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Piola’s approach to the equilibrium problem for bodies with second gradient energies. Part I: First gradient theory and differential geometry

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Abstract

In this study, some pioneering contributions, envisaged in the works of Gabrio Piola, were developed through tools of the modern differential geometry and applied to the second gradient continua. Part I introduced the variational approach for the equilibrium problem according to the first gradient theory and exploited differential geometric perspectives for the present scenario. By prescribing the stationarity of the Lagrangian energy functional, the virtual work equations for a Cauchy’s medium were recovered. The focus was on the deformation process regarded as a diffeomorphism between Riemannian embedded submanifolds, emphasizing the roles of the pullback metrics and of the covariant differentiation. Novel transport formulae were provided for normal and tangent vectors in the neighborhood of a boundary edge. The divergence theorem for curved surfaces with border was revisited, providing remarkable relationships between Lagrangian and Eulerian expressions involving projectors.

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Notes

  1. In materials science, the sharp surface between two homogeneous phases (or interface) is usually distinguished from an intermediate phase (or interphase), possibly existing within a system as the transition zone between two phases.

  2. Piola was born in Milan on 15 July 1794; he died in Giussano on 9 November 1850.

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Correspondence to Roberto Fedele.

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Communicated by Andreas Öchsner.

Dedicated to the 90th birthday of Professor Giulio Maier.

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Properties of surface projectors and their transport formulae

Properties of surface projectors and their transport formulae

In this Appendix, basic properties of the surface projectors and novel formulae for their transport from the Eulerian to the Lagrangian form, were detailed, endowed with short proofs and remarks.

Let us consider the Lagrangian configuration. As well known, at each point \({\mathbf {P}}\) of a curved surface \(\varSigma _{\star }\), one can specify a tangent space \({\mathcal {T}}_{P}\varSigma _{\star }\), including all the vectors tangent at that point to curves drawn over that face, and a normal space \({\mathcal {N}}_{P}\varSigma _{\star }\), i.e. its orthogonal complement with respect to the ambient space. Hence, the ambient space can be expressed as the direct sum of the above spaces, namely \({\mathcal {T}}_{P}\varSigma _{\star }\oplus {\mathcal {N}}_{P}\varSigma _{\star }={\mathcal {R}}^{3}\).

Accordingly, at each point of the same curved surface two linear operators can be defined, apt to project any vector of the ambient space onto the tangential and normal spaces, referred to as the (Lagrangian) tangential and orthogonal projectors and denoted by symbols \([M_{\parallel }]^{A\,}_{\, B}\) and \([M_{\perp }]^{A\,}_{\, B}\), respectively. The same considerations can be repeated for the Eulerian projectors, for which lowercase symbols were adopted.

The projectors possess the following noteworthy properties (in both index and matrix notation):

$$\begin{aligned}&[M_{\parallel }]^{A\,}_{\, B}+[M_{\perp }]^{A\,}_{\, B}=\delta ^{A\,}_{\, B} \,; \quad {\mathbf {M}}_{\parallel }+{\mathbf {M}}_{\perp }=\mathbbm {1}\,;\nonumber \\&[M_{\perp }]^{C\,}_{\, A}=N^{C} \, N_{A}\,;\quad {\mathbf {M}}_{\perp }={\mathbf {N}} \otimes {\mathbf {N}}\,;\nonumber \\&[M_{\parallel }]^{C\,}_{\, A}=\delta ^{C}_{A} -N^{C} \, N_{A}\,;\quad {\mathbf {M}}_{\parallel }= \mathbbm {1}-{\mathbf {N}}\otimes {\mathbf {N}}\,;\nonumber \\&[M_{\parallel }]^{A\,}_{\, B}[M_{\parallel }]^{B\,}_{\, C}=[M_{\parallel }]^{A\,}_{\, C}\,;\quad {\mathbf {M}}_{\parallel }^{2}={\mathbf {M}}_{\parallel }\,; \nonumber \\&[M_{\perp }]^{A\,}_{\, B}[M_{\perp }]^{B\,}_{\, C}=[M_{\perp }]^{A\,}_{\, C}\,;\quad {\mathbf {M}}_{\perp }^{2}={\mathbf {M}}_{\perp }\,; \nonumber \\ \end{aligned}$$
(82)

where the Kronecker symbol \(\delta ^{A}_{\, B}=\mathbbm {1}=g^{A}_{B}\) represents the identity operator, coincident with the mixed form of the metric tensor.

