Lecture 02: Tensor Algebra
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Overview
Steven M. Wise develops second-order tensor algebra by defining tensors as linear maps from a vector space to itself, then showing how their actions and components correspond to familiar matrix operations. Examples including projection, rotation, and reflection lead to coordinate formulas using orthonormal frames, Einstein summation, and dyadic products, culminating in component rules for tensor composition.
Key takeaways
- A second-order tensor is a linear map V → V, and it is fully determined by its action on a basis of V.
- In an orthonormal frame, the tensor action Sv becomes the matrix-vector rule ui = Sij vj, with the repeated index summed.
- Tensor products represent composition: ST applies T first and S second, so order matters and (ST)ij = Sik Tkj.
- For a unit normal e, reflection across the plane perpendicular to e is v − 2(v·e)e and is self-inverse; projection onto that plane is v − (v·e)e and is not invertible.
- The dyadic product satisfies (a⊗b)v = (b·v)a and has components ai bj, while its matrix is the outer product of a column and a row.
- Component calculations require a consistent frame; addition or multiplication of coordinate matrices is meaningful only when the tensors use the same frame.
Chapters
- A second-order tensor on V maps vectors in V back into V and satisfies additivity and scalar homogeneity.
- The zero tensor maps every vector to zero, while the identity tensor maps every vector to itself.
- Two tensors are equal only when their actions agree on every vector, which also implies equality of all components in a frame.
- For a unit vector e, the projection P(v) = (v·e)e extracts the component of v along e.
- A rotation about the e3 axis is specified by its action on the basis vectors: e1 and e2 rotate in their plane while e3 remains fixed.
- Defining a linear tensor on the three basis vectors determines its action on every vector because each vector is a linear combination of that basis.
- For a plane with unit normal e, reflection is T(v) = v − 2(v·e)e.
- Reflecting twice returns the original vector, so the reflection tensor is its own inverse.
- Orthogonal projection onto the plane removes the normal component, giving v − (v·e)e; this projection is not invertible.
- Tensor addition is defined by (S + T)v = Sv + Tv for every vector v.
- Scalar multiplication is defined by (αT)v = α(Tv), and both operations preserve tensor linearity.
- These operations make the set of second-order tensors a real vector space.
- The product ST is defined by (ST)v = S(Tv), so T acts first and S acts on the result.
- Composition produces another second-order tensor because both maps take V into V.
- Tensor multiplication is generally noncommutative, unlike multiplication of scalar components.
- For an orthonormal frame {e1, e2, e3}, the components are Sij = ei·(S ej), giving nine scalar entries.
- The frame must be specified or understood before tensor components have a definite meaning.
- The component indices identify matrix position: i is the row and j is the column.
- Writing v = vj ej and u = Sv gives the component relation ui = Sij vj, with j summed from 1 to 3.
- The derivation uses linearity of S and the dot product, plus orthonormality ei·ej = δij.
- The Kronecker delta contracts a repeated index, transferring an index and eliminating the corresponding sum.
- A free index such as i labels a component, while a repeated dummy index such as j is summed over the three frame directions.
- Dummy indices can be renamed within a term, but free indices must remain consistent on both sides of an equation.
- Checking index structure and tensor type provides a practical consistency test for component calculations.
- The matrix [S] in a frame collects all nine Sij components, with i indexing rows and j indexing columns.
- A matrix transpose interchanges row and column indices, so the transposed components are Sji.
- Component notation is frame-dependent, even though the tensor itself can be described independently of coordinates.
- The jth column of [S] is the coordinate vector of S ej, so the three basis-vector images provide all nine entries.
- For rotation through θ about e3, the matrix has the planar block [[cos θ, −sin θ], [sin θ, cos θ]] and leaves the third axis unchanged.
- The identity tensor has components δij in every orthonormal frame because I ej = ej.
- The dyadic product a⊗b is defined by (a⊗b)v = (b·v)a, making it a second-order tensor.
- In an orthonormal frame, its components are (a⊗b)ij = ai bj, with i and j both free indices.
- The dyadic’s action on a vector reduces to a scalar dot product multiplied by a, which makes its linearity direct to verify.
- The coordinate matrix of a⊗b is the column vector of a multiplied by the row vector bᵀ.
- A 3×1 column times a 1×3 row produces a 3×3 matrix whose entries are ai bj.
- This outer product is distinct from a dot product, which would produce a scalar.
- A tensor can be reconstructed as S = Sij ei⊗ej, with both i and j summed from 1 to 3.
- The proof tests both sides on an arbitrary vector and uses equality of tensor actions on every vector.
- Expanding the vector in the frame and contracting with δij reduces the expression to the components of Sv.
- For tensors expressed in the same frame, the matrix of S + T is [S] + [T].
- Scalar multiplication also commutes with taking components: [αT] = α[T].
- Coordinate operations cannot be combined directly across tensors represented in different frames.
- For U = ST, the component rule is Uij = Sik Tkj, with k summed from 1 to 3.
- A dyadic identity, (a⊗b)(c⊗d) = (b·c)(a⊗d), reduces products of dyadics to a scalar times one dyadic.
- Applying that identity to the dyadic expansions of S and T and contracting the shared index yields the standard matrix multiplication formula.
Transcript excerpt
0:04 All right, no questions. So, we're um we're still in chapter 1. We're now in section 1.3. We talked about very basic uh information about vectors and now we want to talk about second order tensors. So the the big question is what is a second order tensor? It's basically a matrix. That's that's going to be the the the big reveal. I guess a second order tensor on a vector space V. Remember V are the vectors that all that matters is the magnitude and direction. We can pick them up and move them around in the plane. Points don't get moved. Oh, and by the way, so now the if you go
0:53 to canvas, you can find the GitHub repository which contains these lecture notes. So if you want copied so I handw write the notes and then I have uh then I convert them into beamer slides and so both sets are there. You can uh have whichever one you want. And of course I'm following the book. So if you if you have a copy of the book and you follow along with that fairly carefully then you may not have to take uh notes in either case. So I would say take margin notes. Okay. Okay. Back to the back to the slide. So a second order tensor on a vector space V is a linear mapping which takes V into itself. And it uh by linear we mean that t u of
1:45 v is equal to t of u plus t of v for any u and v in the vector space. So the standard linearity requirements plus we need to add the fact that if I take a t acting on a scaler time a vector that's the same as a scalar times the tensor
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Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Steven M. Wise.