# Dispersion relation with damping force

We consider a linear chain of atom connected by springs with constant $$K$$. We have the usual elastic force and we add damping force such that the dispersion relation is:

$$\omega = 2\sqrt \frac K m \sin \left(\frac {qa} 2\right) - \frac{i\Gamma}{2m}$$

I don't know if the expression correct, but for $$\Gamma=0$$ we fall back to the classical expression, so I was satisfied :). But now the problem is that I want to understand what happens if $$q=0$$ or $$q=\frac{\pi}{a}$$ and the above expression results in a complex number, so I don't see what the physical meaning is.

Thanks!

The general solution to displacement for this differential system is given by $$y(t)=y_0e^{\pm i(\omega t+\phi)}$$
And plugging in your form of omega then tells us that the complex part is just the decay related to the damping force present in the system if we take the relation to be negative in the exponent. Otherwise forces $$\Gamma>0$$ if the exponent is positive.
• So there is no difference between q=$0$ and q=$\frac{\pi}{a}$ since there are both complex and with the same "relaxation time"? May 12, 2020 at 19:05
• I reduced the problem to the nearest neighbors and I found $m\ddot{u}_n = Ku_{n-1} + Ku_{n+1} -2Ku_n -\Gamma\dot{u}_n$ where $u_n$ is the displacement for the nth atom in my chain. With the anstaz $\mathbf{u}_{n}=\frac{1}{\sqrt{m}} \mathbf{u}(q)e^{i( \mathbf{q}. \mathbf{r}_n -\omega t)}$. I think my derivation is correct, but maybe I have to consider $-\Gamma\dot{u}_{n-1}$ and $-\Gamma\dot{u}_{n-1}$. I did it and I just find a cosine in the imaginary part. May 12, 2020 at 19:34