# Energy of light vs energy of photon

1. Jan 30, 2013

### xaratustra

How is energy of electromagnetic waves (classical electrodynamics → Poynting vector) related to the photon energy (E=hv) ?

e.g. what is exactly the difference between photons from two radio stations or two laser devices with the same frequency, but different power?

thanks!

2. Jan 30, 2013

### Drakkith

Staff Emeritus
The energy PER PHOTON is given by the equation E=hv, where h is planks constant and v is the frequency of the EM wave. This does not change, no matter how much power you pump out of a laser or radio antenna. If you increase the power of a transmitter or laser by 10x, you increase the number of photons emitted by 10x, not their energy.

3. Feb 1, 2013

### xaratustra

Great! thanks.

4. Feb 1, 2013

### Crazymechanic

Frequency is the measure of a photons energy, but power from a laser or antenna or what not is frequency times (x) photon density.

Like if you compare what one Bruce Lee could do and then imagine Bruce Lee copied himself to ten , their both the same guy with the same abilities just in one case they are far more so they can do more in a given amount of time but the power of each one of them stays the same.

Last edited: Feb 1, 2013
5. Feb 1, 2013

### Drakkith

Staff Emeritus
Crazy I feel your description is going to confuse people. First, transformers work by induction, not by sending out EM radiation. A "larger" transformer doesn't really mean anything. The physical size of the transformer has no direct bearing on the amount of power consumed by it. Number of turns in the coils, spacing, and a dozen other variables are far more important. Just because one transformer is bigger than another doesn't mean that it actually transfers more power than a smaller one.

6. Feb 1, 2013

### Crazymechanic

Well I agree , my analogy wasn't the best one, I'm sorry.
But a larger transformer of the same frequency does produce a stronger EM field hence more photons.
I think we can use a simple analogy that when I built my first amplifier I used a test transformer which was about 100w and 220v at 50hz, then I took one that is rated at 300w and with the second one I had problems of mains hum going through my circuit , that's why I needed to shield or move the transformer further away.
How can you explain this than by a larger more powerful EM field and as long as I know em field is made of photons.?

P.S. Anyways edited my post without the transformer analogy as I don't want to make anyones life more complicated as it already are.

7. Feb 1, 2013

8. Feb 1, 2013

### Drakkith

Staff Emeritus
What do you mean by "larger"?

Ah ok, by "larger" you mean more power. That's fine.

Because as I said a transformer does not use EM radiation to transfer power. You are correct in that the photon is the gauge boson for the EM force, aka it is the force carrier. However if we want to talk about the field being made up of photons then we would have to get into virtual particles and other topics that are far too complex for this thread.

A good rule of thumb is to never take away information from your posts, as that can lead to a confusing thread when people start replying to you and especially when they quote you. Simply put something like "Edit: Ignore this analogy" instead. (But that's just my personal opinion. It's just what I do myself) This thread is actually a good example. People may read the thread, get to my post where I replied to you and then get all confused since you deleted some of yours.

9. Feb 1, 2013

### Crazymechanic

Yes , I agree again it's far better to edit a post before someone has responded , after that just leave it as it was.

Well the transformer was an unnecessary analogy in this thread from the start and I said my mistake for it but it's actually kinda complicated one too because the transformer itself relies on induction as long as we speak of how the current is induced from primary to secondary , but the current itself while passing creates the EM field , without which it couldn't pass in the first place as we know it and the field around the transformer even though in many cases is more of a headache than a benefit is also Em again photons, maybe it would just be wise either from me and from other more experienced forum members to choose when to explain situation in simple classical terms and when to go deeper , I would not have made all this talk if had not made the now infamous analogy about the transformer which dragged me into a little discussion.Well I hope that this info only serves to educate someone who is reading it, as I sometimes learn quite a bit from some of the threads with a lot of arguments here , rather than the ones were someone just asks a question and someone just gives a simple answer even if it doesn't describe the situation fully and then leaves it there.

Last edited: Feb 1, 2013
10. Feb 1, 2013

### Drakkith

Staff Emeritus
No problem. We all have to learn somehow.

