Dual Space Dynamics Worksheet Dual Space Dynamics
Advanced Vector Analysis · Lesson 1
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Conceptual Anchor
"A vector is an arrow; a co-vector is a stack of surfaces." In this problem set, we move beyond the Euclidean comfort of identifying vectors with their duals. We will rigorously define the dual space \(V^*\) and explore how linear functionals act upon the elements of \(V\).
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The Riesz Representation Trap
In finite-dimensional Hilbert spaces, we often use the Riesz Representation Theorem to claim \(V \cong V^*\). However, this isomorphism is not canonical without a fixed inner product.
Let \(V\) be the space of polynomials of degree at most 2. Define a functional \(\phi \in V^*\) such that for any \(p(x) \in V\): \[ \phi(p) = \int_{0}^{1} p(x) dx \]
If we choose the basis \(\mathcal{B} = \{1, x, x^2\}\), find the components of \(\phi\) in the dual basis \(\mathcal{B}^*\).
2
Gradients vs. Velocities
Consider a temperature field \(T(x, y)\) on a metal plate. At a point \(P\), the velocity of a particle is a vector \(\mathbf{v} \in V\), while the temperature gradient \(\nabla T\) is naturally a co-vector \(\omega \in V^*\).
Explain why the "rate of change of temperature experienced by the particle" is properly described by the pairing \(\langle \omega, \mathbf{v} \rangle\) rather than a dot product \(\mathbf{w} \cdot \mathbf{v}\) in a coordinate-independent way.
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Dual Basis Construction
Let \(\{\mathbf{e}_1, \mathbf{e}_2\}\) be a basis for \(\mathbb{R}^2\) such that \(\mathbf{e}_1 = (2, 1)\) and \(\mathbf{e}_2 = (1, 1)\).
Construct the dual basis \(\{\epsilon^1, \epsilon^2\}\) explicitly. Verify the property \(\epsilon^i(\mathbf{e}_j) = \delta^i_j\) and show your work using matrix inversion.
"The dual space is the space of all ways to measure the original space."
Dual Space Slides Defining Dual Spaces
Linear Functionals & The Geometry of Measurement
Graduate Vector Analysis Lesson 01
The Fundamental Question
"Why is the gradient of a function technically not a vector in the same sense as a velocity vector?"
Velocity
Displacement over time. An intrinsic 'arrow' in the space \(V\).
Gradient
A density of surfaces. An operator acting on vectors to produce a scalar.
The Dual Space \(V^*\)
Definition:
The dual space \(V^*\) of a vector space \(V\) over a field \(\mathbb{F}\) is the set of all linear functionals: \[ \phi : V \to \mathbb{F} \]
Linearity: \(\phi(a\mathbf{u} + b\mathbf{v}) = a\phi(\mathbf{u}) + b\phi(\mathbf{v})\)Vector Space: \(V^*\) is itself a vector space with pointwise addition and scalar multiplication.Dimension: For finite-dimensional \(V\), \(\dim(V^*) = \dim(V)\).
Constructing the Dual Basis
Given a basis \(\{\mathbf{e}_1, \dots, \mathbf{e}_n\}\) for \(V\), there exists a unique dual basis \(\{\epsilon^1, \dots, \epsilon^n\}\) for \(V^*\) defined by:
\(\epsilon^i(\mathbf{e}_j) = \delta^i_j\)
Vector Representation:
\[ \mathbf{v} = v^i \mathbf{e}_i \]
Indices of components are contravariant (upper).
Functional Representation:
\[ \phi = \phi_i \epsilon^i \]
Indices of components are covariant (lower).
Visualizing the Dual
If a vector is an arrow , a co-vector is a series of parallel surfaces .
The value \(\phi(\mathbf{v})\) is the number of surfaces pierced by the arrow.
Higher "magnitude" co-vector = denser surfaces.
Dual Surface Stack Vector v
Summary
Vectors exist in \(V\); Dual vectors exist in \(V^*\).
