Variational calculus, tensors, groups, and differential geometry

Four questions about mathematical models

Comparing functions, changing a basis, composing transformations and measuring geometry require different mathematical tools. We will introduce each tool through a supplied exact model and then connect them. The models contain no measurements or inferred laws of motion.

A function assigns an output to each allowed input. Its derivative measures local change, while an integral accumulates values over a domain. A vector has components once a basis is chosen. A matrix acts on a column by multiplying each row by that column and adding the products. These are the local prerequisites used below.

Variational calculus compares whole admissible functions. A tensor can encode a multilinear relationship independently of the basis used to describe it. A group organizes transformations that compose and can be undone. Differential geometry studies smooth local coordinates and geometric quantities such as lengths and curvature. This lesson proves three small cases; general variational problems, arbitrary tensors, continuous groups and general curvature calculations require further study.

Choose a function while preserving the constraints

Let t be a time coordinate on 0 ≤ t ≤ τ, with τ = 2 s. An admissible q is a real position function with a continuous first derivative on this interval, written q ∈ C¹([0, τ]), and fixed endpoints q(0) = 0 m and q(τ) = L = 4 m. Endpoint derivatives are understood one-sided. There is no restriction on intermediate positions or the sign of q'.

Supply J[q] = ∫₀^τ (q'(t))² dt. Brackets emphasize that the input is an entire function. Such a scalar-valued rule on functions is a functional. Here q' has units m/s, so J has units m²/s. This stipulated penalty is the quantity to minimize; no energy, mechanical action or observed trajectory has been supplied.

To vary q, choose any η ∈ C¹([0, τ]) with units m and η(0) = η(τ) = 0 m. For a dimensionless real parameter ε, qε = q + εη remains admissible. We are changing a candidate function while keeping its domain and endpoint values fixed.

The first variation in this direction is δJ[q; η] = (d/dε)J[q + εη] at ε = 0. Expanding the square gives J[q + εη] = J[q] + 2ε∫₀^τ q'η' dt + ε²∫₀^τ (η')² dt, so δJ[q; η] = 2∫₀^τ q'η' dt. All three coefficients have units m²/s. The formula is exact for this functional and these continuous derivatives.

A candidate is stationary when this first variation vanishes for every allowed η. A differentiable local minimum must be stationary when the admissible variations are two-sided as they are here: each function ε ↦ J[q + εη] then has a minimum at ε = 0. Stationarity by itself does not establish a minimum. Other constraints can change which variations are allowed.

Prove the minimum by an exact decomposition

Consider q*(t) = (L/τ)t = (2 m/s)t. For every allowed η, its first variation is 2(L/τ)∫₀^τ η' dt = 2(L/τ)[η(τ) − η(0)] = 0 m²/s. Thus q* is stationary.

Every admissible q can be written q = q* + η with the required zero-endpoint η. Expanding again gives J[q] = L²/τ + 2(L/τ)[η(τ) − η(0)] + ∫₀^τ (η')² dt = 8 m²/s + ∫₀^τ (η')² dt. The final integral is nonnegative, proving a global minimum over the full stated C¹ class.

Equality requires the continuous nonnegative function (η')² to have zero integral. If it were positive anywhere, continuity would give a positive contribution on a nearby interval. Hence η' is zero everywhere, η is constant, and its zero endpoints force η = 0. The minimum is unique.

Change the objective to −J without changing admissibility. The same q* is stationary, but the proved inequality becomes −J[q] ≤ −8 m²/s: q* is now a global maximum. The choice of objective and the additional sign argument determine the conclusion; the word 'stationary' does not.

A bilinear tensor is more than its component array

In a real two-dimensional Euclidean plane, choose perpendicular unit direction vectors e₁ and e₂. Write displacement vectors v and w with scalar components vₓ, vᵧ, wₓ and wᵧ in metres. Supply B(v, w) = 2vₓwₓ + vᵧwᵧ. Its output has units m² and its coefficients are dimensionless. This is a defined weighted bilinear form, with no assigned material or energy interpretation.

Bilinear means linear in either input when the other is fixed: for dimensionless real a and b, B(av + bw, z) = aB(v, z) + bB(w, z), and the corresponding identity holds in the second input. A scalar-valued bilinear form is a covariant tensor of order two, also called type (0, 2). The number two counts its vector inputs; matrix rank is the maximum number of linearly independent columns.

With component columns, B(v, w) = v_oldᵀ C_old w_old, where C_old = [[2, 0], [0, 1]]. A superscript ᵀ means transpose: rows and columns are exchanged. Now use e'₁ = e₂ and e'₂ = −e₁. Put these new basis directions in the columns of S = [[0, −1], [1, 0]], expressed in the old basis.

The same vector has v_old = S v_new, hence v_new = S⁻¹v_old. Substitution into B gives C_new = Sᵀ C_old S and B(v, w) = v_newᵀ C_new w_new. This transformation rule applies to this bilinear form under any invertible basis matrix with the declared convention. Only for an orthogonal S, as here, does S⁻¹ = Sᵀ.

Changing the basis changes both vector components and form components, preserving B on the same inputs. An array without a specified object, inputs and transformation rule is insufficient to identify a tensor. The Euclidean squared norm uses the dot product; our weighted B is a different form. A coordinate change does not permit replacing either form's matrix arbitrarily.

Compose four reversible rotations

Keep the original perpendicular axes and origin fixed on the whole plane. Define the active rotation R(x, y) = (−y, x), where x and y have units m. This changes a vector or point. It differs from describing the same vector in the changed basis. Repetition gives R²(x, y) = (−x, −y), R³(x, y) = (y, −x) and R⁴ = E, where E is the identity map.

Take G = {E, R, R², R³} with composition f ∘ g, meaning apply g first and then f. A group requires closure under its operation, associativity, an identity and an inverse for every element. Commutativity is an additional property, not a group requirement.

For i and j in {0, 1, 2, 3}, Rⁱ ∘ Rʲ = Rᵏ with k the remainder of i + j after division by 4 and R⁰ = E. This proves closure. Composition is associative because both (f ∘ g) ∘ h and f ∘ (g ∘ h) send an input p to f(g(h(p))). E acts as identity on either side. The inverses of E, R, R² and R³ are E, R³, R² and R respectively.

Every element is a power of R, so this is the cyclic group C₄. It happens to be commutative because i + j = j + i. The table lists all sixteen ordered products. These four rotations are a finite group; they do not include every possible rotation angle.

R preserves Euclidean length: (−y)² + x² = x² + y², and the same calculation on differences preserves distances. It need not preserve another supplied form. For v = (3, 4) m, B(v, v) = 34 m², whereas B(Rv, Rv) = B((−4, 3), (−4, 3)) = 41 m². A group of transformations is a symmetry of a chosen rule only when that rule is invariant under the transformations. No conservation law follows from merely listing this group.

Coordinates can vary from place to place

A smooth local coordinate chart labels a region with coordinates and has a smooth inverse there. For our plane, restrict to x > 0 and y > 0. Use r > 0 in metres and 0 < θ < π/2, with θ the dimensionless numerical angle in radians. The formulas x = r cos θ and y = r sin θ invert to r = √(x² + y²) and θ = arctan(y/x). This chart excludes the origin and the coordinate axes and avoids angle ambiguity.

The parameterization p(r, θ) = (r cos θ, r sin θ) has coordinate directions ∂p/∂r = (cos θ, sin θ) and ∂p/∂θ = (−r sin θ, r cos θ). The first is dimensionless and the second has units m. They depend on the location and are independent for r > 0. They are perpendicular here, but the angular coordinate direction has norm r and is not a unit direction.

A tangent vector at one point describes a direction and magnitude in the linear tangent plane at that point. If its coordinate coefficients are δr in metres and dimensionless δθ, its Cartesian components are δp = (∂p/∂r)δr + (∂p/∂θ)δθ. Thus δx = cos θ δr − r sin θ δθ and δy = sin θ δr + r cos θ δθ.

Expanding the Euclidean squared norm cancels the cross terms and gives |δp|² = (δr)² + r²(δθ)². The Riemannian metric used here is a smoothly varying symmetric bilinear form on tangent vectors, assigning positive squared length to every nonzero tangent vector. Symmetric means exchanging its two inputs leaves the value unchanged. Its matrix is G_polar = [[1, 0], [0, r²]], describing the same Euclidean geometry in different coordinates.

The radial coefficient is dimensionless, while the angular coefficient r² has units m² because the coordinate units differ. Both contributions to squared length have units m². The expression ds² = dr² + r²dθ² is the associated line element, evaluated locally along a differentiable curve; it is not an exact finite chord formula. Our C₄ transformations act on the whole plane, while this particular chart labels only the positive quadrant.

Specify which curvature you mean

A circle drawn in a flat plane bends as a curve. For a twice continuously differentiable regular curve, regular means its derivative never vanishes. With arclength s in metres and dimensionless unit tangent T, its curve curvature is κ = |dT/ds|, with units m⁻¹.

On an arc of a circle of radius r₀ = 2 m, use p(θ) = r₀(cos θ, sin θ). Increasing θ gives T = (−sin θ, cos θ), and s = r₀θ up to an additive constant. Hence dT/ds = (−cos θ, −sin θ)/r₀ and κ = 1/r₀ = 0.5 m⁻¹. These are geometric derivatives with respect to angle and arclength, not a supplied time evolution.

Gaussian curvature K_G is a different quantity: intrinsic surface curvature, with units m⁻² when lengths are measured in metres. It is determined by the surface metric, independently of how a surface is embedded in a surrounding space. The Euclidean plane has K_G = 0 m⁻². Expressing it in polar coordinates does not change that value. The circle's nonzero curve curvature therefore does not make its containing plane intrinsically curved.

The varying metric component r² records how much length one angular coordinate change represents at a location. Its variation alone is not a test for intrinsic curvature. Conversely, a tangent vector's norm is an exact local metric calculation even when its coordinate coefficients are written as small finite numbers: those coefficients specify a vector at one base point, not the difference between two finite endpoints.

