MHB Two dimensional rotational Matrix

[cbarker1]

Dear Everybody,

I am trying to learn about the electrodynamics. I am using the textbook, Introduction to Electrodynamics (2nd Ed) by D. J. Griffiths. I am working on the Problem 1.8. The question state:

Prove that the two-dimensional rotation matrix perverse the length of A. (That is, show that $(A_y')^2+(A_z')^2=(A_y)^2+(A_z)^2$)

$\left(\begin{array}{cc} A_y'\\ A_z'\end{array}\right)$=$\left(\begin{array}{cc} cos(\phi) & sin(\phi)\\ -sin(\phi) & cos(\phi) \end{array} \right)$ $\left(\begin{array}{cc} A_y\\A_z\end{array}\right)$

I am struck at the beginning.

 

[topsquark]

Dear Everybody,

I am trying to learn about the electrodynamics. I am using the textbook, Introduction to Electrodynamics (2nd Ed) by D. J. Griffiths. I am working on the Problem 1.8. The question state:

Prove that the two-dimensional rotation matrix perverse the length of A. (That is, show that $(A_y')^2+(A_z')^2=(A_y)^2+(A_z)^2$)

$\left(\begin{array}{cc} A_y'\\ A_z'\end{array}\right)$=$\left(\begin{array}{cc} cos(\phi) & sin(\phi)\\ -sin(\phi) & cos(\phi) \end{array} \right)$ $\left(\begin{array}{cc} A_y\\A_z\end{array}\right)$

I am struck at the beginning.

From the vector equation you have
[math]A_{y'} = cos( \theta ) A_y + sin( \theta ) A_z[/math]
[math]A_{z'} = -sin( \theta ) A_y + cos( \theta ) A_z[/math]

So
[math](A_{y'})^2 + (A_{z'})^2 = \left ( cos( \theta ) A_y + sin( \theta ) A_z \right ) ^2 + \left ( -sin( \theta ) A_y + cos( \theta ) A_z \right ) ^2[/math]

Expand it out (it's not quite as bad as it looks) and you will get
[math](A_{y'})^2 + (A_{z'})^2 = \left ( sin^2 ( \theta ) + cos^2( \theta ) \right ) (A_{y})^2 + \left ( sin^2( \theta ) + cos^2 ( \theta ) \right ) (A_z)^2[/math]

(Hint: The terms in [math]A_y ~ A_z[/math] cancel out.)

Can you finish out the details?

-Dan

 

[HOI]

I hope you mean "preserve", not "perverse"!

 

[cbarker1]

I hope you mean "preserve", not "perverse"!

I meant to preserve.