Indices: Contravariant \(v^i\) vs Covariant \(\phi_i\).
"Next: How do these components change when we rotate, stretch, or skew our perspective?"
Dual Space Answer Key Answer Key Dual Space
Teacher Resource · Lesson 1
1. The Riesz Representation Trap
Goal: Find components of \(\phi(p) = \int_0^1 p(x) dx\) in basis \(\mathcal{B}^* = \{\epsilon^0, \epsilon^1, \epsilon^2\}\) corresponding to \(\mathcal{B} = \{1, x, x^2\}\).
The components \(\phi_i\) are given by the action of \(\phi\) on the basis vectors \(\mathbf{e}_i\):
\(\phi_0 = \phi(1) = \int_0^1 1 dx = [x]_0^1 = 1\)
\(\phi_1 = \phi(x) = \int_0^1 x dx = [\frac{1}{2}x^2]_0^1 = \frac{1}{2}\)
\(\phi_2 = \phi(x^2) = \int_0^1 x^2 dx = [\frac{1}{3}x^3]_0^1 = \frac{1}{3}\)
Thus, \(\phi = \epsilon^0 + \frac{1}{2}\epsilon^1 + \frac{1}{3}\epsilon^2\). Note that these are the covariant components of the functional.
2. Gradients vs. Velocities
Explanation: The pairing \(\langle \omega, \mathbf{v} \rangle\) is the natural application of a linear functional (the gradient) to an element of the vector space (the velocity). This pairing yields a scalar (rate of temperature change) without requiring an auxiliary structure like a metric tensor.
In contrast, the "dot product" \(\mathbf{w} \cdot \mathbf{v}\) implicitly assumes the existence of a metric to map the co-vector \(\omega\) to a vector \(\mathbf{w}\). In non-Euclidean or general curvilinear coordinates, \(\nabla T\) components transform differently than velocity components, making the functional pairing the only coordinate-invariant definition.
3. Dual Basis Construction
Step 1: Arrange basis vectors \(\mathbf{e}_1, \mathbf{e}_2\) as columns of a matrix \(E\):
\(E = \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix}\)
Step 2: The dual basis vectors (as rows of a matrix) are the rows of the inverse matrix \(E^{-1}\). Let \(\det(E) = (2)(1) - (1)(1) = 1\).
\(E^{-1} = \frac{1}{1} \begin{pmatrix} 1 & -1 \\ -1 & 2 \end{pmatrix}\)
Conclusion:
\(\epsilon^1 = [1, -1]\) → Functional acting as \(\epsilon^1(x, y) = x - y\)
\(\epsilon^2 = [-1, 2]\) → Functional acting as \(\epsilon^2(x, y) = -x + 2y\)
Verification: \(\epsilon^1(\mathbf{e}_1) = 1(2) - 1(1) = 1\) (Correct). \(\epsilon^1(\mathbf{e}_2) = 1(1) - 1(1) = 0\) (Correct).
Transformation Laws Worksheet Transformation Laws
Graduate Vector Analysis · Lesson 2
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The Einstein Convention & Index Notation
Contravariant Transformation (Vectors):
\(\bar{v}^i = \frac{\partial \bar{x}^i}{\partial x^j} v^j\)
Transforms inversely to the basis.
Covariant Transformation (Co-vectors):
\(\bar{\omega}_i = \frac{\partial x^j}{\partial \bar{x}^i} \omega_j\)
Transforms with the basis.
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Scaling and Perspective
Consider a transformation where we double the length of our basis vectors: \(\bar{\mathbf{e}}_1 = 2\mathbf{e}_1\) and \(\bar{\mathbf{e}}_2 = 2\mathbf{e}_2\).
(a) Determine the transformation matrix \(L^i_j\) and its inverse \(W^j_i\).
(b) If a velocity vector has components \(v^i = (10, 0)\) in the old basis, what are its components \(\bar{v}^i\) in the new basis? Show your work using the contravariant law.