Keep the object, change and conditions explicit

The four tools connect without becoming interchangeable. Variations compare admissible functions; a passive basis change describes the same tensor and vectors; an active group transformation changes inputs; a coordinate chart describes a region and its tangent directions. A metric is itself a bilinear tensor, now assigned at each point. Its components change with the coordinates while geometric lengths remain the same.

Before using any of these tools, state the domain, admissible inputs, quantity kinds, units and the operation being performed. In the function problem, C¹ regularity and zero-endpoint variations permit the exact proof. In the tensor problem, an invertible basis change is essential. In the chart, r > 0 and the restricted angle range supply a smooth inverse. Curve curvature requires the stated regularity.

A physical conclusion needs a physical model in addition to the mathematics. A functional's extremum does not on its own identify an actual motion; length-preserving rotations do not automatically preserve B; coordinate-dependent coefficients do not on their own diagnose curvature. The next chapter, SI units and modern metrology, connects declared quantities and units to physical measurement.

Worked examples

Try it yourself

1. For the displayed q_a family, set a = −1. Check the endpoints, find the position at t = 1 s, and calculate J[q_a] with units.

   Show the answer and reasoning

   The endpoints are 0 m and 4 m. With a = −1, q_a(1 s) = 1 m and J[q_a] = 32/3 m²/s.

   At u = 1/2, L[u − u(1 − u)] = 4 m(1/2 − 1/4) = 1 m. The factor u(1 − u) still vanishes at both endpoints. Squaring a in J[q_a] gives 8(1 + 1/3) m²/s.

2. On the same full admissible C¹ class, is q*(t) = (2 m/s)t a minimum, maximum or neither for the functional −J? Does its first variation vanish?

   Show the answer and reasoning

   q* is the unique global maximum of −J, with value −8 m²/s, and its first variation vanishes for every admissible direction.

   Negating the exact decomposition gives −J[q] = −8 m²/s − ∫₀^τ (η')² dt. The second term cannot increase the value, and equality forces η = 0. The first variation is the negative of δJ, hence still zero at q*.

3. Take w_old = (1, 2) m with the same S and B. Find w_new and B(w, w) in the new basis, then find B(Rw, Rw) in the old basis. Also compare the Euclidean squared norms.

   Show the answer and reasoning

   w_new = (2, −1) m and B(w, w) = 6 m²; Rw = (−2, 1) m gives B(Rw, Rw) = 9 m². Both Euclidean squared norms are 5 m².

   The passive calculation uses C_new = [[1, 0], [0, 2]], yielding 2² + 2(−1)² = 6 in m². The active calculation keeps C_old and yields 2(−2)² + 1² = 9 in m². The Euclidean form instead adds the two component squares.

4. In C₄, find R³ ∘ R² and the inverse of R³. Apply the composite to (1, 2) m, with the rightmost transformation acting first.

   Show the answer and reasoning

   R³ ∘ R² = R, the inverse of R³ is R, and the composite sends (1, 2) m to (−2, 1) m.

   The exponents add to 5, whose remainder after division by 4 is 1. R³ ∘ R = R⁴ = E on either side. Directly, R² sends (1, 2) to (−1, −2), and R³ then sends that pair to (−2, 1).

5. At r = 3 m and θ = π/4, a tangent vector has δr = 0 m and δθ = 1/10. Give the angular metric coefficient with its unit and calculate the vector's norm.

   Show the answer and reasoning

   The angular metric coefficient is r² = 9 m², and the tangent vector's norm is 3/10 m = 0.3 m.

   The squared norm is 0 m² + (9 m²)(1/10)² = 9/100 m². Taking the positive square root gives 3/10 m. Treating dimensionless δθ as a length would discard the required radial factor.

6. A radius-2 m circular arc lies in the Euclidean plane. Give its curve curvature and the plane's Gaussian curvature with units. Does the varying polar component r² make the plane intrinsically curved?

   Show the answer and reasoning

   The arc has κ = 0.5 m⁻¹; the plane has K_G = 0 m⁻². The varying component r² does not make this plane intrinsically curved.

   The circle's unit tangent changes per arclength, producing 1/r₀. Gaussian curvature concerns the intrinsic surface metric, not the turning of an individual curve. Polar coordinates change its components while preserving the flat Euclidean geometry.

What to carry forward

• A variational problem specifies admissible functions, constraints and a functional before asking for an extremum.
• A zero first variation supplies stationarity; the nonnegative remainder proves the particular unique minimum here.
• A bilinear tensor and its component matrix are distinct: vector and tensor components transform together under a passive basis change.
• C₄ satisfies all four group requirements under composition, but a chosen physical or mathematical rule needs its own invariance check.
• A coordinate-dependent metric converts tangent coefficients into geometric lengths with consistent units.
• Curve curvature and Gaussian surface curvature are different quantities; a curved circle can lie in an intrinsically flat plane.