## the energy of an em radiation

 Quote by apratim.ankur ok, for a single photon the energy E = hv . I get a definite answer. Doesn't this mean that its energy is related to the magnitude of em fields associated with it (as knowledge of the former allows me to calculate the latter or vice-versa)? sorry if it appears stupid, but I am confused....
The number of photons is proportional to the square of the magnitude of the EM fields.
 Recognitions: Gold Member Science Advisor It is sometimes better to consider the Energy rather than the Field because the E and H fields do not always have the same ratio (when not in a Vacuum) and Z0 is different.
 one question is in the diagram itself and there is another one based on that here :- would this induced E-field (in the wires) be equal or related to the E-field component of the incident light as per the diagram? sorry for the crude drawing Attached Thumbnails

 Quote by Darwin123 The photon is not truly "massless". It has a zero rest mass. However, it can have a large relativistic mass. The energy of a photon is never zero. The photon always has a finite energy in any inertial frame. The uncertainty relationship holds to energy and time, not rest mass and time. Photons have a zero rest mass. This does not mean that they have no energy. So your reasoning doesn't really work on photons.
I know all of this, I have a degree in physics :P

First, I don't at all like the term "relativistic mass", it's antiquated, and educators are no longer using it. Mass is invariant under the Lorentz transformation, and it doesn't really shed any light on anything to make up some new form of mass that varies with velocity; there is a more formal explanation for why energy goes to infinity as a massive object approaches c.

Second, nowhere did I say that photons don't have momentum or energy. That would just be silly! You may recall an experimental argument for the uncertainty principle, in which relativistic muons created in the upper atmosphere are detected on the Earth's surface, when they should have decayed long before reaching it. Because the muons are traveling close to c, their measurement of Δt between creation and decay is much different than ours, and their journey to the Earth's surface happens in a time interval, in their reference frame, which is shorter than their decay time in that reference frame. It was explained to me that the infinite range of electromagnetism happens the same way. Photons travel exactly the speed of light. In SR, it doesn't make sense to talk about time and space from the reference frame of a photon, and this is why infinitely propagating virtual photons don't violate the Uncertainty principle. Though, the more I think about this explanation, the less it makes sense, because it seems to violate energy conservation in other frames. Maybe you can work through that one for me.

I understood the rest of your post, but I didn't understand where you were going with it. What does virtual photon polarization have to do with the range of electromagnetism or the range of virtual photons? (Not saying it doesn't have anything to do with it, I'm just curious, and I didn't see the connection in your post.)

 Quote by soothsayer That's not true; Electromagnetism would not have an infinite range if the virtual photons mediating it had a finite range. I was under the impression that the range of the virtual photon should be infinite since it is massless, thus it circumvents the uncertainty principle because the photon has no reference frame or Δt of which to speak of.
Electromagnetism doesn't really have an infinite range. The statement "infinite range" is a qualitative statement meaning "very big". If you preferentially weight the interactions with large momentum transfer, the effective range is much less than infinite.
I can not tell for sure what you mean by infinite range. However, I conjecture that you mean that the scattering total cross section of an electric charge is infinite. However, the total cross section includes interactions with a very low change in momentum. The total cross section is merely the integrated value of the differential cross section over all angles.
The differential cross section of an electric charge is finite, decreasing with scattering angle. The momentum transfer increases with scattering angle. The scattering angle decreases with distance of closest approach to the charge. Therefore, the momentum transfer is actually decreasing with distance of closest approach. If you think of the distance of closest approach as a type of "range", one can easily see that the "range" decreases with momentum transfer. The momentum transfer is proportional to the energy of a photon. So the "range" of the electric field actually decreases with photon energy.
Your visual picture implies that a Coulomb potential has an infinite range because the electric field is nonzero at all finite distances. However, that is equally true of the Yukawa potential (i.e., exponentially decreasing potential). The fact that the meson or gluon has a nonzero rest mass does not change the fact that the Yukawa potential tapers off to zero at infinite distances.
The truth is that almost all force laws present a nonzero force at finite distances. The Yukawa potential results in a finite value for the total cross section. In contrast, the Coulomb potential has an infinite value for total cross section. Both forces decrease with distance and never disappear completely.
Virtual photons with greater energy don't travel as far as virtual photons of low energy because of the uncertainty principle. This corresponds to a differential potential that decreases with scattering angle. Claiming that virtual photons travel forever because of their zero mass is misleading.
In fact, photons have a zero REST mass. The zero rest mass is associated with an infinite value for total cross section of an electric charge. However, the total cross section is not really a range.
One can correlate a "range" with the square root of the differential cross section. However, the differential cross section varies with momentum transfer. Therefore, one could say that the range varies with the momentum of the photon.
The uncertainty you should be considering is not in energy, but momentum. The uncertainty is determined by,
2πΔpΔx≥h
where Δp is the momentum of the virtual photon, Δx is the "range" and h is Planck's constant. Even though the rest mass is zero, Δp does not have to be zero.
The photon should not be called massless since it has a momentum. It should be called "rest massless."
The following link is badly written. It claims that “since the photon has no mass, the coulomb potential has an infinite range.” It does not make it clear what they mean by range and what them mean by mass.
http://en.wikipedia.org/wiki/Virtual_particle
“Some field interactions which may be seen in terms of virtual particles are:
The Coulomb force (static electric force) between electric charges. It is caused by the exchange of virtual photons. In symmetric 3-dimensional space this exchange results in the inverse square law for electric force. Since the photon has no mass, the coulomb potential has an infinite range.
The magnetic field between magnetic dipoles. It is caused by the exchange of virtual photons. In symmetric 3-dimensional space this exchange results in the inverse square law for magnetic force. Since the photon has no mass, the magnetic potential has an infinite range.”