2
Gradient Covariance
Let \(f(x, y) = 3x + y\). The components of the gradient are \(\omega_i = (\partial_x f, \partial_y f) = (3, 1)\).
Using the same scaling transformation from Problem 1 (where the unit of measurement doubled), calculate the new components \(\bar{\omega}_i\). How does this result compare to the velocity transformation in 1(b)? Explain the physical significance.
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Coordinate Mapping
Transform from Cartesian coordinates \((x, y)\) to Polar coordinates \((r, \theta)\), where \(x = r \cos \theta\) and \(y = r \sin \theta\).
Write out the full Jacobian matrix \(\frac{\partial x^j}{\partial \bar{x}^i}\) required to transform a covariant vector field from Cartesian to Polar coordinates.
Transformation Slides Transformations & Notation
Co- vs. Contra-variance and the Einstein Chain
Graduate Vector Analysis Lesson 02
Shrinking the Ruler
If we shrink our basis units by half (a "smaller" ruler), what happens to numerical values?
Velocity Vector:
The numerical value doubles. It transforms inversely.
Temperature Gradient:
The numerical value halves. It transforms with the basis.
Contravariant
Components change contrary to basis change.
Covariant
Components change co-operatively with basis change.
The Einstein Convention
"Any index appearing twice in a term (once upper, once lower) implies a summation over all values of that index."
\(v^i \mathbf{e}_i = \sum_{i=1}^n v^i \mathbf{e}_i\)
Free Indices:
Appear once. Present on both sides of an equation.
Dummy Indices:
Appear twice. Summed over. Can be renamed at will.
Change of Coordinates
Let \(\bar{x}^i = \bar{x}^i(x^1, \dots, x^n)\). The Jacobians are the "exchange rates" between systems.
Contravariant Law
\(\bar{v}^i = \frac{\partial \bar{x}^i}{\partial x^j} v^j\)
Example: Velocity, Position
Covariant Law
\(\bar{\omega}_i = \frac{\partial x^j}{\partial \bar{x}^i} \omega_j\)
Example: Gradients, Normals
The Power of Pairing
When a covariant object and a contravariant object are paired, the transformation Jacobians cancel out!
\(\bar{\omega}_i \bar{v}^i = \omega_j v^j = \text{Scalar (Invariant)}\)
This is why physical laws (scalars) look the same to every observer, even if their "rulers" are different.
Transformation Answer Key Answer Key Transformations
Teacher Resource · Lesson 2
1. Scaling and Perspective
(a) Jacobians: Since \(\bar{\mathbf{e}}_i = 2\mathbf{e}_i\), the transformation is \(\bar{x}^i = \frac{1}{2}x^i\) (coordinate values must be halved to represent the same physical distance with a larger unit).
(b) Velocity Transformation: \(v^i = (10, 0)\).
2. Gradient Covariance
Calculation: Using \(\bar{\omega}_i = W^j_i \omega_j\) with \(W^j_i = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}\) and \(\omega_j = (3, 1)\).
Significance: When the ruler (unit) doubled in size, the velocity components decreased (contravariant), but the gradient components increased (covariant). This is because a gradient represents "change per unit". If the unit is larger, there is more change contained within it. This demonstrates why these objects must be treated differently.
3. Coordinate Mapping (Cartesian to Polar)
Given \(x = r \cos \theta\) and \(y = r \sin \theta\). The Jacobian for covariant transformation is \(W^j_i = \frac{\partial x^j}{\partial \bar{x}^i}\) where \(\bar{x}^1 = r, \bar{x}^2 = \theta\).
\(\begin{pmatrix} \partial_r x & \partial_r y \\ \partial_\theta x & \partial_\theta y \end{pmatrix} = \begin{pmatrix} \cos \theta & \sin \theta \\ -r \sin \theta & r \cos \theta \end{pmatrix}\)
Note: This matrix allows us to convert Cartesian gradient components \(\omega_x, \omega_y\) into polar components \(\omega_r, \omega_\theta\).
