Solving Laplace's Equation with Mass

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Discussion Overview

The discussion revolves around solving a modified form of Laplace's equation, specifically in the context of a hypothetical scenario where a photon has mass. Participants explore the mathematical structure of the equation and potential approaches to derive its solutions.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the equation and indicates that the solutions are linear combinations of e^{\pm\mu r}/r, seeking a method to derive this rather than just verifying it.
  • Another participant suggests expanding the differentials to form a second-order equation and proposes a substitution of the form \(\psi = r \phi\).
  • A follow-up comment echoes the suggestion to expand the differentials and substitute, inquiring about the reasoning behind this approach.
  • A participant explains that factoring out r is a common technique when dealing with differential equations that have solutions resembling e^{kr}, particularly in cylindrical and spherical coordinates.
  • This participant notes that such equations often take the form of 1/r^n d/dr (r^n times something), reflecting the geometry of the problem.

Areas of Agreement / Disagreement

The discussion does not reach a consensus on a specific method to derive the solutions, and multiple approaches are proposed without resolution of which is preferable.

Contextual Notes

Participants do not clarify the assumptions behind the substitution or the implications of treating a photon as having mass, leaving some aspects of the discussion open to interpretation.

homology
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Hi folks, I'm drawing a blank on the following:

[tex] \frac{1}{r^2}\frac{\partial}{\partial r}r^2\frac{\partial \phi}{\partial r}+\mu^2 \phi=0[/tex]

I 'know' the solutions which are linear combinations of [tex]e^{\pm\mu r}/r[/tex] as you can check, but I've been trying to see if I could show the fact (as opposed to just checking that it works, which it does).

Context: The above, as I'm sure you realize, is 'almost' Laplace's equation. "If" a photon had mass (the term [tex]\mu[/tex] would be proportional to it) the above equation would result for a region with no charge.

I'm guessing that the answer is simple so please be vague with your hints (I suppose that's almost asking for a smartass comment) as I'd like to figure it out myself.

Thanks
 
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Expand the differentials so you have a 2nd order equation.

Then substitute [itex]\psi = r \phi[/itex]
 
AlephZero said:
Expand the differentials so you have a 2nd order equation.

Then substitute [itex]\psi = r \phi[/itex]

awesome! was that inspired by experience or a general rule of thumb?
 
You said the solution was something like [itex]e^{kr}/r[/itex]. I know how to solve a DE with solutions like [itex]e^{kr}[/itex] so factoring out the r seemed like a good thing to try :smile:

In cylindrical and spherical coordinates, solutions often have a factor of 1/r or 1/r^2 compared with cartesian coordinates, because they represent something being spread out over a plane or solid angle, compared with a strip of constant width.

The DEs often have the form 1/r^n d/dr (r^n times something) as well.
 

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