11. Feb 1, 2013

### Darwin123

The energy that is transformed in a transformer is not carried by real photons. The energy in a transformer is carried by “virtual photons”. The quantum-mechanical description of near-field systems like transformers are not consistent either with source-free electromagnetic waves or with real photons.

Transformers do not carry “radio waves” in the most precise definition of the phrase “radio waves”. The wavelength of the “waves” in a transformer are not well-defined. Therefore, the momentum of the “radio wave” photons in the transformer can’t be well defined. See the first link that I copied below.

In classical electrodynamics, electromagnetic disturbances can be classified roughly in terms of near-field effects and far-field effects. I won’t go into the mathematical details but the physics relate roughly to how closely associated the electric charges are to the disturbance. The far-field effects include radio waves, light waves, and other long-range waves. These disturbances act independently of the oscillating electric charges that originally generated them. For instance, a static-magnetic field has a very prcise relationship with the electric current in the wire that generates it. The near-field effects include electromagnetic-induction, static-electric fields and static-magnetic fields. These disturbances are highly correlated with the oscillating electric charges that generated them.

In quantum mechanics, the near-field effects correspond to virtual photons while the far-field effects correspond to real photons. Although the de Broglie relationships have a clear interpretation in terms of real particles and far-fields, they have a less clear interpretation in terms of virtual particles. This corresponds to the wave properties of a near-field disturbance not being well defined.

The virtual particles do not have a well-defined energy or momentum. The waves associated with near-field effects do not have a well-defined wavelength or direction.

http://en.wikipedia.org/wiki/Virtual_particle
“Virtual photons are also a major component of antenna near field phenomena and induction fields, which have shorter-range effects, and do not radiate through space with the same range-properties as do electromagnetic wave photons. For example, the energy carried from one winding of a transformer to another, or to and from a patient in an MRI scanner, in quantum terms is carried by virtual photons, not actual photons.”

http://multimedia.biol.uoa.gr/2009-2010/metaptyxiaka/aktinobolies/global neww/2-6-10/emfields.pdf
“ Now what about the energy in static or reactive electromagnetic fields? Quantum physics states that any energy must consist of individual packets, or quanta, but this statement implies that even the static field must consist of particles. In fact, the static field does consist of particles—virtual photons. To explain virtual photons, step further into the strange world of quantum physics.”

12. Feb 1, 2013

### Staff: Mentor

By the way, an interesting historical note is that the answer to this question is what earned Einstein his Nobel prize, not his work on relativity. He measured the photoelectric effect with light of different intensity and frequency. He found that increasing the frequency made more energetic electrons (E=hv) and increasing the power just made a larger number of electrons at the same energy.

13. Feb 2, 2013

### Crazymechanic

@Darwin123 Interesting because I thought that the difference between real or virtual photons was the field with which they associate like virtual ones in a static field like that of a permanent magnet and real ones in a time varying , changing field.

So basically a virtual photon becomes a real one after some time that it has been into existence ,in that case how do we measure a photon from a radio transmitter tower ? Like when were standing on the tower or in a close proximity the field is composed of virtual photons , and when we backup go away the field now is further and the time of the photon to be in existence has been longer so it has become a real one?

Oh and by the way have been wondering about this one , why the photon dies off quicker in a near field like the transformer case than in an antenna , because of it's lower wavelength?

P.S. do we consider wireless internet sitting like 20m from the router a real photon field or virtual? :D I just wanted to know am I sitting in real photon field now or not? :D

Last edited: Feb 2, 2013
14. Feb 2, 2013

### xaratustra

this has become a wonderful thread thanks to you all, I love it!

In fact Darwin123 made a wonderful description, which I can now relate to what I have read in a book. Maybe the answer to Crazymechanic is also here.

Near field effects seem to be the result of the interaction of 2 charges that are in relative constant motion to each other, including velocity = 0. I forgot in which book (either Jackson or Rohrlich), but the near fields are also called velocity fields. This is the domain of the virtual photons, as was mentioned before. The far field effects on the contrary, are sometimes called acceleration fields. Here you need only one charge which is being accelerated. The electromagnetic radiation is then caused by real photons.

