Tensor analysis — scalars, vectors, rank-2 tensors: Exercises

Before we begin, a one-line refresher on the notation this page uses, so no symbol appears unearned.

Level 1 — Recognition

Exercise 1.1

You are handed four $2 \times 2$ grids of numbers. Which of them could be a rank-2 tensor and which cannot possibly be, based only on their definition?

(a) The stress at a point in a solid.
(b) A restaurant menu's price matrix (rows = dishes, columns = day of week).
(c) The Kronecker delta $δ_{j}^{i}$ .
(d) The metric $g_{ij}$ built from arc length $d s^{2} = g_{ij} d x^{i} d x^{j}$ .

Recall Solution

The test is not shape, it is behaviour. A grid is a tensor only if its entries carry a transformation law under a change of coordinates.

(a) Tensor. Stress is defined by how forces act across surfaces; rotate the axes and the components obey $\tilde{T} = R T R^{⊤}$ . ✔
(b) Not a tensor. "Dishes" and "days" are not coordinate axes; there is no rule telling the prices how to change when you rotate anything. It is just a table. ✘
(c) Tensor (isotropic). $\tilde{δ}_{j}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ^{k}}{\partial x ~ ^{j}} = δ_{j}^{i}$ — same in every frame. ✔
(d) Tensor. $d s^{2}$ is an invariant scalar and $d x^{i}$ is contravariant, so $g_{ij}$ is forced to be covariant rank 2 (quotient theorem). ✔

Exercise 1.2

Classify each object by rank and index type: (i) temperature $ϕ$ , (ii) velocity $V^{i}$ , (iii) a gradient $\partial_{i} ϕ$ , (iv) the inertia tensor $I^{ij}$ .

Recall Solution

(i) $ϕ$ — rank 0, a scalar. $n^{0} = 1$ component.
(ii) $V^{i}$ — rank 1, contravariant (upper index, transforms like $d x^{i}$ ).
(iii) $\partial_{i} ϕ$ — rank 1, covariant (lower index, transforms like a gradient with the inverse Jacobian).
(iv) $I^{ij}$ — rank 2, contravariant (two upper indices, $n^{2}$ components).

Level 2 — Application

Exercise 2.1

In flat 2D Cartesian coordinates a contravariant vector has components $V^{k} = (2, 3)$ . Rotate the axes by $θ = 9 0^{\circ}$ using

R = (cos θ sin θ - sin θ cos θ) .

Find the components $\tilde{V}^{i}$ in the new frame. See the figure.

Figure — Tensor analysis — scalars, vectors, rank-2 tensors

Recall Solution

WHAT: we apply the contravariant rule $\tilde{V}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} V^{k} = R^{i}_{k} V^{k}$ , i.e. matrix $R$ times column $V$ . WHY $R$ and not $R^{⊤}$ : the new coordinates $\tilde{x}$ are obtained from the old by $R$ , and a contravariant (upper) index takes the forward Jacobian $\partial \tilde{x} / \partial x$ , which is exactly $R$ . At $θ = 9 0^{\circ}$ : $cos 9 0^{\circ} = 0, sin 9 0^{\circ} = 1$ , so $R = (01 - 1 0)$ .

\tilde{V} = (01 - 1 0) (23) = (- 3 2) .

WHAT IT LOOKS LIKE: the arrow itself never moved (amber in the figure); we only spun the axes a quarter-turn, so its readings change from $(2, 3)$ to $(- 3, 2)$ . The length $2^{2} + 3^{2} = 13$ is unchanged — a rotation preserves length.

Exercise 2.2

A covariant vector has components $W_{k} = (2, 3)$ in the same Cartesian frame. Rotate by the same $θ = 9 0^{\circ}$ . Find $\tilde{W}_{i}$ and confirm it equals the contravariant answer only because the rotation is orthonormal.

Recall Solution

WHAT: covariant rule $\tilde{W}_{i} = \frac{\partial x ^{k}}{\partial x ~ ^{i}} W_{k} = (R^{- 1})^{k}_{i} W_{k}$ , which in matrix form is $\tilde{W} = (R^{- 1})^{⊤} W = R W$ for orthonormal $R$ , since $R^{- 1} = R^{⊤}$ so $(R^{- 1})^{⊤} = R$ .

\tilde{W} = (01 - 1 0) (23) = (- 3 2) .

WHY it matched: for a rotation the inverse Jacobian equals the transpose, so upper and lower vectors move identically. In a non-orthonormal (e.g. polar or skewed) frame they would differ — see Metric tensor and Riemannian geometry.

