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The photon and Electromagnetic field

by Naty1
Tags: electromagnetic, field, photon
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Naty1
#1
Nov3-08, 03:08 PM
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How does an EM field, with both an electric and a magnetic component, each of which "disappears" produce an apparently steady photon?

A point charge has an electromagnetic field it creates...A stationary observer with respect to the charge sees and electric field; an observer in motion sees a magnetic field. But photons, presumably bumps in the EM field, don't wink in and out of existence depending on our motion or the appearance of one field component or the other. .
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tiny-tim
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Nov3-08, 06:20 PM
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Quote Quote by Naty1 View Post
A stationary observer with respect to the charge sees and electric field; an observer in motion sees a magnetic field.
Hi Naty1!

I'm not sure I'm getting the point you're making

if you're saying that the observer in motion will see only a magnetic field, and no electric field, then that's wrong

the field he sees will still be mostly electric (E2 - B2 = E02)

there is no observer for whom the electric field disappears.
Mentz114
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Nov4-08, 04:29 AM
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"Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter. There is no evidence ( nor any way of getting any ) that photons exist in the EM field when it is not interacting with matter.

Speculations about 'free' photons usually lead to apparent contradictions, as evidenced by your question.

atyy
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Nov4-08, 06:24 AM
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The photon and Electromagnetic field

http://www.physicsforums.com/showthread.php?t=251358
Naty1
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Nov4-08, 07:22 AM
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if you're saying that the observer in motion will see only a magnetic field, and no electric field, then that's wrong

there is no observer for whom the electric field disappears
Well that's contrary to another thread here...I don't remember which.....I'll see if I can find it... I clearly remember the (apparent) conclusion because prior to that my understanding was the same as yours...even so the magnetic field seems to appear and disappear...right??

"Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter
Well thats novel.... and difficult for messenger particles to comply with I should think! From Fabric of the Cosmos, Brian Greene:
Photons pass completely unhindered through the Higgs ocean and so have no mass at all....When Schrodinger's equation is applied in a simple situation such as a single, isolated photons passing through a screen with two slits, it gives rise to the interference pattern.....
And what about the polarization of photons, (their spin directions), wave particle duality....photoelectric effect, etc,etc,etc...

Had someone answered I was mixing quantum theory with classicial field theory I would have immediately understood....because there are always anomalies when trying to do that...

Also, the answer may be related to the following:

PARALLEL WORLDS, Michio Kaku, Pg 216
...This symmetry can be made exact if we add monopoles into Maxwell's equations.
The revised Maxwell's equations remain precisely the same if we exchange the electric field with the magnetic field and interchange the electric charge e with the inverse of the magnetic charge g. This means that electricity (if the electric charge is low) is precisely equivalent to magnetism (if the magnetic charge is high). This equivalence is called duality.
(Without monopoles the equations look almost the same.)
Naty1
#6
Nov4-08, 07:28 AM
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I found a quote, but it was used in another thread..still looking for the thread...

Also keep in mind that electric field in one reference frame is a magnetic field in another; so a static (electric) charge on a rotating body will have a perceived magnetic field.
tiny-tim
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Nov4-08, 07:30 AM
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Quote Quote by Naty1 View Post
Well that's contrary to another thread here...I don't remember which.....I'll see if I can find it... I clearly remember the (apparent) conclusion because prior to that my understanding was the same as yours...even so the magnetic field seems to appear and disappear...right??
Hi Naty1!

E2 - B2 is a relativistic invariant.

(btw, so is E.B)

So you can't have E = 0 in one frame and B = 0 in another.

EDIT:
Also keep in mind that electric field in one reference frame is a magnetic field in another; so a static (electric) charge on a rotating body will have a perceived magnetic field.
The second half of that is correct. The first half should say "Also keep in mind that electric field in one reference frame has an additional magnetic field in another"
spidey
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Nov4-08, 07:35 AM
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Quote Quote by Mentz114 View Post
"Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter. There is no evidence ( nor any way of getting any ) that photons exist in the EM field when it is not interacting with matter.

Speculations about 'free' photons usually lead to apparent contradictions, as evidenced by your question.
I am confused. Do u say photon doesn't exist when they are not interacting with matter?
Supposing photon is emitted from sun and reaches earth.So, you say photon exist when it is emitted by Sun and when it interacts with earth.In the mean time, photon doesn't exist between sun and earth.
Jonathan Scott
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Nov4-08, 10:18 AM
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Quote Quote by spidey View Post
I am confused. Do u say photon doesn't exist when they are not interacting with matter?
Supposing photon is emitted from sun and reaches earth.So, you say photon exist when it is emitted by Sun and when it interacts with earth.In the mean time, photon doesn't exist between sun and earth.
A photon is a way of describing a quantized transfer of electromagnetic energy and angular momentum. However, while the energy is in flight, it isn't really a collection of independent particles but rather a flow of waves containing the total amount of energy. When those waves interact with something, they will transfer quantized amounts of energy of the appropriate frequency. If the energy density is very low, then there may be a one to one relationship between photons being emitted and received, but various physical effects such as interference demonstrate that one cannot assume that the emitted and received photons were actually in some sense the same particle.

