mth3008 portfolio cheat sheet

MTH3008 Tensor Analysis, Portfolio Cheat Sheet

Created by William Fayers

Good luck and have fun!! :))

0. Reference Tables & Foundational Material

Suffix Notation Dictionary

Vector/Matrix form	Suffix form	Notes
Dot product $a\cdot b$	$a_j{b}_j$	Scalar (0 free indices)
Cross product $(a\times b{)}_i$	${ϵ}_{ijk}{a}_j{b}_k$	Vector (1 free index, $i$)
Matrix mult. $(AB{)}_{ij}$	$A_{ik}{B}_{kj}$	Rank-2 (2 free indices, $i,j$)
Transpose $(A^T{)}_{ij}$	$A_{ji}$	Swap indices
Trace $\text{Tr}(A)$	$A_{ii}$	Sum over diagonal
Gradient $[\nabla f{]}_i$	$\partial f/\partial x_i$	Scalar → Vector
Divergence $\nabla \cdot u$	$\partial u_j/\partial x_j$	Vector → Scalar
Curl $[\nabla \times u{]}_i$	${ϵ}_{ijk}\partial u_k/\partial x_j$	Vector → Vector
Laplacian ${\nabla }^{2}f$	${\partial }^{2}f/\partial x_j\partial x_j$	Scalar → Scalar (= $\nabla \cdot \nabla f$)
$[\nabla \times (\nabla f){]}_i=0$	${ϵ}_{ijk}{\partial }^{2}f/\partial x_j\partial x_k=0$	Scalar → Vector (always $0$, kill rule)
$[\nabla (\nabla \cdot u){]}_i$	${\partial }^2{u}_j/\partial x_i\partial x_j$	Vector → Vector
$\nabla \cdot (\nabla \times u)=0$	${ϵ}_{ijk}{\partial }^2{u}_k/\partial x_j\partial x_i=0$	Vector → Scalar (always $0$, kill rule)
$[\nabla \times (\nabla \times u){]}_i$	${ϵ}_{ijk}{ϵ}_{klm}{\partial }^2{u}_m/\partial x_j\partial x_l$	Vector → Vector

Identity: $\nabla \times (\nabla \times u)=\nabla (\nabla \cdot u)-{\nabla }^{2}u$, i.e. ${\nabla }^2{u}_i={\partial }^2{u}_i/\partial x_j\partial x_j$.

Key Rules & Identities

1. Index Counting: Dummy indices appear exactly twice per term (summed over 1 to 3). Free indices appear exactly once per term. Every term in an equation must have the exact same free indices.
2. Kronecker Delta ${\delta }_{ij}$:
   • Definition: $1$ if $i=j$, $0$ otherwise.
   • Substitution: ${\delta }_{ij}{a}_j=a_i$.
   • Collapse: ${\delta }_{ij}{\delta }_{jk}={\delta }_{ik}$.
   • Trace: ${\delta }_{ii}=3$.
   • Partial derivative of coordinates: $\frac{\partial x_j}{\partial x_i}={\delta }_{ij}$ (Cartesian coords are independent).
3. Alternating Tensor ${ϵ}_{ijk}$:
   • Definition: ${ϵ}_{ijk}=\{\begin{array}{ll}0& \text{if any of }i,j,k\text{ are equal}\\ +1& \text{if }(i,j,k)=(1,2,3),(2,3,1),\text{or }(3,1,2)\text{ - even permutation}\\ -1& \text{if }(i,j,k)=(1,3,2),(2,1,3),\text{or }(3,2,1)\text{ - odd permutation}\end{array}$.
   • Permutations: ${ϵ}_{ijk}={ϵ}_{jki}={ϵ}_{kij}$ (cyclic, keep sign). ${ϵ}_{ijk}=-{ϵ}_{jik}$ (swap two, flip sign).
4. The $\delta -ϵ$ Identity:
   • ${ϵ}_{ijk}{ϵ}_{klm}={\delta }_{il}{\delta }_{jm}-{\delta }_{im}{\delta }_{jl}$; (first/second X second/third, first/third X second/second)
   • Always cyclically permute so the shared dummy index is in the 3rd position before applying.
5. Symmetry: $S_{ij}=S_{ji}$ (symmetric), $A_{ij}=-A_{ji}$ (antisymmetric). If neither holds, it is neither. For example, ${ϵ}_{ijk}$ is always antisymmetric and mixed partials are always symmetric.
6. The Kill Rule: Symmetric $\times$ Antisymmetric $=0$.
   • If $S_{jk}=S_{kj}$, then ${ϵ}_{ijk}{S}_{jk}=0$.
   • Example: mixed partials commute (${\partial }_j{\partial }_k={\partial }_k{\partial }_j$), so ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f=0$.

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1. Transformation Law — Tensor Character & Rank

Goal: Prove $Q$ is a scalar, vector, or rank-$n$ tensor under coordinate rotation.