In the Eulerian configuration, the orthogonal projector can be expressed as a function of its Lagrangian counterpart by using the transformation formula for the normal vector, as follows

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}&=n^{r}n_{s}=g^{rt}\frac{\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}N_{Q}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert } \frac{\left( {\mathbf {F}}^{-1}\right) _{s}^{V}N_{V}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= \frac{g^{rt}\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}g_{QS}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}} \left[ N^{W}N_{V}\right] \left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=\frac{g^{rt}\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}g_{QS}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}} \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \end{aligned}$$
(83)

The above expression can be further simplified through multiplying the rhs by \(\delta ^{r}_{y}\) and then by expressing the Kronecker symbol as the product of the deformation gradient with its inverse, namely

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}&=\delta ^{r}_{y}\frac{g^{yt}\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}g_{QS}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}} \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=F^{r}_{E}\left( {\mathbf {F}}^{-1}\right) _{y}^{E}\frac{g^{yt}\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}g_{QS}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}} \left[ M_{\perp }\right] ^{Q}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=\frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,g^{yt}\left( {\mathbf {F}}^{-1}\right) _{t}^{Q}g_{QS}\left( {\mathbf {F}}^{-1}\right) _{y}^{E} F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&= \frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,g^{\star \,EQ}g_{QS} F^{r}_{E}\left[ M_{\perp }\right] ^{Q}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&= \frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \end{aligned}$$
(84)

where symbol \(g^{\star \,EQ}\) denotes the pullback metric tensor, with \(g^{\star \,EQ}N_{E}N_{Q}=\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}\). It is worth noting that in Eq. (84) the symbol \(g^{\star \,E}_{S}\) constitutes a slight abuse of notation, as an abbreviated form of \(g^{\star \,EQ}\,g_{QS}\), admissible if its meaning appears clear from the context. In fact, the proper metric tensor apt to lower indices of the contravariant pullback metrics is the doubly covariant pullback tensor \(g^{\star }_{EQ}\), as outlined in Sect. 5. Moreover, since the pullback metrics \(g^{\star \,EQ}\) often provides counterintuitive results with respect to the conventional Lagrangian metric tensor, it is convenient to write explicitly all the terms involved.

Analogously, the Eulerian tangential projector can be expressed as follows

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}&=\delta ^{r}_{s}-n^{r}n_{s}= \delta ^{r}_{s}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=F^{r}_{S}\delta ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=F^{r}_{S}[M_{\parallel }]^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} +F^{r}_{S}[M_{\perp }]^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&=F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}+ \left[ F^{r}_{S}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] \left[ M_{\perp }\right] ^{S}_{V}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V}\,; \end{aligned}$$
(85)

In the above equation, the addends within square brackets provide a contribution purely tangential to the face. In fact, if one expresses the Lagrangian orthogonal projector as the tensor product of the normal vectors, namely \([M_{\perp }]^{S}_{V}=N^{S}N_{V}\), and decomposes additively the Eulerian vector \(F^{r}_{S}N^{S}\) into a tangential and a normal component, the normal component cancel out the second addend within the parentheses. By utilizing the transformation rule for the covariant normal vector \(n_{r}=\left( {\mathbf {F}}^{-1}\right) _{r}^{Q}N_{Q}\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{-1} \), from Eq. (85), one has

