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?


1 Answer 1


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.

  • $\begingroup$ I have already considered this route and came to the same conclusion, but squaring this adds more $dx^2$ and $dy^2$ terms... $\endgroup$ Mar 5, 2015 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... $\endgroup$ Mar 5, 2015 at 8:26

Your Answer

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

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