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Klein-Gordon equation for electro-magnetic field?

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Lojzek
#1
Apr9-08, 05:48 PM
P: 249
Can we imagine electro-magnetic field in vakuum as a massless particle that respects
Klein-Gordon equation (instead of Electromagnetic wave equation)?
It seems to me that both equations are the same, except that the electro-magnetic field can have 2 possible polarizations (then we count them as 2 different particles?).
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pam
#2
Apr9-08, 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.
genneth
#3
Apr10-08, 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.

pellman
#4
Apr11-08, 11:18 AM
P: 582
Klein-Gordon equation for electro-magnetic field?

but the Klein-Gordon equation is a scalar equation. Are the two modes not coupled? They propagate independently each satisfying the (massless) K-G equation?

As a comparison, there is the Dirac equation. The 4 components each individually satisfy the K-G equation as well (for free particles, that is) but they are coupled by the Dirac equation and not independent.
lbrits
#5
Apr11-08, 12:27 PM
P: 410
Quote Quote by pellman View Post
but the Klein-Gordon equation is a scalar equation. Are the two modes not coupled? They propagate independently each satisfying the (massless) K-G equation?

As a comparison, there is the Dirac equation. The 4 components each individually satisfy the K-G equation as well (for free particles, that is) but they are coupled by the Dirac equation and not independent.
All good and true... but you can decouple the Dirac equation by finding it's eigenspinors. This is what you do, e.g., when you perform a mode expansion of the field.

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.
pellman
#6
Apr11-08, 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 de-coupled.
Normouse
#7
Apr11-08, 12:52 PM
P: 40
Quote Quote by Lojzek View Post
Can we imagine electro-magnetic field in vakuum as a massless particle that respects
Klein-Gordon equation (instead of Electromagnetic wave equation)?
It seems to me that both equations are the same, except that the electro-magnetic field can have 2 possible polarizations (then we count them as 2 different particles?).
Out from the dunce corner with my weekly dumb question. Looked up Klein-Gordon equations in wikipedia--nothing. Looked in my McGraw-Hill encyclopedia of Physics--nothing. Kindly generalize in words, no acronyms please, what those(or that) equation(s) is about.
dipstik
#8
Apr11-08, 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
genneth
#9
Apr11-08, 02:33 PM
P: 980
Quote Quote by Normouse View Post
Out from the dunce corner with my weekly dumb question. Looked up Klein-Gordon equations in wikipedia--nothing. Looked in my McGraw-Hill encyclopedia of Physics--nothing. Kindly generalize in words, no acronyms please, what those(or that) equation(s) is about.
[tex]\partial^\mu \partial_\mu \psi + m \psi^2 == 0[/tex]
lbrits
#10
Apr11-08, 06:58 PM
P: 410
Strange, there is plenty of information on the subject. http://en.wikipedia.org/wiki/Klein-Gordon_equation
Mentz114
#11
Apr11-08, 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]]
nrqed
#12
Apr11-08, 07:25 PM
Sci Advisor
HW Helper
P: 2,923
Quote Quote by Mentz114 View Post
Genneth made a small typo - it should read

[tex]\partial^\mu \partial_\mu \psi + m^2 \psi == 0[/tex]

or

[tex](\Box^2 + \mu^2)\psi = 0[/tex]
Mentz made a very minor typo: there is no exponent of two on [tex] \Box [/tex]
Mentz114
#13
Apr11-08, 07:32 PM
PF Gold
P: 4,087
nrqed - for me

[tex]\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2[/tex]
nrqed
#14
Apr11-08, 07:40 PM
Sci Advisor
HW Helper
P: 2,923
Quote Quote by Mentz114 View Post
nrqed - for me

[tex]\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2[/tex]
Ok. It's just that I have never seen this definition used in any book so it's not the standard definition, as far as I know. But I may be wrond and if you know of any book using that notation I would be interested to know.

Regards
Mentz114
#15
Apr11-08, 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
pam
#16
Apr12-08, 05:52 AM
P: 455
I think Gordan spelled his name that way, although Google votes 100-1 the other way.
pam
#17
Apr12-08, 05:58 AM
P: 455
Quote Quote by Mentz114 View Post
nrqed - for me

[tex]\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2[/tex]
Both signs are used in books.
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).
lbrits
#18
Apr12-08, 09:28 AM
P: 410
Quote Quote by pam View Post
I think Gordan spelled his name that way, although Google votes 100-1 the other way.
The Gordon in Klein-Gordon and the Gordan in Clebsch-Gordan are two different people.


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