I am pretty sure that by mass they mean “rest mass” and by range they mean “total cross section.” However, what you think of as range probably has to do with differential cross section than total cross section.
Each virtual photon has a finite "range" that decreases with its momentum. Since the rest mass of a photon has nothing to do with its momentum, the range has nothing to do with its rest mass.
There is a conceptual construct called "relativistic mass" that resolves some these intuitive problems. However, I have been told that it causes more problems then it resolves. So I won't insist on using "relativistic mass." What is important is that a photon has a nonzero momentum.
 I understand everything you try to say, but you lose me when you say that the Yukawa and Coulomb potentials have infinite range even though the photos, gluons, etc that are responsible for the potential do not have infinite range. Does the first derivative of the potential go to zero after a finite amount of distance? In classical electromagnetism, the Coulomb Force is proportional to r-2. This means that the force is nonzero for all finite r. Am I to understand that QED puts a constraint on this force, such that the Coulomb force is zero for some finite r? If not, then how can there be a force at some arbitrarily large r where virtual photons cannot reach due to their limited nature?

 Quote by soothsayer I understand everything you try to say, but you lose me when you say that the Yukawa and Coulomb potentials have infinite range even though the photos, gluons, etc that are responsible for the potential do not have infinite range. Does the first derivative of the potential go to zero after a finite amount of distance?
No. The first derivative with respect to distance from the stationary body does not go to zero for either of these potentials. The first derivative with respect to distance for both potentials is negative all the way out to infinite distance.
Both potentials approach zero in the limit where the distance approaches infinity. However, the force is nonzero for both potentials.

 Quote by soothsayer In classical electromagnetism, the Coulomb Force is proportional to r-2. This means that the force is nonzero for all finite r. Am I to understand that QED puts a constraint on this force, such that the Coulomb force is zero for some finite r?
No. The Coulomb force is never zero at any finite distance. A previous poster (you?) tried to explain this by saying the photon has zero mass. However, he left out something.
The Yukawa force is never zero at any finite distance. However, the mediator of the Yukawa force does have a positive rest mass. So the "range" of the Yukawa potential is limited in the same sense that the "range" of the Coulomb potential is limited. The picture in both cases is that the mediator, a virtual particle, must disappear before it violates an uncertainty relation.
The mediator of the Yukawa potential is mesons for nuclear physics, plasmons for solid state physics, etc. These mediators have a finite rest mass. Because the maximum speed of any object in SR is "c" (speed of light in a vacuum), one can use the uncertainty principle for energy and time to estimate the range of the mediator in a vacuum. The range of the mediator is finite because the Yukawa mediator always has a positive energy. Here, by range I mean the square root of the scattering cross section.
The mediator of the Coulomb potential is photons for electromagnetic theory, gravitons for gravitational theory, and photon-polaritons for solid state theory. These mediators have a zero rest mass, Because the photons have a zero rest mass, one can not use the uncertainty principle for energy and time to estimate the range of the mediator in a vacuum. However, one can use the uncertainty principle for momentum and distance to estimate the range of the mediator in a vacuum.
The uncertainty principle for momentum clearly defines a range for the virtual photon. However, the "range" is infinite because the momentum of a photon can be zero. Here, by range I mean the square root of the total cross section.
 Quote by soothsayer If not, then how can there be a force at some arbitrarily large r where virtual photons cannot reach due to their limited nature?
The same reason there is force at some arbitrarily large r where the Yukawa mediator can not reach due to its limited nature. There is a finite "possibility" that a virtual particle will have have a small enough energy or momentum to reach the distance, r.
The uncertainty principle has an arbitrary constant determined by the implicit precision of the measurement. The constant usually given (h) is for a measurement precise up to the standard deviation of measurements. For measurements precise up to any multiple (n) of the standard deviation, the constant has a higher value (nh).