In the first case, as was mentioned in this thread, the motion of charged particles in the near fields are strongly correlated. In the transformer analogy, the electrons in the secondary winding and in the primary winding know about each other. That is why the power that is consumed in the secondary must be provided by the primary, according to the ratio of their turns.

Acting as a primary winding of an imaginary transformer, a radio station radiates a certain amount of power no matter if there is an imaginary secondary winding somewhere in the world that receives it or not. Whether your charge is located in the near field or far field depends mainly on the frequency. For "mains hum" of a high power transmission line, near field is several kilometers, for a GHz "horn antenna" a couple of mm.

Now, maybe the answer to Crazymechanic is the following:

If your charge is situated in the near field of and accelerating source, then I think you have a superposition of velocity and acceleration fields there, so that there are both virtual photons and real photons in this region. But only real photons can actually propagate, unless they are stopped by doing work on your charge.

what do you think?

A related thing is: What about gravitational fields, are we most of the times in the "near field" of gravitational sources?

cheers!

15. Feb 2, 2013

### Crazymechanic

good post @xaratustra , well the transformer /radio antenna was a pretty good analogy , well the electrons from secondary to primary don't know for sure about each other I guess only through the magnetic field, it's like an elastic wire and two cars towing each other, you don't know what the first is doing but you can feel the effect.

I don't know about the near field or far field of gravity but I know that it gets weaker by distance just like em field by the square root of r.
the difference would be that em field carries energy but gravity does not according to how it is described in GR, I think of gravity like a property of mass which it exerts on the matrix of space.
Still there are some questions in my last post which I kinda would wait for more explanation from another source until i can fully coupe with them.
About the when a virtual photon become a real one well in the power line example how do we know were the near field ends, as the time a photon has been in existence can be measured by the speed of the photon and the distance it has traveled, so if the near field of lower frequency power line is many km which it is then where does the near field end and why doesn't the photon go any further.?

Last edited: Feb 2, 2013
16. Feb 2, 2013

### Darwin123

Exactly!

Most of the phenomena that we currently associate with gravity is in the "near field"! The near field includes all orbital phenomena, as well as scattering. Therefore, the gravitons that are exchanged in orbital phenomena are "virtual gravitons" rather than "real gravitons".

Astronomers are looking for "gravitational waves". These are far field phenomena. The gravitons associated with these gravity waves would be real gravitons, not virtual gravitons.

I would like to get away from gravitons for a while because of a well known conflict. It is well known that there are some contradictions between general relativity and quantum mechanics. Therefore, I am a little uncomfortable talking about "real gravitons" versus "virtual gravitons". Lets get back to the electromagnetic field, which we know doesn't contradict relativity.

This is one of those situations where I strongly recommend learning the classical limits well before examining the quantum mechanics too closely. I recommend that you look real closely at how a transformer is supposed to work in terms of classical electrodynamics. The quantum electrodynamics analysis is an extrapolation of the classical electrodynamics analysis.

The far field has important constraints on the direction of the fields. Outside the radio antennae, the electric field of a plane wave has to be perpendicular to magnetic field of the plane wave. Therefore, what we call a radio wave has an electric field that is perpendicular to the magnetic field. This is a far field disturbance. The radio wave propagates independently of the orientation of the antennae that generated it. This type of wave has a real photon associated with it.

The near field has some of these constraints lifted. Inside a radio antennae (I mean in the empty space between the metal wires), the electric field and the magnetic field may have pointed in nearly the same direction. Thus, the disturbance inside the antennae is not a true radio wave. The disturbance does not propagate independently of the orientation of the antennae. The photons associated with this disturbance close to the antennae is more virtual.

You probably can guess that the distinction between real and virtual photons isn't always sharp. So there is some truth to what you said about a virtual photon "becoming" a real photon. However, discussing this issue can lead to even more problems. We can address the problem of graded definitions when we have some idea of what the distinction is.

Note: the phrase "gravitational wave" is used for several unrelated phenomena. Meteorologists talk about "gravity waves" which referred to an vertically oscillating air mass. Oceanographers talk about gravity waves that move on the surface of the oceans, where the restoring force is gravity. Neither are these are the "gravity waves" that astronomers are seeking.

There is no problem with the physics of these different "gravity waves". It is just a matter of the jargon used by scientists in different specialties.