Metric Tensor Worksheet Metric Tensor Dynamics
Graduate Vector Analysis · Lesson 3
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The Bridge Between V and V*
Without a metric tensor \(g_{\mu\nu}\), vectors and co-vectors are separate entities. The metric defines geometry: length, angle, and the ability to map one space to the other.
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The Non-Orthogonal Metric
Consider a basis \(\{\mathbf{e}_1, \mathbf{e}_2\}\) where \(|\mathbf{e}_1|=1, |\mathbf{e}_2|=2\), and the angle between them is \(60^\circ\).
Calculate the components of the metric tensor \(g_{ij} = \mathbf{e}_i \cdot \mathbf{e}_j\).
Find the inverse metric \(g^{ij}\). What physical role does this inverse play?
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Raising and Lowering
Let a vector \(\mathbf{v}\) have contravariant components \(v^i = (2, -1)\).
Using the metric tensor from Problem 1, lower the index to find the covariant components \(v_i\).
Calculate the magnitude of the vector squared \(|\mathbf{v}|^2 = g_{ij} v^i v^j\). Verify that this is equal to \(v_i v^i\).
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Polar Metric
In polar coordinates \((r, \theta)\), the line element is \(ds^2 = dr^2 + r^2 d\theta^2\).
Identify the components of the metric \(g_{ij}\) in this system. Explain why the metric is no longer a constant across the manifold.
Metric Tensor Slides The Metric Tensor
The Geometry of Measurement
Graduate Vector Analysis Lesson 03
The Missing Link
"How do we calculate the length of a vector if our coordinate axes are neither orthogonal nor normalized?"
Distance
\(ds^2 = g_{\mu\nu} dx^\mu dx^\nu\)
Angle
\(\cos \theta = \frac{g_{ij} u^i v^j}{|u||v|}\)
Mapping
\(V \xrightarrow{g} V^* \xrightarrow{g^{-1}} V\)
The Metric Matrix
The components of the metric tensor are the inner products of the basis vectors:
\(g_{ij} = \mathbf{e}_i \cdot \mathbf{e}_j\)
Symmetry: \(g_{ij} = g_{ji}\)Positive Definite: In Euclidean signatures.Orthonormal Basis: \(g_{ij} = \delta_{ij}\) (The Identity).
60° e2 e1
Non-orthogonal basis vectors generate a non-identity metric.
Raising & Lowering Indices
The metric tensor acts as a musical sharp or flat, moving components between the primal space and dual space.
Lowering (V → V*)
\(v_i = g_{ij} v^j\)
"Contract with the first index of the metric."
Raising (V* → V)
\(v^i = g^{ij} v_j\)
"Use the inverse metric components."
The Magnitude Invariant
Regardless of the basis or the index position, the "actual" physical length squared is:
\(L^2 = g_{ij} v^i v^j = v_i v^i\)
The pairing of a vector with its own dual counterpart defines its identity in the manifold.
Metric Tensor Answer Key Answer Key Metric Tensor
Teacher Resource · Lesson 3
1. The Non-Orthogonal Metric
Given \(|\mathbf{e}_1|=1, |\mathbf{e}_2|=2, \theta=60^\circ\):
\(g_{11} = \mathbf{e}_1 \cdot \mathbf{e}_1 = 1^2 = 1\)
\(g_{22} = \mathbf{e}_2 \cdot \mathbf{e}_2 = 2^2 = 4\)
\(g_{12} = g_{21} = |\mathbf{e}_1||\mathbf{e}_2|\cos(60^\circ) = (1)(2)(0.5) = 1\)
Metric matrix \(g_{ij} = \begin{pmatrix} 1 & 1 \\ 1 & 4 \end{pmatrix}\). Inverse metric \(g^{ij}\):
\(\det(g) = (1)(4) - (1)(1) = 3\). Thus, \(g^{ij} = \frac{1}{3} \begin{pmatrix} 4 & -1 \\ -1 & 1 \end{pmatrix}\).