$$\begin{aligned}&\left[ F^{b}_{S}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] \nonumber \\&\quad =F^{b}_{S}N^{S}N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, g^{tj}\left( {\mathbf {F}}^{-1}\right) _{t}^{E}\left( {\mathbf {F}}^{-1}\right) _{j}^{P}g_{PS} F^{r}_{E}N^{S}N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\nonumber \\&\quad =F^{b}_{S}N^{S}n_{s}\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert -g^{tj}\delta _{t}^{r}n_{j}n_{s}\nonumber \\&[m_{\parallel }]^{r}_{v}F^{v}_{S}N^{S}n_{s}\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert =+[m_{\perp }]^{r}_{v}F^{v}_{S}N^{S}n_{s}\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert -n_{r}n_{s}\nonumber \\&\quad =[m_{\parallel }]^{r}_{v}F^{v}_{S}N^{S}N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} +\underbrace{\frac{n^{r}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }n_{s}\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert -n_{r}n_{s}}_{=0}= [m_{\parallel }]^{r}_{v}F^{v}_{S}[M_{\perp }]^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \end{aligned}$$
(86)

Thus, one can write Eq. (85) in a shorter but implicit form, as follows

$$\begin{aligned}&\left[ m_{\parallel }\right] ^{r}_{s}=F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}+ \left[ m_{\parallel }\right] ^{r}_{b}\left[ F^{b}_{S}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] \end{aligned}$$
(87)

One can easily check that the proposed transport rules are compatible with the idempotence of the projectors. Preliminarily we notice that, due to the idempotence, for any vector \(w^{s}\) one has

$$\begin{aligned}&\left[ m_{\parallel }\right] ^{t}_{r}\left( \left[ m_{\parallel }\right] ^{r}_{s}w^{s}\right) = \left[ m_{\parallel }\right] ^{t}_{r}\left( w^{s}_{\parallel }\right) \,; \end{aligned}$$
(88)

being \(w^{s}_{\parallel }\) the tangential projection of \(w^{s}\). Thereafter, considering that the outer projector acts on a purely tangential vector and thus, the addend of Eq. (85) including the orthogonal projector vanishes, one finds

$$\begin{aligned}&=F^{t}_{R}\left[ M_{\parallel }\right] ^{R}_{Q}\left( {\mathbf {F}}^{-1}\right) _{r}^{Q}\left( F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s}+ \left[ m_{\parallel }\right] ^{r}_{b}\left[ F^{b}_{S}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] w^{s}\right) \nonumber \\&=F^{t}_{R}\left[ M_{\parallel }\right] ^{R}_{Q}\left( \delta _{S}^{Q}\left[ M_{\parallel }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s}+ \left[ \left( {\mathbf {F}}^{-1}\right) _{r}^{Q}\left[ m_{\parallel }\right] ^{r}_{b}F^{b}_{S}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] w^{s}\right) \nonumber \\&=F^{t}_{R}\left[ M_{\parallel }\right] ^{R}_{V} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} w^{s} +\left[ m_{\parallel }\right] ^{t}_{r}\left[ m_{\parallel }\right] ^{r}_{b} F^{b}_{S}N^{S}\left[ \left( {\mathbf {F}}^{-1}\right) _{s}^{V}N_{V}w^{s}\right] \nonumber \\&=F^{t}_{R}\left[ M_{\parallel }\right] ^{R}_{V} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} w^{s} +\left[ m_{\parallel }\right] ^{t}_{b}\left[ F^{b}_{S}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s}\right] \end{aligned}$$
(89)

