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Follow up to this question: Why is there no physical interpretation to non-real potentials in classical electromagnetism?.

Is there anything that the imaginary quantities in math map onto in the physical realm? For example, Silly Goose cites here a solution to Laplace's equation for an imaginary potential and claims that Griffiths dismisses this solution by saying it's nonphysical. In this case my question specifies to "what, if anything, do these imaginary solutions map on to, or represent" in reality. Here (What does imaginary number maps to physically?) Anna V discusses how complex numbers are tools, but I am wondering more specifically if there is anything that this solution to Laplace's equation represents and if we are missing something by dismissing this solution?

Wondering in general, not just for potential, but potential is a good example to start with I suppose.

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  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/560748/2451 , physics.stackexchange.com/q/76595/2451 , physics.stackexchange.com/q/11396/2451 and links therein. $\endgroup$
    – Qmechanic
    Jul 24, 2022 at 6:28
  • $\begingroup$ To reopen this post (v2) consider to make title and body less broad and more focused. $\endgroup$
    – Qmechanic
    Jul 24, 2022 at 6:30
  • $\begingroup$ You sometimes see the imaginary part of a physical constant associated with loss. E.g., the complex dielectric function has a real part that is related to the usual index of refraction and has an imaginary part that is related to the absorption (when you absorb you lose beam energy and turn it into heat energy). This is related to a complex potential for $\vec D$ since $\vec D = \epsilon \vec E$. But, anyways, as written this question seems to be too broad and unclear exactly what you want from an answer. $\endgroup$
    – hft
    Jul 25, 2022 at 4:05
  • $\begingroup$ You also often see a complex "self energy" associated with quasiparticles. In this case the imaginary part might be related to the mean free path for inelastic scattering. Etc. $\endgroup$
    – hft
    Jul 25, 2022 at 4:07
  • $\begingroup$ inspirehep.net/literature/99841 $\endgroup$ Aug 1, 2022 at 15:15

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I think the only answer in general is precisely Anna V's statement that complex numbers are tools. In many contexts they can mean many different things.

For the application you mentioned in electrostatics, complex numbers are useful because the real and imaginary parts of any complex differentiable function satisfy Laplace's equation. That is, if $f:\mathbb{C}\to\mathbb{C}$ is complex differentiable, then $\nabla^2 \text{Re}(f(x+yi))=0$ and $\nabla^2 \text{Im}(f(x+yi))=0$. So you can imagine this is an incredibly powerful tool for use in 2 dimensions. For example, if $f(z)=z^2$, then $f(x+yi)=x^2-y^2+i(2xy)$ and we get the harmonic functions $x^2-y^2$ and $xy$ for free. For $\sin(z)$ we get the harmonic functions $\sin(x)\cosh(y)$ and $\cos(x)\sinh(y)$, and we could spend all day coming up with harmonic functions and inventing practice problems around them. Here's one: "find the potential in $-\infty<x<\infty$ and $-1\le y\le 1$, with boundary conditions $\phi(x,1)=\cos(x)$ and $\phi(x,-1)=-\cos(x)$." We immediately know from the previous sentences that $\phi(x,y)=\text{Im}(\sin(x+y i))=\cos(x)\sinh(y)$ satisfies this.

What does $\text{Re}(\sin(x+y I))=\cosh(y)\sin(x)$ represent? It turns out that lines of constant values of that are field lines! Here I plot the equipotentials as contours of $\cos(x)\sinh(y)$, and the blue dashed lines are contours of $\cosh(y)\sin(x)$.

field lines for phi(x,1)=cos(x), phi(x,-1)=-cos(x) Mathematica source code

This is because of the very interesting property of complex differentiable functions: if $f(x+y i)=u(x+y i)+i v(x+y i)$, then I've already said $\nabla^2 u=0$ and $\nabla^2 v=0$, but we also have the very interesting relation $\partial_x u=\partial_y v$ and $\partial_y u=-\partial_x v$. In other words, the gradient of $u$ is equal to the gradient of $v$ rotated ninety degrees!

For an example with fluid flows, see this wikipedia article for Potential flow in two dimensions. The real part of the complex function is called the velocity potential, and the imaginary part is called the stream function. Lines of constant imaginary part are called streamlines.

