Parametric representation of a vector electric field

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SUMMARY

The discussion focuses on the parametric representation of a vector electric field, specifically addressing the equations for the components E_x and E_y. Participants identified errors in the expressions, particularly regarding the cosine function's properties and the handling of negative signs. The correct expressions are E_x = A_x cos(kz - ωt + φ_x) and E_y = A_y cos(r). The conversation emphasizes the importance of accurately applying trigonometric identities to simplify equations in electromagnetic theory.

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  • Understanding of vector electric fields
  • Familiarity with trigonometric identities, particularly cosine properties
  • Basic knowledge of electromagnetic wave equations
  • Ability to manipulate parametric equations
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Homework Statement



See Attachment

Homework Equations



None I can think of

The Attempt at a Solution



I'm fairly certain that phi_yx is zero
Also I tried factoring out the cos and splitting up the equation into it's respective components, but to no avail. Am I even going about this correctly?
 

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The last bit is a typo. It should be \phi_{xy} = \phi_x.

I'm not sure what you mean by "splitting up the equation into its respective components" since the equation you're supposed to show is true is a scalar equation.
 
Thanks for the reply Vela! Actually I think I figured it out. Maybe you or someone else can check my work? I assumed that it the phi's were not typos, but they could very well be. I went ahead and attached my work.
 

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Your expressions for Ex and Ey are slightly wrong. For example,
E_x = A_x\cos(kz-\omega t+\phi_x) = A_x\cos(-r+\phi_x) = A_x\cos(r-\phi_x)since cosine is an even function. You have an extra negative signs floating around.

You must have made a mistake somewhere in the middle because the cross term has the wrong sign. The sign doesn't flip the way you did it because, again, \cos (-\theta)=\cos\theta.

You actually made it more complicated by keeping the E's and A's around. You know that
\begin{align*}
\frac{E_x}{A_x} &= \cos(\phi_x-r) \\
\frac{E_y}{A_y} &= \cos(r) \\
\end{align*}so you could have just written the lefthand side in terms of cosines and then shown it simplifies down to the righthand side.
 
Ah that I forgot about that cosine property and I found a convenient sign mistake half way through the problem. Thanks Vela!
 

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