Alternating linear chain of masses

In summary, the problem involves a chain of atoms connected by identical springs, with atoms of mass m alternating with atoms of mass M. The Lagrangian is set up and the equations of motion for the x and y coordinates are derived. The dispersion relation w(q) is determined and the dispersion relation in the long wavelength limit (q→0) is found. The potential energy is also discussed and the correct equations for the derivatives are provided.
  • #1
jncarter
49
0

Homework Statement


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 traveling wave solutions of the form
[itex]x_{n} = Xe^{i(qna-wt)}[/itex] and [itex] y_{n} = Ye^{i(qna-wt)}[/itex]​
(c) Find the dispersion relation in the long wavelength limit (q→0) and determine the speed of sound.

Homework Equations


[itex]x^{0}_{n+1}-x^{0}_{n} = a [/itex] where x0 denotes equilibrium position.
[itex]y^{0}_{n+1}-y^{0}_{n} = a [/itex]
[itex]x^{0}_{n}-y^{0}_{n} = \frac{a}{2} [/itex]
[itex]T = \frac{1}{2} m \Sigma \dot{x}^{2}_{n} + \frac{1}{2} M \Sigma \dot{y}^{2}_{n}[/itex]

The Attempt at a Solution


Now, here is where I run into a problem; how to express the potential energy? I started with [itex]U = \frac{1}{2} k \Sigma (y_{n+1} - y_{n})^{2} + (x_{n+1} - x_{n})^{2}[/itex]. 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 [itex]U = \frac{1}{2} k \Sigma (x_{n}-y_{n})^{2}+(y_{n+1}-x_{n})^{2}[/itex] And the equations of motion would then be
[itex]m\ddot{x}_{n} = -\frac{\partial U}{\partial y_{n+1}} - -\frac{\partial U}{\partial y_{n}}[/itex]
[itex]M\ddot{y}_{n} = -\frac{\partial U}{\partial x_{n+1}} - -\frac{\partial U}{\partial x_{n}}[/itex]​
or
[itex] m \ddot{x}_{n} = -2kx_{n} +k(y_{n+1}+y_{n}) [/itex]
[itex] m \ddot{y}_{n} = -2ky_{n} +k(x_{n+1}+x_{n}) [/itex]​

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!
 
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  • #2
jncarter said:
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 [itex]U = \frac{1}{2} k \Sigma (x_{n}-y_{n})^{2}+(y_{n+1}-x_{n})^{2}[/itex] And the equations of motion would then be
[itex]m\ddot{x}_{n} = -\frac{\partial U}{\partial y_{n+1}} - -\frac{\partial U}{\partial y_{n}}[/itex]
[itex]M\ddot{y}_{n} = -\frac{\partial U}{\partial x_{n+1}} - -\frac{\partial U}{\partial x_{n}}[/itex]​
or
[itex] m \ddot{x}_{n} = -2kx_{n} +k(y_{n+1}+y_{n}) [/itex]
[itex] m \ddot{y}_{n} = -2ky_{n} +k(x_{n+1}+x_{n}) [/itex]​
Your Physics intuition works well, the potential is correct. Take care with the derivatives.




ehild
 
  • #3
When I took the derivatives more carefully I ended up with the same x-equation, but the y-equation come out as
[itex] m\ddot{y_{n}} = 2ky_{n} - k(x_{n} + x_{n+1}) [/itex]
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
That y equation is not correct yet...
I write out the terms containing yn and xn: it is easier to find the derivatives.

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

ehild
 
  • #5


Your approach to the problem seems correct. The potential energy should indeed depend on the neighboring particles, as each atom is affected by the forces from its neighboring atoms. Therefore, your expression for the potential energy in terms of x and y coordinates is correct.

One thing to note is that in your equations of motion, you have written the partial derivatives of the potential energy with respect to the positions of the neighboring atoms. This is correct, but it may be easier to express the potential energy in terms of the displacements from equilibrium (xn and yn) as you have done in the first part of the problem. This will make the equations of motion simpler to solve.

Overall, your approach seems correct and you are on the right track. Keep going and see if you can solve for the dispersion relation and speed of sound in the long wavelength limit. Good luck!
 

FAQ: Alternating linear chain of masses

What is an alternating linear chain of masses?

An alternating linear chain of masses is a physical system that consists of a series of masses connected by springs, with alternating masses being equal and the springs between them having equal stiffness. This results in a repeating pattern of mass-spring-mass-spring, and so on.

What are the properties of an alternating linear chain of masses?

The properties of an alternating linear chain of masses include the mass of each individual mass, the stiffness of each spring, and the length of each spring. These properties determine the dynamics and behavior of the system.

How does an alternating linear chain of masses behave?

An alternating linear chain of masses behaves in a complex manner, with each mass and spring influencing the movement of the others. The system can exhibit a variety of behaviors, including oscillations, waves, and energy transfer.

What are some applications of an alternating linear chain of masses?

Alternating linear chains of masses have numerous applications in various fields of science and engineering. They are used to model and study the behavior of physical systems, such as crystals, polymers, and molecules. They also have practical applications in designing structures and materials with specific mechanical properties.

What are the limitations of an alternating linear chain of masses model?

While the alternating linear chain of masses model is useful for studying the behavior of physical systems, it has some limitations. For instance, it assumes that the masses are point particles and the springs are ideal, which may not accurately represent real-world systems. Additionally, it does not take into account external forces or environmental factors that may affect the system.

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