How can I solve this tough integral equation for a dipole magnetic field?

[vibe3]

Hi, I have an integral equation for which I'd appreciate any tips! The equation is:

<br /> B^2(\vec{r}) = \int_{\vec{r}}^{\infty} \left[ (\vec{B}(\vec{s}) \cdot \nabla) \vec{B}(\vec{s}) \right] \cdot \vec{ds}<br />

The path of integration can be chosen arbitrarily, starting at some point r, and ending at infinity (where B = 0). B can be thought of as a dipole magnetic field.

I've tried expanding the integrand into its components along and normal to the field lines, and choosing ds to be a path always normal to the field lines. This yields the following equation:

<br /> B^2(\vec{r}) = -\int_{\vec{r}}^{\infty} \frac{B^2(\vec{s})}{R_c(\vec{s})} ds<br />

where R_c is the radius of curvature of the field line at the point s. But here it seems difficult to continue without numerical assistance. Anyone have any ideas?

 

[Mute]

I would try to solve that numerically before you put any great effort into getting an (approximate) analytical expression. Your integral equation has the trivial solution B = 0, and that might be your only solution. It's worth your time to set up a numerical solution anyways, because there may not be a nice, easy to get analytical expression.

However, if I were to try to see if I could find a closed-form expression, then because the variable r only appears as part of the limits of the integral, at least in the simplified expression, you might be able to differentiate both sides and derive a differential equation for B^2. (I would do B^2 rather than B because your simplified integral equation is linear in B^2 but non-linear in B). You still might not be able to solve the equation, though, and I'm not sure how nice everything will turn out because it's a multi-dimensional integral.

 

[Ugly American]

I have a solution, but mine is off from your's by a sign and a factor of 2. They key is the vector identity

\nabla (\vec{A}\cdot\vec{B}) = \vec{A}\times(\nabla\times\vec{B}) + \vec{B}\times(\nabla\times\vec{A}) + (\vec{A}\cdot\nabla)\vec{B} + (\vec{B}\cdot\nabla)\vec{A}.

We also use the fact that \vec{B} is the magnetic field from a dipole (which I assume is not osscilating). This implies that

\nabla\times\vec{B} = \displaystyle\frac{\partial \vec{E}}{\partial t} + \vec{J} = 0,

since the fields are constant in time \left(\displaystyle\frac{\partial\vec{E}}{\partial t} = 0 \right) and there is no current \left( \vec{J}=0 \right). This alows us to rewrite the right hand side of your equation as

\displaystyle\int_{\vec{r}}^{\infty} \left[\left(\vec{B}(\vec{s})\cdot\nabla\right)\vec{B} (\vec{s}) \right] \cdot d\vec{s} = \displaystyle\int_{\vec{r}}^{\infty} \left[ \frac{\nabla B^2}{2} - \vec{B}\times(\underbrace{\nabla\times\vec{B}}_0) \right]\cdot d\vec{s} = \frac{B^2(\infty)}{2}-\frac{B^2(\vec{r})}{2} = -\frac{B^2(\vec{r})}{2},

where we have used the Gradient Theorem and the fact that the dipole field goes to zero at infinity.

 

[vibe3]

Thanks for your solution, UA, but unfortunately there is a current which is not 0. My original statement was an attempt to solve the integral which comes up in the physical problem which I didn't describe very well. The actual problem is as follows.

Consider a dipole field \vec{B}_0, where plasma is then introduced to the system and allowed to reach an equilibrium state. In this equilibrium state, there will be a diamagnetic current which satisfies
<br /> \nabla P = \vec{J} \times \vec{B}<br />
where P = nkT is the plasma pressure, J is the current, and B is the total field:
<br /> \vec{B} = \vec{B}_0 + \vec{B}_d<br />
and \vec{B}_d is the field due to the diamagnetic current J. Using
<br /> \vec{J} = \frac{1}{\mu_0} \nabla \times \vec{B}<br />
as well as the identity you quoted, the equation becomes
<br /> \nabla \left( P + \frac{B^2}{2 \mu_0} \right) = \frac{1}{\mu_0} \left( \vec{B} \cdot \nabla \right) \vec{B}<br />
Dotting both sides by a path length \vec{ds} and then integrating along some path starting at the point of interest \vec{r} and ending at infinity where all the fields go to 0, allows us to use the gradient theorem on the LHS of the equation:
<br /> P(\vec{r}) + \frac{B^2(\vec{r})}{2\mu_0} = -\frac{1}{\mu_0} \int_{\vec{r}}^{\infty} (\vec{B}(\vec{s}) \cdot \nabla) \vec{B}(\vec{s}) \cdot \vec{ds}<br />
which was my original equation, except I left out the pressure P term for simplicity.

I think that this equation cannot in general be solved for \vec{B}, but I would be happy if there was a way to use perturbation theory to solve for the scalar correction to the magnitude of B. IE: if
<br /> B = B_0 + b<br />
where b is the perturbation/correction to the dipole field magnitude due to the diamagnetic current J. I would be happy with either a solution for b where you can neglect b^2 terms, or a full solution for \vec{B}

If it helps, the following identity is true:
<br /> (\vec{B} \cdot \nabla) \vec{B} = \frac{d}{db} \left( \frac{B^2}{2} \right) \hat{b} - \frac{B^2}{R_c} \hat{n}<br />
where \hat{b} is a unit vector along the field line, \hat{n} is normal to the field line and directed anti-radially, and R_c is the radius of curvature of the field line. Because of the derivative in the first term, I thought at first it might be best to integrated along a field line rather than integrate to infinity. This makes it possible to solve the integral, but then you end up with a solution dependent on the second point you choose in the integral which doesn't seem right.

 

[Antiphon]

Use the "method of moments" aka the method of weighted residuals.

 

[vibe3]

Use the "method of moments" aka the method of weighted residuals.

I don't understand how that will help...?