However, it's quite common to describe the flow of electromagnetic waves as a stream of photons anyway in cases where the distinction doesn't really matter.
Naty1
#10
Nov4-08, 10:32 AM
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Jonathan posted..

However, while the energy is in flight, it isn't really a collection of independent particles but rather a flow of waves containing the total amount of energy. When those waves interact with something, they will transfer quantized amounts of energy of the appropriate frequency.
I'm not sure I agree entirely with the first sentence; however it's largely immaterial as we are into the wave (classical field theory) versus particle (quantum mechanical) viewpoint....The difficulty is akin to the double slit experiment...electromegnatic waves in some situations appear to be continuous and wavelike, in other situations discrete quantum like particle effects are manifested....it's a good concept to remember, but has been debated for about 90 years...
DrGreg
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Nov4-08, 11:47 AM
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Quote Quote by Mentz114 View Post
"Photons" only exist at the moment they are emitted or absorbed i.e. when they interact with matter. There is no evidence ( nor any way of getting any ) that photons exist in the EM field when it is not interacting with matter.

Speculations about 'free' photons usually lead to apparent contradictions, as evidenced by your question.
It's true that we can measure photons only by emitting them or absorbing them. But to conclude that they "do not exist" between those events seems a very perverse use of the English language.

You say that there is no evidence that they exist, but the evidence is that they are emitted (i.e. come into existence!) and are absorbed (i.e. cease to exist!).

I think what you meant to say is that we can't tell exactly where a photon is between its creation and destruction. Quantum theory tells us the photon takes all possible routes -- so it can be in many places at once -- and the best we can do is describe it via a wavefunction.
Naty1
#12
Nov4-08, 12:18 PM
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Quantum theory tells us the photon takes all possible routes -- so it can be in many places at once -- and the best we can do is describe it via a wavefunction.
Now that I can buy into!!!
Naty1
#13
Nov4-08, 12:22 PM
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Here's a quote from wikipedia,
http://en.wikipedia.org/wiki/Electromagnetic_field

which supports Tiny tim's post....so I'm back to my original understanding...thanks tim...

In the past, electrically charged objects were thought to produce two types of field associated with their charge property. An electric field is produced when the charge is stationary with respect to an observer measuring the properties of the charge and a magnetic field (as well as an electric field) is produced when the charge moves (creating an electric current) with respect to this observer. Over time, it was realized that the electric and magnetic fields are better thought of as two parts of a greater whole the electromagnetic field.
(my boldface added)
Naty1
#14
Nov4-08, 12:26 PM
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So nobody has answered my original question yet...If a characteristic of the EM field is observer dependent, why not the photon?

Rather odd...but maybe because the photon only moves at lightspeed, that locks in the characteristic which might otherwise change....
tiny-tim
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Nov4-08, 01:26 PM
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Quote Quote by Naty1 View Post
How does an EM field, with both an electric and a magnetic component, each of which "disappears" produce an apparently steady photon?
Quote Quote by Naty1 View Post
So nobody has answered my original question yet...If a characteristic of the EM field is observer dependent, why not the photon?

Rather odd...but maybe because the photon only moves at lightspeed, that locks in the characteristic which might otherwise change....
Hi Naty1!

The EM fields which produce photons have E = B (and E.B = 0) (I think: )

and if E = B in one frame, then E = B in every frame

so, unlike an electron, a "photon-field" will look the same in any frame.
Mentz114
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Nov4-08, 04:13 PM
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Quote Quote by DrGreg View Post
It's true that we can measure photons only by emitting them or absorbing them. But to conclude that they "do not exist" between those events seems a very perverse use of the English language.

You say that there is no evidence that they exist, but the evidence is that they are emitted (i.e. come into existence!) and are absorbed (i.e. cease to exist!).

I think what you meant to say is that we can't tell exactly where a photon is between its creation and destruction. Quantum theory tells us the photon takes all possible routes -- so it can be in many places at once -- and the best we can do is describe it via a wavefunction.
(my bolding).

I think that strengthens my case. The wave function for a photon is not a Schroedinger equation in the normal sense. I quote from (1)

This choice of normalization reflects the fact that a photon has no mass that can be localized at a point; rather it has only helicity and energy, and the energy cannot strictly be localized at a point. The square of the wave function gives the probability density for energy, not particle location
So the wave function does not describe a particle, even in the loosest sense.

(1)"The Maxwell wave function of the photon" M. G. Raymer and Brian J. Smith.
In proc. SPIE conference Optics and Photonics, The Nature of Light: What is a Photon?" (San Diego, Aug. 2005)
jtbell
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Nov4-08, 05:04 PM
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[QUOTE=tiny-tim;1943254]The EM fields which produce photons have E = B (and E.B = 0) (I think: )

Electromagnetic waves have E = cB (more precisely [itex]|\vec E| = c |\vec B|[/itex] at all points.
tiny-tim
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Nov4-08, 05:12 PM
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Quote Quote by jtbell View Post
Electromagnetic waves have E = cB (more precisely [itex]|\vec E| = c |\vec B|[/itex] at all points.
Hi jtbell!

As you've probably guessed, I learnt my electromagnetism with c = 1, and I keep forgetting to put the c back in.

hmm then I suppose it's E2 - c2B2 which is invariant?


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