Lecture 05: Fourth-Order Tensors
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Overview
Steven M. Wise develops the coordinate representation of fourth-order tensors in three dimensions, where 81 components act on a second-order tensor by contracting two indices. He then establishes dyadic bases and symmetry conditions, and derives the isotropic linear elasticity law C(A) = λ tr(A) I + 2μ sym(A) for an operator that maps symmetric tensors to symmetric tensors and annihilates skew-symmetric tensors.
Key takeaways
- In three dimensions, a fourth-order tensor has 81 components, and its action on a second-order tensor is expressed as Uᵢⱼ = CᵢⱼₖₗTₖₗ, summing over k and l.
- The dyadic products eᵢ ⊗ eⱼ ⊗ eₖ ⊗ eₗ form a basis for fourth-order tensors, giving an expansion with 81 coefficients.
- Major symmetry swaps the index pairs, Cᵢⱼₖₗ = Cₖₗᵢⱼ; right and left minor symmetries swap indices within the input and output pairs, respectively.
- Isotropy requires a tensor function to behave consistently under every rotation Q, using the transformed input QAQᵀ.
- An isotropic linear operator that maps symmetric tensors to symmetric tensors and annihilates skew-symmetric tensors has the form C(A) = λ tr(A)I + 2μ sym(A).
- To prove tensor identities, Wise repeatedly tests both sides on an arbitrary vector or tensor; equality of their actions on every input establishes equality of the operators.
Chapters
- In three dimensions, a fourth-order tensor C has 3⁴ = 81 scalar components Cᵢⱼₖₗ, defined relative to a frame.
- C maps a second-order tensor T to another second-order tensor U, with components Uᵢⱼ = Cᵢⱼₖₗ Tₖₗ.
- The repeated indices k and l are summed from 1 to 3, leaving i and j as the free indices of U.
- Wise derives the component formula from the frame basis and linearity, emphasizing that tensor operations must match the types of their inputs and outputs.
- Wise defines a fourth-order tensor C by its action C(T) = AT, where A and T are second-order tensors composed by multiplication.
- Since composition is linear and returns a second-order tensor, this action meets the requirements for a fourth-order tensor.
- Evaluating C on the dyadic basis tensor eₖ ⊗ eₗ gives components Cᵢⱼₖₗ = Aᵢₖ δₗⱼ.
- The Kronecker delta δₗⱼ arises from (eₖ ⊗ eₗ)eⱼ = δₗⱼ eₖ.
- An equivalent component formula uses the tensor inner product: Cᵢⱼₖₗ = (eᵢ ⊗ eⱼ) : C(eₖ ⊗ eₗ).
- Wise first shows that Aᵢⱼ = (eᵢ ⊗ eⱼ) : A, using the dyadic basis coordinates δᵢₖδⱼₗ.
- For fixed k and l, C(eₖ ⊗ eₗ) is a second-order tensor, so its inner product with eᵢ ⊗ eⱼ extracts its i,j component.
- The derivation illustrates how component identities follow from basis expansions and tensor inner products rather than index manipulation alone.
- For vectors a, b, c, and d, the four-factor dyadic product is defined through its action on a second-order tensor T.
- Its action is (a ⊗ b ⊗ c ⊗ d)T = (c · (T d))(a ⊗ b), producing a scalar times a second-order tensor.
- The definition supplies an output for every T, while linearity follows from the linearity of tensor-vector multiplication and the dot product.
- Wise uses this action-based definition to verify that the four-factor dyadic product is a fourth-order tensor.
- For a frame with three basis vectors, the products eᵢ ⊗ eⱼ ⊗ eₖ ⊗ eₗ provide 81 candidate basis elements for fourth-order tensors.
- Wise establishes the expansion C = Cᵢⱼₖₗ(eᵢ ⊗ eⱼ ⊗ eₖ ⊗ eₗ), with each of the four indices summed from 1 to 3.
- The proof applies the proposed expansion to an arbitrary second-order tensor T and uses Uᵢⱼ = CᵢⱼₖₗTₖₗ to recover U = C(T).
- Because the expansion has the same action as C for every T, the operators are equal; linear independence is identified as the remaining basis property.
- Major symmetry means A : C(B) = C(A) : B for all second-order tensors A and B, making C self-adjoint under the tensor inner product.
- Right minor symmetry means C(B) = C(sym B) for every B, so C ignores the skew-symmetric part of its input.
- Left minor symmetry means C(A) = C(sym A) in the corresponding inner-product identity, restricting the operator on the output-side argument.
