Smallest relative velocity driving a two-stream instability

The physical picture is a two-stream system of cold electrons and ions (i.e. $T_i=T_e=0$) with realtive velocities $V_i$ and $V_e$. The dispersion relation obtained is $$D(\omega)=1-\frac{\omega_{pi}^2}{(\omega-kV_i)^2}-\frac{\omega_{pe}^2}{(\omega-kV_e)^2}=0$$ where $\omega_{pi}$, $\omega_{pe}$ respectevely correspond to electron and ion plasma frequency.

I am asked to give the minimum relative velocity $V_e-V_i$ that may drive an instability. What I know is that there can be an instability if $D(\omega)<0$, because such situation give two real and two imaginary roots, the second case give exponential solutions if $\text{Im}(\omega)>0$.

The first thing to do is use a different reference frame to simplify things by going into the ion rest frame. Thus, the dispersion relation goes to: $$D\left( \omega \right) = 1 - \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} - \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 0 \tag{1}$$ where $\mathbf{V}_{o} = \mathbf{V}_{e} - \mathbf{V}_{i}$. We can further simplify things by rewriting $D\left( \omega \right) = 1 - F\left( \omega \right)$, where this new term is given by: $$F\left( \omega \right) = \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} + \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 1 \tag{2}$$
We can see that $F\left( \omega \right)$ has two poles at $\omega = 0$ and $\omega = \mathbf{k} \cdot \mathbf{V}_{o}$ and a minimum at $\partial F/\partial \omega = 0$, given by: \begin{align} \frac{ \partial F }{ \partial \omega } & = - \frac{ 2 }{ \omega } \left( \frac{ \omega_{pi} }{ \omega } \right)^{2} - - \frac{ 2 }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \left( \frac{ \omega_{pe} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right) } \right)^{2} = 0 \tag{3a} \\ & = \frac{ \omega_{pi}^{2} }{ \omega^{3} } + \frac{ \omega_{pe}^{2} }{ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right)^{3} } = 0 \tag{3b} \\ \omega_{pe}^{2} \ \omega^{3} & = - \omega_{pi}^{2} \ \left( \omega - \mathbf{k} \cdot \mathbf{V}_{o} \right)^{3} \tag{3c} \end{align} After we make a few substitutions (i.e., $\zeta = \tfrac{\omega}{k \ V_{o}}$ and $\alpha = \tfrac{\omega_{pi}^{2}}{\omega_{pe}^{2}}$) and assume everything is one-dimensional (i.e., $\mathbf{k}$ is parallel to $\mathbf{V}_{o}$), Equation 3c can be reduced to: $$\zeta^{3} + \alpha \ \left( \zeta - 1 \right)^{3} = 0 \tag{4}$$ The three roots of Equation 4 are messy but when we note that if this is a proton-electron plasma, then $\alpha = \tfrac{m_{e}}{m_{p}}$, where $m_{s}$ is the mass of species $s$. Thus, we can expand in a Taylor series for small $\alpha$ to find more simplified results. We can also take advantage of the fact that we are looking for the real part of the frequency, so we can rearrange Equation 4 to get: $$\zeta^{3} = \alpha \left( 1 - \zeta \right)^{3} \tag{5}$$ If we take the cubic root of Equation 5, we can solve for $\zeta$ to find: \begin{align} \zeta & = \frac{ \alpha^{1/3} }{ 1 + \alpha^{1/3} } \tag{6a} \\ & = \frac{ 1 }{ 1 + \alpha^{-1/3} } \tag{6b} \\ \omega_{sol} & = \frac{ k \ V_{o} }{ 1 + \left( \frac{ m_{p} }{ m_{e} } \right)^{1/3} } \tag{6c} \end{align} where we replaced our normalized parameters with the original inputs.
To find the threshold for instability, we use the results in Equation 6c and impose an additional constraint that $F\left( \omega_{sol} \right) > 1$, which gives us: \begin{align} F\left( \omega_{sol} \right) & = \left( \frac{ \omega_{pe} }{ k \ V_{o} } \right)^{2} \left[ 1 + \left( \frac{ m_{e} }{ m_{p} } \right)^{1/3} \right]^{3} > 1 \tag{7a} \\ \left( k \ V_{o} \right)^{2} & < \omega_{pe}^{2} \ \left[ 1 + \left( \frac{ m_{e} }{ m_{p} } \right)^{1/3} \right]^{3} \tag{7b} \end{align}
• I'm not quite sure of this. The $\omega_\text{sol}$ is supposed to be imaginary and positive in order to give a growing instability. But how can one see that from your solution? – Saavestro Jun 30 '17 at 1:43