Confusion regarding the scalar potential

In summary, the conversation discusses the determination of whether the magnetic field \mathbf{B} in cylindrical coordinates has a scalar potential or not. It is shown that because the domain is not simply connected, \mathbf{B} does not have a scalar potential. However, there is a contradiction when considering a function \phi(\varphi) = \frac{I\mu_0}{2\pi}\varphi, which implies that \mathbf{B} does have a scalar potential. The issue lies in the discontinuity of \phi at \varphi = 0 and \varphi = 2\pi, and it is suggested that physically acceptable fields are always periodic in \varphi with period 2\
  • #1
lampCable
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Homework Statement


Consider the following in cylindrical coordinates [itex]\rho,\varphi,z[/itex]. An electric current flows in an infinitely long straight cylindrical wire with the radius [itex]R[/itex]. The magnetic field [itex]\mathbf{B}[/itex] outside of the thread is [tex]\textbf{B}(\textbf{r})=\frac{I\mu_0}{2\pi}\frac{\mathbf{e}_{\varphi}}{\rho}, ρ>R.[/tex] We want to determine wether [itex]\mathbf{B}[/itex] has a scalar potential or not.

Homework Equations



The Attempt at a Solution


Since the domain is not simply connected, there is no use showing that [itex]\nabla\times\mathbf{B}=\mathbf{0}[/itex] for [itex]ρ>R[/itex]. Therefore, we must show that for all points [itex]P[/itex] and [itex]Q[/itex] the line integral of [itex]\mathbf{B}[/itex] from [itex]P[/itex] to [itex]Q[/itex] is independent of path, since this implies that [itex]\mathbf{B}[/itex] has a scalar potential.

Consider a circle, [itex]C[/itex], that is concentric with the cylindrical wire with radius [itex]R_1[/itex] such that [itex]R_1>R[/itex]. Line integration of [itex]\mathbf{B}[/itex] over the closed curve [itex]C[/itex] yields [tex]\oint_C\mathbf{B}⋅d\mathbf{r} = [d\mathbf{r} = R_1d\varphi\mathbf{e}_{\varphi}] = \frac{I\mu_0}{2\pi}\int^{2\pi}_0\mathbf{e}_{\varphi}⋅\mathbf{e}_{\varphi}d\varphi = I\mu_0 \neq 0.[/tex] But this says that [itex]\mathbf{B}[/itex] is dependent of path, and therefore [itex]\mathbf{B}[/itex] does not have a scalar potential.

However, if we now consider the function [tex]\phi(\varphi) = \frac{I\mu_0}{2\pi}\varphi,[/tex] we observe that [tex]\nabla\phi = \frac{1}{\rho}\frac{I\mu_0}{2\pi}\mathbf{e}_\varphi = \mathbf{B}.[/tex] But this says that [itex]\mathbf{B}[/itex] is independent of path, and therefore [itex]\mathbf{B}[/itex] has a scalar potential.

And so, we end up with a contradiction. Since I am very confident that the line integral of [itex]\mathbf{B}[/itex] over [itex]C[/itex] is correct, I assume that there is something in the last part that is wrong, but I cannot find the error.
 
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  • #2
Your problem here is that [itex]\varphi[/itex] lies between [itex]0[/itex] and [itex]2\pi[/itex], but both [itex]\varphi = 0[/itex] and [itex]\varphi = 2\pi[/itex] represent the same physical half-plane. Thus if [itex]\phi = k\varphi[/itex], [itex]k \neq 0[/itex], then there is a half-plane where [itex]\phi[/itex] is discontinuous: from one side [itex]\phi \to 0[/itex] and from the other [itex]\phi \to 2k\pi[/itex]. This is a problem for the existence of [itex]\nabla \phi[/itex] *everywhere* in [itex]\rho > R[/itex].

To avoid this problem, physically acceptable scalar or vector fields are always periodic in [itex]\varphi[/itex] with period [itex]2\pi[/itex].

(The other way to avoid this is to admit multivalued potentials, in which case yes, [itex]\nabla (k\varphi) = (k/\rho)\mathbf{e}_{\varphi}[/itex].)
 
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Related to Confusion regarding the scalar potential

What is the scalar potential?

The scalar potential is a mathematical concept used in physics to describe the potential energy of a system. It is a scalar quantity, meaning it has only magnitude and no direction.

What causes confusion about the scalar potential?

The concept of the scalar potential can be confusing because it is often described using abstract mathematical language that is not easily understood by non-experts. Additionally, its application in different areas of physics can vary, leading to further confusion.

How is the scalar potential different from other potentials?

The scalar potential is different from other potentials, such as the vector potential and the gravitational potential, in that it is a scalar quantity and does not have a direction associated with it. It is also dependent on the position of an object in space, rather than its velocity or mass.

What is the significance of the scalar potential in physics?

The scalar potential is an important concept in physics because it helps to describe the behavior of systems and the relationships between different physical quantities. It is used in many areas of physics, including electromagnetism, quantum mechanics, and fluid dynamics.

How can one better understand the scalar potential?

To better understand the scalar potential, it is helpful to have a strong foundation in mathematics, particularly calculus and vector calculus. It can also be useful to study specific applications of the scalar potential in different areas of physics, as well as seeking out explanations and visualizations from reputable sources.

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