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PHY450H1S. Relativistic Electrodynamics Lecture 24 (Taught by Prof. Erich Poppitz). Non-relativistic electrostatic Lagrangian.

Posted by peeterjoot on March 30, 2011

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Covering chapter 5 \S 37, and chapter 8 \S 65 material from the text [1].

Covering pp. 181-195: the Lagrangian for a system of non relativistic charged particles to zeroth order in $(v/c)$: electrostatic energy of a system of charges and .mass renormalization.

A closed system of charged particles.

Consider a closed system of charged particles $(m_a, q_a)$ and imagine there is a frame where they are non-relativistic $v_a/c \ll 1$. In this case we can describe the dynamics using a Lagrangian only for particles. i.e.

\begin{aligned}\mathcal{L} = \mathcal{L}( \mathbf{x}_1, \cdots, \mathbf{x}_N, \mathbf{v}_1, \cdots, \mathbf{v}_N)\end{aligned} \hspace{\stretch{1}}(2.1)

If we work t order $(v/c)^2$.

If we try to go to $O((v/c)^3$, it’s difficult to only use $\mathcal{L}$ for particles.

This can be inferred from

\begin{aligned}P = \frac{2}{3} \frac{e^2}{c^3} {\left\lvert{\dot{d}{\mathbf{d}}}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.2)

because at this order, due to radiation effects, we need to include EM field as dynamical.

Start simple

Start with a system of (non-relativistic) free particles

\begin{aligned}S \end{aligned}

So in the non-relativistic limit, after dropping the constant term that doesn’t effect the dynamics, our Lagrangian is

\begin{aligned}\mathcal{L}(\mathbf{x}_a, \mathbf{v}_a) = \frac{1}{{2}} \sum_a m_a \mathbf{v}_a^2 - \frac{1}{{8}} \frac{m_a \mathbf{v}_a^4}{c^2}\end{aligned} \hspace{\stretch{1}}(3.3)

The first term is $O((v/c)^0)$ where the second is $O((v/c)^2)$.

Next include the fact that particles are charged.

\begin{aligned}\mathcal{L}_{\text{interaction}} = \sum_a \left( \cancel{q_a \frac{\mathbf{v}_a}{c} \cdot \mathbf{A}(\mathbf{x}_a, t)} - q_a \phi(\mathbf{x}_a, t) \right)\end{aligned} \hspace{\stretch{1}}(3.4)

Here, working to $O((v/c)^0)$, where we consider the particles moving so slowly that we have only a Coulomb potential $\phi$, not $\mathbf{A}$.

HERE: these are NOT ‘EXTERNAL’ potentials. They are caused by all the charged particles.

\begin{aligned}\partial_i F^{i l} = \frac{4 \pi}{c} j^l = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(3.5)

For $l = \alpha$ we have have $4 \pi \rho \mathbf{v}/c$, but we won’t do this today (tomorrow).

To leading order in $v/c$, particles only created Coulomb fields and they only “feel” Coulomb fields. Hence to $O((v/c)^0)$, we have

\begin{aligned}\mathcal{L} = \sum_a \frac{m_a \mathbf{v}_a^2}{2} - q_a \phi(\mathbf{x}_a, t)\end{aligned} \hspace{\stretch{1}}(3.6)

What’s the $\phi(\mathbf{x}_a, t)$, the Coulomb field created by all the particles.

\paragraph{How to find?}

\begin{aligned}\partial_i F^{i 0} = \frac{4 \pi}{c} = 4 \pi \rho\end{aligned} \hspace{\stretch{1}}(3.7)

or

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \rho = - \boldsymbol{\nabla}^2 \phi \end{aligned} \hspace{\stretch{1}}(3.8)

where

\begin{aligned}\rho(\mathbf{x}, t) = \sum_a q_a \delta^3 (\mathbf{x} - \mathbf{x}_a(t))\end{aligned} \hspace{\stretch{1}}(3.9)

This is a Poisson equation

\begin{aligned}\Delta \phi(\mathbf{x}) = \sum_a q_a 4 \pi \delta^3(\mathbf{x} - \mathbf{x}_a)\end{aligned} \hspace{\stretch{1}}(3.10)

(where the time dependence has been suppressed). This has solution

\begin{aligned}\phi(\mathbf{x}, t) = \sum_b \frac{q_b}{{\left\lvert{\mathbf{x} - \mathbf{x}_b(t)}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.11)

This is the sum of instantaneous Coulomb potentials of all particles at the point of interest. Hence, it appears that $\phi(\mathbf{x}_a, t)$ should be evaluated in 3.11 at $\mathbf{x}_a$?

