Why is the curl of H zero in a plasma and where does the factor of me come from?

[PhDeezNutz]

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.

 

[jedishrfu]

In answer to the meaning of curl

https://betterexplained.com/articles/vector-calculus-understanding-circulation-and-curl/
This article uses the pinwheel in the flow visualization and if it spins then the curl is not zero.

Khan Academy and mathispower4u.com have videos you can search for that solve curl problems in a variety of vector fields using line integrals to verify the curl calculation.

 

[PhDeezNutz]

I think it has something to do with assuming the exponential form of $\vec{E}$ and thus $\vec{V}$ hence their derivatives will be scalar multiples of each other.

Let me take a crack at it. (And the thing is I don't even think we need to invoke Ampere's Law, at least not directly)

$$\vec{E} = \frac{Nq}{\sigma} \vec{V}$$

$$\frac{\partial \vec{E}}{\partial t} = \frac{Nq}{\sigma} \frac{\partial \vec{V}}{\partial t} $$

Assuming the form

$$\vec{E} = \vec{E_0} e^{i\left(\vec{k} \cdot {x} - \omega t \right)}$$ the time derivative is merely $- i \omega \vec{E}$

So we have$$-i \omega \vec{E} = \frac{Nq}{\sigma} \frac{\partial \vec{V}}{\partial t}$$

$$\frac{\partial \vec{V}}{\partial t} = \frac{-i \omega \sigma}{Nq} \vec{E} $$

Multiply through by $m_e$ we get$$ m_e \frac{\partial \vec{V}}{\partial t} = m_e \frac{-i \omega \sigma}{Nq} \vec{E} $$

Which is close but not quite the answer I'm looking for

The answer I'm looking for is

$$m_e \frac{\partial \vec{V}}{\partial t} = - q \vec{E}$$

(Which is reasonable since it's essentially saying Force = Force)

I guess somehow

$$m_e \frac{-i \omega \sigma}{Nq} = q$$

But I'm not seeing it.

Any ideas? Thanks for the help in advanced.

Also the factor of $i$ is concerning, I don't see how to get rid of it. It seems inevitable when taking the time derivative of a wave form.

 

[jedishrfu]

As an aside, I just found this 3blue1brown video on curl and divergence that can help to understand the meaning behind these vector field operations: