# Plasma in magnetic field in equilibrium

Hello,

Starting from the equilibrum equation of a steady-state plasma in a magnetic field:
$$(1) \nabla P = \vec{J} \times \vec{B}$$
where $P = nkT$ is the plasma pressure, $\vec{J}$ is the current and $\vec{B}$ is the magnetic field, I am trying to solve for the magnetic field $\vec{B}$ in terms of a known pressure $P$.

Using $\vec{J} = \frac{1}{\mu_0} \nabla \times \vec{B}$, and the identity $\vec{B} \times (\nabla \times \vec{B}) + (\vec{B} \cdot \nabla) \vec{B} = \nabla( B^2/2)$, the first equation becomes
$$(2) \nabla \left( P + \frac{B^2}{2 \mu_0} \right) = \frac{1}{\mu_0} (\vec{B} \cdot \nabla) \vec{B}$$
Dotting both sides by a path length $\vec{ds}$ and integrating from the point of interest $\vec{r}$ to some other point $p_2$, and then using the gradient theorem gives
$$(3) P(p_2) + \frac{B^2(p_2)}{2 \mu_0} - P(\vec{r}) - \frac{B^2(\vec{r})}{2 \mu_0} = \frac{1}{\mu_0} \int_{\vec{r}}^{p_2} \left[ (\vec{B} \cdot \nabla) \vec{B} \right] \cdot \vec{ds}$$
The integrand of (3) can be expressed as
$$(4) \frac{1}{\mu_0} (\vec{B} \cdot \nabla) \vec{B} = \hat{b} \frac{\partial}{\partial l} \left( \frac{B^2}{2 \mu_0} \right) - \hat{n} \frac{B^2}{\mu_0 R_c}$$
where $l$ is a coordinate along a magnetic field line, $\hat{b}$ is a unit vector tangent to the field line, $\hat{n}$ is a unit vector normal to the field line and anti-radial, and $R_c$ is the radius of curvature of the field line (see for example Plasma physics and controlled nuclear fusion By Kenrō Miyamoto, eq 6.7).

By choosing the path of integration either in the $\hat{b}$ direction or $\hat{n}$ direction perhaps we can get a solution.

CHOICE 1: $\vec{ds} = \hat{n} ds$

Here, we will choose $p_2$ to be infinity, where we have the advantage that all quantities will go to 0. So (3) becomes
$$(5) P(\vec{r}) + \frac{B^2(\vec{r})}{2 \mu_0} = \frac{1}{\mu_0} \int_{\vec{r}}^{\infty} \frac{B^2}{R_c} ds$$
Unfortunately, it does not seem that there is a closed form solution to this integral.

CHOICE 2: $\vec{ds} = \hat{b} dl$

Here, we are integrating along a field line, but there does not seem to be an obvious choice for $p_2$, but the integral will be trivial. Leaving $p_2$ in place for now, we find that (3) becomes
$$(6) P(p_2) + \frac{B^2(p_2)}{2 \mu_0} - P(\vec{r}) - \frac{B^2(\vec{r})}{2 \mu_0} = \frac{B^2(p_2)}{2 \mu_0} - \frac{B^2(\vec{r})}{2 \mu_0}$$
which unfortunately leads to $P(\vec{r}) = P(p_2)$ which isn't very helpful. Interestingly, if we integrate the other way on the field line, so that $\vec{ds} = -\hat{b} dl$ we get a completely different solution:
$$(7) B^2(\vec{r}) = B^2(p_2) - \mu_0 (P(\vec{r}) - P(p_2))$$
which has the unfortunate property that the solution depends on which point $p_2$ we pick. Can anyone shed some light on where I'm going wrong with these integrals? Many thanks