On this page, we will work with expressions of the form
∂ u ‾ ∂ v ‾ \begin{aligned} \frac{\partial \underline{\boldsymbol{ u}}}{\partial \underline{\boldsymbol{ v}}} \end{aligned} ∂ v ∂ u that is, differentiation a tensor valued expression wrt. to a tensor. In this case, u ‾ = u i e ‾ i \underline{\boldsymbol{ u}}=u_i\underline{\boldsymbol{ e}}_{ i} u = u i e i v ‾ = v i e ‾ i \underline{\boldsymbol{ v}}=v_i \underline{\boldsymbol{ e}}_{ i} v = v i e i constant orthonormal coordinate systems, we use that
∂ u ‾ ∂ v ‾ = ∂ u i ∂ v j e ‾ i ⊗ e ‾ j \begin{aligned} \frac{\partial \underline{\boldsymbol{ u}}}{\partial \underline{\boldsymbol{ v}}} = \frac{\partial u_i}{\partial v_j} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} \end{aligned} ∂ v ∂ u = ∂ v j ∂ u i e i ⊗ e j Here, constant implies that the base vectors are constant in space. Hence, it suffices to differentiate the coefficients as the derivative of each base vector is zero. In general, for an N N N y = y i 1 i 2 ⋯ i N e ‾ i 1 ⊗ e ‾ i 2 ⊗ ⋯ ⊗ e ‾ i N \boldsymbol{y} = y_{i_1 i_2 \cdots i_N}\underline{\boldsymbol{ e}}_{ i_1}\otimes\underline{\boldsymbol{ e}}_{ i_2}\otimes\cdots\otimes\underline{\boldsymbol{ e}}_{ i_N} y = y i 1 i 2 ⋯ i N e i 1 ⊗ e i 2 ⊗ ⋯ ⊗ e i N M M M x = x j 1 j 2 ⋯ j M e ‾ j 1 ⊗ e ‾ j 2 ⊗ ⋯ ⊗ e ‾ j M \boldsymbol{x} = x_{j_1 j_2 \cdots j_M}\underline{\boldsymbol{ e}}_{ j_1}\otimes\underline{\boldsymbol{ e}}_{ j_2}\otimes\cdots\otimes\underline{\boldsymbol{ e}}_{ j_M} x = x j 1 j 2 ⋯ j M e j 1 ⊗ e j 2 ⊗ ⋯ ⊗ e j M
∂ y ∂ x = [ ∂ y i 1 i 2 ⋯ i N ∂ x j 1 j 2 ⋯ j M ] e ‾ i 1 ⊗ e ‾ i 2 ⊗ ⋯ ⊗ e ‾ i N ⊗ e ‾ j 1 ⊗ e ‾ j 2 ⊗ ⋯ ⊗ e ‾ j M \begin{aligned} \frac{\partial \boldsymbol{y}}{\partial \boldsymbol{x}} &= \left[\frac{\partial y_{i_1 i_2 \cdots i_N}}{\partial x_{j_1 j_2 \cdots j_M}}\right] \underline{\boldsymbol{ e}}_{ i_1}\otimes\underline{\boldsymbol{ e}}_{ i_2}\otimes\cdots\otimes\underline{\boldsymbol{ e}}_{ i_N}\otimes\underline{\boldsymbol{ e}}_{ j_1}\otimes\underline{\boldsymbol{ e}}_{ j_2}\otimes\cdots\otimes\underline{\boldsymbol{ e}}_{ j_M} \end{aligned} ∂ x ∂ y = [ ∂ x j 1 j 2 ⋯ j M ∂ y i 1 i 2 ⋯ i N ] e i 1 ⊗ e i 2 ⊗ ⋯ ⊗ e i N ⊗ e j 1 ⊗ e j 2 ⊗ ⋯ ⊗ e j M Furthermore, since we can consider each free coefficient, e.g., u 1 u_1 u 1 Consider the following in 2d: The tensor
b \boldsymbol{ b} b is a function of the tensor
a \boldsymbol{ a} a , such that
b ( a ) = a a \boldsymbol{ b}(\boldsymbol{ a}) = \boldsymbol{ a}\boldsymbol{ a} b ( a ) = a a (i.e.
b i j = a i n a n j b_{ij} = a_{in}a_{nj} b ij = a in a nj ). We would like to differentiate
a b \boldsymbol{ a}\boldsymbol{ b} a b wrt.
a \boldsymbol{ a} a .
