# EM radiation and momentum

1. May 10, 2010

### snoopies622

I've read that even before the 20th century physicists realized that light carries momentum, and that - although experiment is the ultimate arbiter in science - one can arrive at this conclusion by studying Maxwell's equations alone. If this is the case, could someone give me an outline of the reasoning employed?

2. May 11, 2010

### clem

In the interaction of EM fields with matter, the momentum of the matter is not conserved.
Conservation of momentum is restored by lattrilbuting momentum to the EM field.

3. May 11, 2010

### snoopies622

Thanks, clem. I'll give that some thought..

4. May 11, 2010

### matonski

The following argument is taken from the Feynman Lectures on Physics, Vol 1 sec 34-9.

Light consists of transverse electric and magnetic fields. If you shine light on a surface, the charges in the surface will move in the direction of the electric field. Since this direction is perpendicular to the magnetic field, the charges will experience a magnetic force in the direction that the light is traveling. Thus, the surface will receive momentum due to the light.

In detail, the force due the the magnetic field is.

$$F = qvB$$

Since $B = E/c$, in a light wave,

$$F = qv \frac{E}{c}$$

However, $q E$ is the electric force on the charge and force times velocity is the work done by the field. Thus, let $W$ be the energy in the field. Then,

$$F = \frac{1}{c}\frac{dW}{dt}$$.

Thus, since $F = dp/dt$

$$p = \frac{W}{c}$$

In other words, the momentum carried by light is equal to its energy divided by c.

5. May 11, 2010

### Creator

Hi Snoops;
Apparently you are looking for the Maxwell derivation (rather than the quantum explanation) and I'll assume you want the explanation for LINEAR momentum , (and not angular momentum), which results in momentum (pressure) given by the equation of the time averaged Poynting Vector, <S> divided by c.
P = <S>/c

That is an excellent question, and not commonly appreciated.
The first thing to recognize is that a formal derivation for the momentum (pressure) is found by Integrating the average Maxwell stress tensor over the area.

So the first thing to do is find that tensor according to Maxwell.

It can be done by equating the mechanical force equation of a single charge (f = qE + qv x B) to that of the electromagnetic field distribution using Maxwell's eqns....(which satisfies the continuity (charge conservation) equation, Div J = - d(rho)/dt ).

See equation 13.5 here....in an excellent formal derivation I found here in a good Optics text by Hecht:

Begin on page 420 with the section "Maxwell Stress Tensor" for the initial set-up in vacuo, and continue thru pages 421 -423 for the develpment, espec. eqn. 13:10. And then see its application on page 423-424 for finding the radiation pressure...(eqn.13:21)

And of course, experimental verification was done by Nichols and Hull (in 1901): See here (pg 315):

Creator

"The Works of the Lord are great, studied by all who have pleasure therein". - Inscribed in the Archway of the Door leading to James Clerk Maxwell Cavendish Laboratory - Cambridge

Last edited: May 11, 2010
6. May 12, 2010

### snoopies622

Thank you both for filling in some of the mathematical details!

Unfortunately in the Novotny-Hecht Google books entry, "pages 424 to 427 are not shown in this preview", but my local university physics library may have a copy of the text.

7. May 12, 2010

### Staff: Mentor

Here is my favorite page on the conservation of momentum in EM:
http://farside.ph.utexas.edu/teaching/em/lectures/node91.html

And nearby is my favorite page on the conservation of energy in EM:
http://farside.ph.utexas.edu/teaching/em/lectures/node89.html

8. May 12, 2010

### snoopies622

Looks interesting DaleSpam, thanks!

9. May 12, 2010

### dx

Just a thought in relation to this: The interaction of the EM field with matter actually involves the recognition that the EM field must also be considered to be matter, i.e. particles. Fundamentally, the exchange of momentum between radiation and matter, like an electron, is not continuous, as the field picture would seem to suggest. The exchange is actually in the form of particle collisions between the electrons and so-called photons.

10. May 22, 2010

### dan_b_

I have a question about the e-m explanation of radiation pressure.
As I understand it, when an e-m wave with low frequency strikes a material with a much higher resonance frequency, the displacement of the electron relative to the atom will be in phase with the electric field (well opposite phase if you account for the negative charge).
When the electric field incident on an atom is at the maximum value of the wave cycle
(ie. E(t) = Eo=Emax) , the velocity of the vibrating electron would be zero (at a turning point). The magnetic field, though also a maximum at this time (t), would produce no force on the electron at this moment.
If one explores the force produced by the magnetic field on the electron a small time before time t (e.g. 10% of a cycle earlier, when the vibrating electron is in motion) one finds that the electron (say moving in the +y direction) will experience a lateral force, say in the direction of the incident radiation. But if one compares this to the force experienced a short moment after time t (e.g. 10% of a cycle later) the electron is moving in the opposite direction (having passed the turning point). The B field still has the same sign but the electron's velocity has reversed - the lateral force produced by the B-field must also be reversed. As I see it radiation pressure is not predicted - the lateral force on the matter
would be oscillatory. The net force produced would average to zero over a cycle.

