Radial force balance

Next, consider the force balance in $\hat{\mathbf{R}}$ direction. The $R$ component of the force balance equation (43) is written

$$J_{\phi} B_Z - J_Z B_{\phi} = \frac{\partial P}{\partial R}$$ (49)

Using the expressions of the current density and magnetic field [Eqs. (5), (38), and (41)], equation (49) is written

$$- \frac{1}{R} \triangle^{\ast} \Psi \frac{1}{R} \frac{\partial \P... ...rac{\partial g}{\partial R} \frac{g}{R} = \mu_0 \frac{\partial P}{\partial R} .$$ (50)

Using the fact that $P$ and $g$ are a function of only $\Psi$, i.e., $P = P (\Psi)$ and $g = g (\Psi)$, Eq. (50) is written

$$- \frac{1}{R} \triangle^{\ast} \Psi \frac{1}{R} \frac{\partial \P... ...ial R} \frac{g}{R} = \mu_0 \frac{d P}{d \Psi} \frac{\partial \Psi}{\partial R},$$ (51)

which can be simplified to

$$\triangle^{\ast} \Psi = - \mu_0 R^2 \frac{d P}{d \Psi} - \frac{d g}{d \Psi} g,$$ (52)

i.e.,

$$\frac{\partial^2 \Psi}{\partial Z^2} + R \frac{\partial}{\partial... ...i}{\partial R} \right) = - \mu_0 R^2 \frac{d P}{d \Psi} - \frac{d g}{d \Psi} g.$$ (53)

Equation (53) is known as Grad-Shafranov (GS) equation.

[Note that the $Z$ component of the force balance equation is written

$$\begin{array}{l} J_R B_{\phi} - J_{\phi} B_R = \frac{\partial P}... ...st} \Psi = - \mu_0 R^2 \frac{d P}{d \Psi} - \frac{d g}{d \Psi} g \end{array}$$

which turns out to be identical with the Grad-Shafranov equation. This is not a coincidence. The reason is that the force balance equation has been satisfied in three different directions, namely, in the $\hat{\ensuremath{\boldsymbol{\phi}}}$, $\hat{\mathbf{R}}$, and $\mathbf {B}$ direction, and thus it must be satisfied in all directions.]