# Alternating linear chain of masses

1. Nov 7, 2011

### jncarter

1. The problem statement, all variables and given/known data
A chain of atoms are connected by identical springs of force constant k. Suppose teh atoms of mass m alternate with atoms of mass M. Thus the crystal consists of a sequence ... MkmkMkmMkmk ... which is the periodic repetition of unit cells Mkmk. The size of the unit cell is a (this is the equilibrium distance between two neighboring m atoms).
(a) In the nth unit cell let xn and yn be the longitudinal displacement from equilibrium of the atoms m and M respectively. Set up the Lagrangian and derive the equations of motion for the x and y coordinates.
(b)Determine the dispersion relation w(q). HINT: consider travelling wave solutions of the form
$x_{n} = Xe^{i(qna-wt)}$ and $y_{n} = Ye^{i(qna-wt)}$​
(c) Find the dispersion relation in the long wavelength limit (q→0) and determine the speed of sound.

2. Relevant equations
$x^{0}_{n+1}-x^{0}_{n} = a$ where x0 denotes equilibrium position.
$y^{0}_{n+1}-y^{0}_{n} = a$
$x^{0}_{n}-y^{0}_{n} = \frac{a}{2}$
$T = \frac{1}{2} m \Sigma \dot{x}^{2}_{n} + \frac{1}{2} M \Sigma \dot{y}^{2}_{n}$

3. The attempt at a solution
Now, here is where I run into a problem; how to express the potential energy? I started with $U = \frac{1}{2} k \Sigma (y_{n+1} - y_{n})^{2} + (x_{n+1} - x_{n})^{2}$. Then I realized that this makes the problem trivial and not all that interesting. My physics intuition tells me that the x's and y's must be coupled because the potential energy should depend on the neighboring particles. I think that this gives $U = \frac{1}{2} k \Sigma (x_{n}-y_{n})^{2}+(y_{n+1}-x_{n})^{2}$ And the equations of motion would then be
$m\ddot{x}_{n} = -\frac{\partial U}{\partial y_{n+1}} - -\frac{\partial U}{\partial y_{n}}$
$M\ddot{y}_{n} = -\frac{\partial U}{\partial x_{n+1}} - -\frac{\partial U}{\partial x_{n}}$​
or
$m \ddot{x}_{n} = -2kx_{n} +k(y_{n+1}+y_{n})$
$m \ddot{y}_{n} = -2ky_{n} +k(x_{n+1}+x_{n})$​

I just want to know if I have over-thought things, or if I am on the right track with the second potential. Thanks for any suggestions!

2. Nov 8, 2011

### ehild

Your Physics intuition works well, the potential is correct. Take care with the derivatives.

ehild

3. Nov 10, 2011

### jncarter

When I took the derivatives more carefully I ended up with the same x-equation, but the y-equation come out as
$m\ddot{y_{n}} = 2ky_{n} - k(x_{n} + x_{n+1})$
My professor pointed out that this is not a restoring force. The yn term needs to be negative, but he says the potential looks like it's correct. The professor doesn't tend to do the homework out himself until after I go talk to him, and usually just because I've gone to talk to him about a problem. Anyway, now I'm working on the other parts of the problem and plan to sort out the issue of the potential by integrating what I know must be true.

I also want to thank ehild for continuing to answer my questions. I know I have been posting frequently and I do appreciate the help. During my undergraduate career I was far from the top of my class. That's not to say I did poorly, just that my peers were stellar. At my current school; however, I tend to be ahead on the homework and now have few people to discuss the problems with until after I have already finished the homework. Receiving help here has been very useful.

4. Nov 10, 2011

### ehild

That y equation is not correct yet...
I write out the terms containing yn and xn: it is easier to find the derivatives.

$$U=k/2(....(y_{n-1}-x_{n-1})^2+(x_{n}-y_{n-1})^2+(y_{n}-x_{n})^2+(x_{n+1}-y_{n})^2....)$$
$$\frac{\partial U}{\partial x_{n}}=k((x_{n}-y_{n-1})-(y_{n}-x_{n}))=k(2x_{n}-y_{n-1}-y_{n})$$
$$\frac{\partial U}{\partial y_{n}}=k((y_{n}-x_{n})-(x_{n}-x_{n+1}))=k(2y_{n}-x_{n}-x_{n+1})$$
$$m\ddot x_{n}=-2kx_n+k(y_{n-1}+y_{n})$$
$$M\ddot y_{n}=-2ky_n+k(x_{n}+x_{n+1})$$

ehild