My point was that if you take the mathematical theory literally, the virtual photons in an electromagnetic interaction have to vanish at sufficiently large distances. The virtual photons have a zero rest mass, but they still have to disappear. The virtual photons travel much farther then a virtual particle with positive rest mass.

I never had a conceptual problem with this because I never took the "virtual particle" model as literal truth in a classical sense. I think of "virtual particles" as a diagram technique to make Fourier analysis that much easier. However, I understand why one may need a heuristic picture to understand "virtual particles".
If one takes the "virtual particles" literally then one should expect the virtual particles to spontaneously disappear at some distance by definition. If one takes "virtual particles" literally, then "real particles" do not spontaneously disappear by definition. Real=Persistent.
Further, there are two uncertainty principles that one should always consider. Uncertainty in energy and time are what determines the "range" of virtual particles with a positive rest mass. Uncertainty in momentum and distance are what determines the "range" of particles with a zero rest mass.

Ok, I think that clears up some of my confusion. Especially the part where you say:

 There is a finite "possibility" that a virtual particle will have have a small enough energy or momentum to reach the distance, r.
I hadn't thought about it that way, but that makes a lot of sense now.

Still, it seems like the Coulomb potential would drop off faster then r^2, in that case, because of an increasing number of vanishing virtual particles and a decreasing virtual particle density due to the spherical configuration of forces, which alone should drop off like r^2.
 Short comment on the range of the fields: coulomb fields fall off as 1/R^2 while radiation fields fall off as 1/R so as you take the limit to infinity, the radiation field dominates. Regarding the field of a single photon, it's best to think of it like the double slit. Each individual photon does obey maxwells equations. They can be considered to be the dynamical equations of a photon. But once you detect a photon (yes, by registering a tiny electric or magnetic field), that field ceases to exist everywhere else. In this sense it never really existed like the classical field typically does even though the probability of finding the photon obeys the energy relation of the classical fields defined over all space. It is the collapse of the wave function phenomenon but with photons. It *is* true that a single higher energy photon would register a larger electric/magnetic field than a longer wavelength/lower energy quantum when detected.

 Quote by soothsayer Ok, I think that clears up some of my confusion. Especially the part where you say: I hadn't thought about it that way, but that makes a lot of sense now. Still, it seems like the Coulomb potential would drop off faster then r^2, in that case, because of an increasing number of vanishing virtual particles and a decreasing virtual particle density due to the spherical configuration of forces, which alone should drop off like r^2.
The Coulomb potential falls of as r, not r^2. The Coulomb force falls off as r^2. So the Coulomb potential falls off a lot slower than the Coulomb force.
The relationship that determines the range of the photons is the uncertainty in momentum times the uncertainty in distance (i.e., range). So the range in the case of the electromagnetic potential is determined by the momentum. So the virtual photons carry momentum, not energy.
Classically, the change in momentum is equal to the impulse which is force times distance. So the flux of virtual photons actually determines the force. The flux through a spherical surface is proportional to the inverse square. So the force is inverse square.
The Coulomb potential is a measure of the "energy" that is carried by the virtual photons. However, the photons have no rest mass. So there is no lower limit to the energy of a virtual photon. So the Coulomb potential extends way past the Coulomb force.
In the quantum mechanical calculations, the projection operators are summed over both momentum states and polarization states. There is no summation over the energy states because the energy of the virtual photon is totally unknown. The momentum of the virtual photon is determined by distance, but the energy of the virtual photon is unknown.
Note that in the case of real photons, the intensity varies as the inverse square. That is because in the case of real photons, the photons carry a fixed momentum and a fixed energy. So both the energy flux and the momentum flux of real photons vary as inverse square. Since the intensity is proportional to both energy flux and momentum flux, the intensity decreases as inverse square.
The virtual photon has a momentum that decreases with distance due to the uncertainty principle of momentum. However, the energy of the virtual photon isn't determined by anything.
Thanks! I didn't understand that part before. Now, some of the Feynmann rules make sense!
 Yeah, whoops, I meant to say the Coulomb Force falls as r^2, not the Coulomb potential. Thanks!