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In this paper online, the author models the deformation of a medium as a diffeomorphism $ \mathbb{R}^3 \rightarrow \mathbb{R}^3$ as given by:

$$ y^i \mapsto x^i(y)=y^i + u^i(x) $$

as given by equation (1). The diffeomorphism induces a transformation of the metric

$$ g_{ij}(x) = \frac{\partial y^k}{\partial x^i} \frac{\partial y^l}{\partial x^j} \delta_{ij}$$

which is just the push forward of $\delta_{ij}$ under the diffeomorphism, as shown in equation (5).

It is stated that after the diffeomorphism, which transforms the metric from $\delta_{ij}$ to $g_{ij}$, the geodesics of the material will become curved because the metric $g_{ij}$ is non-trivial. Therefore, sound waves through the medium will now take curved paths as they are postulated to follow geodesics. However, this seems completely bizarre to me. A diffeomorphism is equivalent to a change of coordinates, so the geodesics of $g_{ij}$ will be the same as the geodesics of $\delta_{ij}$, which are straight lines, not curved. It is just now the geodesic equation will look a bit more complicated because we are working in a general curvilinear coordinate system. In fact, both metrics are flat because the curvature is invariant under diffeomorphisms, so I assume this is another reason to argue that the geodesics will be straight lines too?

My question

How can one describe a deformation of a material, something which physically affects the density of the material and the paths sound waves travel, as a diffeomorphism, something which does not change the manifold structure and can be viewed as a change of coordinates so should be unphysical?

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… as a diffeomorphism, something which does not change the manifold structure and can be viewed as a change of coordinates so should be unphysical …

Diffeomorphism does not have to be the change of coordinates. It can also have nontrivial physical meaning, as it has here. This is because the Euclidean space under consideration here has an elastic material occupying it. And when Euclidean space is mapped onto itself this is also accompanied by the displacement and deformation of that material. So when we write: $$ y^i \mapsto x^i(y)=y^i + u^i(x) $$ the physical interpretation is that the physically small element of the material that initially occupied the neighborhood around position $y$ is now occupying the neighborhood around $x(y)$. The displacement of this material element is given by $u$ and its deformation is described by the tensor of small deformations $\epsilon$. Note, that $y^i$ and $x^i$ are the Cartesian coordinates of the initial and final placements of the elastic substance, and so the straight lines in initial state generally would not be mapped to the straight lines in the final placement.

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  • $\begingroup$ Yes I understand that under diffeomorphism, the initial geodesics will be warped as they are dragged along with the transformation, i.e. if $\phi$ is the diffeo and $\gamma$ is the geodesic, then $ \gamma’ = \phi \circ \gamma$ is not necessarily a geodesic. However, the set of geodesics of the Euclidean metric $\delta_{\mu \nu}$ will surely be the same as the geodesics of the transformed metric $g_{\mu \nu}$ because $g_{\mu \nu}$ is the push forward transformation, equivalent to a change of coordinates, so nothing should have changed. $\endgroup$
    – Hermitian_hermit
    Commented Jul 28, 2020 at 8:30
  • $\begingroup$ ... this is because if $g$ has components $g_{\mu \nu}$ in the coordinate basis $\{ e_\mu \}$, then $\phi_* g$ has the components $g_{\mu \nu}$ in the coordinate basis $\{\phi_* e_\mu\}$, so the metrics are equivalent so one would expect their geodesics to be equivalent. $\endgroup$
    – Hermitian_hermit
    Commented Jul 28, 2020 at 8:34
  • $\begingroup$ nothing should have changed would have been true if there was nothing else inside the manifold. But there is substance there: the elastic medium. So the geodesic of the new metric would be passing through the same atoms as in the initial configuration. That is the equivalence. But this sequence of atoms is no longer arranged in a straight line of the Euclidean space in the final configuration. $\endgroup$
    – A.V.S.
    Commented Jul 28, 2020 at 10:58
  • $\begingroup$ I do not understand why the geodesics of the new metric will be curved, this is the essence of my confusion. $\endgroup$
    – Hermitian_hermit
    Commented Jul 29, 2020 at 10:49
  • $\begingroup$ Once again, they would be curved with respect to the metric of Euclidean space of classical mechanics. They would be straight lines with respect to $g_{ij}(x)$ (which is not the metric on Euclidean space). Example: take the line $y^1=y^2=0$ (geodesic of initial state) and apply the following deformation: $u^1 = a \sin x^3$, $u^2=u^3=0$ (pure shear). Under the deformation the initial line becomes $x^1=a \sin x^3,$ $x^2=0$, which is not a straight line. $\endgroup$
    – A.V.S.
    Commented Jul 29, 2020 at 14:13

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