# Displacements-strain to stress-momentum formulation in Linear Elasticity

I was going through this paper (page 2) where they describe a duality between a Fracton theory and linear elasticity in 2 dimensions. Here, he starts with the action for elasticity, given by $$S=\int d^{2} x d t \frac{1}{2}\left[\left(\partial_{t} u^{i}\right)^{2}-C^{i j k \ell} u_{i j} u_{k \ell}\right]$$

He then introduces Hubbard Stratonovich fields $$\pi$$ and $$\sigma$$ which are like the momentum and stress fields respectively. \begin{aligned} S=\int d^{2} x d t [\frac{1}{2} C_{i j k t}^{-1} \sigma^{i j} \sigma^{k \ell}-\frac{1}{2} \pi^{i} \pi_{i}-\sigma^{i j}\left(\partial_{i} \tilde{u}_{j}+u_{i j}^{(s)}\right)+\pi^{i} \partial_{t}(\tilde{u}_{i}+u_{i}^{(s)})] \end{aligned}

and claims that we can "integrate out" the diplacement field $$u$$ to get the constraint given below, which the momentum transport equation. $$\partial_{t} \pi^{i}-\partial_{j} \sigma^{i j}=0$$

I understand that this is a classical system and I think the exponential of the action is being integrated (over imaginary time(?)) over $$u$$ in the first equation to get the partition function of the system. But I do not understand how they are integrating out $$u$$ to get the constraint. I'm guessing we should get a delta functional like $$\delta(\partial_{t} \pi^{i}-\partial_{j} \sigma^{i j})$$ out when we integrate out $$u$$ but I don't see how that comes.

Here $$u_{ij}$$ are the derivatives of the displacement field. Assuming you are in a Minkowski space-time:

$$u_{ij} := u_{i;j} = \partial_j u_i$$

the Lagrangian density $$\mathcal{L}$$ is:

$$\mathcal{L} := \frac{1}{2}\left[\rho\left(\partial_{t} u^{i}\right)^{2}-C^{i j k \ell} u_{i j} u_{k \ell}\right], \qquad S = \int_{\mathbb{R}^3\times\mathbb{R}} [\mathcal{L}\ \text{d}V]\text{d}t$$

So the Euler-Lagrange equations for the Lagrangian density $$\mathcal{L}$$ are:

$$\frac{\partial}{\partial t}\left(\frac{\partial \mathcal{L}}{\partial (\partial_t u_i)}\right) + \frac{\partial}{\partial x^j}\left(\frac{\partial \mathcal{L}}{\partial u_{ij}}\right) = \frac{\partial \pi_i}{\partial t} -\frac{\partial}{\partial x_j}\left(C^{ijk\ell}u_{k\ell}\right)$$

but by definition:

$$\pi_i := \rho\partial_t u_i, \qquad \sigma^{ij} :=C^{ijk\ell}u_{k\ell}$$