I'm trying to understand how the following equation:
$-\nabla P + j\ \times\ B + pg = 0 $
Reduces to:
$\frac{P_0}{L_0}=\frac{B_0^2}{\mu_0 \rho_0}$
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Sign up to join this communityI'm trying to understand how the following equation:
$-\nabla P + j\ \times\ B + pg = 0 $
Reduces to:
$\frac{P_0}{L_0}=\frac{B_0^2}{\mu_0 \rho_0}$
Why is the magnetic field magnitude squared?
(I rephrased your question as I assume this is what you were asking)
In MHD, one generally ignores the displacement current density such that: $$ \mathbf{j} = \frac{ 1 }{ \mu_{o} } \nabla \times \mathbf{B} \tag{0} $$ then we have the following: $$ \mathbf{j} \times \mathbf{B} = \frac{ 1 }{ \mu_{o} } \left( \nabla \times \mathbf{B} \right) \times \mathbf{B} \tag{1} $$ From vector calculus identities, we can show that: $$ \mathbf{j} \times \mathbf{B} = \frac{ 1 }{ \mu_{o} } \left[ \left( \mathbf{B} \cdot \nabla \right) \mathbf{B} - \frac{1}{2} \nabla \left( \mathbf{B} \cdot \mathbf{B} \right) \right] = \frac{ \left( \mathbf{B} \cdot \nabla \right) \mathbf{B} }{ \mu_{o} } - \nabla \left( \frac{ B^{2} }{ 2 \ \mu_{o} } \right) \tag{2} $$
Where do these terms come from?
(I rephrased your question as I assume this is what you were asking)
The factor of $L_{o}^{-1}$ comes from the assumption that $\nabla \ Q \propto L^{-1} \ Q$, where $L^{-1}$ is some characteristic scale length and $Q$ is the relevant parameter (i.e., you assume all $Q$ are proportional to an exponential function of height/altitude). In your particular problem, it seems the notation is $L_{o}$, which I am guessing represents an atmospheric scale height or some reference altitude at which the mass density is $rho_{o}$.