∂ a i m b m j ( a ) ∂ a k l = = ∂ [ a i 1 b 1 j ( a ) + a i 2 b 2 j ( a ) ] ∂ a k l , ( Expand dummy index summation ⇒ 2 4 individual scalar expressions, one for each i , j , k , l ) = ∂ a i 1 b 1 j ( a ) ∂ a k l + ∂ a i 2 b 2 j ( a ) ∂ a k l , ( Product rule ) = a i 1 ∂ b 1 j ( a ) ∂ a k l + ∂ a i 1 ∂ a k l b 1 j ( a ) + a i 2 ∂ b 2 j ( a ) ∂ a k l + ∂ a i 2 ∂ a k l b 2 j ( a ) , ( Chain rule ) = a i 1 ∂ a 1 n a n j ∂ a k l + δ i k δ 1 l b 1 j ( a ) + a i 2 ∂ a 2 n a n j ∂ a k l + δ i k δ 2 l b 2 j ( a ) = a i 1 [ a 1 n ∂ a n j ∂ a k l + ∂ a 1 n ∂ a k l a n j ] + δ i k δ 1 l b 1 j ( a ) + a i 2 [ a 2 n ∂ a n j ∂ a k l + ∂ a 2 n ∂ a k l a n j ] + δ i k δ 2 l b 2 j ( a ) = a i 1 [ a 1 n δ n k δ j l + δ 1 k δ n l a n j ] + δ i k δ 1 l b 1 j ( a ) + a i 2 [ a 2 n δ n k δ j l + δ 2 k δ n l a n j ] + δ i k δ 2 l b 2 j ( a ) = a i 1 [ a 1 k δ j l + δ 1 k a l j ] + δ i k δ 1 l b 1 j ( a ) + a i 2 [ a 2 k δ j l + δ 2 k a l j ] + δ i k δ 2 l b 2 j ( a ) = a i m [ a m k δ j l + δ m k a l j ] + δ i k δ m l b m j ( a ) , ( Identify as summation, reinstate dummy indices ) = a i m [ a m k δ j l + δ m k a l j ] + δ i k b l j ( a ) \begin{aligned} &\frac{\partial a_{im}b_{mj}(\boldsymbol{ a})}{\partial a_{kl}} =\\ &= \frac{\partial \left[a_{i1}b_{1j}(\boldsymbol{ a}) + a_{i2}b_{2j}(\boldsymbol{ a})\right]}{\partial a_{kl}}, \quad \left(\begin{matrix}\text{Expand dummy index summation} \Rightarrow 2^4 \text{ individual}\\ \text{scalar expressions, one for each }i,j,k,l\end{matrix}\right)\\ &= \frac{\partial a_{i1}b_{1j}(\boldsymbol{ a})}{\partial a_{kl}} + \frac{\partial a_{i2}b_{2j}(\boldsymbol{ a})}{\partial a_{kl}}, \quad (\text{Product rule})\\ &= a_{i1}\frac{\partial b_{1j}(\boldsymbol{ a})}{\partial a_{kl}} + \frac{\partial a_{i1}}{\partial a_{kl}}b_{1j}(\boldsymbol{ a}) + a_{i2}\frac{\partial b_{2j}(\boldsymbol{ a})}{\partial a_{kl}} + \frac{\partial a_{i2}}{\partial a_{kl}}b_{2j}(\boldsymbol{ a}), \quad (\text{Chain rule}) \\ &= a_{i1}\frac{\partial a_{1n}a_{nj}}{\partial a_{kl}} + \delta_{ik}\delta_{1l} b_{1j}(\boldsymbol{ a}) + a_{i2}\frac{\partial a_{2n}a_{nj}}{\partial a_{kl}} + \delta_{ik}\delta_{2l} b_{2j}(\boldsymbol{ a}) \\ &= a_{i1}\left[a_{1n}\frac{\partial a_{nj}}{\partial a_{kl}}+\frac{\partial a_{1n}}{\partial a_{kl}}a_{nj}\right] + \delta_{ik}\delta_{1l} b_{1j}(\boldsymbol{ a}) + a_{i2}\left[a_{2n}\frac{\partial a_{nj}}{\partial a_{kl}}+\frac{\partial a_{2n}}{\partial a_{kl}}a_{nj}\right] + \delta_{ik}\delta_{2l} b_{2j}(\boldsymbol{ a}) \\ &= a_{i1}\left[a_{1n}\delta_{nk}\delta_{jl}+\delta_{1k}\delta_{nl}a_{nj}\right] + \delta_{ik}\delta_{1l} b_{1j}(\boldsymbol{ a}) + a_{i2}\left[a_{2n}\delta_{nk}\delta_{jl}+\delta_{2k}\delta_{nl}a_{nj}\right] + \delta_{ik}\delta_{2l} b_{2j}(\boldsymbol{ a})\\ &= a_{i1}\left[a_{1k}\delta_{jl}+\delta_{1k}a_{lj}\right] + \delta_{ik}\delta_{1l} b_{1j}(\boldsymbol{ a}) + a_{i2}\left[a_{2k}\delta_{jl}+\delta_{2k}a_{lj}\right] + \delta_{ik}\delta_{2l} b_{2j}(\boldsymbol{ a}) \\ &= a_{im}\left[a_{mk}\delta_{jl}+\delta_{mk}a_{lj}\right] + \delta_{ik}\delta_{ml} b_{mj}(\boldsymbol{ a}), \quad \left( \begin{matrix} \text{Identify as summation,} \\ \text{reinstate dummy indices} \end{matrix} \right)\\ &= a_{im}\left[a_{mk}\delta_{jl}+\delta_{mk}a_{lj}\right] + \delta_{ik} b_{lj}(\boldsymbol{ a}) \end{aligned} ∂ a k l ∂ a im b mj ( a ) = = ∂ a k l ∂ [ a i 1 b 1 j ( a ) + a i 2 b 2 j ( a ) ] , ( Expand dummy index summation ⇒ 2 4 individual scalar expressions, one for each i , j , k , l ) = ∂ a k l ∂ a i 1 b 1 j ( a ) + ∂ a k l ∂ a i 2 b 2 j ( a ) , ( Product rule ) = a i 1 ∂ a k l ∂ b 1 j ( a ) + ∂ a k l ∂ a i 1 b 1 j ( a ) + a i 2 ∂ a k l ∂ b 2 j ( a ) + ∂ a k l ∂ a i 2 b 2 j ( a ) , ( Chain rule ) = a i 1 ∂ a k l ∂ a 1 n a nj + δ ik δ 1 l b 1 j ( a ) + a i 2 ∂ a k l ∂ a 2 n a nj + δ ik δ 2 l b 2 j ( a ) = a i 1 [ a 1 n ∂ a k l ∂ a nj + ∂ a k l ∂ a 1 n a nj ] + δ ik δ 1 l b 1 j ( a ) + a i 2 [ a 2 n ∂ a k l ∂ a nj + ∂ a k l ∂ a 2 n a nj ] + δ ik δ 2 l b 2 j ( a ) = a i 1 [ a 1 n δ nk δ j l + δ 1 k δ n l a nj ] + δ ik δ 1 l b 1 j ( a ) + a i 2 [ a 2 n δ nk δ j l + δ 2 k δ n l a nj ] + δ ik δ 2 l b 2 j ( a ) = a i 1 [ a 1 k δ j l + δ 1 k a l j ] + δ ik δ 1 l b 1 j ( a ) + a i 2 [ a 2 k δ j l + δ 2 k a l j ] + δ ik δ 2 l b 2 j ( a ) = a im [ a mk δ j l + δ mk a l j ] + δ ik δ m l b mj ( a ) , ( Identify as summation, reinstate dummy indices ) = a im [ a mk δ j l + δ mk a l j ] + δ ik b l j ( a ) The same result is achieved without expanding the dummy indices \textcolor{blue}{ \text{dummy indices}} dummy indices
