Prove: scalar product of 2 1-order tensors is 0-order tensor
\[ \begin{align} u^{\prime}_{i} v^{\prime}_{i} &= L_{ij} u_{j} L_{ik} v_{k} \\ &= L_{ij} L_{ik} u_{j} v_{k} \\ &= \delta_{jk} u_{j} v_{k} \\ &= u_{j} v_{j} \\ \end{align} \]
Suppose electrostatic potential \(\phi\), its component \(E_{i} = - \frac{\partial \phi}{\partial x_{i}}\).
Prove \(\vec{E}\)is 1-order tensor.
\[ \begin{align} E^{\prime}_{i} &= (-\frac{\partial \phi}{\partial x_{i}})^{\prime} \\ &= - \frac{\partial \phi_{\prime}}{\partial x^{\prime}_{i}} \\ &= - \frac{\partial \phi}{\partial x^{\prime}_{i}} \\ &= - \frac{\partial \phi}{\partial x_{j}} \frac{\partial x_{j}}{\partial x^{\prime}_{i}} \\ &= L_{ij} E_{j} \\ \end{align} \]
Definition
\[ \begin{equation} T^{\prime}_{ij} = L_{ik} L_{jl} T_{kl} \end{equation} \]
Given the definition, it’s natural to display 2-order tensor in matrix form.
Note: it’s similar to linear operator(???), but it requires both 2 subscript refer to the same coordinate system.
Definition
\[ \begin{equation} T_{ij} = u_{i} v_{j} \end{equation} \]
denoted as \(\vec{T} = \vec{u} \bigotimes \vec{v}\), then
\[ \begin{align} \vec{T} &= u_{i} \vec{e_{i}} \bigotimes u_{j} \vec{e_{j}} \\ &= u_{i} v_{j} \vec{e_{i}} \bigotimes \vec{e_{j}} \\ &= T_{ij} \vec{e_{i}} \bigotimes \vec{e_{j}} \\ \end{align} \]
Prove \(T_{ij}\) is 2-order tensor.
\[ \begin{align} T_{ij} &= u^{\prime}_{i} v^{\prime}_{j} \\ &= L_{ik} u_{k} L_{jl} v_{l} \\ &= L_{ik} L_{jl} u_{k} v_{l} \\ &= L_{ik} L_{jl} T_{kl} \\ \end{align} \]
Definition
\[ \begin{equation} T_{ij} = \frac{\partial v_{i}}{\partial x_{j}} \end{equation} \]
denoted as \(\vec{T} = \nabla \vec{V}\) (???)
Prove it’s a 2-order tensor
\[ \begin{align} T^{\prime}_{ij} &= \frac{\partial v^{\prime}_{i}}{\partial x^{\prime}_{j}} \\ &= \frac{\partial (L_{ik} v_{k})}{\partial x_{l}} \frac{\partial x_{l}}{\partial x^{\prime}_{j}} \\ &= L_{ik} \frac{\partial v_{k}}{\partial x_{l}} L_{jl}(???) \\ &= L_{ik} L_{jl} T_{kl} \\ \end{align} \]
Addition
\[ \begin{align} S_{ij} = b_{ij} + c_{ij} \end{align} \]
Contraction
\[ \begin{align} W_{k} \equiv d_{iik} \end{align} \]
means contracting \(i\) and \(j\). N-order tensor becomes N-2 order tensor. \(W_{k}\) is 1-order tensor.
Tensor products
\[ \begin{align} d_{ijk} = u_{i} b_{jk} \end{align} \]
Inner products
\[ \begin{align} d_{ilm} = b_{ij} c_{jlm} \end{align} \]
N-order and M-order tensors lead to N+M-2 order tensor. (???)
Division
No such tensor operation.
Gradients
\[ \begin{align} \nabla & \equiv \vec{e}_{i} \frac{\partial}{\partial x_{i}} \\ h_{kij} &= \nabla \vec{b} = \frac{\partial b_{ij}}{\partial x_{k}} \\ \end{align} \]
It’s similar to tensor product of \(\nabla, \vec{b}\).
Divergence
\[ \begin{align} \nabla \cdot \vec{b} = \vec{e}_{j} \frac{\partial b_{ij}}{\partial x_{i}} \end{align} \]
Take inner product of \(\nabla, \vec{b}\). (???)