Physical Role: The inverse metric \(g^{ij}\) allows for the mapping of covariant components back to contravariant components (raising indices).
2. Raising and Lowering
Lowering Indices: \(v^j = (2, -1)\).
Magnitude squared:
3. Polar Metric
From \(ds^2 = dr^2 + r^2 d\theta^2\), the metric components are \(g_{rr} = 1, g_{\theta\theta} = r^2\), and \(g_{r\theta} = g_{\theta r} = 0\).
\(g_{ij} = \begin{pmatrix} 1 & 0 \\ 0 & r^2 \end{pmatrix}\)
Explanation: The metric depends on the coordinate \(r\). This is because the physical "width" of a unit change in \(\theta\) (an angle) grows linearly with the distance from the origin. In curvilinear coordinates, the "exchange rate" between coordinate change and physical distance is spatially varying.
Higher Rank Tensor Worksheet Tensor Mechanics Case Study
Graduate Vector Analysis · Lesson 4
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Multilinear Mapping
A tensor of rank \((p, q)\) is a multilinear map that takes \(p\) co-vectors and \(q\) vectors as inputs to produce a scalar. In physical systems, tensors describe how internal forces (stress) relate to the orientation of surfaces.
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Cauchy Stress Relationship
The Cauchy stress tensor \(\sigma\) is a rank-2 tensor (\(1, 1\) or \(0, 2\) depending on representation) that maps a surface normal vector \(\mathbf{n}\) to a traction vector \(\mathbf{t}\): \[ t^i = \sigma^i_j n^j \]
(a) If \(\sigma^i_j\) is a constant tensor, and we rotate our coordinate system, how do the components \(\sigma^i_j\) transform? Write the transformation law using two Jacobian matrices.
(b) Explain why the stress tensor must be symmetric (\(\sigma_{ij} = \sigma_{ji}\)) for a body in static equilibrium (Hint: Consider angular momentum).
2
Constructing Ranks
Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in \(V\). The tensor product \(\mathbf{T} = \mathbf{u} \otimes \mathbf{v}\) is a rank \((2, 0)\) tensor with components \(T^{ij} = u^i v^j\).
Prove that the contraction of this tensor product \(C(T) = g_{ij} T^{ij}\) is equal to the inner product \(\mathbf{u} \cdot \mathbf{v}\). Show all index steps.
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The Riemann Preview
A tensor \(A_{ijk}\) is completely anti-symmetric if swapping any two indices negates the component (e.g., \(A_{ijk} = -A_{jik}\)).
In a 3-dimensional space, how many independent components does a completely anti-symmetric rank-3 tensor have? List them or justify your answer.
Higher Rank Slides Generalizing Tensors
Multilinear Maps & Physical Systems
Graduate Vector Analysis Lesson 04
Beyond Rank-1
"How can we describe a physical quantity like stress that has a magnitude and a direction, but acts on a surface with its own specific orientation?"
Vector
One direction. Rank 1.
Tensor
A relationship between directions. Rank N.
The Modern Definition
A tensor \(T\) of rank \((p, q)\) is a multilinear map:
\(T : \underbrace{V^* \times \dots \times V^*}_{p} \times \underbrace{V \times \dots \times V}_{q} \to \mathbb{F}\)
Rank (1, 0):
\(T(\phi) = \phi(\mathbf{v})\) → A Vector.
Rank (0, 1):
\(T(\mathbf{v}) = \phi(\mathbf{v})\) → A Co-vector.
Rank (0, 2):
\(T(\mathbf{u}, \mathbf{v})\) → The Metric Tensor.
Rank (1, 1):
\(T(\phi, \mathbf{v})\) → A Linear Transformation.
The Stress Tensor \(\sigma\)
In a solid, the force per unit area (traction) depends on the orientation of the cut.
\(t^i = \sigma^{ij} n_j\)
Diagonal elements \(\sigma^{ii}\): Normal stress.
