Recasting integrals from Lagrangian to Eulerian frame Working on a research problem in the continuum mechanics of fluids. For clarity, uppercase will be used for tensors in the reference configuration, and corresponding spatial items will be in lowercase.
For a moving body, define a map $\Phi:(X,t)\to{(x,t)}\;$ from a reference configuration to spatial configuration, whose deformation gradient is $F=\nabla\Phi$ with the determinant $J=\textrm{Det }F$.
What is the correct approach to convert a surface integral in reference configuration to one in spatial configuration? I have come across conflicting information in multiple sources: e.g. for a reference surface $A$ with area element $dS$,
(a) Using a change of variables (integration by substitution)
i.e. $\int_A T\;dS = \int_a J t\;ds$ where $a=\Phi(A)$, $ds=\Phi(dS)$ and $t=\Phi(T)$ for an arbitrary tensor $T$ in the reference configuration.
(b) Applying Piola transforms.
Certainly these approaches give very different answers unless I am missing something?
The actual integrals I am trying to convert to spatial integrals have the general forms:
$$I_1 = \int_A (N\cdot{J^k}t^k(F^{-\top})^kN)\;(U^{k+1}{\cdot}N)\;dS$$
$$I_2 = \int_A (U^{k+1}{\cdot}N)(V^{k+1}{\cdot}N)\;dS$$
$$I_3 = \int_A (U^{k+1}{\cdot}V^{k+1})\;dS$$
where $N$ is the unit normal to surface area element $dS$, $t$ is a symmetric second order tensor (in the spatial frame), and $U$ and $V$ are vectors in the reference frame. The superscripts $k$ and $k+1$ refer to two separate times.
Please any experienced insights would be very welcome.
 A: BOTTOM LINE UP FRONT: Accounting can make a huge difference in the speed of understanding this material. For example, always note whether a function is a function on the reference configuration or on the deformed configuration. If you work with indices, raised and lowered indices and Einstein's summation convention are very helpful.



My references are Chapter 2 of Non-Linear Elastic Deformations by R. W. Ogden and the Continuum mechanics/Volume change and area change section of Wikiversity.



I will re-write some of the notation to make some expressions simpler. For example, I will use the sumbol $\Phi_t$ for the map that transforms a body (or volume of fluid) at time 0 to its configuration at time $t$. Then for each $t$, $\Phi_t$ is a (smooth?) map from one 3-dimensional manifold to another, and for fixed $t$ we write $\mathbf{x} = \Phi_t(\mathbf{X})$.

Let the reference configuration of the body/volume of fluid be $\mathcal{B}$, a subset of $\mathbb{R}^3$. This body consists of material points. At time $t$, those same material points occupy $\Phi_t(\mathcal{B})$ (also a subset of $\mathbb{R}^3$). $\Phi_t$ is a map from $\mathcal{B}$ onto $\Phi_t(\mathcal{B})$, but if $\mathbf{T}$ is a tensor field on $\mathcal{B}$, it is incorrect to claim that the corresponding field $\mathbf{t}$ on $\Phi_t(\mathcal{B})$ is $\Phi_t(\mathbf{T})$. It's not quite correct to claim that $\mathbf{t}(\mathbf{x}) = \mathbf{T}(\Phi_t^{-1}(\mathbf{x}))$, as one is constructed from vectors attached to the point $\mathbf{x}\in\Phi_t(\mathcal{B})$ while the other is constructed from vectors attached to the point $\Phi^{-1}(\mathbf{x})\in\mathcal{B}$.
There are different transformation rules for different kinds of tensor fields as well as for differential forms such as the volume form $dV$, area form $d\mathbf{S}$, and line form $d\mathbf{\ell}$.