Laplacian
\[ \begin{align} \nabla^{2} = \nabla \cdot \nabla = \frac{\partial^{2}}{\partial x_{i} \partial x_{i}} \end{align} \]
Taylor series
Let \(\vec{b}(\vec{x})\) is a smooth second-order tensor field, then the value of \(\vec{b}\) at position \(\vec{y} + \vec{r}\) can be obtained as
\[ \begin{align} b_{ij}(\vec{y} + \vec{r}) = b_{ij}(\vec{y}) + \big ( \frac{\partial b_{ij}}{\partial x_{k}} \big )_{y} r_{k} + \frac{1}{2!} \big ( \frac{\partial^{2} b_{ij}}{\partial x_{k} \partial x_{l}} \big )_{y} r_{k} r_{l} + \ldots \end{align} \]
Gauss’s theorem
Let \(\mathcal{A}\) be a piecewise smooth closed orientable surface that encloses a volume \(\mathcal{V}\), and let \(\vec{n}\) denote the outward-pointing unit normal on \(\mathcal{A}\). Then
\[ \begin{align} \iiint_{\mathcal{V}} \frac{\partial b_{ij}}{\partial x_{k}} dV = \iint_{\mathcal{A}} b_{ij} n_{k} dA \end{align} \]
Vector cross product
The alternating symbol defined by
\[ \begin{align} \varepsilon_{ijk} &= 1, \quad (i,j,k) \quad cyclic; \\ &= -1, \quad (i,j,k) \quad anticyclic; \\ &= 0, \quad otherwise \\ \end{align} \]
Cyclice orderings are 123, 231, and 312; anticyclic orderings are 321, 132, and 213, otherwise two or more of the suffixes are the same.
\[ \begin{align} \vec{r} &= \vec{u} \times \vec{v} = det \begin{vmatrix} \vec{e}_{1} & \vec{e}_{2} & \vec{e}_{3} \\ u_{1} & u_{2} & u_{3} \\ v_{1} & v_{2} & v_{3} \\ \end{vmatrix} \\ &= \varepsilon_{ijk} \vec{e}_{i} u_{j} v_{k} \\ \end{align} \]
Cross product and curl are not 1-order tensor, they are pseudovectors, And the direction of them is determined by the right handed rule.
2 main kinds
\(\int_{C} \phi d \vec{r}\)
Since Cartesian unit vectors are of constant magnitude and direction, hence
\[ \begin{align} \int \phi \vec{i} dx = \vec{i} \int \phi dx \end{align} \]
Then
\[ \begin{align} \int_{C} \phi d \vec{r} = \vec{i} \int_{C} \phi(x,y,z) dx + \vec{j} \int_{C} \phi(x,y,z) dy + \vec{k} \int_{C} \phi(x,y,z) dz \end{align} \]
\(\int_{C} \vec{a} \cdot d \vec{r}\)
\[ \begin{align} \int_{C} \vec{a} \cdot d \vec{r} &= \int_{C} \Big ( a_{x} \vec{i} + a_{y} \vec{j} + a_{z} \vec{k} \Big ) \cdot \Big ( dx \vec{i} + dy \vec{j} + dz \vec{k} \Big ) \\ &= \int_{C} \Big ( a_{x} dx + a_{y} dy + a_{z} dz \Big ) \\ \end{align} \]
Helmholtz’s theorem, also known as the fundamental theorem of vector calculus, states that vector field in 3D can be resolved into the sum of the irrotational(curl-free) vector field and a solenoidal(divergence-free) vector field. That’s known as the Helmholtz decomposition.
\[ \begin{align} \vec{F} = \nabla \phi + \nabla \times \vec{psi} \end{align} \]
\(\nabla \phi\) is irrotational part since \(\nabla \times \nabla \phi = 0\). \(\nabla \times \vec{psi}\) is solenoidal part since \(\nabla \cdot (\nabla \times \vec{\psi}) = 0\).
Footnotes
[fn:1] Riley K F, Hobson M P, Bence S J. Mathematical methods for physics and engineering: a comprehensive guide[M]. Cambridge University Press, 2006.