Klein-Gordon equation for electro-magnetic field?

In summary, the Klein-Gordon equation is the equation for a massless particle in a vacuum, and the Clebsch-Gordan equation is the equation for a massless particle in a material.
  • #1
Lojzek
249
1
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?).
 
Physics news on Phys.org
  • #2
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
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
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.
 
  • #5
pellman said:
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.
 
  • #6
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.
 
  • #7
Lojzek said:
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.
 
  • #8
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 each other. something like the laplacian of E equals the E double dot, where E is the electric field.

or something
 
Last edited:
  • #9
Normouse said:
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]
 
  • #11
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]]
 
Last edited:
  • #12
Mentz114 said:
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]
:wink:
 
  • #13
nrqed - for me

[tex]\square^2 \equiv -\frac{d^2}{dt^2} + \nabla^2[/tex]
 
  • #14
Mentz114 said:
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
 
  • #15
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
I think Gordan spelled his name that way, although Google votes 100-1 the other way.
 
  • #17
Mentz114 said:
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).
 
Last edited:
  • #18
pam said:
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.
 
  • #19
Here is a more direct question about this topic:

Is relativistic quantum mechanic also supposed to describe the dynamics of photons? Or are the similarities between equations concidental?
 
  • #20
pam said:
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).

Thanks for sending the details of that operator but I've been seeing such equations for years. Are there any physical/intuitive ideas associated with this one?
 
  • #21
Normouse,
the square is called the D'Alembertian operator. More info here -

en.wikipedia.org/wiki/D'Alembertian

It should strictly be written like this ( note the factor c^2)

[tex]\square^2 \equiv +\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2[/tex]

in which case, (as has been stated above)

[tex](\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2)\psi = 0[/tex] means that [tex]\psi[/tex] is a field whose quanta travel at c.
 
  • #22
Normouse said:
Thanks for sending the details of that operator but I've been seeing such equations for years. Are there any physical/intuitive ideas associated with this one?

You might say so :smile:

The whole first part of my book is devoted to this equation...
for instance chapter 1:

http://physics-quest.org/Book_Chapter_EM_basic.pdf


Regards, Hans
 
Last edited:
  • #23
lbrits said:
The Gordon in Klein-Gordon and the Gordan in Clebsch-Gordan are two different people.
Thank you. I now recall that I knew that once.
 
  • #24
Lojzek said:
Here is a more direct question about this topic:

Is relativistic quantum mechanic also supposed to describe the dynamics of photons? Or are the similarities between equations concidental?
RQM does describe the kinematics and dynamics of photons.
There are very few coincidences in physics.
 
  • #25
pam said:
RQM does describe the kinematics and dynamics of photons.
There are very few coincidences in physics.
This seems strange to me, since a straight wave in quantum mechanics implies uniform probability density and consequently uniform energy distribution (1 photon=h*f energy), while a straight electromagnetic wave implies energy distribution proportional to E^2, which oscilates.
 
  • #26
Good news

Mentz114 said:
Normouse,
the square is called the D'Alembertian operator. More info here -



It should strictly be written like this ( note the factor c^2)

[tex]\square^2 \equiv +\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2[/tex]

in which case, (as has been stated above)

[tex](\frac{1}{c^2}\frac{d^2}{dt^2} - \nabla^2)\psi = 0[/tex] means that [tex]\psi[/tex] is a field whose quanta travel at c.
Wow! A wonderfully clear and straight answer with actual intuition and physical feel. (VERY UNUSUAL for you physicists--careful your physicist card might be in jeopardy((just kidding))). Thanks Mentz114.
 
  • #27
Hans de Vries said:
You might say so :smile:

The whole first part of my book is devoted to this equation...
for instance chapter 1:




Regards, Hans

Books sure have improved since I was an undergraduate. I hope to "line by line" it soon. Thanks for that site(with your book on it). I wish I didn't have to work.
 
  • #28

1. What is the Klein-Gordon equation for electro-magnetic field?

The Klein-Gordon equation is a relativistic wave equation that describes the behavior of spinless particles, such as electromagnetic waves. It is a combination of the Maxwell's equations for the electromagnetic field and the Klein-Gordon equation for a free particle.

2. How is the Klein-Gordon equation derived?

The Klein-Gordon equation is derived by applying special relativity principles to the Schrödinger equation. This results in a second-order partial differential equation that describes the propagation of spinless particles, including electromagnetic waves.

3. What are the physical implications of the Klein-Gordon equation?

The Klein-Gordon equation has several important physical implications. Firstly, it shows that electromagnetic waves are quantized and have discrete energy levels. It also predicts the existence of antiparticles, which have opposite charge and spin to their corresponding particles.

4. How does the Klein-Gordon equation relate to quantum electrodynamics?

The Klein-Gordon equation is a fundamental equation in quantum electrodynamics, which is the theory that describes the interaction between electrically charged particles and the electromagnetic field. It is used to describe the behavior of spinless particles in the presence of electromagnetic fields.

5. What are the limitations of the Klein-Gordon equation?

The Klein-Gordon equation has several limitations. It does not account for the spin of particles, which is a crucial aspect of quantum mechanics. It also does not take into account the effects of relativity at high energies, which is necessary for describing the behavior of particles at the subatomic level. As such, the Klein-Gordon equation is often superseded by more advanced equations, such as the Dirac equation.

Similar threads

  • Quantum Physics
Replies
2
Views
1K
  • Quantum Physics
Replies
23
Views
577
Replies
9
Views
721
Replies
18
Views
1K
  • Quantum Physics
Replies
13
Views
708
Replies
45
Views
3K
Replies
1
Views
745
Replies
7
Views
1K
  • Quantum Physics
Replies
4
Views
915
Replies
6
Views
1K
Back
Top