Analogously, for the orthogonal projector one has

$$\begin{aligned}&[m_{\perp }]^{r}_{s}[m_{\perp }]^{s}_{t}=[m_{\perp }]^{r}_{t}\,;\nonumber \\&\quad \frac{g^{\star \,E}_{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}[M_{\perp }]^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\, \frac{g^{\star \,D}_{G}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{s}_{D}[M_{\perp }]^{G}_{W}\left( {\mathbf {F}}^{-1}\right) _{t}^{W}\nonumber \\&\quad =\frac{g^{\star \,E}_{S}\,g^{\star \,D}_{G}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{4}} \delta ^{V}_{D}F^{r}_{E}[M_{\perp }]^{S}_{V}[M_{\perp }]^{G}_{W}\left( {\mathbf {F}}^{-1}\right) _{t}^{W}\nonumber \\&\quad =\frac{g^{\star \,E}_{S}\,g^{\star \,D}_{G}N_{D}N^{G}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{4}} F^{r}_{E}N^{S}N_{W}\left( {\mathbf {F}}^{-1}\right) _{t}^{W}\nonumber \\&\quad =\frac{g^{\star \,E}_{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}} F^{r}_{E}[M_{\perp }]^{S}_{W}\left( {\mathbf {F}}^{-1}\right) _{t}^{W}\,; \end{aligned}$$
(90)

The examples which follow provide some insight into the use of surface projectors and are especially suitable to characterize the moving frame vectors over a boundary face (namely \({\mathbf {B}}\),\({\mathbf {T}}\),\({\mathbf {N}}\)) and their transformation.

(i) If we multiply both the projectors by the covariant normal vector expressed through its transport rule, we expect that the orthogonal projection coincides with the same vector, whilst the tangential projection vanishes. By formulae

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}n_{r}&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{\left( {\mathbf {F}}^{-1}\right) _{r}^{D}N_{D}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\delta _{E}^{D}N_{D} \frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\nonumber \\&=\frac{g^{\star \,E}_{S}N_{E}N^{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\nonumber \\&=\frac{N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }=n_{s}\,; \end{aligned}$$
(91)

and, by using the very last expression in Eq. (86), one has

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}n_{r}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) n_{r}=0\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{\left( {\mathbf {F}}^{-1}\right) _{r}^{D}N_{D}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= \nonumber \\&=\frac{g^{\star \,E}_{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ N^{S}N_{E} \delta ^{D}_{V}N_{D}-\delta ^{D}_{E}N_{D}N^{S}N_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= 0\,; \end{aligned}$$
(92)

If instead we consider the Eulerian vector \({\tilde{n}}^{s}=F^{s}_{Q}N_{Q}\), we expect that both its projections do not vanish. In fact, one has

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}{\tilde{n}}^{s}&=n^{r}n_{s}{\tilde{n}}^{s}=n^{r}n_{s}F^{s}_{Q}N_{Q}\nonumber \\&=n^{r}\frac{\left( {\mathbf {F}}^{-1}\right) _{s}^{W}N_{W}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }F^{s}_{Q}N^{Q}= n^{r}\frac{\delta _{Q}^{W}N_{W}N^{Q}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }=\frac{n^{r}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\,; \end{aligned}$$
(93)

The same result is attained by utilizing Eq. (85), namely

$$\begin{aligned}&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} F_{Q}^{s}N_{Q}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\delta _{Q}^{V}N^{Q}=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}N^{S}\nonumber \\&=\frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}F^{r}_{E}N^{S}g^{ts} \left( {\mathbf {F}}^{-1}\right) _{t}^{E}\left( {\mathbf {F}}^{-1}\right) _{s}^{O}g_{OS} =\frac{1}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\delta ^{r}_{t}g^{ts} \left( {\mathbf {F}}^{-1}\right) _{s}^{O}N_{O}\,=\frac{n^{r}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\,; \end{aligned}$$
(94)

