# How do you reconcile EM fields with frequency of light?

1. Apr 13, 2009

### benbenny

I know that a charged particle emits photons which are the "carriers" of the electric and magnetic fields, and that these photons, interacting with another charged particle, cause an attractive force or a repulsive force.

I also know that examining the amount of energy in the photons - which are the quantas of light radiation - is basically synonymous with looking at the frequency of the light waves.

BUT I also know that the same type of electromagnetic radiation can be emmited as heat radiation such as the lamp in my room.

I cannot understand how these concepts work together.

I guess my first questions would be: Does the light coming from the lamp in my room carry E and B field force potential? i.e if there was a floating electron right here would it be affected by these photons?

If not, then why not? how do you characterize the difference between these heat emitted photons and the "type" of photons that are the carriers of the E and B forces?

If yes, then please explain how to reconcile culoumbs law with this. i.e how do you represent a force inversely related to distance squared with photons? And is there a relation between the quanta of energy of the photons ( the frequency) and the strength of the EM field?

Assuming that these photons that are emmited from my lamp can not affect a charged particle - then how do you characterize the difference between them and the "type" of photons that are the carriers of the E and B forces?

Thanks for any input.

2. Apr 13, 2009

### Staff: Mentor

These are virtual photons, not real photons. Look in the Quantum Physics forum here and you'll find lots of threads discussing virtual photons and how they're different from real photons. (Sorry, I'm too busy figuring out my taxes right now to try to give a more long-winded answer. :grumpy:)

3. Apr 13, 2009

### benbenny

ah ok. Thanks. now I know what to search for.

4. Apr 14, 2009

### Matterwave

Also, Coulomb's law is an electro-statics law. Electric fields can also be caused by changing magnetic fields not just static charges. So, I don't think you can really do with Coulomb's law when dealing with EM waves.

5. Apr 14, 2009

### atyy

Yes.

Yes.

Light comes from Maxwell's equations. Coulomb's law is what you get when Maxwell's equations are applied to static charges.

6. Apr 14, 2009

### Born2bwire

The heat radiation you are talking about is the same electromagnetic radiation as visible light with the only difference being the frequency. On the point of the strength of the field, the energy of the photon is dependent upon the frequency but the energy of the EM wave is dependent upon the amplitude of the wave. The difference here is that the EM wave is born out of classical EM theory (though you can relate it back to photons using QED). If we have a source that is radiating two EM waves at different frequencies but at the same field amplitudes, then the difference in terms of the photons is that the lower frequency case has a higher rate of photons being emitted. This can actually introduce a "granularity" in the photons for a given wave. A high frequency wave can exhibit more granularity than a low frequency wave of the same energy due to the difference in the rate of photons.

7. Apr 14, 2009

### benbenny

ah. ok.
so frequency of the wave is related to the rate of photons AND the quanta of energy in each photon as given by planck's formula?
you are talking about "rate" of photon emission which is different than number of photons i.e intensity of the light?
so you are talking about 2 waves at same amplitude but different frequencies. what about same frequency but different amplitudes? how do you explain that with photons?

I think that this is all explained by "virtual photons" which jtbell mentioned above and which ive started reading about elsewhere.

thanks a bunch

8. Apr 14, 2009

### Staff: Mentor

No, the amplitude of a classical electromagnetic wave is related to the number of photons in the wave.

[added] Oops, for a fixed amplitude, the number of photons does depend on the frequency. See post #11 below. I was thinking of the energy in the wave depending only on the amplitude.

An electromagnetic wave falling perpendicularly onto a surface of area A for a time interval $\Delta t$ delivers energy to the surface:

$$E = \frac{1}{2} \epsilon_0 c E_0^2 A \Delta t$$

where $E_0$ is the amplitude of the electric field in the wave. This corresponds to a certain number N of photons via E = Nhf.

With electromagnetic radiation (waves) you're dealing with real photons. Virtual photons enter into things like the electrostatic Coulomb force.

Last edited: Apr 14, 2009
9. Apr 14, 2009

### benbenny

10. Apr 14, 2009

### Naty1

benbnny:
You've asked a lot of good questions...try reading wikipedia under PHOTONS..there is a lot to consider.

regarding Coulomb's law, the inverse square ratio results from the surface of a sphere A = 4(pi)r2...so if you visualizes lines of electromagnetic force (or equivalently numbers of photons) originating from a point charge and passing through a pair of concentric spheres, the concentration of the lines (lines per unit area) varies as the inverse square of the radius....in other words photons will be less concentrated passing through the larger sphere....and that's a result of our three dimensional world...in higher dimensions, force drops off more rapidly...

More generally,
photons sure are curious: they are not electromagnetic, not carrying any charge, so they pass through electric and magnetic fields without deflection....yet gravitational potential will change their direction...hence accelerating them...but they do NOT carry the gravitational force (gravitons do that)....seems strange.....
are they affected by the strong and weak nuclear forces??..I'm not sure...

Also, it worth noting that Coulomb's law does NOT hold rigorously at relativistic speeds...only if charged particles do not move with respect to each other is it precise...

Equally curious: every electric charge gives rise to an electric field, yet a moving electric field gives rise also to a magnetic field...How do the photons know this??... Is a photon from a static field different than one from a field in motion?? I'm not sure...

11. Apr 14, 2009

### Staff: Mentor

Yes.

The area is there because if you double the area of the surface (think of a sheet of paper that you hold up, facing the sun), you double the number of joules of energy that hit the surface in the given time interval $\Delta t$.

That's because the two equations are really for two different purposes. The first one tells you how much energy you get out of an EM wave with a given amplitude $E_0$. The second one tells you (after rearranging it to N = E/hf) how many photons that corresponds to. You can combine the two equations to get

$$N = \frac{\epsilon_0 c E_0^2 A \Delta t}{2hf}$$

which tells you how many photons hit that same surface in the same time interval.

It isn't. The energy in a classical EM wave does not depend on the frequency, only on the amplitude.

Note that the higher the frequency, the larger the energy of a single photon. This means that for a constant amplitude (and therefore constant energy), higher frequency EM waves correspond to fewer photons.

In classical EM, you can distinguish between two "kinds" of fields: radiation fields (EM waves) which carry energy and momentum from one place to another, and non-radiation fields, which do not. The Coulomb field produced by a stationary electric charge is a non-radiation field. So is the magnetic field produced by a stationary magnet, or a steady electric current. Radiation fields correspond to real photons. Non-radiation fields correspond to virtual photons.

Last edited: Apr 14, 2009