- These are distinct properties: a fourth-order tensor may have major symmetry without either minor symmetry.
- Major symmetry is equivalent to Cᵢⱼₖₗ = Cₖₗᵢⱼ, which swaps the two index pairs while preserving order within each pair.
- Right minor symmetry gives Cᵢⱼₖₗ = Cᵢⱼₗₖ, swapping the indices in the input pair.
- Left minor symmetry gives Cᵢⱼₖₗ = Cⱼᵢₖₗ, swapping the indices in the output pair.
- Wise derives the major-symmetry identity by moving C across the tensor inner product, then recommends applying the same method to the minor-symmetry identities.
- For a function G from second-order tensors to second-order tensors, isotropy is defined by QG(A)Qᵀ = G(QAQᵀ) for every rotation Q.
- The transformed tensor QAQᵀ represents A under a rotated frame, so isotropy expresses independence from orientation.
- For a scalar-valued function G, the corresponding condition is G(A) = G(QAQᵀ).
- Wise notes that Qᵀ is the inverse of Q and uses this fact to rearrange the tensor-valued isotropy condition into an equivalent form.
- For isotropic G that maps symmetric tensors to symmetric tensors, G(A) can be represented using A and its three scalar invariants.
- The first invariant is tr(A), the third is det(A), and the second is the intermediate invariant defined from A.
- The invariant values do not depend on the frame used to calculate them, allowing coefficient functions to depend on the invariant vector rather than orientation.
- Wise also gives a separate representation for symmetric, invertible inputs, while noting that the general representation is cited rather than proved.
- The fourth-order operator C is assumed isotropic, maps symmetric tensors to symmetric tensors, and satisfies C(W) = 0 for every skew-symmetric tensor W.
- The isotropic representation for symmetric inputs, combined with linearity, restricts the allowable terms to a constant multiple of tr(H)I and a constant multiple of H.
- Applying C to the zero tensor forces the constant identity-only term to vanish.
- Decomposing any A into sym(A) plus its skew-symmetric part yields C(A) = λ tr(A)I + 2μ sym(A), with two scalar constants λ and μ.
- Wise proves S(A ⊗ B) = (SA) ⊗ B by applying each side to an arbitrary vector V.
- The key dyadic action is (A ⊗ B)V = (B · V)A, which turns both compositions into (B · V)SA.
- The exercises also ask students to verify S(A ⊗ B) = A ⊗ (SᵀB) and (A ⊗ B)ᵀ = B ⊗ A.
- Testing equal actions on every V establishes equality of the resulting second-order tensors.
- Wise checks that the second invariant is unchanged when a tensor is represented in two different frames.
- The trace of S is already known to be frame-invariant, so any expression formed from that trace alone, including its square, is invariant.
- For the remaining term, the rotated representation of S² is AᵀS²A; taking traces uses the same frame-change argument applied to S.
- Since both parts of the second-invariant expression are frame-invariant, their difference is also frame-invariant.
- Wise closes with one additional exercise and points students to the posted slides for worked examples and practice.
- The exercises reinforce dyadic products, summation-convention calculations, and invariant checks before the homework.
- The lecture ends at about 69 minutes with an invitation for questions.
Transcript excerpt
0:04 All right. So, welcome back to continuum mechanics. This is class number five. So, we're finished today with tensor algebra. Um and so, I want to talk more about fourth-order tensors, as as especially the last section of the last set of slides that we didn't get to cover well. I decided to put those in as a first section for this set of slides. So, we were we introduced uh what we meant what what we meant by fourth-order tensors. They these are linear mappings that take a second-order tensor to a second-order tensor, okay? So, you can you can grab their coordinates and you can do a lot of nice things with them in um in coordinate frames, and so that's
0:51 what we want to talk about here. So, suppose that C is a fourth-order tensor, and suppose that F is a frame. The 81 components of C are following numbers. They're scalars. Components are always scalars, and and components always have to be with respect to some frame. So, the 81 coordinates are C I J K L for I J K and L between 1 and 3. Right? So, there's three choices for each one, so 3 * 3 * 3 * 3 81. And they are defined um by this equation. C I J K L is E I dotted with a vector.
1:40 Okay, what is that vector? The vector is this two tensor acting on the vector, which gives a vector. This four tensor acting on this two tensor gives that two tensor inside. Okay, so everything has to pass the what I call a smell test, so that you're comparing apples to apples. I have to When I do a dot product, I have to be dotting a vector with a vector. So, I have a vector here, and so
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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.