- #1
PhDeezNutz
- 693
- 440
- Homework Statement
- I'm following my instructor's notes and they read as follows
"In a plasma all the electrons are free. Then
$$\vec{J_f} = \sigma \vec{E} = Nq\vec{V} $$
and
$$m_e \frac{\partial \vec{V}}{\partial t} = - q \vec{E}$$
I don't see how this is possible unless we assume
$$\nabla \times \vec{H} = \vec{0}$$
per Maxwell's Equations. And I have no idea why that would physically substantiated.
- Relevant Equations
- $$\nabla \times \vec{H} = \vec{J_f} + \epsilon \frac{\partial \vec{E}}{\partial t}$$
If we assume ##\nabla \times \vec{H} = \vec{0}## (again I have no idea why this would be true)
$$\vec{0} = \sigma \vec{E} + \epsilon \frac{\partial \vec{E}}{\partial t}$$
$$\vec{0} = \sigma \vec{E} + \epsilon Nq\frac{\partial \vec{V}}{\partial t}$$
$$-\sigma \vec{E} = \epsilon Nq\frac{\partial \vec{V}}{\partial t}$$
multiply through by ##\frac{q}{\sigma}##
$$-q\vec{E} = \frac{\epsilon N q^2}{\sigma} \frac{\partial \vec{V}}{\partial t}$$
I guess my question is two fold
1) Why is ##\nabla \times \vec{H} = \vec{0}## physically or mathematically ?
2) Where does the factor of ##m_e## come from?
As always any help is appreciated.
$$\vec{0} = \sigma \vec{E} + \epsilon \frac{\partial \vec{E}}{\partial t}$$
$$\vec{0} = \sigma \vec{E} + \epsilon Nq\frac{\partial \vec{V}}{\partial t}$$
$$-\sigma \vec{E} = \epsilon Nq\frac{\partial \vec{V}}{\partial t}$$
multiply through by ##\frac{q}{\sigma}##
$$-q\vec{E} = \frac{\epsilon N q^2}{\sigma} \frac{\partial \vec{V}}{\partial t}$$
I guess my question is two fold
1) Why is ##\nabla \times \vec{H} = \vec{0}## physically or mathematically ?
2) Where does the factor of ##m_e## come from?
As always any help is appreciated.