Pre-Quantum Electrodynamics
Definition; Gauge Choicesems.ms.vp.A
- PM 6.3
- Gr 5.4.1
Since \({\boldsymbol \nabla} \cdot {\bf B} = 0\) in magnetostatics, following Helmholtz's theorem (see c_m_vf_helm) we can write
Substituting this in Ampère's law and using curlcurl gives
\[ {\boldsymbol \nabla} \times {\bf B} = {\boldsymbol \nabla} \times ({\boldsymbol \nabla} \times {\bf A} ) = {\boldsymbol \nabla} ({\boldsymbol \nabla} \cdot {\bf A}) - {\boldsymbol \nabla}^2 {\bf A} = \mu_0 {\bf J} \]
In the case of electrostatics, we could add any constant to electrostatic potential without affecting the physics. Here, in magnetostatics, we can add any curlless function (said otherwise: the gradient of any scalar field) to the vector potential, without changing the magnetic field. Making a specific choice is called making a gauge choice. For example, we can always choose to eliminate the divergence of \({\bf A}\), giving us the example of the
Coulomb gauge
\[ {\boldsymbol \nabla} \cdot {\bf A} = 0. \tag{CoulG}\label{CoulG} \]
Proof: suppose our starting \({\bf A}_0\) is not divergenceless. We add \({\boldsymbol \nabla} \lambda\) to the vector potential, so \({\bf A} = {\bf A}_0 + {\boldsymbol \nabla} \lambda\). Then, \[ {\boldsymbol \nabla} \cdot {\bf A} = {\boldsymbol \nabla} \cdot {\bf A}_0 + {\boldsymbol \nabla}^2 \lambda. \] The scalar field then obeys a Poisson-like equation, \[ {\boldsymbol \nabla}^2 \lambda = -{\boldsymbol \nabla} \cdot {\bf A}_0, \] whose solution we know how to find, because it's mathematically exactly the same as solving the Poisson equation 🐟 with p_vcd. Therefore, provided \({\boldsymbol \nabla} \cdot {\bf A}_0\) goes to zero at infinity, \[ \lambda ({\bf r}) = \frac{1}{4\pi} \int d\tau' \frac{{\boldsymbol \nabla}' \cdot{\bf A}_0 ({\bf r}')}{|{\bf r} - {\bf r}'|}. \] (exercise: prove this in one line using the Laplacian and Lap1or).
Under this gauge choice, Ampère's law becomes
Again this is precisely like the Poisson equation, with one separate equation for each component. For currents falling off sufficiently rapidly at infinity, we thus have
\[ {\bf A} ({\bf r}) = \frac{\mu_0}{4\pi} \int d\tau' \frac{J({\bf r}')}{|{\bf r} - {\bf r}'|} \tag{A_CoulG}\label{A_CoulG} \]
For line and surface currents,
\[ {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int dl' \frac{{\bf I ({\bf r}')}}{|{\bf r} - {\bf r}'|}, \tag{A_CoulG_l}\label{A_CoulG_l} \]
\[ {\bf A}({\bf r}) = \frac{\mu_0}{4\pi} \int da' \frac{{\bf K ({\bf r}')}}{|{\bf r} - {\bf r}'|}. \tag{A_CoulG_s}\label{A_CoulG_s} \]
Example: infinite solenoid
Task: find the vector potential of an infinite solenoid with \(n\) turns pet unit length, radius \(R\) and current \(I\).
Solution: one cannot use Amp_CoulG since the current extends to infinity (comment: you are encouraged to check that the integral converges anyway, by combining the integrals for \(z > 0\) and \(z < 0\) into one).
Nice trick: notice that
\[ \oint d{\bf l} \cdot {\bf A} = \int d{\bf a} \cdot ({\boldsymbol \nabla} \times {\bf A}) = \int d{\bf a} \cdot {\bf B} = \Phi. \] This is reminiscent of Ampère's law in integral form Amp_int, \[ \oint d{\bf l} \cdot {\bf B} = \mu_0 I_{enc}. \] Mathematically, it's the same equation! Replacement: \({\bf B} \rightarrow {\bf A}\) and \(\mu_0 I_{enc} \rightarrow \Phi\). And to paraphrase Feynman's lectures: the same equations have the same solutions.
Use symmetry: vector potential can only be circumferential. Using an amperian loop at a radius \(r\) inside the solenoid, and the fact that the field inside a solenoid is \(\mu_0 n I\) (as per Btor), we get \[ \oint d{\bf l} \cdot {\bf A} = A (2\pi r) = \int d{\bf a} \cdot {\bf B} = \mu_0 n I (\pi r^2), \] so
\[ {\bf A} = \frac{\mu_0 n I}{2} r ~\hat{\boldsymbol \varphi}, \hspace{1cm} r < R. \] For an amperian loop outside, the flux is always \(\mu_0 n I (\pi R^2)\), so
\[ {\bf A} = \frac{\mu_0 n I}{2} \frac{R^2}{r} \hat{\boldsymbol \varphi}, \hspace{1cm} r > R. \] Exercise: check that \({\boldsymbol \nabla} \times {\bf A} = {\bf B}\) and that \({\boldsymbol \nabla} \cdot {\bf A} = 0\).
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Created: 2024-02-27 Tue 10:31