Let $F$ be a smooth-enough real-valued function on $\Phi_t(\mathcal{B})$. How to express the integral of $F$ over material points $\mathbf{x}\in\Phi_t(\mathcal{B})$ as an integral over material points $\mathbf{X}\in\mathcal{B}$ (the reference configuration)?
\begin{equation}
\int_{\Phi_t(\mathcal{B})}F(\mathbf{x})dv(\mathbf{x})
=
\int_{\mathcal{B}}F(\Phi_t(\mathbf{X}))J(\mathbf{X})dV(\mathbf{X}),
\end{equation}
where $J$ is the absolute value of the determinant of the deformation gradient $\mathbf{F}$:
\begin{equation}
J ~=~ \left|\det\mathbf{F}\right| ~=~ \left|\det\left(\frac{\partial\mathbf{x}}{\partial\mathbf{X}}\right)\right|.
\end{equation}
Note that we must integrate $F\circ\Phi_t$ (not $F$) over $\mathcal{B}$ because $F$ is not a function of the points $\mathbf{X}\in\mathcal{B}$. $(F\circ\Phi_t)(\mathbf{X})$ is equal to the value of $F$ at the "deformed" material point $\mathbf{x} = \Phi_t(\mathbf{X})$.
This shows the transformation of the volume form: $dv \longleftrightarrow JdV$. I don't write that they are equal because they are forms on different spaces. Many authors are not bothered by this technicality, but I think it is worth noting.
I am torn over whether to claim that $J$ is a function on $\mathcal{B}$ or on $\Phi_t(\mathcal{B})$, since $\Phi_t$ is a diffeomorphism between these two sets, allowing us to change our minds if we need to do so.
Note, however, I do like to write $dV(\mathbf{X})$ (as opposed to just $dV$) and $dv(\mathbf{x})$ (as opposed to just $dv$) for accounting. Another accounting principle will help remember the side of equation that has $J$. Consider dimensions in $\mathcal{B}$ and in $\Phi_t(\mathcal{B})$ as distinct. Then the dimensions of $J$ are
\begin{equation}
\left[J\right]
~=~ \left[\det\left(\frac{\partial\mathbf{x}}{\partial\mathbf{X}}\right)\right]
~=~ \frac{\textrm{length$^3$ in $\Phi_t(\mathcal{B})$}}{\textrm{length$^3$ in $\mathcal{B}$}}.
\end{equation}
Hence, $J$ has the same dimensions as the non-rigorous expression $dv/dV$, and we conclude $dv \longleftrightarrow JdV$.



The analogous transformation for the area form is called Nanson's formula:
\begin{equation}
\mathbf{n}da \longleftrightarrow (\mathbf{F}^{-\mathsf{T}}\cdot\mathbf{N})dA,
\end{equation}
where


*

*$da$ is the differential area in $\Phi_t(\mathcal{B})$,

*$\mathbf{n}$ is the unit normal vector attached to $da$,

*$dA$ is the differential area in $\mathcal{B}$,

*$\mathbf{N}$ is the unit normal vector attached to $dA$,

*$\mathbf{F}^{-\mathsf{T}}$ is the inverse of the transpose of $\mathbf{F}$.





It might be a good time to note that $\mathbf{F}$ is a two-point tensor field, and that  $\mathbf{F}(\mathbf{X})$ maps from the tangent space at $\mathbf{X}\in\mathcal{B}$ to the tangent space at $\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B})$. In Cartesian co-ordinates,
\begin{equation}
\mathbf{F} = \frac{\partial x^i}{\partial X^j}\mathbf{e}_i\otimes\mathbf{E}^j.
\end{equation}
$\mathbf{F}^{-1}$ maps in the opposite direction, but $\mathbf{F}^{-\mathsf{T}} = (\mathbf{F}^{-1})^{\mathsf{T}}$ maps in the opposite direction from that$^{\dagger}$, so it maps from the tangent space at $\mathbf{X}\in\mathcal{B}$ to the tangent space at $\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B})$.


*

*$\mathbf{F}(\mathbf{X}):(\textrm{tangent space at}~\mathbf{X}\in\mathcal{B})\to(\textrm{tangent space at}~\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B}))$

*$\mathbf{F}^{-1}(\mathbf{x}):(\textrm{tangent space at}~\mathbf{x}\in\Phi_t(\mathcal{B}))\to(\textrm{tangent space at}~\Phi_t^{-1}(\mathbf{x})\in\mathcal{B})$

*$\mathbf{F}^{-\mathsf{T}}(\mathbf{X}):(\textrm{tangent space at}~\mathbf{X}\in\mathcal{B})\to(\textrm{tangent space at}~\Phi_t(\mathbf{X})\in\Phi_t(\mathcal{B}))$
For the sake of accounting, note that the units of a particular entry of $\mathbf{F}$ are
\begin{equation}
\left[F^i_j\right] ~=~ \left[\frac{\partial x^i}{\partial X^j}\right] ~=~ \frac{\textrm{length in $\Phi_t(\mathcal{B})$}}{\textrm{length in $\mathcal{B}$}}.
\end{equation}


