“Proof” of Alfven’s Theorem of Flux Freezing

SOURCE FOR CONTENT: Choudhuri, A.R., 2010. Astrophysics for Physicists. Ch. 8. 

In the previous post we saw the consequences of different regimes of the magnetic Reynolds’ number under which either diffusion or advection of the magnetic field dominates. In this post, I shall be doing a “proof” of Alven’s Theorem of Flux Freezing. (I hesitate to call it a proof since it lacks the mathematical rigor that one associates with a proof.) Also note in this post, I will be working with the assumption of a high magnetic Reynolds number.

Alfven’s Theorem of Flux Freezing: Suppose we have a surface S located within a plasma at some initial time t_{0}. From the theorem it is known that the flux of the associated magnetic field is linked with surface S by

\displaystyle \int_{S}\textbf{B}\cdot d\textbf{S}. (1)

At some later time t^{\prime}, the elements of plasma contained within S at t_{0} move to some other point and will constitute some different surface M. The magnetic flux, linked to M at t^{\prime} by

\displaystyle \int_{M}\textbf{B}\cdot d\textbf{M}, (2)

from which we may mathematically state the theorem as

\displaystyle \int_{S}\textbf{B}\cdot d\textbf{S}=\int_{M}\textbf{B}\cdot d\textbf{M}. (3)

If we know that the magnetic field evolves in time in accordance to the induction equation we may express Eq.(3) as

\displaystyle \frac{d}{dt}\int_{S}\textbf{B}\cdot d\textbf{S}=0. (4)

To confirm that this is true, we note the two ways magnetic flux may change as being due to either (1) some intrinsic variability of the magnetic field strength or (2) movement of the surface. Therefore, either way it follows that

\displaystyle \frac{d}{dt}\int_{S}\textbf{B}\cdot d\textbf{S}=\int_{S}\frac{\partial \textbf{B}}{\partial t}\cdot d\textbf{S}+\int_{S}\textbf{B}\cdot \frac{d}{dt}(d\textbf{S}). (5)

Now, consider again the two surfaces. Let us suppose now that M is displaced some distance relative to S. Further, let us also suppose that this displacement occurs during a time interval t^{\prime}=t_{0}+\delta t. Additionally, if we imagine a cylinder formed by projecting a circular cross-section from one surface to the other, we may consider its length to be \delta l with area given by the cross product: -\delta t \textbf{v}\times \delta\textbf{l}. Moreover, since we know that the area of integration is a closed region we see that the integral vanishes (goes to 0). Thus, we may write the difference

\displaystyle d\textbf{M}-d\textbf{S}-\delta \oint \textbf{v}\times \delta \textbf{l}=0 (6).

Recall the definition for a derivative, we may apply it to the second term on the right hand side of Eq.(5) to get

\displaystyle \frac{d}{dt}(d\textbf{S})=\lim_{\delta t\rightarrow 0}\frac{d\textbf{M}-d\textbf{S}}{\delta t}=\oint \textbf{v}\times \delta \textbf{l}. (7)

Thus the term becomes

\displaystyle \int_{S}\textbf{B}\cdot \frac{d}{dt}(d\textbf{S})=\int \oint \textbf{B}\cdot (\textbf{v}\times \delta\textbf{l})=\int \oint (\textbf{B}\times \textbf{v})\cdot \delta l. (8)

Since the integrals that exist interior to the boundary of the surface (call it path C) vanish and recall Stokes’ theorem

\displaystyle \oint_{\partial \Omega}\textbf{F}\cdot d\textbf{r}=\int\int_{\Omega} (\nabla \times \textbf{F})\cdot d\Omega,

and applying it to Eq.(8) we arrive at

\displaystyle \frac{d}{dt}\int_{S}\textbf{B}\cdot d\textbf{S}=\oint_{\partial S}(\textbf{B}\times\textbf{v}) \cdot \delta \textbf{l}=\int\int_{S}\bigg\{\nabla \times (\textbf{B}\times\textbf{v})\bigg\}\cdot d\textbf{S}. (9)

Recall that we are dealing with high magnetic Reynolds number, if we use the corresponding form of the induction equation in Eq.(9) we arrive at

\displaystyle \frac{d}{dt}\int_{S}\textbf{B}\cdot d\textbf{S}=\int\int_{S}d\textbf{S}\cdot \bigg\{ \frac{\partial \textbf{B}}{\partial t}-\nabla\times (\textbf{v}\times \textbf{B})\bigg\}= 0. (10)

Thus, this completes the “proof”.

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