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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.)]

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.

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.