Transformation Rules:

• Coordinate transform: $x_i^{\prime }=L_{ij}{x}_j$. Inverse: $x_i=L_{ji}{x}_j^{\prime }$.
  • Rotation (Cartesian): $T_{i_1\cdots i_n}^{\prime }=L_{i_1{j}_1}\cdots L_{i_n{j}_n}{T}_{j_1\cdots j_n}$ (one $L$ per free index).
  • General coordinates: $T^{\prime i_1^{\prime }\cdots i_p^{\prime }}{}_{j_1^{\prime }\cdots j_q^{\prime }}=L^{i_1^{\prime }}{}_{k_1}\cdots L^{i_p^{\prime }}{}_{k_p}{L}^{l_1}{}_{j_1^{\prime }}\cdots L^{l_q}{}_{j_q^{\prime }}{T}^{k_1\cdots k_p}{}_{l_1\cdots l_q}$, where $L^{i^{\prime }}{}_j=\frac{\partial x^{\prime i^{\prime }}}{\partial x^j}$ (contravariant) and $L^j{}_{i^{\prime }}=\frac{\partial x^j}{\partial x^{\prime i^{\prime }}}$ (covariant).
• Rotation matrix properties: $L_{ij}{L}_{kj}={\delta }_{ik}$ and $L_{ji}{L}_{jk}={\delta }_{ik}$.
• Chain rule: $\frac{\partial }{\partial x_i^{\prime }}=L_{ji}\frac{\partial }{\partial x_j}$.

Method:

1. Write the primed quantity (e.g., $(a\cdot b{)}^{\prime }=a_i^{\prime }{b}_i^{\prime }$).
2. Substitute the transformation law for each piece (e.g., $a_i^{\prime }=L_{ij}{a}_j$).
3. Simplify if possible (e.g., with chain rule or partial differentiation like product rule).
4. Group the $L$ matrices together.
5. Collapse $L_{ij}{L}_{kj}$ into ${\delta }_{ik}$ if possible.
6. Use ${\delta }_{ik}$ to collapse dummy indices and recover the unprimed quantity.

Key Results:

• Contraction reduces rank: If $T_{ij}$ is a rank-2 tensor, $T_{ii}$ is a scalar. Proof: $T_{ii}^{\prime }=L_{ia}{L}_{ib}{T}_{ab}={\delta }_{ab}{T}_{ab}=T_{aa}$.
• Outcome: Single term with $n$ $L$s for $n$ free indices → rank-$n$ tensor. Extra terms that don’t cancel → not a tensor.

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2. Suffix Notation — Simplification & Proofs

2a. Simplifying $\delta$ and $ϵ$ Expressions

Goal: Reduce suffix expressions like ${ϵ}_{ijk}{ϵ}_{klm}$ or ${\delta }_{ij}{ϵ}_{ijk}$.

Method:

• Repeated $ϵ$ index: If an $ϵ$ has two identical indices (e.g. ${\delta }_{ij}{ϵ}_{ijk}={ϵ}_{iik}$), the term equals $0$.
• Product of two $ϵ$s:
  1. Identify the shared dummy index (e.g., $k$).
  2. Cyclically permute to move the shared index to the 3rd position of both $ϵ$s (Rule 3 in §0).
  3. Apply the $\delta$–$ϵ$ identity (Rule 4 in §0).
  4. Expand brackets and collapse $\delta$s (${\delta }_{ab}{c}_b=c_a$, ${\delta }_{aa}=3$).
• Standard deductions:
  • ${ϵ}_{ijk}{ϵ}_{ilm}={\delta }_{jl}{\delta }_{km}-{\delta }_{jm}{\delta }_{kl}$
  • ${ϵ}_{ijk}{ϵ}_{ijm}=2{\delta }_{km}$
  • ${ϵ}_{ijk}{ϵ}_{ijk}=6$

2b. Suffix Algebra Proofs (“Show that…“)

Goal: Prove vector/matrix identities (e.g., $(AB{)}^T=B^T{A}^T$, $a\times b=-b\times a$) or symmetry properties.

Method:

1. Convert LHS into suffix notation. Assign one free index (e.g., $i$) for vectors, two ($i,j$) for matrices.
2. Convert RHS into suffix notation.
3. Relabel dummy indices in LHS to match RHS (you can change any $j$ to $k$ as long as you change both copies in the term).
4. Reorder scalar variables freely to group them properly.
5. Convert back to vector/matrix notation.

Factoring scalars: In suffix form, any sub-expression that is already a scalar (no free indices, e.g. $a_k{c}_k=\mathbf{a}\cdot \mathbf{c}$) can be factored out of a term freely.

Example ($C_{ik}=A_i{B}_k-A_k{B}_i$ is antisymmetric):

• Evaluate transpose/swap: $C_{ki}=A_k{B}_i-A_i{B}_k$.
• Factor minus sign: $=-(A_i{B}_k-A_k{B}_i)=-C_{ik}$.

2c. $\nabla$ Operator Identities & Radial Functions

Goal: Prove vector calculus identities using suffix notation.

