User:Eml4500.f08.lulz.abcd/HW4

Prove $$\mathbf{\tilde{f}^{(e)}}=\mathbf{\tilde{T}^{(e)}{f}^{(e)}}$$
$$\mathbf{\tilde{f}^{(e)}}=\mathbf{\tilde{T}^{(e)}{f}^{(e)}}$$

$$\mathbf{\tilde{T}^{(e)}}=\begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0\\ m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & -m^{(e)}\\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}$$

Since we know that $$l^{(e)} = cos(\theta)^{(e)}$$, $$m^{(e)} = sin(\theta)^{(e)}$$, and that the angle is 30 degrees, we can substitute values and compare the transformed matrix output to the values previously calculated. Confirmation will prove the relation.

$$\mathbf{\tilde{T}^{(1)}}=\begin{bmatrix} \sqrt{3}/2 & -0.5 & 0 & 0\\ 0.5 & \sqrt{3}/2 & 0 & 0\\ 0 & 0 & \sqrt{3}/2 & -0.5\\ 0 & 0 & 0.5 & \sqrt{3}/2 \end{bmatrix}$$

$$\mathbf{\tilde{T}^{(1)}f^{(1)}}=\begin{bmatrix} \sqrt{3}/2 & -0.5 & 0 & 0\\ 0.5 & \sqrt{3}/2 & 0 & 0\\ 0 & 0 & \sqrt{3}/2 & -0.5\\ 0 & 0 & 0.5 & \sqrt{3}/2 \end{bmatrix}\begin{bmatrix} -4.378\\ -2.5622\\ 4.4378\\ 2.5622 \end{bmatrix}=\mathbf{\tilde{f}^{(1)}}$$

$$\mathbf{\tilde{f}^{(1)}}=\begin{bmatrix} -5.1243\\ 0\\ 5.1243\\ 0 \end{bmatrix}$$

This confirms as expected and thus the proof is confirmed.

Prove that $$ \mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$
$$ \mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$

We already know the values from previous work for $$\mathbf{k^{(e)}}$$, $$\mathbf{\tilde{k}^{(e)}}$$, $$\mathbf{\tilde{T}^{(e)}}$$, and $$\mathbf{\tilde{T}^{(e)^T}}$$ is trivial to find:

$$ \mathbf{\tilde{T}^{(e)^T}} = \begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)} \end{bmatrix}$$

$$\mathbf k^{(e)} = \tilde \mathbf T^{(e)^{T}}\tilde \mathbf k^{(e)}\tilde \mathbf T^{(e)}$$

$$\mathbf k^{(e)} = k^{(e)}\begin{bmatrix} l^{(e)} & -m^{(e)} & 0 & 0\\ m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & -m^{(e)}\\ 0 & 0 & m^{(e)} & l^{(e)} \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 & 0\\ 0 & 0 & 0 & 0\\ -1 & 0 & 1 & 0\\ 0 & 0 & 0 & 0\end{bmatrix}\begin{bmatrix} l^{(e)} & m^{(e)} & 0 & 0\\ -m^{(e)} & l^{(e)} & 0 & 0\\ 0 & 0 & l^{(e)} & m^{(e)}\\ 0 & 0 & -m^{(e)} & l^{(e)}\end{bmatrix} = \begin{bmatrix} (l^{(e)})^{2} & l^{(e)}m^{(e)}  &  -(l^{(e)})^{2}  &  -l^{(e)}m^{(e)}   \\ l^{(e)}m^{(e)} &  (m^{(e)})^{2}  &  -l^{(e)}m^{(e)}  &  -(m^{(e)})^{2}   \\ -(l^{(e)})^{2} & -l^{(e)}m^{(e)}  &  (l^{(e)})^{2}  &  l^{(e)}m^{(e)} \\ -l^{(e)}m^{(e)} &  -(m^{(e)})^{2}  &  l^{(e)}m^{(e)}  &  (m^{(e)})^{2} \\ \end{bmatrix} = \mathbf k^{(e)}$$