So that's for 2 dimensional electrostatics, and you can have similar statements for 2d fluid flows.

The use case of analyzing voltages in time dependent circuits $V(t)$ is going to have a totally different explanation. There we analyze the complex valued $\tilde V(\omega)=\int dt e^{-i\omega t}V(t)$, and the magnitude and phase of $\tilde{V}$ describe how the circuit behaves at a certain frequency $\omega$. So that's a totally different discussion and I'd have to fall back on the explanation, "complex numbers are useful tools."

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  • $\begingroup$ Most excellent and clear! +1 -NN $\endgroup$ Jul 31, 2022 at 20:53
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I think you will find this very usefull https://www.youtube.com/watch?v=T647CGsuOVU&list=PLiaHhY2iBX9g6KIvZ_703G3KJXapKkNaF watch the entire series, is really well done!

To the point.

The confusion begins by calling a certain set of numbers (those that belong to ℝ) as "real" numbers.

In reality what is "real" and what is not, is a philosophical question, and thus, it sits outside of the practice of mathematics.

"Real" numbers are not more "real" than any other number... Natural numbers are real numbers too. And of course, imaginary numbers (and complex numbers) are also... real numbers too.

Or in other words... All numbers are equally real as the "real" ones and equally imaginary as the "imaginary" ones.

Imaginary numbers are DEFINITELY as valid and as real and as usefull and as "physical" as any other number. They are NOT mathematical hacks, or "tools" that we use in a "shut up and calculate" fashion.

They make PERFECT sense in mathematical logic, and they are the DIRECT extension of the function of powers and square roots.

Imaginary numbers are basically numbers that sit at right angles from the "real" number line. So everytime we use a system of 2 axes, we knowingly or unknowingly use imaginary and complex numbers. They are hardwired into our physical reality and they are not at all weird or spooky.

The fact that $\sqrt{-1}$ exists, is not weird at all... We don't square a "real" number to get -1, we square a number that sits at right angles from the real number line, its not weird that when we do square it, it gives -1. Its just the next logical step after we have defined powers and square roots.

Its school that has made us all a great disservice, by not teaching us imaginary logic correctly, and made us all believe that somehow we can introduce all kinds of nonesense into math - we cant.

Imaginary numbers are not nonesense, they are valid numbers, as valid as any other number that you can think off.

Everytime you use a graph of 2 axes, I don't know if you use vectors or polar forms or whatever, you are still using imaginary numbers even withought knowing it.

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    $\begingroup$ This perspective seems to dodge the question that was being asked. Whenever I measure a quantity in the lab, the result is always real; I never measure a potential difference of $10i \text{ V}$ between two points. Even if imaginary numbers are "just as valid" as real numbers, physical results are only ever real; and the OP is asking whether there is a valid interpretation of the numbers that never seem to show up on our laboratory devices. $\endgroup$ Aug 4, 2022 at 13:01
  • $\begingroup$ This is out of convention thought... Here imgur.com/a/gKXSTIr we have axes X and Z. And we have replaced "real" numbers with "right" numbers and "imaginary" numbers with "up" numbers... So why should the right numbers be any diffrent from the up ones? Or why can a room have a "15 right" temperature but not a "15 up" temperature? Its all just a convention that happened naturally as mathematics evolved into the centuries. – – $\endgroup$
    – Nuke
    Aug 8, 2022 at 17:06
  • $\begingroup$ You seem to be saying that what we view as a "real" number and what we view as an "imaginary" number is arbitrary, and that we could switch all real numbers with imaginary numbers and it wouldn't make a difference. But that doesn't seem to be true at all: there are imaginary numbers whose square is real, but there are no real numbers whose square is imaginary. $\endgroup$ Aug 8, 2022 at 17:11
  • $\begingroup$ Yes because that's what the function of square roots and powers do. They take you from one number line and transfer you to the other at right angles. In Special Relativity, my speed through spacetime, can be 60+80i. I'm traveling at 60% the speed of light in space and 80% the speed of light in time - both numbers indicate a very real and physical observation. The fact that things move and age... We don't "use" complex numbers in special relativity, but we do use other things like metrics and differential geometry that is complex analysis, just hidden... $\endgroup$
    – Nuke
    Aug 9, 2022 at 17:30

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