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KleinGordon equation for electromagnetic field? 
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#1
Apr908, 05:48 PM

P: 249

Can we imagine electromagnetic field in vakuum as a massless particle that respects
KleinGordon equation (instead of Electromagnetic wave equation)? It seems to me that both equations are the same, except that the electromagnetic field can have 2 possible polarizations (then we count them as 2 different particles?). 


#2
Apr908, 06:59 PM

P: 455

Yes. The EM wave equation for A^\mu is the KG equation of a massless particle.
Only having two polarizations is due to the masslessness. 


#3
Apr1008, 05:52 AM

P: 980

However, note that because we do not observe longitudinal modes (which are allowed by the KG eqns) you have to include a constraint into the solution.



#4
Apr1108, 11:18 AM

P: 582

KleinGordon equation for electromagnetic field?
but the KleinGordon equation is a scalar equation. Are the two modes not coupled? They propagate independently each satisfying the (massless) KG equation?
As a comparison, there is the Dirac equation. The 4 components each individually satisfy the KG equation as well (for free particles, that is) but they are coupled by the Dirac equation and not independent. 


#5
Apr1108, 12:27 PM

P: 410

Every free relativistic field will satisfy the KG equation, as it merely expresses [tex]p^2 = m^2[/tex]. Note that [tex]\mathbf{E}[/tex] and [tex]\mathbf{B}[/tex] also obey the massless Klein Gordon equation. I think to get the KG equation for [tex]A[/tex] you need to be in the Lorenz gauge. 


#6
Apr1108, 12:49 PM

P: 582

Ok. I get it. The components of A are in general coupled by the Maxwell equations. It is just by selecting a special gauge they are decoupled.



#7
Apr1108, 12:52 PM

P: 40




#8
Apr1108, 01:11 PM

P: 99

klien gordon is kind of a relativistic schodinger equation. it uses energy squared is equal to the (momentum times the speed of light in vaccuum) squared plus the mass squared times the speed of light in vacuum to the fourth power, i think. and turns everything into waves and operators.
the electro magnetic equations can be found by comparing maxwells equations to eachother. something like the laplacian of E equals the E double dot, where E is the electric field. or something 


#9
Apr1108, 02:33 PM

P: 980




#10
Apr1108, 06:58 PM

P: 410

Strange, there is plenty of information on the subject. http://en.wikipedia.org/wiki/KleinGordon_equation



#11
Apr1108, 07:22 PM

PF Gold
P: 4,087

Genneth made a small typo  it should read
[tex]\partial^\mu \partial_\mu \psi + m^2 \psi == 0[/tex] or [tex](\square^2 + \mu^2)\psi = 0[/tex] [nrqed you're too quick  I'm still editing  d*mn latex. It depends on your definition of [tex]\square[/tex]] 


#12
Apr1108, 07:25 PM

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#13
Apr1108, 07:32 PM

PF Gold
P: 4,087

nrqed  for me
[tex]\square^2 \equiv \frac{d^2}{dt^2} + \nabla^2[/tex] 


#14
Apr1108, 07:40 PM

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Regards 


#15
Apr1108, 09:44 PM

PF Gold
P: 4,087

nrqed,
it's true it is used differently in various texts. Itzykson&Zuber use your definition, but the Wiki page uses mine. I guess I&Z are more authoritative. Best to define just what one means by it, I guess. M 


#16
Apr1208, 05:52 AM

P: 455

I think Gordan spelled his name that way, although Google votes 1001 the other way.



#17
Apr1208, 05:58 AM

P: 455

The opposite sign [tex]\square^2 \equiv +\frac{d^2}{dt^2}  \nabla^2[/tex] is winning out recently, because it corresponds to the metric (1,1,1,1). 


#18
Apr1208, 09:28 AM

P: 410




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