# Why is there no physical interpretation to non-real potentials in classical electromagnetism?

In Griffiths 4ed, he solves Laplace's equation and omits one of the solutions because it is imaginary. What are the grounds on which imaginary solutions or values can be marked as "non-physical"? And, how can this marking be done rigorously?

I mean to ask this question in the context of the whole of physics, not just classical E&M.

• physics.stackexchange.com/q/471063 Commented Jul 24, 2022 at 4:22
• reddit.com/r/AskPhysics/comments/3qnaoj/… Commented Jul 24, 2022 at 4:23
• physics.stackexchange.com/q/262899 Commented Jul 24, 2022 at 4:36
• these dont exactly answer your question, but good places to look for discussion? i myself don't know - I feel like it's probably that we use numbers to describe things in a way that is useful. i am eager to see what the answers are. Commented Jul 24, 2022 at 4:40
• Thank you for these resources! @Relativisticcucumber Commented Jul 24, 2022 at 5:22

I'm not sure whether I've ever seen a book explicitly discuss this, but the thing with complex solutions is that they are a way of solving two problems at once because they get "easier" together. To give an example, I'll use the driven harmonic oscillator. Suppose you want to solve $$\ddot{x} + \omega_0^2 x = F_0 \cos(\omega t). \tag{1}$$ This equation is difficult to solve. A different, but similar problem, is $$\ddot{y} + \omega_0^2 y = F_0 \sin(\omega t). \tag{2}$$ This is also difficult to solve. However, suppose $$z(t)$$ is a complex function satisfying $$\ddot{z} + \omega_0^2 z = F_0 e^{i\omega t}. \tag{3}$$ Then this is often easier to deal with (because exponentials are easier to differentiate and integrate than trigonometric functions), but there is something interesting: let us write $$z = x + i y$$ for real functions $$x$$ and $$y$$. We have $$(\ddot{x} + i \ddot{y}) + \omega_0^2(x + i y) = F_0 \big(\cos(\omega t) + i \sin(\omega t)\big). \tag{4}$$

Now, for a complex equation to be satisfied, the equation must hold independently for the real and imaginary parts. If we pick the real part of Eq. (4), we retrieve Eq. (1)! Hence, the real part of $$z(t)$$ is the solution to the problem we initially considered.

What if we pick the imaginary part? Then we get Eq. (2)! We solved two problems at once. The reason to pick this or that part is a matter of which problem we originally wanted.

While I didn't check the specific case Griffiths uses for the Laplace equation, this is fairly often what we do when solving (linear) equations using complex numbers in Physics. We bring two problems together, solve then together, and then split them at the end.

For situations in which you do not have a non-homogeneous term to make this splitting clear, one usually gets a real solution by imposing real initial conditions. For example, if we solved Eq. (4) with $$F_0 = 0$$ using complex numbers, at the end of the day we'd pick real initial conditions because, in the lab, we measure the initial conditions as being real numbers. The solution to the imaginary part will then be identically zero due to the imposition that its initial conditions vanish (assuming the equation to be sufficiently well-behaved).