Question about Classical Transport Theory

Question

With a distribution function of the form $f=f_{0} + \vec{v} \cdot \vec{g}$, one can obtain the current density. My question is about $\vec{g}$; we assume a general solution to $\vec{g}$ of the form $\vec{g}= \alpha \vec{E}+\beta \vec{B} + \gamma(\vec{B} \times \vec{E}) $. By substitution into $$\frac{\mu}{c}(\vec{v} \times \vec{B})\cdot \vec{g} + \vec{v} \cdot \vec{g} = e \tau \frac{\partial f_{0}}{\partial \epsilon} \vec{E} \cdot \vec{v}$$ and by equating both sides and equating the coefficients of $\vec{v} \cdot \vec{E}$, $\vec{v} \cdot \vec{B}$, and $\vec{v} \cdot (\vec{B} \times \vec{E})$ we are supposed to get $$\vec{g} = \frac{\partial f_{0}}{\partial \epsilon} \frac{e \tau}{1+ (\omega_{c}\tau)^{2}}[\vec{E}+(\omega_{c}\tau)^{2}(\hat{z}\cdot\vec{E})\hat{z}+\omega_{c}\tau(\vec{E} \times \hat{z})]$$

My question is how? I'm afraid this isn't apparent to me. I don't know where to start to try and prove this!

Answer 1

When we substitute the expression for $\vec{g}$ in the equation you wrote, we can do a number of simplifications, using identities that involve the cross product. The first term contains a dot product of $\vec{v} \times \vec{B}$ with $\vec{g}$, in which we can use:

$$ (\vec{v} \times \vec{B}) \cdot \vec{B} = 0 \\ (\vec{v} \times \vec{B}) \cdot \vec{E} = (\vec{B} \times \vec{E}) \cdot \vec{v} \\ (\vec{v} \times \vec{B}) \cdot (\vec{B} \times \vec{E}) = (\vec{v} \cdot \vec{B})(\vec{B} \cdot \vec{E}) - B^2 ~\vec{v} \cdot \vec{E}$$

After the substitution we get:

$$ \frac{\mu}{c} \left[ \alpha \vec{v} \cdot (\vec{B} \times \vec{E}) + \gamma (\vec{B} \cdot \vec{E}) (\vec{v} \cdot \vec{B}) -\gamma B^2 ~ \vec{v} \cdot \vec{E} \right] +\alpha ~ \vec{v} \cdot \vec{E} + \beta ~ \vec{v} \cdot \vec{B} + \gamma ~ \vec{v} \cdot(\vec{B} \times \vec{E}) = e \tau \frac{\partial f_0}{\partial \epsilon} \vec{v} \cdot \vec{E} $$

We can now bring all the terms to the left side, and take out the common factor $\vec{v}$:

$$ \vec{v} \cdot \left\{ \frac{\mu}{c} \left[ \alpha (\vec{B} \times \vec{E}) + \gamma (\vec{B} \cdot \vec{E}) \vec{B} -\gamma B^2 ~ \vec{E} \right] +\alpha ~ \vec{E} + \beta ~ \vec{B} + \gamma ~ \vec{B} \times \vec{E} - e \tau \frac{\partial f_0}{\partial \epsilon} \vec{E} \right\} =0 $$

Since this equation holds for all $\vec{v}$, we know that the expression in the curly brackets needs to vanish. Assuming $\vec{E}$ and $\vec{B}$ are not parallel, the three vectors $\vec{E}$, $\vec{B}$ and $\vec{B} \times \vec{E}$ are linearly independent, so if a combination of them vanishes then each coefficient in the combination needs to vanish. In this case:

$$ -\frac{\mu}{c} \gamma B^2 +\alpha -e \tau \frac{\partial f_0}{\partial \epsilon}=0 ~~~~~ \text{(the coefficient of $\vec{E}$)} \\ \frac{\mu}{c} \gamma ~ \vec{B} \cdot \vec{E} + \beta =0 ~~~~~ \text{(the coefficient of $\vec{B}$)} \\ \frac{\mu}{c} \alpha + \gamma =0 ~~~~~ \text{(the coefficient of $\vec{B} \times \vec{E}$)}$$

We have a system of three equations we can solve for $\alpha$, $\beta$ and $\gamma$. After solving we substitute the solution in the expression for $\vec{g}$, and using the definition of $\omega_c$ and the fact that $\vec{B} = B \hat{z}$ we get the desired result (I leave these last steps to you).