Can we define the zero potential at an imaginary point?

Question

Consider a force field defined as

$$\vec{F}(x) = \left(\frac{A}{x^2}-B\right)\hat{i}\space$$

where $A, B$ are positive constants. We want to get the potential energy function for this field. We can integrate this,

$$U(x_f)-U(x_i) = -\int_{x_i}^{x_f} Fdx= \left(\frac{A}{x_f}+Bx_f\right)-\left(\frac{A}{x_i}+Bx_i\right)$$

Now, we want to define a zero potential. Can we do so at $x=\sqrt{-\frac{A}{B}}$? (which is an imaginary point) Because putting this into the equation makes the last term on the RHS zero giving a simpler expression.

Specifically, is it valid for us to define the zero of potential energy function at an imaginary point? Does talking about imaginary numbers even make sense when talking about the potential of a force field?

Answer 1

There's no such thing as an imaginary point. In other words, you can certainly plug imaginary numbers into a formula, but those imaginary numbers don't represent points in space, and thus the results you get will not represent the conditions at any actual point.

However, you don't need an actual point in space for this purpose. The potential energy field only needs to satisfy $\vec{F}(\vec{x}) = -\vec{\nabla}U(\vec{x})$, and that definition allows you to include an arbitrary constant in the potential. The constant doesn't actually need to be the value of the potential at a physical point. (As a corollary, there is no need to have $U=0$ anywhere.) So you can write

$$U(x) = \frac{A}{x} + Bx + C$$

and then arbitrarily choose the value of $C$, for example $C = 0$; you don't need to set $C$ to the value of $U$ at some point, as you did in your question.

Answer 2

It is up to you where to define the zero potential (potential energy undefined up to a constant, so by adding any constant, the zero becomes anywhere you want it), but let's consider the point where the potential itself is minimum and subtract this value so that the potential is everywhere positive except at it's minimum (just a convention) $$\vec{F} = \left(\frac{A}{x^2} - B\right)\hat{x}$$ $$U(x) - U(x_0) = \left( \frac{A}{x} + Bx\right) - \left( \frac{A}{x0} + Bx_0\right)$$ This potential is positive everywhere for $x>0$. The minimum of this potential happens at $$\frac{dU(x)}{dx} = B - \frac{A}{x^2}\implies x_\text{min} = \sqrt{\frac{B}{A}}$$ Assuming you're only interested in the region $x>0$

Now $U(x_\text{min}) = \sqrt{\frac{B}{A}}B\left(1+\frac{A^2}{B^2} \right)$, and you can redefine your potential to be $U(x)\rightarrow U(x) - U(x_\text{min})$, in which case your potential is indeed zero now at $x_\text{min}$

Answer 3

There was a mistake in the Integration: $- \int_{x_i}^{x_f} F dx = [\frac{A}{x}+\frac{1}{2}Bx^2]_{x_i}^{x_f}$.

When you set the potential difference $U(x_f)-U(x_i)$ to Zero, you will get an $x$ that is always a real number; you will have $x = (\frac{-2A}{B})^{\frac{1}{3}}$.

Usually, imaginary potentials are unphysical; every measurement of physical quantities must be represented in form of real numbers.