∂ a i m b m j ( a ) ∂ a k l = = a i m ∂ b m j ( a ) ∂ a k l + ∂ a i m ∂ a k l b m j ( a ) = a i m ∂ a m n a n j ∂ a k l + δ i k δ m l b m j ( a ) = a i m [ a m n ∂ a n j ∂ a k l + ∂ a m n ∂ a k l a n j ] + δ i k b l j ( a ) = a i m [ a m n δ n k δ j l + δ m k δ n l a n j ] + δ i k b l j ( a ) = a i m [ a m k δ j l + δ m k a l j ] + δ i k b l j ( a ) \begin{aligned} &\frac{\partial a_{i\textcolor{blue}{ m}}b_{\textcolor{blue}{ m}j}(\boldsymbol{ a})}{\partial a_{kl}} =\\ &= a_{i\textcolor{blue}{ m}}\frac{\partial b_{\textcolor{blue}{ m}j}(\boldsymbol{ a})}{\partial a_{kl}} + \frac{\partial a_{i\textcolor{blue}{ m}}}{\partial a_{kl}}b_{\textcolor{blue}{ m}j}(\boldsymbol{ a})\\ &= a_{i\textcolor{blue}{ m}}\frac{\partial a_{\textcolor{blue}{ mn}}a_{\textcolor{blue}{ n}j}}{\partial a_{kl}} + \delta_{ik}\delta_{\textcolor{blue}{ m}l} b_{\textcolor{blue}{ m}j}(\boldsymbol{ a})\\ &= a_{i\textcolor{blue}{ m}}\left[a_{\textcolor{blue}{ mn}}\frac{\partial a_{\textcolor{blue}{ n}j}}{\partial a_{kl}}+\frac{\partial a_{\textcolor{blue}{ mn}}}{\partial a_{kl}}a_{\textcolor{blue}{ n}j}\right] + \delta_{ik} b_{lj}(\boldsymbol{ a})\\ &= a_{i\textcolor{blue}{ m}}\left[a_{\textcolor{blue}{ mn}}\delta_{\textcolor{blue}{ n}k}\delta_{jl}+\delta_{\textcolor{blue}{ m}k}\delta_{\textcolor{blue}{ n}l}a_{\textcolor{blue}{ n}j}\right] + \delta_{ik} b_{lj}(\boldsymbol{ a})\\ &= a_{i\textcolor{blue}{ m}}\left[a_{\textcolor{blue}{ m}k}\delta_{jl}+\delta_{\textcolor{blue}{ m}k}a_{lj}\right] + \delta_{ik} b_{lj}(\boldsymbol{ a}) \end{aligned} ∂ a k l ∂ a i m b m j ( a ) = = a i m ∂ a k l ∂ b m j ( a ) + ∂ a k l ∂ a i m b m j ( a ) = a i m ∂ a k l ∂ a mn a n j + δ ik δ m l b m j ( a ) = a i m [ a mn ∂ a k l ∂ a n j + ∂ a k l ∂ a mn a n j ] + δ ik b l j ( a ) = a i m [ a mn δ n k δ j l + δ m k δ n l a n j ] + δ ik b l j ( a ) = a i m [ a m k δ j l + δ m k a l j ] + δ ik b l j ( a ) And for completeness, this is a 2 ⊗ ‾ I + a ⊗ ‾ a T + I ⊗ ‾ b T \boldsymbol{ a}^2 \overline{\otimes} \boldsymbol{ I} + \boldsymbol{ a}\overline{\otimes}\boldsymbol{ a}^{\mathrm{T}} + \boldsymbol{ I}\overline{\otimes}\boldsymbol{ b}^{\mathrm{T}} a 2 ⊗ I + a ⊗ a T + I ⊗ b T
If we consider a = f ( x ) = x b \boldsymbol{ a} = f(x) = x\boldsymbol{ b} a = f ( x ) = x b
∂ a ∂ x = ∂ x b i j ∂ x e ‾ i ⊗ e ‾ j = b i j e ‾ i ⊗ e ‾ j = b \begin{aligned} \frac{\partial \boldsymbol{ a}}{\partial x} = \frac{\partial x b_{ij}}{\partial x} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} = b_{ij} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} = \boldsymbol{ b} \end{aligned} ∂ x ∂ a = ∂ x ∂ x b ij e i ⊗ e j = b ij e i ⊗ e j = b because b i j b_{ij} b ij x x x