Level 3 — Analysis

Exercise 3.1

Take the symmetric stress tensor $T = (3113)$ (MPa). Show explicitly that its trace $T^{i}_{i}$ is invariant under the $4 5^{\circ}$ rotation $R = \frac{1}{2} (11 - 1 1)$ , by computing $\tilde{T} = R T R^{⊤}$ and taking its trace.

Recall Solution

WHAT: rotate both slots. $R^{⊤} = \frac{1}{2} (1 - 1 11)$ . First $T R^{⊤} = \frac{1}{2} (3113) (1 - 1 11) = \frac{1}{2} (2 - 2 44)$ . Then

\tilde{T} = R (T R^{⊤}) = \frac{1}{2} (11 - 1 1) (2 - 2 44) = \frac{1}{2} (4008) = (2004) .

Trace before: $3 + 3 = 6$ . Trace after: $2 + 4 = 6$ . ✔ Invariant. WHY: trace is a contraction $T^{i}_{i}$ ; contraction glues a forward and inverse Jacobian into $δ_{k}^{l}$ , so the transformation factors cancel and the result is a genuine scalar. WHAT IT LOOKS LIKE: the off-diagonal $1$ (shear) vanished — we landed on the principal axes (see figure). The rotation only re-labelled; the physics (trace $6$ , eigenvalues $2, 4$ ) is untouched.

Exercise 3.2

Split $T = (5425)$ into its symmetric part $S_{ij}$ and antisymmetric part $A_{ij}$ , and verify $T = S + A$ .

Recall Solution

Use $S_{ij} = \frac{1}{2} (T_{ij} + T_{j i})$ and $A_{ij} = \frac{1}{2} (T_{ij} - T_{j i})$ .

S = \frac{1}{2} [(5425) + (5245)] = (5335), A = \frac{1}{2} [(5425) - (5245)] = (01 - 1 0) .

Check: $S + A = (5425) = T$ . ✔ WHY it matters: symmetry is coordinate-independent — a symmetric tensor stays symmetric in every frame, so this split survives any rotation.

Level 4 — Synthesis

Exercise 4.1

Prove the quotient theorem in a simple case: suppose $A^{ij}$ is an unknown grid, but for every covariant vector $V_{j}$ the object $B^{i} \equiv A^{ij} V_{j}$ is a genuine contravariant vector. Show $A^{ij}$ must transform as a contravariant rank-2 tensor.

Recall Solution

WHAT we know: $B^{i}$ is contravariant, so $\tilde{B}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} B^{k}$ , and $V_{j}$ is covariant, so $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ . Write the same equation in both frames:

\tilde{A}^{ij} \tilde{V}_{j} = \tilde{B}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} B^{k} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} V_{m} .

Rewrite $V_{m}$ using the inverse of the covariant rule. Since $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ , the free index $j$ on the left already carries $\tilde{V}_{j}$ ; substitute $\tilde{V}_{j} = \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m}$ on the left:

\tilde{A}^{ij} \frac{\partial x ^{m}}{\partial x ~ ^{j}} V_{m} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} V_{m} .

WHY the next step works: this holds for every $V_{m}$ , so we may strip $V_{m}$ from both sides (equate coefficients):

\tilde{A}^{ij} \frac{\partial x ^{m}}{\partial x ~ ^{j}} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} A^{k m} .

Solve for $\tilde{A}^{ij}$ by hitting both sides with $\frac{\partial x ~ ^{n}}{\partial x ^{m}}$ and summing; since $\frac{\partial x ^{m}}{\partial x ~ ^{j}} \frac{\partial x ~ ^{n}}{\partial x ^{m}} = δ_{j}^{n}$ (chain rule), the left collapses to $\tilde{A}^{in}$ :

\tilde{A}^{in} = \frac{\partial x ~ ^{i}}{\partial x ^{k}} \frac{\partial x ~ ^{n}}{\partial x ^{m}} A^{k m} .

That is exactly the contravariant rank-2 law. So $A^{ij}$ was a tensor all along — we proved it without ever inspecting its entries. This is the litmus test referenced by the parent note.

Exercise 4.2

Compute the metric $g_{ij}$ in polar coordinates $(r, θ)$ starting from Cartesian $x = r cos θ, y = r sin θ$ , using $g_{ij} = \frac{\partial x ^{a}}{\partial u ^{i}} \frac{\partial x ^{a}}{\partial u ^{j}}$ (sum over Cartesian $a$ , with $u^{1} = r, u^{2} = θ$ ).

