In the first chapter of Altland and Simons (2nd Edition), pg. 34, there is the following exercise:

Consider the one-dimensional elastic chain discussed in Section 1.1. Convince yourself that the continuity equation of the system assumes the form $\partial_t \int_{x_1}^{x_2}dx \rho(x,t) = \partial_t(\phi(x_2,t)-\phi(x_1,t))$, where $\rho(x,t)$ is the local density of the medium. Use this result to show that $\rho = \partial_x\phi$. Show that the momentum $\int dx (\mathrm{particle~density})\times(\mathrm{velocity}) = \int dx\rho\partial_t\phi$ carried by the system coincides with the Noether momentum.

Rather than clutter this question with several additional lines, the 1D harmonic chain of section 1.1 is defined as in this question: Taylor expansion in classical 1D harmonic chain (classical field theory)

Now, my question is, why does $\rho = \partial_x\phi$ and not $\rho = -\partial_x\phi$?

Equivalently, why, in the first equation of the exercise, is the right hand side $\partial_t(\phi(x_2,t)-\phi(x_1,t))$ and not $\partial_t(\phi(x_1,t)-\phi(x_2,t))$?

Given that $\phi(x,t)$ is a displacement field, I would think that the particle density in the region $[x_1,x_2]$ should go up as the particles in the chain are displaced toward the region rather than away from it, i.e. density would rise for $\partial_t \phi(x_1,t) > 0$ (the opposite of what is stated in the exercise).

I have looked online, but found no one else who has talked about the particle density of the 1D harmonic chain, despite the number of books/lecture notes on the 1D harmonic chain.

Also, I have the rest of the question figured out; I just don't see how Altland and Simons got the sign that they did. Every way I think about it, I get the opposite sign from theirs.



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