a simple derivation of magnetic reconnection, part 2
In the
first part of this derivation and
an erratum, we used 5 equations to derive the magnetic field
B around a current-carrying rod.
In this part of the derivation, we will express that magnetic field in both cylindrical and Cartesian coordinates.
In part 3, we will show that the magnetic field around four current-carrying rods reproduces both of the figures in Dungey's 1958 paper, figure 3a in the survey paper by Yamada et al, and the still figure in Wikipedia's
current article on magnetic reconnection.
In part 4, I will describe a simple variation of
the experiment I've been suggesting to Michael Mozina and prove that the topology of the magnetic field changes during that experiment. As Yamada et al explain in their appendix, that change in the topology of the magnetic field is what we mean by magnetic reconnection.
[size=+1]
Notation for B[/size]
When we write ∇∙
B, we're thinking of
B as a function of points in 3-space. When we write d
B/dt or ∂
B/∂t, we're thinking of
B as a function of time.
As will become painfully clear if you read
Michael Mozina's posts, that contextual overloading of
B can confuse people. Let's take a minute to review the notation.
We usually speak of
B as the magnetic field, which means it's a function from points in 3-space (which we write as R
3) to vectors of magnetic flux density (which we also write as R
3). In that kind of context,
B is a function from R
3 to R
3, and it would be correct to write
B ∈ (R3 ➝ R3)
where (R
3 ➝ R
3) is the set of all functions from R
3 to R
3.
When we speak of
B as a function from time to magnetic fields, it would be correct to write
B ∈ (R ➝ (R3 ➝ R3))
We may also regard
B as a function of four variables (x, y, z, and t), so it would be correct to write
B ∈ ((R3 × R) ➝ R3)
Because (R ➝ (R
3 ➝ R
3)) is isomorphic to ((R
3 × R) ➝ R
3), these last two meanings for
B are consistent.
I'll continue to use the traditional notation (instead of
lambda calculus, which would be more precise), but everyone should keep in mind that context determines whether we're thinking of
B as a function of space or time.
From
part 1, here are the equations we'll need as we move forward:
[size=+1]
Equation 3 (integral form of Ampère's law)[/size]
[latex]
\[
\oint_C \hbox{{\bf B}} \cdot d \hbox{{\bf l}} =
\int_S \nabla \times \hbox{{\bf B}} \cdot \hbox{{\bf n}} \; da =
\mu_0 \int_S \hbox{{\bf J}} \cdot \hbox{{\bf n}} \; da
\]
[/latex]
[size=+1]
Equation 4 (magnitude of magnetic field around a current-carrying rod)[/size]
In cylindrical coordinates:
[latex]
\[
| \hbox{{\bf B}} | = | \hbox{{\bf B}} (r, \theta, z, t) | = | \hbox{{\bf B}} (t) (r, \theta, z) | = \frac{\mu_0}{2 \pi} \frac{I(t)}{r}
\]
[/latex]
where those first two equalities illustrate the context-dependent overloading of
B.
[size=+1]
Equation 5 (superposition)[/size]
[latex]
\[
\hbox{{\bf B}} = \sum_{i} \hbox{{\bf B}}_i
\]
[/latex]
[size=+1]
Magnetic field around a single rod (cylindrical coordinates)[/size]
From equations 3 and 4, the magnetic field around a single current-carrying rod positioned at r=0 is
[latex]
\[
\hbox{{\bf B}} = \hbox{{\bf B}} (r, \theta, z, t) = \hbox{{\bf B}} (t) (r, \theta, z) = \frac{\mu_0}{2 \pi} \frac{I(t)}{r} \hbox{{\bf e}}_\theta
\]
[/latex]
where
eθ is the unit vector in the θ direction.
[size=+1]
Magnetic field around a single rod (Cartesian coordinates)[/size]
In part 3, we'll derive the magnetic field for two and then four current-carrying rods. They won't all be positioned at r=0. To derive the magnetic field around rods positioned elsewhere, Cartesian coordinates will be more convenient. From the equation above for cylindrical coordinates, the magnetic field around a single current-carrying rod positioned at the origin of a Cartesian coordinate system is
[latex]
\[
\begin{align*}
\hbox{{\bf B}} &= \hbox{{\bf B}} (x, y, z, t) = \hbox{{\bf B}} (t) (x, y, z) \\
&= \frac{\mu_0}{2 \pi} \frac{I(t)}{\sqrt{x^2+y^2}}
\left( - \frac{y}{\sqrt{x^2+y^2}} \, \hbox{{\bf e}}_x + \frac{x}{\sqrt{x^2+y^2}} \, \hbox{{\bf e}}_y \right) \\
&= \frac{\mu_0}{2 \pi} \frac{I(t)}{x^2+y^2}
\left( - y \, \hbox{{\bf e}}_x + x \, \hbox{{\bf e}}_y \right)
\end{align*}
\]
[/latex]
where
ex and
ey are the unit vectors in the x and y directions.
Everything we've done so far is found within standard textbooks on electromagnetism. In part 3, we'll go beyond freshman-level textbooks by reproducing the magnetic fields shown in Dungey's figures and in Yamada et al's figure 3a.
)
!
.
Where does the kinetic energy come from at two ZERO points in two magnetic fields RC? Dungey used ELECTRICAL DISCHARGES.