Method:

1. Write the expression fully in suffix form (see dictionary in §0).
2. Use the product rule on derivatives: ${\partial }_i(A_j{B}_k)=({\partial }_i{A}_j)B_k+A_j({\partial }_i{B}_k)$.
3. If you hit ${ϵ}_{ijk}{\partial }_j{\partial }_{k}f$, it immediately equals $0$ (kill rule, Rule 5 in §0).

Radial Functions $f(r)$:

• Definition: $r=(x_j{x}_j{)}^{1/2}$.
• Derivative of $r$: $\frac{\partial r}{\partial x_i}=\frac{x_i}{r}$.
• Derivative of $f(r)$: $[\nabla f(r){]}_i=f^{\prime }(r)\frac{x_i}{r}$.

2d. Determinant Identities

Goal: Prove matrix determinant rules using the alternating tensor.

Formula: ${ϵ}_{pqr}|M|={ϵ}_{ijk}{M}_{pi}{M}_{qj}{M}_{rk}$, where $p,q,r$ index rows and $i,j,k$ index columns of $M$.

• Row expansion (set $(p,q,r)=(1,2,3)$): $|M|={ϵ}_{ijk}{M}_{1i}{M}_{2j}{M}_{3k}$.
• Column expansion (set $(i,j,k)=(1,2,3)$): ${ϵ}_{pqr}|M|=M_{p1}{M}_{q2}{M}_{r3}$.

Method:

• To get $6|M|$: Multiply both sides by ${ϵ}_{pqr}$. The LHS becomes ${ϵ}_{pqr}{ϵ}_{pqr}|M|=6|M|$.
• To prove $|M^T|=|M|$: Write out the formula for $|M^T|$, swap the indices on the $M$s, then relabel the dummy indices to match the original formula.

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3. Bases, Components & Geometry

3a. Dual Bases & Components (Covariant/Contravariant)

Given: A basis $e_1,e_2,e_3$.

Find Dual Basis $e^1,e^2,e^3$:

1. Compute the three cross products: $e_2\times e_3$, $e_3\times e_1$, $e_1\times e_2$.
   • Cross product: $\left(\begin{array}{c}a\\ b\\ c\end{array}\right)\times \left(\begin{array}{c}d\\ e\\ f\end{array}\right)=\left(\begin{array}{c}\left|\begin{array}{cc}b& e\\ c& f\end{array}\right|\\ -\left|\begin{array}{cc}a& d\\ c& f\end{array}\right|\\ \left|\begin{array}{cc}a& d\\ b& e\end{array}\right|\end{array}\right)$ (cover 1st, 2nd but neg., 3rd).
2. Compute volume $V=e_1\cdot (e_2\times e_3)$.
3. Divide: $e^1=\frac{e_2\times e_3}{V}$, $e^2=\frac{e_3\times e_1}{V}$, $e^3=\frac{e_1\times e_2}{V}$.

Find Components:

• Covariant: $A_i=A\cdot e_i$ (dot with the original basis).
• Contravariant: $A^i=A\cdot e^i$ (dot with the dual basis).

Raising/Lowering via Metric:

• Covariant metric $g_{ij}=e_i\cdot e_j$. Contravariant metric $g^{ij}=e^i\cdot e^j$ (matrix inverse of $g_{ij}$).
• Raise/lower: $A_i=g_{ij}{A}^j$, $A^i=g^{ij}{A}_j$.
• Orthogonal shortcut: If $g_{ij}=0$ for $i\ne j$, then $A_i=g_{ii}{A}^i$ and $g^{ii}=1/g_{ii}$.

3b. Metric Tensor, Basis Vectors & Arc Length

Goal: Find metric components and scale factors from coordinate definitions.

Find Basis $e_i$:

1. Write position vector $r$ in terms of the new coordinates $(x^1,x^2,x^3)$ and Cartesian unit vectors $i_1,i_2,i_3$.
2. Differentiate: $e_1=\frac{\partial r}{\partial x^1}$, $e_2=\frac{\partial r}{\partial x^2}$, $e_3=\frac{\partial r}{\partial x^3}$.

Find Metric Tensor $g_{ij}$:

• Compute $g_{ij}=e_i\cdot e_j$ as a $3\times 3$ matrix. Off-diagonals all $0$ → orthogonal.

Arc Length & Scale Factors:

• Scale factors: $h_i=\sqrt{g_{ii}}$ (or $h_i=|e_i|$).
• Arc length element (orthogonal): $d{s}^2=h_1^2(d{x}^1{)}^2+h_2^2(d{x}^2{)}^2+h_3^2(d{x}^3{)}^2$.

Note: write both of these if asked to “describe the arc length element in terms of the metric coefficients”, or similar.

3c. Expansion Coefficients / Finding $L_{m^{\prime }}^n$

Goal: Find the coefficients to transform between bases.

Method:

• Given new basis vectors $e_m^{\prime }$ in terms of Cartesian $i_n$.
• The expansion coefficient is $L_{m^{\prime }}^n=e_m^{\prime }\cdot i_n$ (just extract the $n$-th component of $e_m^{\prime }$).
• Transform covariant components: $B_i^{\prime }=L_{i^{\prime }}^j{B}_j$.