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Photon is electromagnetic field, right? 
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#19
Dec1212, 05:24 AM

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http://en.wikipedia.org/wiki/Field_%28physics%29 http://en.wikipedia.org/wiki/Electric_charge 


#20
Dec1212, 06:02 AM

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#21
Dec1212, 06:23 AM

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http://prola.aps.org/abstract/PR/v50/i2/p115_1 


#22
Dec1212, 06:33 AM

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#23
Dec1212, 06:36 AM

P: 68

The point is, when you ask whether there is electric field present at some point in space, it is just about the same as asking whether there is electric charge there, so in that sense they are synonyms, and you can see those two articles use the two terms interchangeably. 


#24
Dec1212, 07:36 AM

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#25
Dec1212, 07:50 AM

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#26
Dec1212, 08:07 AM

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Thanks
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#27
Dec1212, 08:28 AM

P: 68

Maxwell's equations do not explain how can photon have electric and magnetic field and yet we can not bend a beam of light by external electric or magnetic fields. Actually polarization should answer this, if we can change polarization then that would be the way to influence these fields, right? Perhaps we can not bend a beam of light, but if we can change the plane of em fields oscillation, well that's pretty good too. 


#28
Dec1212, 08:41 AM

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#29
Dec1212, 09:01 AM

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[tex]\nabla \cdot E_{\gamma} = 0[/tex] [tex]\nabla \times E_{\gamma} = \frac{\partial B_{\gamma}}{\partial t}[/tex] [tex]\nabla \cdot B_{\gamma} = 0[/tex] [tex]\nabla \times B_{\gamma} = \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t}[/tex] However, instead of sending that beam of light through the vacuum, we send it through some system of charges and currents that have already created some fields [itex]E[/itex] and [itex]B[/itex]. These fields must satisfy Maxwell's Equations with charges and currents. [tex]\nabla \cdot E = \frac{\rho}{\epsilon_0}[/tex] [tex]\nabla \times E = \frac{\partial B}{\partial t}[/tex] [tex]\nabla \cdot B = 0[/tex] [tex]\nabla \times B = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t}[/tex] Here, [itex]\rho[/itex] and [itex]J[/itex] are charge and current densities, respectively. We wish to demonstrate that this does not alter the trajectory of the light beam. In other words, that the new electric field is simply [itex]E + E_{\gamma}[/itex] and new magnetic field is simply [itex]B + B_{\gamma}[/itex]. We show that by demonstrating that these sums satisfy the Maxwell's equations with the same charge and current densities. [tex]\nabla \cdot (E + E_{\gamma}) = \nabla \cdot E + \nabla \cdot E_{\gamma} = \frac{\rho}{\epsilon_0} + 0 = \frac{\rho}{\epsilon_0}[/tex] [tex]\nabla \times (E + E_{\gamma}) = \nabla \times E + \nabla \times E_{\gamma} = \frac{\partial B}{\partial t} \frac{\partial B_{\gamma}}{\partial t} = \frac{\partial(B + B_{\gamma})}{\partial t}[/tex] [tex]\nabla \cdot (B + B_{\gamma}) = \nabla \cdot B + \nabla \cdot B_{\gamma} = 0 + 0 = 0[/tex] [tex]\nabla \times (B + B_{\gamma}) = \nabla \times B + \nabla \times B_{\gamma} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial E}{\partial t} + \mu_0 \epsilon_0 \frac{\partial E_{\gamma}}{\partial t} = \mu_0 J + \mu_0 \epsilon_0 \frac{\partial (E + E_{\gamma})}{\partial t}[/tex] The equations are satisfied. Therefore, the electromagnetic field of a beam of light simply adds on top of electromagnetic fields already present without any kind of distortion. In other words, photons are not affected by electric or magnetic fields. Are we finally done with your "charged photon" nonsense? 


#30
Dec1212, 09:12 AM

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#31
Dec1212, 09:15 AM

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You aren't just confused on a couple of definitions. You have no clue what you are talkinga bout at all. You need to read a textbook. That was probably the most helpful suggestion you got.



#32
Dec1212, 09:41 AM

PF Gold
P: 3,207

Barry, you seem to miss the point that K^2 tried so hard to show you.
There can be an electric field without any charge. An EM wave is just this. For example the light coming from stars contain no charge at all, yet it is made of an electromagnetic field. All the light is like this. You can take a region in space with an electric field and find no charge. 


#33
Dec1212, 09:55 AM

P: 260

There does not need to be any charges for there to be an electric field. Maxwell's Equations tell us that a changing magnetic field will also produce an electric field. Similarly, you don't need any currents for there to be a magnetic field; a changing electric field will produce a magnetic field. Essentially what happens in an EM wave is that a changing electric field produces a changing magnetic field which produces a changing electric field which... This allows the electric and magnetic fields to propagate without any charges or currents. I'll even go through how this is derived, step by step: As you've been shown numerous times, Maxwell's Equations in a vacuum are: [tex]\nabla \cdot E=0[/tex] [tex]\nabla \times E = \frac{\partial B}{\partial t}[/tex] [tex]\nabla \cdot B=0[/tex] [tex]\nabla \times B = \frac{1}{c^2} \frac{\partial E}{\partial t}[/tex] Now, using the following "curl of the curl" identity: [itex]\nabla \times \nabla \times A=\nabla(\nabla\cdot A)\nabla^2A[/itex] [tex]\nabla \times \nabla \times E=\nabla(0)\nabla^2E=\nabla \times \frac{\partial B}{\partial t}=\frac{\partial }{\partial t}[\nabla \times B]=\frac{\partial }{\partial t}[\frac{1}{c^2} \frac{\partial E}{\partial t}][/tex] Simplifying you get: [tex]\nabla^2E=\frac{1}{c^2}\frac{\partial^2 E}{\partial t^2}[/tex] You can apply the same identity to the curl of the magnetic field to get the following: [tex]\nabla^2B=\frac{1}{c^2}\frac{\partial^2 B}{\partial t^2}[/tex] The solution to both of these differential equations is a sinusoidal wave which moves at a velocity c. You accuse K^2 of being condescending, yet he's absolutely correct. You have no idea how basic electrodynamics works. What you need to do is stop pretending you know what you're talking about and go pick up a textbook. You're arguing about things you don't understand. 


#34
Dec1212, 10:14 AM

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#35
Dec1212, 10:25 AM

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#36
Dec1212, 10:37 AM

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http://en.wikipedia.org/wiki/Superposition_principle 


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