under the influence of thermal fluctuations can be deduced from simple thermodynamic arguments. We consider an elastic line of length $L$ and line energy ${\cal T}$ pinned at both ends. The initial line direction is identified with the $x$ direction of a Cartesian coordinate system. The fundamental modes of the line are given by $y_n(x) = A_n \sin(n \pi x/L)$ where $n$ is a positive integer number. The associated energy is $$E_n = \frac{{\cal T}}{2} \int \left(\frac{\partial y_n(x)}{\partial x} \right)^2 dx = n^2 \frac{{\cal T} \pi^2}{4 L} A_n^2.$$ From the equipartition theorem it then follows that $E_n = k_BT/2$, hence $$A_n^2 = \frac{2k_B T}{ {\cal T}L} \left(\frac{L}{\pi n}\right)^2,
\label{eq:intens}$$ i.e., the square mode amplitude (’intensity’) is expected to be proportional to temperature and inversely proportional to the square of the wave number. Conversely, from measurements of the mode amplitude the line energy ${\cal T}$ can be deduced as $${\cal T} =  \frac{2 k_B T}{L} \frac{1}{(\pi n A_n)^{2}}
\label{eq:linetens}$$ For the case of dislocations, this provides an easy check to see whether a line tension approximation is warranted or whether corrections must be taken into account.

Defining the mode wave vector as $q = n \pi/L$ and noting that the phases 