Off-diagonal \(\sigma^{ij}\): Shear stress.
Balance of torques → \(\sigma^{ij} = \sigma^{ji}\).
Internal Forces
σ11
σ12
σ13
σ21
σ22
σ23
σ31
σ32
σ33
Higher Rank Transformations
Each index transforms according to its position (upper vs lower) using its own Jacobian.
\(\bar{T}^{ij}_k = \frac{\partial \bar{x}^i}{\partial x^l} \frac{\partial \bar{x}^j}{\partial x^m} \frac{\partial x^n}{\partial \bar{x}^k} T^{lm}_n\)
"A tensor is not just an array of numbers; it is an object that transforms correctly."
Higher Rank Answer Key Answer Key Higher Rank Tensors
Teacher Resource · Lesson 4
1. Cauchy Stress Relationship
(a) Transformation Law: For a rank (1, 1) tensor \(\sigma^i_j\), the law is:
\(\bar{\sigma}^i_j = \frac{\partial \bar{x}^i}{\partial x^k} \frac{\partial x^l}{\partial \bar{x}^j} \sigma^k_l\)
(b) Symmetry for Static Equilibrium: If \(\sigma_{ij}\) were not symmetric, the internal shear forces would create a net non-zero torque on an infinitesimally small volume element. For the element to remain in static equilibrium (zero angular acceleration), the sum of moments must be zero, which requires \(\sigma_{12} = \sigma_{21}\), etc.
2. Tensor Products & Contraction
Proof:
Definition: \(T^{ij} = u^i v^j\).
Contraction: \(C(T) = g_{ij} T^{ij} = g_{ij} u^i v^j\).
By the index lowering property: \(g_{ij} u^i = u_j\).
Substitute: \(C(T) = u_j v^j\).
By definition of the inner product in dual space pairing: \(u_j v^j = \mathbf{u} \cdot \mathbf{v}\).
This shows that contraction is the tensor generalization of the dot product.
3. The Riemann Preview (Anti-symmetry)
Independent Components: In 3D, for a rank-3 completely anti-symmetric tensor \(A_{ijk}\), the indices must all be distinct for a non-zero component. If any two indices are the same (e.g., \(A_{112}\)), anti-symmetry implies \(A_{112} = -A_{112} \implies A_{112} = 0\).
The only distinct combinations are permutations of \(\{1, 2, 3\}\). There is only one independent component (e.g., \(A_{123}\)). All other non-zero components are \(\pm A_{123}\) based on the parity of the permutation.
This is related to the Levi-Civita symbol \(\epsilon_{ijk}\).
Non Euclidean Transport Worksheet Manifold Vector Transport
Graduate Vector Analysis · Lesson 5
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Non-Euclidean Curvature
In curved spaces, "straight lines" are geodesics and "constant direction" requires parallel transport. Here, we apply the covariant derivative \(\nabla_\mu\) and Christoffel symbols \(\Gamma^\sigma_{\mu\nu}\) to understand how vector quantities evolve across a manifold.
1
The Triangle Walk
Imagine a vector at the North Pole of a unit sphere pointing along the \(0^\circ\) longitude line. Transport it "parallelly" to the equator, then $90^\circ$ along the equator, then back to the North Pole.
Sketch the final orientation of the vector compared to its initial orientation. Explain why the final direction has rotated even though we kept it "parallel" at every step.
Sketch and Explanation Space
2
Correcting the Gradient
The ordinary partial derivative \(\partial_\mu v^\nu\) of a vector field is not a tensor. We must add a correction term involving the Christoffel symbols.
\(\nabla_\mu v^\nu = \partial_\mu v^\nu + \Gamma^\nu_{\mu\lambda} v^a\)
Using the polar metric \(g_{rr} = 1, g_{\theta\theta} = r^2\), calculate the non-zero Christoffel symbols. (Hint: Use \(\Gamma^k_{ij} = \frac{1}{2} g^{kl} (\partial_i g_{lj} + \partial_j g_{li} - \partial_l g_{ij})\)).
