Editing Green's function (section)

===Units===
While it does not uniquely fix the form the Green's function will take, performing a [[dimensional analysis]] to find the units a Green's function must have is an important sanity check on any Green's function found through other means. A quick examination of the defining equation,
<math display="block"> L G(x, s) = \delta(x - s), </math>
shows that the units of <math>G</math> depend not only on the units of <math>L</math> but also on the number and units of the space of which the position vectors <math>x</math> and <math>s</math> are elements. This leads to the relationship:
<math display="block"> [[G]] = [[L]]^{-1} [[d x]]^{-1}, </math>
where <math>[[G]]</math> is defined as, "the physical units of {{nowrap|<math>G</math>"}}{{Explain | reason=What does this mean? This is ungoogleable and without a description this section is meaningless. | date = October 2024}}, and <math>dx</math> is the [[volume element]] of the space (or [[spacetime]]).

For example, if <math>L = \partial_t^2</math> and time is the only variable then:
<math display="block">\begin{align}[]
[[L]] &= [[\text{time}]]^{-2}, \\[1ex]
[[dx]] &= [[\text{time}]],\ \text{and} \\[1ex]
[[G]] &= [[\text{time}]].
\end{align}</math>
If {{nowrap|<math>L = \square = \tfrac{1}{c^2}\partial_t^2 - \nabla^2</math>,}} the [[d'Alembert operator]], and space has 3 dimensions then:
<math display="block">\begin{align}[]
[[L]] &= [[\text{length}]]^{-2}, \\[1ex]
[[dx]] &= [[\text{time}]] [[\text{length}]]^3,\ \text{and} \\[1ex]
[[G]] &= [[\text{time}]]^{-1} [[\text{length}]]^{-1}.
\end{align}</math>