As for the tangential projection, one has

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}{\tilde{n}}^{s}&= \left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) F^{s}_{Q}N^{Q}= F^{r}_{Q}N^{Q}-n^{r}n_{s}F^{s}_{Q}N^{Q}\nonumber \\&=F^{r}_{Q}N^{Q}-n^{r}\frac{ \left( {\mathbf {F}}^{-1}\right) _{s}^{D}N_{D}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }F^{s}_{Q}N^{Q}= F^{r}_{Q}N^{Q}-n^{r}\frac{\delta _{Q}^{D}N_{D}N^{Q}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= {\tilde{n}}^{r}-\frac{n^{r}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\,; \end{aligned}$$
(95)

or alternatively, through Eq. (85),

$$\begin{aligned}&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} F^{s}_{Q}N^{Q}\nonumber \\&\quad =\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}\delta ^{V}_{Q}N^{Q}-F^{r}_{E}[M_{\perp }]^{S}_{V}\delta ^{V}_{Q}N^{Q}\right\} \nonumber \\&\quad =\frac{g^{\star \,E}_{S}N^{S}N_{E} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{V}\right\} - \frac{g^{\star \,E}_{S}N^{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{E}\right\} \nonumber \\&\quad ={\tilde{n}}^{r}- \frac{n^{r}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\,; \end{aligned}$$
(96)

(ii) If we consider the vector \(t^{r}=F^{r}_{Q}T^{Q}\Vert {\mathbf {F}}{\mathbf {T}}\Vert ^{-1} \), we expect that its tangential projection equals the same vector, whilst the orthogonal projection vanishes. In fact,

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}t^{s}&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{F^{s}_{Q}T^{Q}}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\delta _{Q}^{V}T^{Q} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }\nonumber \\&=\frac{g^{\star \,E}_{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,N^{S}\underbrace{N_{V}T^{V}}_{=0}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }=0\,; \end{aligned}$$
(97)

whilst, for the tangential projection, through Eq. (86) one finds

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}t^{s}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) t^{s}=t^{r}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} \frac{F^{s}_{Q}T^{Q}}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \delta _{Q}^{V}T_{Q} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}T^{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}T^{V}\right\} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }= \nonumber \\&=\frac{g^{\star \,E}_{S}N^{S}N_{E}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}T^{V}\right\} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }= \frac{F^{r}_{V}T^{V}}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }=t^{r}\,; \end{aligned}$$
(98)

On the contrary, if we consider the Eulerian vector \(\breve{t}_{r}=\left( {\mathbf {F}}^{-1}\right) _{r}^{Q}T_{Q}\), we expect that both its projections do not vanish. As for the orthogonal component, one finds

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}\breve{t}_{r}&=n^{r}n_{s}\breve{t}_{r}=\langle {\mathbf {n}}, \breve{{\mathbf {t}}}\rangle n_{s}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \left( {\mathbf {F}}^{-1}\right) _{r}^{Q}T_{Q}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\delta ^{Q}_{E}T_{Q}\nonumber \\&=\frac{g^{\star \,E}_{S}N^{S}T_{E}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}= \langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {T}}\rangle _{g}\,\frac{n_{s}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= \langle {\mathbf {n}},\breve{{\mathbf {t}}}\rangle _{g}\,n_{s}\,; \end{aligned}$$
(99)

whilst for the tangential projection one has

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}\breve{t}_{r}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) \breve{t}_{r}=\breve{t}_{s}-\langle {\mathbf {n}}, \breve{{\mathbf {t}}}\rangle _{g} n_{s}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} \left( {\mathbf {F}}^{-1}\right) _{r}^{Q}T_{Q}= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ \delta ^{Q}_{V}T_{Q}N^{S}N_{E}-\delta ^{Q}_{E}T_{Q}[M_{\perp }]^{S}_{V}\right\} = \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ T_{V}N^{S}N_{E}-T_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V}= \nonumber \\&=\left( {\mathbf {F}}^{-1}\right) _{s}^{V}T_{V} -\frac{g^{\star \,E}_{S}T_{E}N^{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,\left( {\mathbf {F}}^{-1}\right) _{s}^{V}N_{V}\nonumber \\&=\breve{t}_{s}\,-\langle {\mathbf {F}}^{-T}{\mathbf {T}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g} \frac{ n_{s}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }=\breve{t}_{s}\,-\langle {\mathbf {F}}^{-T}{\mathbf {T}},{\mathbf {n}}\rangle _{g} n_{s}; \end{aligned}$$
(100)

