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For my field theory class I am trying to build the Lagrangian for the following system. Consider a 2D square lattice where the nearest and next-nearest neighbor interactions are modeled by springs with spring constant $C$. Let $u_1(\mathbf{x},t)$ and $u_2(\mathbf{x},t)$ be the displacement fields in the $x_1$ and $x_2$ direction respectively.

I need to show that the lagrangian density for this system is given by

$$\mathscr{L}=\frac{1}{2}\left\{\sum_i\rho\left(\frac{\partial u_i}{\partial t}\right)^2-\lambda\left(\sum_i\frac{\partial u_i}{\partial x_i}\right)^2\right\}$$

The first term is easy, and the terms containing $(\frac{\partial u_i}{\partial x_i})^2$ I can also get by evaluating the added energy by the horizontal and vertical springs and approximating their change of length $\Delta l_i=a\frac{\partial u_i}{\partial x_i}$ and taking the continuum limit which gets rid of the $a$. But I am unable to show that the change in length of the diagonal spring is proportional to $2a^2\frac{\partial u_1}{\partial x_1}\frac{\partial u_2}{\partial x_2}$.

My best attempt so far is using the fact that an expansion of the potential $V(x_1,x_2)$ for small changes $\delta x_i$ gives (with reference potential being 0)

$$V\approx \frac{1}{2}\frac{\partial^2 V}{\partial x_1^2}\delta x_1^2+\frac{1}{2}\frac{\partial^2 V}{\partial x_2^2}\delta x_2^2+\frac{\partial^2 V}{\partial x_1 x_2}\delta x_1 \delta x_2$$

and setting all the derivatives of the potential equal to $C$ if I want this potential to represent the combined energy of one vertical, horizontal and diagonal spring, but I am not satisfied with this solution, since I just assume that all the double derivatives equal $C$. I feel that my calculation is rather sloppy and cutting corners, even though it might lead to the correct solution. Is there a more illustrative way to build this lagrangian?

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Doesn't the result follow from a simple diagram? If you change the X,Y position by a small amount dx, dy, the change in length of the hypotenuse is simply

$$\begin{align} &= \sqrt{(a+dx)^2 + (a+dy)^2} - \sqrt{a^2 + a^2}\\ &= \sqrt{2a^2 + 2a dx + 2a dy + ...} - \sqrt{2a^2}\\ &= a\sqrt{2}\left(1 + \frac{dx}{a} + \frac{dy}{a} - 1\right)\\ &= \sqrt{2}\left(dx + dy\right) \end{align}$$

The resulting change in energy comes from the square of this, which is where you get the terms you had in your derivation.

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  • $\begingroup$ I have already considered this route and came to the same conclusion, but squaring this adds more $dx^2$ and $dy^2$ terms... $\endgroup$ – Gehaktmolen Mar 5 '15 at 8:19
  • $\begingroup$ And shouldn't the first order terms in the expansion be $dx_i/2a$? I just woke up, so may be wrong... $\endgroup$ – Gehaktmolen Mar 5 '15 at 8:26

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