# How are body deformations modeled in Lagrangian mechanics?

With rigid-body systems, we choose a finite number of generalized coordinates to model a system, i.e. a pendulum. However, I've read that deformable bodies like elastomers have "infinite" degrees of freedom. How does one write the Lagrange's equation for such systems? Is there an integral of infinitesimal volume?

• You would need a field theory description, (classical) field theory is just (classical) mechanics for objects with infinite (continuous) degrees of freedom. Commented May 11, 2020 at 15:27

As far as I know, in continuum mechanics a body is discribed by a fixed orientable Riemannian manifold $$B$$ together with a (time-dependent) embedding in space, e.g. the $$\mathbb{R}^3$$: $$X_t:B\rightarrow\mathbb{R}^3.$$ You could then define a free Lagrangian as a integral over the model body $$B$$: $$L(X)=\int_B \frac{1}{2} \rho \langle\partial_t X, \partial_t X\rangle dV$$ with $$\rho:B\rightarrow\mathbb{R}$$ the density and $$\langle\cdot,\cdot\rangle$$ the scalar product of $$\mathbb{R}^3$$. This one could be seen as some kind of continuum limit of a system with $$N$$ free particles. Now you could add a potential function $$U$$ depending on the field $$X$$. For example if you want to model a (isotropic) rubber band a common choice would be $$U=-\frac{c}{2}g^{ij}\langle\partial_iX,\partial_jX\rangle$$ where $$g^{ij}$$ are the coefficients of the (co)metric on $$B$$ and $$c$$ is some quantity measuring how easy the body can be deformed. The Lagrangian than takes the form $$L(X)=\int_B\frac{1}{2}\left( \rho\langle\partial_t X,\partial_t X\rangle - g^{ij} \langle\partial_i X,\partial_j X\rangle \right).$$ Put in words, the potential $$U$$ is the deformation energy, see for example here. It measures how much the embedding $$X$$ deforms the geometry of $$B$$, thus the stress and strain of the body. The lowest energy configurations of $$U$$ are so-called harmonic maps, for example for a 1D rubber band these are giving a homogeneously strained straight line.