(iii) If we consider the Eulerian vector \({\tilde{b}}^{r}=F^{r}_{Q}B^{Q}\), being \(B^{Q}\) the Lagrangian vector normal to the border edge which belongs to the tangent plane, we expect that the tangential projection equals the same vector, whilst its orthogonal projection trivially vanishes. In fact, through the transformation of the covariant normal vector one finds

$$\begin{aligned}&[m_{\perp }]^{r}_{s}{\tilde{b}}^{s}=n^{r}n_{s}{\tilde{b}}^{s}=n^{r} \frac{\left( {\mathbf {F}}^{-1}\right) ^{Q}_{s}N_{Q}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }\,F^{s}_{Q}B^{Q}=0\,; \end{aligned}$$
(101)

and the same result is provided through Eq. (85), namely

$$\begin{aligned}&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} F^{s}_{Q}B^{Q}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\delta _{Q}^{V}B^{Q}\nonumber \\&=\frac{g^{\star \,E}_{S}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,F^{r}_{E}N^{S}N_{V}B^{V}=0\,; \end{aligned}$$
(102)

As for the tangential projection, one finds

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}{\tilde{b}}^{s}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) {\tilde{b}}^{s}={\tilde{b}}^{r} -n^{r}n_{s}{\tilde{b}}^{s}={\tilde{b}}^{r}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} F^{s}_{Q}B^{Q}= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \delta _{Q}^{V}B_{Q}= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}B^{V}N^{S}N_{E}-F^{r}_{E}N^{S}N_{V}B^{V}\right\} \frac{1}{\Vert {\mathbf {F}}{\mathbf {T}}\Vert }= \nonumber \\&=\frac{g^{\star \,E}_{S}N^{S}N_{E}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}B^{V}\right\} = F^{r}_{V}B^{V}={\tilde{b}}^{r}\,; \end{aligned}$$
(103)

If we consider the Eulerian vector \(\breve{b}_{r}=\left( {\mathbf {F}}^{-1}\right) _{r}^{Q}B_{Q}\), both its projections do not vanish. As for the orthogonal component, one obtains

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}\breve{b}_{r}&=n^{r}n_{s}\breve{b}_{r}=\langle {\mathbf {n}}, \breve{{\mathbf {b}}}\rangle n_{s}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \left( {\mathbf {F}}^{-1}\right) _{r}^{Q}B_{Q}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\delta ^{Q}_{E}B_{Q}\nonumber \\&=\frac{g^{\star \,E}_{S}N^{S}B_{E}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,N_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}= \langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {B}}\rangle _{g}\,\frac{n_{s}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }= \langle {\mathbf {n}},\breve{{\mathbf {b}}}\rangle _{g}\,n_{s}\,; \end{aligned}$$
(104)

whilst for the tangential component one finds

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}\breve{b}_{r}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) \breve{b}_{r}=\breve{b}_{s}-\langle {\mathbf {n}}, \breve{{\mathbf {b}}}\rangle _{g} n_{s}\nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ F^{r}_{V}N^{S}N_{E}-F^{r}_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V} \left( {\mathbf {F}}^{-1}\right) _{r}^{Q}B_{Q}= \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ \delta ^{Q}_{V}B_{Q}N^{S}N_{E}-\delta ^{Q}_{E}B_{Q}[M_{\perp }]^{S}_{V}\right\} = \nonumber \\&=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, \left\{ B_{V}N^{S}N_{E}-B_{E}[M_{\perp }]^{S}_{V}\right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V}= \nonumber \\&=\left( {\mathbf {F}}^{-1}\right) _{s}^{V}B_{V} -\frac{g^{\star \,E}_{S}T_{E}N^{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\,\left( {\mathbf {F}}^{-1}\right) _{s}^{V}N_{V}\nonumber \\&=\breve{b}_{s}\,-\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g} \frac{ n_{s}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }=\breve{b}_{s}\,-\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {n}}\rangle _{g} n_{s}; \end{aligned}$$
(105)