Let's first consider the differentiating a tensor wrt. itself. For a first-order tensor, we have
∂ u ‾ ∂ u ‾ = ∂ u i ∂ u j e ‾ i ⊗ e ‾ j ∂ u i ∂ u j = δ i j ∂ u ‾ ∂ u ‾ = I \begin{aligned} \frac{\partial \underline{\boldsymbol{ u}}}{\partial \underline{\boldsymbol{ u}}} &= \frac{\partial u_i}{\partial u_j} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j}\\ \frac{\partial u_i}{\partial u_j} &= \delta_{ij} \\ \frac{\partial \underline{\boldsymbol{ u}}}{\partial \underline{\boldsymbol{ u}}} &= \boldsymbol{ I} \end{aligned} ∂ u ∂ u ∂ u j ∂ u i ∂ u ∂ u = ∂ u j ∂ u i e i ⊗ e j = δ ij = I As ∂ u i / ∂ u j \partial u_i/\partial u_j ∂ u i / ∂ u j i = j i=j i = j i ≠ j i\neq j i = j
If we now consider a 2nd order tensor, we have
∂ a ∂ a = ∂ a i j ∂ a k l e ‾ i ⊗ e ‾ j ⊗ e ‾ k ⊗ e ‾ l ∂ a i j ∂ a k l = δ i k δ j l ∂ a ∂ a = I \begin{aligned} \frac{\partial \boldsymbol{ a}}{\partial \boldsymbol{ a}} &= \frac{\partial a_{ij}}{\partial a_{kl}} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j}\otimes\underline{\boldsymbol{ e}}_{ k}\otimes\underline{\boldsymbol{ e}}_{ l} \\ \frac{\partial a_{ij}}{\partial a_{kl}} &= \delta_{ik}\delta_{jl}\\ \frac{\partial \boldsymbol{ a}}{\partial \boldsymbol{ a}} &= \textbf{\textsf{ I}} \end{aligned} ∂ a ∂ a ∂ a k l ∂ a ij ∂ a ∂ a = ∂ a k l ∂ a ij e i ⊗ e j ⊗ e k ⊗ e l = δ ik δ j l = I ∂ a i j / ∂ a k l \partial a_{ij}/\partial a_{kl} ∂ a ij / ∂ a k l i = k i=k i = k j = l j=l j = l ∂ a i j / ∂ a k l = δ i k δ j l \partial a_{ij}/\partial a_{kl}=\delta_{ik}\delta_{jl} ∂ a ij / ∂ a k l = δ ik δ j l
To consider a more complicated example, we look at
∂ [ v ‾ a ] ∂ v ‾ = ∂ v k a k i ∂ v j e ‾ i ⊗ e ‾ j ∂ v k a k i ∂ v j = ∂ v k ∂ v j a k i = δ k j a k i = a j i ∂ [ v ‾ a ] ∂ v ‾ = a T \begin{aligned} \frac{\partial \left[\underline{\boldsymbol{ v}}\boldsymbol{ a}\right]}{\partial \underline{\boldsymbol{ v}}} &= \frac{\partial v_k a_{ki}}{\partial v_j} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} \\ \frac{\partial v_k a_{ki}}{\partial v_j} &= \frac{\partial v_k}{\partial v_j} a_{ki} = \delta_{kj} a_{ki} = a_{ji} \\ \frac{\partial \left[\underline{\boldsymbol{ v}}\boldsymbol{ a}\right]}{\partial \underline{\boldsymbol{ v}}} &= \boldsymbol{ a}^{\mathrm{T}} \end{aligned} ∂ v ∂ [ v a ] ∂ v j ∂ v k a ki ∂ v ∂ [ v a ] = ∂ v j ∂ v k a ki e i ⊗ e j = ∂ v j ∂ v k a ki = δ kj a ki = a ji = a T If we consider y = f ( a ) = a : a y = f(\boldsymbol{ a}) = \boldsymbol{ a}:\boldsymbol{ a} y = f ( a ) = a : a