$\dagger$ One might have guessed this by observing that the formula includes $\mathbf{F}^{-\mathsf{T}}$ acting a vector attached to a point of $\mathcal{B}$.
A: I suspect that you hope to find a formula involving $\mathbf{F}$, but the deformation gradient is used primarily to re-write integrals over deformed configurations as integrals over the reference configuration. Expressing everything in terms of the reference configuration is what makes the Lagrangian approach to solid mechanics so convenient.
Again, let $\mathcal{B}$ be the portion of $\mathbb{R}^3$ occupied by the reference configuration, so that $\Phi_t(\mathcal{B})$ is the set that those (deformed) material points occupy at time $t$.
We use the symbol $\partial\mathcal{B}$ for the boundary of $\mathcal{B}$ and $\partial\Phi_t(\mathcal{B})$ for the boundary of $\Phi_t(\mathcal{B})$.
I think that your $I_3$ is
\begin{equation}
\int_{\partial\mathcal{B}}\mathbf{U}(\Phi_{t_{k+1}}(\mathbf{X}))\cdot\mathbf{V}(\Phi_{t_{k+1}}(\mathbf{X}))dS(\mathbf{X}).
\end{equation}
It would be more helpful to know what the vector fields $\mathbf{U}$ and $\mathbf{V}$ are, as we usually integrate fields whose immediate argument is $\mathbf{X}$ (as opposed to $\mathbf{x} = \Phi_t(\mathbf{X})$) and use transformations such as Nanson's formula to allow us to integrate functions of $\mathbf{X}$ over $\mathcal{B}$.




The most general I can get without venturing into differential forms is to note that we assume there are parametrizations of the material points in the reference configuration. Since the surface $\partial\mathcal{B}$ is a 2-dimensional surface, there should be (at least locally) some real variables $s_1$ and $s_2$ (for example, the angles $\theta$ and $\phi$ on a spherical surface) such that a point on that surface can be described as $\mathbf{X}(s_1,s_2)$. Let $F$ be a smooth-enough function on $\partial\mathcal{B}$. Then in a small area around $\mathbf{X}(s_1,s_2)$ the differential area is
\begin{equation}
dA(\mathbf{X}(s_1,s_2)) = \left\Arrowvert\frac{\partial\mathbf{X}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{X}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2,
\end{equation}
where we take the norm of the cross-product of the partial derivates of $\mathbf{X}(s_1,s_2)$. The integral of a function $F$ over a (perhaps small) patch would be the iterated integral
\begin{equation}
\int_{a}^{b}\int_{c}^{d}F(\mathbf{X}(s_1,s_2))\left\Arrowvert\frac{\partial\mathbf{X}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{X}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2.
\end{equation}

But the co-ordinates $s_1$ and $s_2$ also parametrize the material points on the surface of the deformed body:
\begin{equation}
\mathbf{x}(s_1,s_2) = \Phi_t(\mathbf{X}(s_1,s_2)).
\end{equation}
Integrating a function $f$ on part of the surface of the deformed body has the same form:
\begin{equation}
\int_{a}^{b}\int_{c}^{d}f(\mathbf{x}(s_1,s_2))\underbrace{\left\Arrowvert\frac{\partial\mathbf{x}}{\partial s_1}(s_1,s_2)\times\frac{\partial\mathbf{x}}{\partial s_2}(s_1,s_2)\right\Arrowvert ds_1 ds_2}_{da(\mathbf{x}(s_1,s_2))}.
\end{equation}
I could re-write this using facts such as
\begin{equation}
\frac{\partial\mathbf{x}}{\partial s_i}(s_1,s_2) = \mathbf{F}(\mathbf{X}(s_1,s_2))\cdot\frac{\partial\mathbf{X}}{\partial s_i}(s_1,s_2),
\end{equation}
but that would change this to an integral over the reference configuration, which is not what you seek.




Do you have a parametrization of the reference configuration and some specific form for the kinds of deformations under consideration? Those details could make this discussion much more concrete.
