## PHY450H1S. Relativistic Electrodynamics Lecture 26 (Taught by Prof. Erich Poppitz). Radiation reaction force for a dipole system.

Posted by peeterjoot on April 11, 2011

# Reading.

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

Covering pp. 181-195: (182-189) [Tuesday, Mar. 29]; the EM potentials to order (190-193); the “Darwin Lagrangian. and Hamiltonian for a system of nonrelativistic charged particles to order and its many uses in physics (194-195) [Wednesday, Mar. 30]

Next week (last topic): attempt to go to the next order – radiation damping, the limitations of classical electrodynamics, and the relevant time/length/energy scales.

# Recap.

A system of N charged particles closed system and nonrelativistic, . In this case we can incorporate EM effects in a Largrangian *ONLY* involving particles (EM field not a dynamical DOF). In *general case*, this works to , because at system radiation effects occur.

In a specific case, when

we can do that (meaning use a Lagrangian with particles only) to because of specific symmetries in such a system.

The Lagrangian for our particle after the gauge transformation is

Next time we’ll probably get to the Lagrangian for the entire system. It was hinted that this is called the Darwin Lagrangian (after Charles Darwin’s grandson).

We find for whole system

This is the Darwin Lagrangian (also Charles). The Darwin Hamiltonian is then

# Incorporating radiation effects as a friction term.

To obvious problem due to radiation (system not closed). We’ll incorporate radiation via a function term in the EOM

Again consider the dipole system

or

gives

(because there’s no radiation).

The energy radiated per unit time averaged per period is

We’ll modify the EOM

Employing an integration factor we have

or

Observe that the last expression, force times velocity, has the form of power

So we can make an identification with the time rate of energy lost by the system due to radiation

Average over period both sides

We demand this last equality, by requiring the energy change rate to equal that of the dipole power (but negative since it is a loss) that we previously calculated.

**Claim:**

**Proof:**

We need to show

We have

We first used . The first integral above is zero since the derivative of is also periodic, and vanishes when integrated over the interval.

We can therefore write

Our “frictional” correction is the radiation reaction force proportional to the third derivative of the position.

Rearranging slightly, this is

where is the “classical radius” of the electron. In our frictional term we have , the time for light to cross the classical radius of the electron.

There are lots of problems with this. One of the easiest is with . Then we have

with solution

where

This is a *self accelerating system*! Note that we can also get into this trouble with , but those examples are harder to find (see: [2]).

FIXME: borrow this text again to give that section a read.

The sensible point of view is that this third term () should be taken seriously only if it is small compared to the first two terms.

# References

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

[2] D.J. Griffith. *Introduction to Electrodynamics*. Prentice-Hall, 1981.

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