diff --git a/_psets/4.md b/_psets/4.md index feeec7d22e4cab20451f1530e407deb8051a4449..8fadaed01bb3b893f4b6e2f5f71c0028c56314b1 100644 --- a/_psets/4.md +++ b/_psets/4.md @@ -5,10 +5,39 @@ title: Problem Set 4 ## (6.1) {:.question} -Prove the BAC–CAB rule $$A \times (B \times C) = B(A \cdot B) - C (A \cdot B)$$ by writing it out in +Prove the BAC–CAB rule $$A \times (B \times C) = B(A \cdot C) - C (A \cdot B)$$ by writing it out in the summations convention, and use it to show that $$\nabla \times (\nabla \times E) = \nabla (\nabla \cdot E) - \nabla^2 E$$. +I'm going to use the [Levi-Civita symbol](https://en.wikipedia.org/wiki/Levi-Civita_symbol). I'm +also going to use $$i_+$$ to represent $$(i + 1) \text{ mod } 3$$ for any coordinate index $$i$$. So +if $$i$$ is 2, then $$i_+ = 3$$ and $$i_{+ +} = 1$$. + +The $$i$$th coordinate of $$A \times (B \times C)$$ is + +$$ +\begin{align*} +\epsilon_{i j k} A_j \epsilon_{k l m} B_l C_m +&= A_{i_+} \epsilon_{i_{+ +} l m} B_l C_m - A_{i_{+ +}} \epsilon_{i_+ l m} B_l C_m \\ +&= A_{i_+} (B_i C_{i_+} - B_{i_+} C_i) - A_{i_{+ +}} (B_{i_{+ +}} C_i - B_i C_{i_{+ +}}) \\ +&= A_i B_i C_i + A_{i_+} B_i C_{i_+} + A_{i_{+ +}} B_i C_{i_{+ +}} - \\ +&\phantom{x=} A_i B_i C_i - A_{i_+} B_{i_+} C_i - A_{i_{+ +}} B_{i_{+ +}} C_i \\ +&= B_i A_j C_j - C_i A_j B_j +\end{align*} +$$ + +which is the $$i$$th coordinate of $$B(A \cdot C) - C (A \cdot B)$$. + +If we take $$\nabla$$ to be a three-vector of partial differentiation operators, then this shows +that + +$$ +\begin{align*} +\nabla \times (\nabla \times E) +&= \nabla (\nabla \cdot E) - E (\nabla \cdot \nabla) \\ +&= \nabla (\nabla \cdot E) - \nabla^2 E +\end{align*} +$$ ## (6.2)