∂ y ∂ a = ∂ a k l a k l ∂ a i j e ‾ i ⊗ e ‾ j = [ ∂ a k l ∂ a i j a k l + a k l ∂ a k l ∂ a i j ] e ‾ i ⊗ e ‾ j = [ δ k i δ l j a k l + a k l δ k i δ l j ] e ‾ i ⊗ e ‾ j = [ a i j + a i j ] e ‾ i ⊗ e ‾ j = 2 a i j e ‾ i ⊗ e ‾ j = 2 a \begin{aligned} \frac{\partial y}{\partial \boldsymbol{ a}} &= \frac{\partial a_{kl}a_{kl}}{\partial a_{ij}} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j}\\ &= \left[\frac{\partial a_{kl}}{\partial a_{ij}} a_{kl} + a_{kl} \frac{\partial a_{kl}}{\partial a_{ij}}\right]\underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} \\ &= \left[\delta_{ki}\delta_{lj} a_{kl} + a_{kl} \delta_{ki}\delta_{lj}\right]\underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} \\ &= \left[a_{ij} + a_{ij}\right]\underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} = 2a_{ij}\underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} = 2\boldsymbol{ a} \end{aligned} ∂ a ∂ y = ∂ a ij ∂ a k l a k l e i ⊗ e j = [ ∂ a ij ∂ a k l a k l + a k l ∂ a ij ∂ a k l ] e i ⊗ e j = [ δ ki δ l j a k l + a k l δ ki δ l j ] e i ⊗ e j = [ a ij + a ij ] e i ⊗ e j = 2 a ij e i ⊗ e j = 2 a Some operations wrt. the coordinates are so common that they have their own name and notation. The concept of a gradient, ∇ f \nabla f ∇ f f ( x ‾ ) f(\underline{\boldsymbol{ x}}) f ( x )
grad ( f ) = ∂ f ∂ x ‾ = ∇ i f ( x ‾ ) e ‾ i \begin{aligned} \text{grad}(f) = \frac{\partial f}{\partial \underline{\boldsymbol{ x}}} = \nabla_i f(\underline{\boldsymbol{ x}}) \underline{\boldsymbol{ e}}_{ i} \end{aligned} grad ( f ) = ∂ x ∂ f = ∇ i f ( x ) e i And we will define the vector operator ∇ ‾ \underline{\boldsymbol{ \nabla}} ∇
∇ ‾ = ∇ i e ‾ i = ∂ ∂ x i e ‾ i \begin{aligned} \underline{\boldsymbol{ \nabla}} = \nabla_i \underline{\boldsymbol{ e}}_{ i} = \frac{\partial }{\partial x_{i}} \underline{\boldsymbol{ e}}_{ i} \end{aligned} ∇ = ∇ i e i = ∂ x i ∂ e i The gradient of higher order tensors is then possible to express as, e.g., v ‾ ⊗ ∇ ‾ \underline{\boldsymbol{ v}}\otimes\underline{\boldsymbol{ \nabla}} v ⊗ ∇ a ⊗ ∇ ‾ \boldsymbol{ a}\otimes\underline{\boldsymbol{ \nabla}} a ⊗ ∇
