frac{1}{{2}} \text{MeV} \sim \text{MeV}.\end{aligned} \hspace{\stretch{1}}(3.16)

At these frequencies it is possible to create e^{+} and e^{-} pairs. A theory where the number of particles (electrons and positrons) is NOT fixed anymore is required. An estimate of this frequency, where these effects have to be considered is possible.

PICTURE: different length scales with frequency increasing to the left and length scales increasing to the right.

\begin{itemize}
\item 10^{-13} \text{cm}, r_e = e^2/m c^2. LHC exploration.
\item 137 \times 10^{-13} \text{cm}, \hbar/m_e c \sim \lambda/2\pi, the Compton wavelength of the electron. QED and quantum field theory.
\item (137)^2 \times 10^{-13}  \text{cm} \sim 10^{-10} \text{cm}, Bohr radius. QM, and classical electrodynamics.
\end{itemize}

here

\begin{aligned}\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c } = \frac{1}{{137}},\end{aligned} \hspace{\stretch{1}}(3.17)

is the fine structure constant.

Similar to the distance scale restrictions, we have field strength restrictions. A strong enough field (Electric) can start creating electron and positron pairs. This occurs at about

\begin{aligned}e E \lambda/2\pi \sim 2 m_e c^2 \end{aligned} \hspace{\stretch{1}}(3.18)

so the critical field strength is

\begin{aligned}E_{\text{crit}} &\sim \frac{m_e c^2 }{\lambda/2\pi e}  \\ &\sim \frac{m_e c^2 }{\hbar e} m_e c  \\ &\sim \frac{m_e^2 c^3}{\hbar e}\end{aligned}

Is this real?

Yes, with a very heavy nucleus with some electrons stripped off, the field can be so strong that positron and electron pairs will be created. This can be observed in heavy ion collisions!

References

[1] L.D. Landau and E.M. Lifshitz. The classical theory of fields. Butterworth-Heinemann, 1980.

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