1. Apr 21, 2012

### Philalethes

Hello,

In electrostatics, classical mechanics predicts that a charged particle's action-at-a-distance force (Coulomb's law) on other charges is accompanied by an equal and opposite reaction force on the particle.

Classically, radiant electromagnetic energy is a self-propagating electromagnetic field; therefore, must there likewise be a reaction force on this radiant energy resulting from its action on charged particles? If so, how would this reaction manifest? Would this mean a decrease in energy of the radiant EM energy? Does it slow down? Change its path? Is it explainable without significant explanation of special relativity or any QED?

Thanks,
Philalethes

2. Apr 21, 2012

### Bob S

You may be interesed in a simulation of the field lines of a moving point charge, using the Cal Tech simulation tool at http://www.cco.caltech.edu/~phys1/java/phys1/MovingCharge/MovingCharge.html
The first panel shows the field lines of a charge moving at a velocity β = 0.5 from the left to the right. The field lines are straight, and slightly compressed toward the 90 degree line (due to length contraction along direction of motion). The lines move out away from the charge at β=1 in the reference frame of the observer.

In the second panel, a charge moving at β = 0.5 is suddenly decelerated to β = 0.225. This produced a kink in the field lines, which also radiates out away from the moving charge at β = 1 in the reference frame of the observer. Any observer at a distance $\ell$ from the particle will see a retarded signal at delayed time $\delta t = \ell / c \space$.

So the E field lines for a constant velocity charge are a longitudinal E field (i.e., radial), while the E field in the kink of the decelerated charge is transverse. Does this mean that the lost energy is being radiated as a pulsed TEM field?

Here is the caption to the applet
When the charge moves at relativistic speed, the electric field is concentrated near the pole, and consequently the field lines are shifted. The field lines always point to where the charge is at that instant, if we are within the current sphere of information. If that charge has changed speed or direction within a time t < r/c, where r is the distance away, and c is the speed of light, we will not know that charge has accelerated, and the field lines will still point to where the charge would be if it hadn't changed speed or direction.

Notice That when the charge is accelerated, because the field lines must be continuous, it is forced in a direction almost perpendicular to the the direction of propogation. As time goes on, the line becomes more and more perpendicular, the horizontal component increasing faster than the vertical component. Associated with this electric field is a magnetic field, perpendicular to the electric field and the direction of propogation, which describes light.

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3. Apr 21, 2012

### Bob S

The classical radiation (radiated power) from a decelerating charged particle is given by
$$\frac{dW}{dt}=-\frac{e^2 \dot v^2}{6 \pi \epsilon_o c^3}$$
where $\dot v$ is the acceleration. This radiation moves away from the charged particle at β = 1.
See Panofsky and Phillips Classical Electricity and Magnetism page 301.

4. Apr 21, 2012

### Philalethes

Thank you Bob, for pointing me toward the Larmor formula. I found this document from the Wikipedia article about it. In section 5.2, he uses the example of blue light scattering in the atmosphere as a demonstration of energy conservation as light passes by an atom, but declines to explain the exact mechanism behind the decrease in energy of the light! (See attachment screenshot of the PDF)

I'm looking over your first post again to see if the answer is contained therein....

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5. Apr 21, 2012

### Bob S

Kinematically, scattering light off of an atom is just like Thomson scattering off of a single electron, which is the classical limit of Compton scattering. Look specifically at the derivation of this formula for the wavelength shift in http://en.wikipedia.org/wiki/Compton_scattering
$$\lambda'-\lambda=\frac{h}{mc}\left(1-\cos\theta \right)$$
where m is the mass of the electron and θ is the scattering angle. In Rayleigh scattering, the recoil mass is the entire atom, so let's cosider scattering at 90 degrees. We have
$$\lambda'-\lambda=\frac{h}{Mc}=\frac{hc}{Mc^2}$$
Using hc = 4.136x 10-15 eV-sec x 3 x 1010cm/sec = 1.24 x 10-4 eV -cm, and Mc2=1.3 x 1010 eV for a nitrogen atom, we get

$$\lambda'-\lambda=\frac{hc}{Mc^2} = \frac{1.24 x 10^{-4} eV \cdot cm}{1.3x10^{10} eV}=9.5 x 10^{-15}cm = 9.5 x 10^{-7} Angstroms$$
So there is a wavelength shift, but it is very small.

Incidentally, the author of the paper you referenced implied that the nucleus was moving and radiating. It is actually the electron cloud that is moving and radiating.

6. Apr 21, 2012

### Philalethes

Thanks! Thomson scattering describes what I was looking for.

Best,
Philalethes