How do you know the vector potential inside a solenoid only depends on r?

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Discussion Overview

The discussion revolves around the vector potential inside a solenoid, specifically addressing why it is argued to depend only on the perpendicular distance from the axis of the solenoid (denoted as \(\rho\)) and why it is believed to have only an azimuthal component (\(\hat{\phi}\)). The scope includes theoretical reasoning and symmetry arguments in electromagnetism.

Discussion Character

  • Exploratory, Technical explanation, Conceptual clarification

Main Points Raised

  • One participant questions the symmetry arguments used to conclude that the vector potential depends only on \(\rho\) and has only an azimuthal component.
  • Another participant suggests that the form of the vector potential is constrained by the surface integral of the current density, leading to only an azimuthal component being possible.
  • A different participant notes that the geometry of the solenoid is invariant in the \(z\) and \(\phi\) dimensions, implying that any properties of the system must also be invariant.
  • One participant elaborates that the radial symmetry of the charge and current configuration in a cylindrical solenoid means that if the vector potential had angular dependence, it would imply a distinction in direction that contradicts the symmetry of the system.

Areas of Agreement / Disagreement

Participants express a general agreement on the idea that the vector potential must exhibit radial symmetry due to the nature of the solenoid's charge and current distribution. However, the discussion includes varying levels of understanding and interpretation of the symmetry arguments, indicating that some aspects remain contested or unclear.

Contextual Notes

Participants do not provide explicit definitions of symmetry or the conditions under which the vector potential is derived, leaving some assumptions and mathematical steps unresolved.

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How can you argue, by symmetry, that the vector potential inside a solenoid depends only on [itex]\rho[/itex], the perpendicular distance from the axis of the solenoid? And how can you argue that there is only a [itex]\hat{\phi}[/itex] (azimuthal) component of the vector potential, such that [itex]\vec{A}[/itex] takes the form

[tex]\vec{A}(\vec{r}) = A_\phi(\rho) \hat{\phi}[/tex]

My E&M teacher is fond of talking about how we can "use symmetry arguments" to reach these conclusions without actually providing the arguments for us. It's a little frustrating...what symmetry arguments is he talking about?
 
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Ok, I think I understand why [itex]A[/itex] has only an azimuthal component. The equation for the vector potential reads
[tex] \vec{A}(\vec{r}) = \frac{\mu_0}{4 \pi} \int\int \frac{\vec{K}(\vec{r}') dA}{|\vec{R}|},[/tex]
and since [itex]\vec{K}(\vec{r}) = \frac{N}{L} I \hat{\phi}[/itex] (where N is the # of turns and L is the length), and we can't obtain other components from performing the surface integral, we're stuck with only a [itex]\phi[/itex] component for the vector potential.

Is that right?!
 
Basically, the geometry of the problem is invariant in the z hat and phi hat dimensions. There is no change in the problem with respect to these variables and dimensions. If the entire system is invariant in a dimension, then any other properties of the system must also be invariant.
 
To put it another way: the vector potential has to be determined by the physical charge and current configuration, right? And in a cylindrical solenoid, that configuration exhibits radial symmetry - that is, you can rotate it around its axis, and the charge/current configuration looks just the same as it did before.

Now suppose the vector potential did depend on [itex]\phi[/itex]. Whatever kind of angular dependence you can imagine, it provides some way to distinguish one direction from another... there has to be some "feature" of the vector potential that points in some particular direction, for it to have angular dependence. Let's say that you start by orienting the solenoid so that this feature points upwards (i.e. in a chosen direction which we'll designate as "up"). Then rotate the solenoid around by 90 degrees. Now the feature of the vector potential points to the left. But the charge and current distribution generating the vector potential still looks the same! How can the same distribution create two different vector potentials, one with this "feature" pointing up and one with it pointing to the left?

(The answer, of course, is that it can't, which is why we know that the vector potential must be radially symmetric)
 

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