1
$\begingroup$

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.

Improve this question
$\endgroup$

2 Answers 2

Reset to default
0
$\begingroup$

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.

Improve this answer
$\endgroup$
4
  • $\begingroup$ So there is no difference between q=$0$ and q=$\frac{\pi}{a}$ since there are both complex and with the same "relaxation time"? $\endgroup$
    – Amateur Physicist
    Commented May 12, 2020 at 19:05
  • $\begingroup$ By the way, do you know if this expression is correct because I don't find any source to compare it on the internet? $\endgroup$
    – Amateur Physicist
    Commented May 12, 2020 at 19:08
  • $\begingroup$ What’s the differential equation you’re considering? $\endgroup$
    – Superfast Jellyfish
    Commented May 12, 2020 at 19:19
  • $\begingroup$ 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. $\endgroup$
    – Amateur Physicist
    Commented May 12, 2020 at 19:34
0
$\begingroup$

Couple of comments:

  1. $\Gamma$ cannot be a force, since $[\omega] = T^{-1}$. It is rather a mass-scaled decay rate.

  2. $\Gamma$ is included in the dispersion relation independently of $q$. Therefore, every mode of the system "relaxes" with the same rate. I don't see how this is consistent with the equation of motion $$m\ddot{u}_n = K(u_{n+1}-u_n) + K(u_{n-1}-u_n) - \Gamma \dot{u}_n$$ OP has proposed in the comments.

  3. The proposed Newtons equation of motion would lead rather to the dispersion relation $$\omega = \frac{i\Gamma}{2m} \pm \sqrt{\left(\frac{i\Gamma}{2m}\right)^2 +\frac{K}{m}\sin^2(ka)}.$$

  4. The physical content of the dispersion relation, which is a time-independent quantity telling you how much energy a certain mode has, loses its original meaning in the context of Hamiltonian mechanics as soon as we start talking about non-conservative systems. What we can talk about instead is about the transient dynamics, which as Superfast Jellyfish pointed out has a trivial exponential decay on top of oscillation.

  5. While the spring constants have a microscopic origin, what is the microscopic origin of the damping? A proper description would require you to include a bath (keywords: Brownian motion, Langevin dynamics and Caldeira-Legett model for the quantum version of Brownian motion).

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.