The original problem comes from plasma physics but my issue is purely mathematical. I encounter the following function
$$\epsilon (k, \omega) = 1 - \frac 1 2 \left[ \frac{\omega_p^2}{(\omega-kv_0)^2} + \frac{\omega_p^2}{(\omega+kv_0)^2}\right]$$.
I am aimed at
- Finding the range of $k$ for which $\omega_p$ is imaginary.
- Determining what $k_{max}$ is.
- Determining what $\text{Im}(\omega_{max})$
This is how I am tackling it
- OK, it should be simply about finding $k$ for which $\epsilon(k, \omega) = 0$ yields an imaginary solution. However I have two issues with it: 1) This is not an easy looking second order equation to solve 2) I should get two solutions (at least one of them being potentially imaginary) and not a "range". How could actually solve it?
- This should be simply about finding for what $k$ values does $\frac{\partial \epsilon}{\partial k} = 0$ hold, do you agree?
- I am confused with this one; Is this simply about finding the function $\epsilon (k, \omega_{max})$ i.e. $\text{Im}(\omega_{max}) = \epsilon (k, \omega_{max})$?
If you need more context please let me know.
EDIT 0
$$\frac{\omega_p^2}{(\omega-kv_0)^2} + \frac{\omega_p^2}{(\omega+kv_0)^2} = 2 \Rightarrow \omega_p^2\left( \frac{\omega^2+2kv_0+k^2v_0^2+\omega^2-2kv_0+k^2v_0^2}{(\omega^2-k^2v_0^2)^2} \right) = 2$$
$$2\omega_p^2\left( \frac{(\omega^2+k^2v_0^2)}{(\omega^2-k^2v_0^2)^2} \right)=2 \Rightarrow \omega_p^2 = \frac{(\omega^2-k^2v_0^2)^2}{\omega^2+k^2v_0^2}$$
Since $(\omega^2-k^2v_0^2)^2$ is always positive, $\omega_p$ will be imaginary if $$\omega^2+k^2v_0^2<0 \Rightarrow k^2v_0^2 < -\omega^2 \Rightarrow k < \pm i\frac{\omega}{v_0}$$
So the answer is $k < \pm i\frac{\omega}{v_0}$
EDIT 1
We can determine what $k_{max}$ is via $\frac{\partial \epsilon}{\partial k} = 0$
$$\frac{\partial \epsilon}{\partial k} = \frac{\omega_p^2k}{(\omega-kv_0)^3}-\frac{\omega_p^2k}{(\omega+kv_0)^3} = 0 \Rightarrow \omega_p^2k(\omega+kv_0)^3 - \omega_p^2k(\omega-kv_0)^3 = 0$$
Solving for $k$ yields $k_{max}$