As ∇ ‾ \underline{\boldsymbol{ \nabla}} ∇ To clarify what operand the gradient is acting on in a larger expression, it can be necessary to enclose the entire expression in brackets:
a b ⊗ ∇ ‾ \boldsymbol{ a} \boldsymbol{ b}\otimes\underline{\boldsymbol{ \nabla}} a b ⊗ ∇ Not clear if the gradient is acting on b \boldsymbol{ b} b a b \boldsymbol{ a}\boldsymbol{ b} a b
a [ b ⊗ ∇ ‾ ] \boldsymbol{ a}\left[ \boldsymbol{ b}\otimes\underline{\boldsymbol{ \nabla}}\right] a [ b ⊗ ∇ ] b \boldsymbol{ b} b
[ a b ] ⊗ ∇ ‾ \left[\boldsymbol{ a}\boldsymbol{ b}\right]\otimes\underline{\boldsymbol{ \nabla}} [ a b ] ⊗ ∇ a b \boldsymbol{ a}\boldsymbol{ b} a b
c [ a b ] ⊗ ∇ ‾ \boldsymbol{ c} \left[\boldsymbol{ a}\boldsymbol{ b}\right]\otimes\underline{\boldsymbol{ \nabla}} c [ a b ] ⊗ ∇ Not clear if gradient is acting on a b \boldsymbol{ a}\boldsymbol{ b} a b c [ a b ] \boldsymbol{ c}\left[\boldsymbol{ a}\boldsymbol{ b}\right] c [ a b ]
c [ [ a b ] ⊗ ∇ ‾ ] \boldsymbol{ c}\left[ \left[\boldsymbol{ a}\boldsymbol{ b}\right]\otimes\underline{\boldsymbol{ \nabla}}\right] c [ [ a b ] ⊗ ∇ ] a b \boldsymbol{ a}\boldsymbol{ b} a b
In some cases, brackets are required also for regular expression, e.g. C = A : [ a ⊗ ‾ b ] ≠ D = [ A : a ] ⊗ ‾ b \textbf{\textsf{ C}}=\textbf{\textsf{ A}}:\left[\boldsymbol{ a}\overline{\otimes}\boldsymbol{ b}\right] \neq \textbf{\textsf{ D}}=\left[\textbf{\textsf{ A}}:\boldsymbol{ a}\right]\overline{\otimes}\boldsymbol{ b} C = A : [ a ⊗ b ] = D = [ A : a ] ⊗ b C i j k l = A i j m n a m k b n l ≠ D i j k l = A i k m n a m n b j l \textsf{ C}_{ ijkl}=\textsf{ A}_{ ijmn}a_{mk}b_{nl}\neq\textsf{ D}_{ ijkl}=\textsf{ A}_{ ikmn}a_{mn}b_{jl} C ijk l = A ijmn a mk b n l = D ijk l = A ikmn a mn b j l ∇ ‾ \underline{\boldsymbol{ \nabla}} ∇
The divergence, div ( v ) \text{div}(\boldsymbol{ v}) div ( v ) ∇ ‾ \underline{\boldsymbol{ \nabla}} ∇
Divergence of 1st order tensor: v ‾ ⋅ ∇ ‾ \underline{\boldsymbol{ v}}\cdot\underline{\boldsymbol{ \nabla}} v ⋅ ∇
Divergence of 2nd order tensor: a ⋅ ∇ ‾ \boldsymbol{ a}\cdot\underline{\boldsymbol{ \nabla}} a ⋅ ∇
Divergence of higher order tensors is not common. As for the gradient, brackets are crucial to ensure that we know which operand (tensor) ∇ ‾ \underline{\boldsymbol{ \nabla}} ∇