However 3.11 becomes infinite due to contributions of the a-th particle itself. Solution to this is to drop the term, but let’s discuss this first.

Let’s talk about the electrostatic energy of our system of particles.

\begin{aligned}\mathcal{E} &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \left(\mathbf{E}^2 + \cancel{\mathbf{B}^2} \right) \\ &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \mathbf{E} \cdot (-\boldsymbol{\nabla} \phi) \\ &= \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \left( \boldsymbol{\nabla} \cdot (\mathbf{E} \phi) - \phi \boldsymbol{\nabla} \cdot \mathbf{E} \right) \\ &= -\frac{1}{{8 \pi}} \oint d^2 \boldsymbol{\sigma} \cdot \mathbf{E} \phi + \frac{1}{{8 \pi}} \int d^3 \mathbf{x} \phi \boldsymbol{\nabla} \cdot \mathbf{E} \\ \end{aligned}

The first term is zero since $\mathbf{E} \phi$ for a localized system of charges $\sim 1/r^3$ or higher as $V \rightarrow \infty$.

In the second term

\begin{aligned}\boldsymbol{\nabla} \cdot \mathbf{E} = 4 \pi \sum_a q_a \delta^3(\mathbf{x} - \mathbf{x}_a(t))\end{aligned} \hspace{\stretch{1}}(3.12)

So we have

\begin{aligned}\sum_a \frac{1}{{2}} \int d^3 \mathbf{x} q_a \delta^3(\mathbf{x} - \mathbf{x}_a) \phi(\mathbf{x})\end{aligned} \hspace{\stretch{1}}(3.13)

for

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a q_a \phi(\mathbf{x}_a)\end{aligned} \hspace{\stretch{1}}(3.14)

Now substitute 3.11 into 3.14 for

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a \frac{q_a^2}{{\left\lvert{\mathbf{x} - \mathbf{x}_a}\right\rvert}} + \frac{1}{{2}} \sum_{a \ne b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.15)

or

\begin{aligned}\mathcal{E} = \frac{1}{{2}} \sum_a \frac{q_a^2}{{\left\lvert{\mathbf{x} - \mathbf{x}_a}\right\rvert}} + \sum_{a < b} \frac{q_a q_b}{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.16)

The first term is the sum of the electrostatic self energies of all particles. The source of this infinite self energy is in assuming a \underline{point like nature} of the particle. i.e. We modeled the charge using a delta function instead of using a continuous charge distribution.

Recall that if you have a charged sphere of radius $r$

PICTURE: total charge $q$, radius $r$, our electrostatic energy is

\begin{aligned}\mathcal{E} \sim \frac{q^2}{r}\end{aligned} \hspace{\stretch{1}}(3.17)

Stipulate that rest energy $m_e c^2$ is all of electrostatic origin $\sim e^2/r_e$ we get that

\begin{aligned}r_e \sim \frac{e^2}{m_e c^2}\end{aligned} \hspace{\stretch{1}}(3.18)

This is called the classical radius of the electron, and is of a very small scale $10^{-13} \text{cm}$.

As a matter of fact the applicability of classical electrodynamics breaks down much sooner than this scale since quantum effects start kicking in.

Our Lagrangian is now

\begin{aligned}\mathcal{L}_a = \frac{1}{{2}} m_a \mathbf{v}_a^2 - q_a \phi(\mathbf{x}_a, t)\end{aligned} \hspace{\stretch{1}}(3.19)

where $\phi$ is the electrostatic potential due to all \underline{other} particles, so we have

\begin{aligned}\mathcal{L}_a = \frac{1}{{2}} m_a \mathbf{v}_a^2 - \sum_{a \ne b} \frac{q_a q_b }{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.20)

and for the system

\begin{aligned}\mathcal{L} = \frac{1}{{2}} \sum_a m_a \mathbf{v}_a^2 - \sum_{a < b} \frac{q_a q_b }{{\left\lvert{\mathbf{x}_a - \mathbf{x}_b}\right\rvert}}\end{aligned} \hspace{\stretch{1}}(3.21)

This is THE Lagrangian for electrodynamics in the non-relativistic case, starting with the relativistic action.

What’s next?

We continue to the next order of $v/c$ tomorrow.

References

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