3
Path of Least Resistance
The geodesic equation \(\frac{d^2 x^\mu}{ds^2} + \Gamma^\mu_{\nu\rho} \frac{dx^\nu}{ds} \frac{dx^\rho}{ds} = 0\) describes the motion of a vector quantity being transported along its own direction.
In flat Euclidean space (Cartesian), the Christoffel symbols are zero. What does the geodesic equation simplify to, and what physical path does this represent?
Non Euclidean Slides Non-Euclidean Transport
Curvature, Connection, & Geodesics
Graduate Vector Analysis Lesson 05 (Capstone)
The Problem of Direction
"If you walk strictly 'north' on a sphere and return to where you started, why is your final facing direction different from your initial one?"
In curved space, "parallel" is no longer global. We need a way to connect the tangent spaces at different points.
Parallel Transport
Transporting a vector \(\mathbf{v}\) along a curve \(\gamma(s)\) such that its covariant derivative along the curve is zero.
\(\frac{Dv^\mu}{ds} = \frac{dv^\mu}{ds} + \Gamma^\mu_{\nu\rho} v^\nu \frac{dx^\rho}{ds} = 0\)
The rotation of the vector depends on the Holonomy of the path.
Curved Connection
Christoffel Symbols \(\Gamma^\sigma_{\mu\nu}\)
These are NOT tensors. They represent the "fictitious forces" or coordinate artifacts of the manifold curvature.
Purpose
To correct the partial derivative so that the result is once again a tensor.
Symmetry
\(\Gamma^\sigma_{\mu\nu} = \Gamma^\sigma_{\nu\mu}\) for torsion-free connections (Levi-Civita).
"The connection tells the tangent spaces how to talk to each other."
The Universal Path
A geodesic is a curve whose velocity vector is parallel-transported along itself.
\(\ddot{x}^\mu + \Gamma^\mu_{\nu\rho} \dot{x}^\nu \dot{x}^\rho = 0\)
In the absence of external forces, objects move along geodesics. This is the foundation of Einstein's General Relativity.
Non Euclidean Answer Key Answer Key Vector Transport
Teacher Resource · Lesson 5
1. The Triangle Walk
Explanation: The final vector at the North Pole will be rotated by \(90^\circ\) relative to the starting vector.
This occurs because the tangent space "rotates" as it moves along the curved manifold. Parallel transport keeps the vector's relationship to the local tangent space as "constant" as possible, but because the sphere's surface is curved, the path taken matters. The total rotation angle is equal to the integral of the Gaussian curvature over the area enclosed by the path (Gauss-Bonnet Theorem).
2. Correcting the Gradient (Christoffel Calculation)
Given \(g_{rr}=1, g_{\theta\theta}=r^2\). Inverse metric: \(g^{rr}=1, g^{\theta\theta}=1/r^2\).
\(\Gamma^r_{\theta\theta} = \frac{1}{2}g^{rr}(\partial_\theta g_{r\theta} + \partial_\theta g_{\theta r} - \partial_r g_{\theta\theta}) = \frac{1}{2}(1)(-2r) = -r\)
\(\Gamma^\theta_{r\theta} = \Gamma^\theta_{\theta r} = \frac{1}{2}g^{\theta\theta}(\partial_r g_{\theta\theta}) = \frac{1}{2}(1/r^2)(2r) = 1/r\)
All other \(\Gamma^k_{ij} = 0\). These terms account for the "spreading" of coordinate lines as radius increases.
3. Path of Least Resistance
In Cartesian space, \(\Gamma = 0\). The equation becomes: \[ \frac{d^2 x^\mu}{ds^2} = 0 \]
The solution is \(x^\mu(s) = a^\mu s + b^\mu\), which is the parametric equation for a straight line . This confirms that in flat space, geodesics are straight lines. In curved space, the \(\Gamma\) term acts as a "correction" that keeps the path as straight as possible given the constraints of the surface.