The curl of a vector field, v ‾ ( x ‾ ) \underline{\boldsymbol{ v}}(\underline{\boldsymbol{ x}}) v ( x )
curl ( v ‾ ) = − v ‾ × ∇ ‾ = − ∂ v i ∂ x j ε i j k e ‾ k \begin{aligned} \text{curl}(\underline{\boldsymbol{ v}}) &= - \underline{\boldsymbol{ v}}\times\underline{\boldsymbol{ \nabla}} = - \frac{\partial v_i}{\partial x_j} \varepsilon_{ijk} \underline{\boldsymbol{ e}}_k \end{aligned} curl ( v ) = − v × ∇ = − ∂ x j ∂ v i ε ijk e k This operation is common in fluid mechanics to find the rotation of a velocity field, v ‾ \underline{\boldsymbol{ v}} v
It is also possible to define the curl for higher order tensors. Here we use the definition from Rubin (2000) :
curl ( a ) = − a × ∇ ‾ = − ∂ a i j ∂ x k ε j k l e ‾ i ⊗ e ‾ l \begin{aligned} \text{curl}(\boldsymbol{ a}) &= - \boldsymbol{ a}\times\underline{\boldsymbol{ \nabla}} = - \frac{\partial a_{ij}}{\partial x_k} \varepsilon_{jkl} \underline{\boldsymbol{ e}}_i \otimes \underline{\boldsymbol{ e}}_l \end{aligned} curl ( a ) = − a × ∇ = − ∂ x k ∂ a ij ε jk l e i ⊗ e l which is the same as for vectors. An important property of the curl of a the gradient of a vector is
− [ u ‾ ⊗ ∇ ‾ ] × ∇ ‾ = 0 \begin{aligned} - \left[ \underline{\boldsymbol{ u}}\otimes\underline{\boldsymbol{ \nabla}}\right]\times\underline{\boldsymbol{ \nabla}} = \boldsymbol{ 0} \end{aligned} − [ u ⊗ ∇ ] × ∇ = 0 Actually, there are many different definitions of the curl for higher order tensors in the literature. For the curl of a second order tensor,
curl ( a ) \text{curl}(\boldsymbol{ a}) curl ( a ) , it can be written as
curl ( a ) = ε o p j ∂ a i p ∂ x o e ‾ i ⊗ e ‾ j \begin{aligned} \text{curl}(\boldsymbol{ a}) = \varepsilon_{opj} \frac{\partial a_{ip}}{\partial x_o} \underline{\boldsymbol{ e}}_{ i}\otimes\underline{\boldsymbol{ e}}_{ j} \end{aligned} curl ( a ) = ε o p j ∂ x o ∂ a i p e i ⊗ e j In the different variations, it could have the opposite sign,
a \boldsymbol{ a} a could be transposed, or the result could be transposed. In many use cases, the sign, and whether or not the result is transposed, is not critical. However, definitions that have
a \boldsymbol{ a} a transposed do not fulfill the important identity that
− [ u ‾ ⊗ ∇ ‾ ] × ∇ ‾ = 0 - \left[ \underline{\boldsymbol{ u}}\otimes\underline{\boldsymbol{ \nabla}}\right]\times\underline{\boldsymbol{ \nabla}} = \boldsymbol{ 0} − [ u ⊗ ∇ ] × ∇ = 0 and should be avoided!