Since the transport formulae are available for both the covariant and contravariant components of the edge normal vector, we can proceed with a further check. As for the orthogonal projection, one finds

$$\begin{aligned}{}[m_{\perp }]^{r}_{s}b_{r}&=[m_{\perp }]^{r}_{s}\left\{ \left( {\mathbf {F}}^{-1}\right) _{r}^{R}B_{R}- \frac{\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}}{\langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}} \left( {\mathbf {F}}^{-1}\right) _{r}^{R}N_{R}\right\} \frac{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }{\Vert J^{-1}{\mathbf {F}}{\mathbf {T}}\Vert }\nonumber \\&=\left\{ \langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {B}}\rangle _{g}\,\frac{n_{s}}{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }- \frac{\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}}{\langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}}\left( {\mathbf {F}}^{-1}\right) _{r}^{R}N_{R} \right\} \frac{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }{\Vert J^{-1}{\mathbf {F}}{\mathbf {T}}\Vert }\,=0; \end{aligned}$$
(106)

whilst for the tangential component one has

$$\begin{aligned}{}[m_{\parallel }]^{r}_{s}b_{r}&=\left( \delta ^{r}_{s}-[m_{\perp }]^{r}_{s}\right) \left\{ \left( {\mathbf {F}}^{-1}\right) _{r}^{R}B_{R}- \frac{\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}}{\langle {\mathbf {F}}^{-T}{\mathbf {N}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}} \left( {\mathbf {F}}^{-1}\right) _{r}^{R}N_{R}\right\} \frac{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }{\Vert J^{-1}{\mathbf {F}}{\mathbf {T}}\Vert }\nonumber \\&=\left\{ \left( {\mathbf {F}}^{-1}\right) _{r}^{R}B_{R}\,-\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {n}}\rangle _{g} n_{s}- 0 \right\} \frac{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert }{\Vert J^{-1}{\mathbf {F}}{\mathbf {T}}\Vert }\,=b_{r}; \end{aligned}$$
(107)

(iv) The following relationships, trivial to prove, are worth considering:

$$\begin{aligned}&\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}{\mathbf {N}}\rangle _{g}=g^{s}_{r}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}B_{V}F^{r}_{M}N^{M}= \delta ^{M}_{V}B_{V}N^{M}=0\,;\nonumber \\&\langle {\mathbf {F}}^{-T}{\mathbf {T}},{\mathbf {F}}{\mathbf {N}}\rangle _{g}=g^{s}_{r}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}T_{V}F^{r}_{M}N^{M}= \delta ^{M}_{V}T_{V}N^{M}=0\,;\nonumber \\ \end{aligned}$$
(108)

since the orthogonality of the above Eulerian vectors was not evident. But, for a generic tangent map (diverse from an isometric transformation), we expect

$$\begin{aligned}&\langle {\mathbf {F}}^{-T}{\mathbf {B}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}\ne 0\,;\nonumber \\&\langle {\mathbf {F}}^{-T}{\mathbf {T}},{\mathbf {F}}^{-T}{\mathbf {N}}\rangle _{g}\ne 0\,;\nonumber \\&\langle {\mathbf {F}}^{-T}{\mathbf {T}},{\mathbf {F}}^{-T}{\mathbf {B}}\rangle _{g}\ne 0\,;\nonumber \\&\langle {\mathbf {F}}{\mathbf {T}},{\mathbf {F}}{\mathbf {B}}\rangle _{g}\ne 0\,;\nonumber \\ \end{aligned}$$
(109)