Recall Solution

WHY this formula: flat space has $d s^{2} = d x^{2} + d y^{2}$ ; expressing $d x, d y$ in terms of $d r, d θ$ and collecting gives $g_{ij}$ automatically. The Jacobian columns are the tangent vectors. Partial derivatives:

\frac{\partial x}{\partial r} = cos θ, \frac{\partial y}{\partial r} = sin θ, \frac{\partial x}{\partial θ} = - r sin θ, \frac{\partial y}{\partial θ} = r cos θ .

Now dot the columns:

g_{r r} = cos^{2} θ + sin^{2} θ = 1, g_{r θ} = cos θ (- r sin θ) + sin θ (r cos θ) = 0,

g_{θ θ} = (- r sin θ)^{2} + (r cos θ)^{2} = r^{2} .

g_{ij} = (10 0 r^{2}) \Rightarrow d s^{2} = d r^{2} + r^{2} d θ^{2} .

WHAT IT LOOKS LIKE: moving $d θ$ at radius $r$ sweeps an arc of length $r d θ$ , so the $θ$ -slot "costs" $r^{2}$ — that is the $r^{2}$ in $g_{θ θ}$ . See Covariant derivative and Christoffel symbols for what this curvature of coordinates forces next.

Level 5 — Mastery

Exercise 5.1

Show that the ordinary derivative $\partial_{k} V^{i}$ is not a tensor, by transforming it and exhibiting the offending extra term. State what object fixes it.

Recall Solution

WHAT: transform $V^{i}$ first, then differentiate. We have $\tilde{V}^{i} = \frac{\partial x ~ ^{i}}{\partial x ^{a}} V^{a}$ . Now differentiate with respect to $\tilde{x}^{k}$ , using the chain rule $\frac{\partial}{\partial x ~ ^{k}} = \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial}{\partial x ^{b}}$ (product rule on the two $x$ -dependent factors):

\frac{\partial V ~ ^{i}}{\partial x ~ ^{k}} = \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial x ~ ^{i}}{\partial x ^{a}} \frac{\partial V ^{a}}{\partial x ^{b}} + extra, non-tensorial \frac{\partial x ^{b}}{\partial x ~ ^{k}} \frac{\partial ^{2} x ~ ^{i}}{\partial x ^{b} \partial x ^{a}} V^{a} .

WHY it breaks: a true mixed rank-2 tensor would show only the first term (two Jacobians hitting $\partial_{b} V^{a}$ ). The second term carries a second derivative of the coordinate map. In curved or curvilinear coordinates that term is nonzero, so $\partial_{k} V^{i}$ does not transform as a tensor. The fix: the covariant derivative $\nabla_{k} V^{i} = \partial_{k} V^{i} + Γ^{i}_{k m} V^{m}$ adds the Christoffel symbol $Γ$ , whose own transformation supplies an equal-and-opposite extra term, cancelling the offender and restoring tensor character.

Exercise 5.2

The Einstein field equations $G_{μν} = \frac{8 π G}{c ^{4}} T_{μν}$ equate two rank-2 tensors. Explain, using only what you proved above, why both sides must be tensors of the same type for the equation to be physically meaningful, and what would go wrong otherwise.

Recall Solution

Argument (three earned facts):

From Ex 2.1–2.2, tensor components change frame-to-frame with Jacobian factors; only the transformation-consistent combination is coordinate-independent.
A law of physics must not depend on which coordinates a human scribbled — the same conclusion in every frame.
If $G_{μν}$ transformed one way and $T_{μν}$ another, an equation true in one frame would become false after a coordinate change (the two sides would pick up different Jacobian factors and no longer match). Therefore both sides are covariant rank-2 tensors: $G_{μν}$ (curvature of spacetime) and $T_{μν}$ (stress–energy). Their equality is frame-independent — see General Relativity — Einstein field equations and, for the source side, Stress and strain tensors (continuum mechanics). This is the ultimate payoff of the slogan "it's a tensor if it transforms like one."

Recall Self-test summary

One-line invariant each level tested ::: L1 shape ≠ tensor; L2 one index → one Jacobian; L3 trace/eigenvalues are similarity invariants; L4 quotient theorem needs "for all V"; L5 ordinary derivative fails, covariant derivative fixes it. Why is $T^{i}_{i}$ frame-independent? ::: Contraction pairs a forward and inverse Jacobian into $δ_{k}^{l}$ , cancelling the transformation factors — the result is a scalar. What extra object rescues $\partial_{k} V^{i}$ ? ::: The Christoffel symbol $Γ^{i}_{k m}$ inside the covariant derivative $\nabla_{k} V^{i}$ .