MHD Fluid: Difference between revisions

Revision as of 14:19, 11 February 2024

\rho \left({\frac {Du}{Dt}}\right)=-\nabla P+J\times B+\eta \nabla ^{2}u

{\frac {\partial B}{\partial t}}=\nabla \times (u\times B-\eta \nabla B)

E+u\times B=\eta J

\rho \left({\frac {D\varepsilon }{Dt}}\right)=-P\nabla \cdot u+\nabla \cdot (k\nabla T)+\eta J^{2}

{\frac {D\rho }{Dt}}+\rho \nabla \cdot u=0

@@ Line 1: / Line 1: @@
-\(E=mc^2\)
-\(E=mc^2\)
-$$\frac{1}{2} mv^2$$
-<math>E=mc^2</math>
-This is an example of a simple inline equation: <math>E=mc^2</math>.
-For a more complex block-style equation:
+<math display="block">
+\rho \left(\frac{Du}{Dt}\right) = -\nabla P + J \times B + \eta \nabla^2 u
+</math>
+<math display="block">
+\frac{\partial B}{\partial t} = \nabla \times (u \times B - \eta \nabla B)
+</math>
+<math display="block">
+E + u \times B = \eta J
+</math>
+<math display="block">
+\rho \left(\frac{D\varepsilon}{Dt}\right) = -P \nabla \cdot u + \nabla \cdot (k \nabla T) + \eta J^2
+</math>
 <math display="block">
-\frac{1}{2} mv^2
+\frac{D\rho}{Dt} + \rho \nabla \cdot u = 0
 </math>