(v) It can be useful to arrange the transport relationships for the two surface projectors in the form of a system of equations:

$$\begin{aligned} \left\{ \begin{aligned}&[m_{\perp }]^{r}_{s}=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V} \\&{[m_{\parallel }]}^{r}_{s}=F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}+ \left[ F^{r}_{S}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}\right] \left[ M_{\perp }\right] ^{S}_{V} \end{aligned}\right. \end{aligned}$$
(110)

Hence, to characterize the generic (contravariant) Eulerian vector with respect to the boundary face, one can multiply by it both sides of the above equations, visualizing simultaneously the results in both the Lagrangian and the Eulerian configuration. For instance, let us consider a vector \(w^{s}\) orthogonal to the face, such that it results \([m_{\parallel }]^{r}_{s}w^{s}=0\) in the Eulerian configuration. We have

$$\begin{aligned} \left\{ \begin{aligned}&[m_{\perp }]^{r}_{s}w^{s}=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s} \\&{[m_{\parallel }]}^{r}_{s}w^{s}=0=\left\{ F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}+ \left[ F^{r}_{S}\left[ M_{\perp }\right] ^{S}_{V}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}\right] \right\} \left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s} \end{aligned}\right. \end{aligned}$$
(111)

By the position \(W^{V}=\left( {\mathbf {F}}^{-1}\right) _{s}^{V}w^{s}\), one finds

$$\begin{aligned} \left\{ \begin{aligned}&[m_{\perp }]^{r}_{s}w^{s}=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}W^{V} \\&{[m_{\parallel }]}^{r}_{s}w^{s}=0=\left\{ F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}W^{V}+ \left[ F^{r}_{S}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\right] \left[ M_{\perp }\right] ^{S}_{V}W^{V}\right\} \end{aligned}\right. \end{aligned}$$
(112)

It is clear that, since vector \(w^{s}\) is orthogonal to the face in the Eulerian configuration, when transported to the Lagrangian configuration by \({\mathbf {F}}^{-1}\) (\(\forall \,{\mathbf {F}}\), \(J=\mathrm {det}\left( {\mathbf {F}}\right) >0\)) it possesses also a tangential component, namely \(\left[ M_{\parallel }\right] ^{S}_{V}W^{V}\ne {\mathbf {0}}\). However, such a tangential component through Eq. (112) must give rise to a contribution equal opposite to that generated by its normal component \(\left[ M_{\perp }\right] ^{S}_{V}W^{V}\).

On the contrary, when we consider a vector tangent to the same face of the Eulerian configuration, one has

$$\begin{aligned} \left\{ \begin{aligned}&[m_{\perp }]^{r}_{s}w^{s}=0=\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\left[ M_{\perp }\right] ^{S}_{V}W^{V} \\&{[m_{\parallel }]}^{r}_{s}w^{s}=\left\{ F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}W^{V}+ \left[ F^{r}_{S}-\frac{g^{\star \,E}_{S} }{\Vert {\mathbf {F}}^{-T}{\mathbf {N}}\Vert ^{2}}\, F^{r}_{E}\right] \left[ M_{\perp }\right] ^{S}_{V}W^{V}\right\} \\&\quad =F^{r}_{S}\left[ M_{\parallel }\right] ^{S}_{V}W^{V}=F^{r}_{S}W^{S}=w^{s}\,; \end{aligned}\right. \end{aligned}$$
(113)

Thus, the image of such a vector through the inverse tangent map (\(\forall \, {\mathbf {F}}\), \(J>0\)) must possess a purely tangential component in the Lagrangian configuration, namely \(\left[ M_{\perp }\right] ^{S}_{V}W^{V}={\mathbf {0}}\).

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Fedele, R. Piola’s approach to the equilibrium problem for bodies with second gradient energies. Part I: First gradient theory and differential geometry. Continuum Mech. Thermodyn. 34, 445–474 (2022). https://doi.org/10.1007/s00161-021-01064-6

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