Derivation of the Maxwell and Proca Field Equations

  • #1
topsquark
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I am back to my writing desk and I was looking up different and (hopefully) relatively "basic" derivations of the field equations.

I found a nice little derivation of the Proca (and Maxwell) equations by Gersten (1998, PRL, 12, 291-298 is the reference, but I got it from the web so what I have may not be the full paper). It starts is a similar way to the Dirac derivation
##E^2-p^2=0##

And now factor this as:
##(E 1_3 - \textbf{p} \cdot \textbf{S} ) (E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} - m^2 \boldsymbol{\psi} = 0##

where ##1_3## is the 3 x 3 identity matrix, p is the 3-momentum, S is a set of three 3 x 3 matrices to be determined, and ##\boldsymbol{\psi}## is a 3 component vector of spacetime functions.

To solve this, Gersten added a term:
##(E 1_3 - \textbf{p} \cdot \textbf{S} ) (E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} + \textbf{p} ( \textbf{p} \cdot \boldsymbol{\psi} ) - m^2 \boldsymbol{\psi} = 0##

and required that
##\textbf{p} \cdot \boldsymbol{\psi} = i m^2 \phi##

and
##(E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} = i m^2 \textbf{A}##

Now, I'm mainly okay with this. We are adding a term and supplying 4 functions to take up the slack for it that we can use to "fix" the correct solution. And I like the simplicity of it.

But, if you take this down to m = 0 (the Maxwell field) you get
##(E 1_3 - \textbf{p} \cdot \textbf{S} ) (E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} + \textbf{p} ( \textbf{p} \cdot \boldsymbol{\psi} ) = 0##

with
##\textbf{p} \cdot \boldsymbol{\psi} = 0##

and
##(E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} = 0##

Now we've lost all of our freedom to "fix" the solution. Clearly, this should generate a solution, but we should be able to pick any extra term to add to the energy equation (so long as its compatible with the second equation, of course.) How does this process guarantee the correct solution to the Maxwell equations?

Is there a way to a priori justify adding this term?

-Dan
 
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  • #2
topsquark said:
##(E 1_3 + \textbf{p} \cdot \textbf{S} ) \boldsymbol{\psi} = i m^2 \textbf{A}##
What is the definition of ##\textbf{A}##?
 
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  • #3
renormalize said:
What is the definition of ##\textbf{A}##?
A would be the 3-space part of the vector potential. In this method, it is defined what it is required to be by the solution of the system of equations. (Being a spin 1 particle, it will, of course, have a gauge condition to work right. I'm going to mention this after I derive the separate wave equations, but not going to go into it, as I feel there are better ways to analyze that problem later on into my work.)

At this level, I see it pretty much as a "dummy" variable, whose value is initially a sliding scale to "fix" the solution to be the Proca field equations. I've seen it done before: to my mind, the most blatant example is in the Ferrari solution to the quartic equation: an undetermined quantity z is introduced into the solution to force one side to be a perfect square. That condition sets a value of z (determined by solving a cubic) and that value is then set for the rest of the solution.

-Dan

Addendum: To be a bit more clear:
Replace E and p with the usual operators. This gives
##\nabla \boldsymbol{\psi} = -m^2 \phi##

##\dfrac{\partial \boldsymbol{\psi}}{\partial t} - i \nabla \times \boldsymbol{\psi} = m^2 \textbf{A}##

This leads to:
##\boldsymbol{\psi} = - \dfrac{\partial \textbf{A}}{\partial t} - i \nabla \times \textbf{A} - \nabla \phi##

We then set ##\boldsymbol{\psi} = \textbf{E} - i \textbf{B}##, E and B real, which leads to the Proca equations.

-Dan
 
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  • #5
vanhees71 said:
The correct citation is

Gersten, A. Maxwell’s Equations as the One-Photon Quantum Equation. Found Phys Lett 12, 291–298 (1999). https://doi.org/10.1023/A:1021648704289
Thank you. I didn't have the title. Unfortunately, as much as I'd like to look at the full article, I can't afford to buy it.

-Dan

Addendum: Ouch! That smarts. Apparently I didn't give the correct Journal, either. (A typo on my part.) Thanks for the catch!
 
  • #7
vanhees71 said:
A similar approach to the Proca equation can be found here:

https://arxiv.org/abs/0901.3300
For the record, that's the paper I have. But it does not explain why we choose the form ##\textbf{p} \cdot \boldsymbol{\psi}##. I mean, it does give the correct answer, but is there any logic to why we should choose this form? Is there a Physical reason this term should be suspected to be 0?

-Dan
 
  • #8
I find this paper not very convincing. The entire thing goes back to the 19th century and the so-called Riemann-Silberstein vector. For electrodynamics, you find the modern argument, using the relativistic formulation of electrodynamics to show that this vector ##\vec{F}=\vec{E}+\mathrm{i} \vec{B}## leads to a representation of the Lorentz group as ##\text{SO}(3,\mathbb{C})##, in my manuscript, p.70.

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
 
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  • #9
vanhees71 said:
I find this paper not very convincing. The entire thing goes back to the 19th century and the so-called Riemann-Silberstein vector. For electrodynamics, you find the modern argument, using the relativistic formulation of electrodynamics to show that this vector ##\vec{F}=\vec{E}+\mathrm{i} \vec{B}## leads to a representation of the Lorentz group as ##\text{SO}(3,\mathbb{C})##, in my manuscript, p.70.

https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf
Thank you! I'm getting over a flu, so it might be a couple of days before I can fully read through this.

-Dan
 
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