I have read Poyting vector analyses that generate the radiation pressure, but they never seem to discuss the instantaneous relationships of the phases of interest.

Is there something wrong with my thinking? Is there an article that accounts for the phase
relations in the radiation pressure analysis? I would be most grateful if I could get the help to "get over this one"!

11. May 23, 2010

### snoopies622

Geez dan_b_, that's an interesting question. Here are a couple thoughts:

1. When the electrical field is pushing the electron in one direction while it's still going in the opposite direction, it isn't doing any work on the electron, but instead the electron is doing worth on it.

2. A vibrating electron creates an electromagnetic wave of its own. Maybe this wave somehow interferes with the original wave to cancel the backward push on the electron and not the forward push.

But I suppose only a complete mathematical analysis - using potentials and all - would really get to the bottom of the matter.

12. May 23, 2010

### matonski

I think the argument works at resonance, where the phase shift is 90 degrees and the light is fully absorbed. Away from resonance, the pressure is less because less of the light is being absorbed. Thus, the amount of light being absorbed and the momentum transferred are kept consistent.

13. May 23, 2010

### dan_b_

Snoopies - thanks for your response. It sounds like your second idea involves considering the effect of the interaction of a secondary wave (produced by the vibrating electron) and the incident wave. If the absorber was a single atom or maybe an antenna, then I guess the secondary wave would follow the dipole radiation distribution. I guess the interference with the incident wave would lead to something that "in the wings" looks like dipole radiation, but in the forward and backward directions may differ due to the interference. I'll have to give this some thought. Maybe I'll think about Rayleigh scatter too, while I'm at it.

Matonski - Thanks also for your reply. I see your point - at resonance the driving force is in phase with the electron's vibration velocity (as opposed to the low driving frequency limit, where the force and displacement are in phase). This would lead to a forward push on the electron (i.e. in the direction of the wave) throughout the entire wave cycle. But I still wonder what happens far from resonance. Do you think the energy of the wave is "preferentially absorbed" only during the portion of the wave cycle that would lead to a transfer of momentum in the forward direction?

14. May 23, 2010

### dan_b_

I think maybe my assumption that the displacement of the vibrating electron is completely in phase with the incident wave was too extreme. For low frequency radiation, although the sinusoidal displacement function lags the electric field by a very small amount, this amount can't be ignored. Although the absorber would experience an ocillating force parallel to the direction of the incident wave, the slight phase difference between driving force and responding vibration displacement leads to a slight difference in the amount of forward and backwards push by the wave. I am not sure if this would be due to a difference between the peak values of the push and pull forces or if it would be due to a small difference in the amount of time that the push and pull forces act (hmmm). Either way, there must be a small dominance by the effects of the forward push force versus the backwards pull force leading to a small amount of radiation pressure in the direction of the incident radiation.

15. May 23, 2010

### sophiecentaur

That appears not to agree with the fact that, when light is reflected, there is twice the momentum change which occurs when it's absorbed.

(just to add confusion):uhh:

16. May 23, 2010

### matonski

Perhaps its because reflection is absorption plus radiation. So you then have to consider the field of the reflected wave as well as the incoming one.

17. May 23, 2010

### sophiecentaur

That would give the right answer.!!

18. May 23, 2010

### snoopies622

Now I'm confused. I thought the assumption was that the electron's speed lags behind its acceleration (the electric field) by ninety degrees, and that the displacement lags behind the speed by another ninety degrees. This would make the electron's velocity and the (magnetic) force work to push the electron forward half the time and backward during the other half.

19. May 23, 2010

### dan_b_

The assumption was the low driving frequency limit, where f(wave) << f(resonance).
Then the electron's transverse displacement will be in phase with the transverse force from the electric field. The electron's transverse velocity would be 270 degrees behind the force and displacement (or 90 degrees ahead, if you prefer). The acceleration would be in phase with the force and transverse displacement.

I think this would lead to longitudinal vibration of the electron, i.e the radiation pressure will alternate from forward to backward twice per cycle. But this would lead to no overall transfer of longitudinal momentum to the atom. Maybe this tells us that the magnetic field's ability to couple with the vibrating electron/atom vanishes as the driving frequency decreases. If so, this might tell us more. The amount of energy absorbed from the electric and magnetic components of the fields must be proportional (E/B = c, and any unused, scattered portion of the wave must have E and B components in this proportion so the absorbed amounts must also be proportional to one another). Consequently, if the magnetic field can't couple to the oscillator, then neither can the electric field. The wave is forward scattered, and the absorption cross section tends to zero. Maybe this represents a classical explanation of why the absorption of an electromagnetic wave must tend to zero in the low frequency limit. What do you think?

20. May 24, 2010

### snoopies622

I'm afraid I still don't understand the set-up. If the frequency of an applied external force and that of a particle's resonance due to a built-in restoring force are different, wouldn't that make the particle's motion the sum of two sinusoidal waves instead of just a simple sinusoidal wave?