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.
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 ‾ × ∇ ‾ \begin{aligned} \text{curl}(\underline{\boldsymbol{ v}}) &= - \underline{\boldsymbol{ v}}\times\underline{\boldsymbol{ \nabla}} \end{aligned} curl ( v ) = − v × ∇ 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 × ∇ ‾ \begin{aligned} \text{curl}(\boldsymbol{ a}) &= - \boldsymbol{ a}\times\underline{\boldsymbol{ \nabla}} \end{aligned} curl ( a ) = − a × ∇ 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!