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Finding the length of the major and minor axis of ellipse.

  1. Jan 19, 2013 #1
    [tex]\vec E\;=\; \hat x E_{x0}\cos(\omega {t} -kz)\;+\;\hat y E_{y0}\cos(\omega{t}-kz+\delta)[/tex]

    For z=0, this is a vector that trace out an ellipse with time t.

    I want to

    1) to verify that using the definition of differential calculus, we can find the length of the major and minor axis by finding the maximum and minimum of the the derivative of the [itex]\vec E[/itex]. This is by taking the derivative of [itex]\vec E[/itex] and let the derivative equal to zero to get the minimum and maximum value.

    2) Find a way to get the maximum and minimum. I don't know how to do this. This is my work:

    [tex]\frac {\partial \vec E}{\partial t}\;=\; -\hat x \omega E_{x0}\sin(\omega {t} -kz)\;-\;\hat y E_{y0}\sin(\omega{t}-kz+\delta)[/tex]

    For [itex]\frac {\partial \vec E}{\partial t}\;=\;0[/itex], both components of x and y has to be zero. I don't know how to do this. Please help.

    thanks

    Alan
     
  2. jcsd
  3. Jan 19, 2013 #2

    I like Serena

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    Hi Alan!

    It's not ##\partial \vec E \over \partial t## that has to be zero.

    What you are looking for are the 4 points where ##\vec E## is perpendicular to its tangent vector.
    Those points correspond to the long axis and the short axis.
    The tangent vector is ##\partial \vec E \over \partial t##.
    Their dot product has to be zero.
    So:

    $$\vec E \cdot {\partial \vec E \over \partial t} = 0$$

    $$(\hat x E_{x_0}\cos(\omega {t} -kz)+\hat y E_{y_0}\cos(\omega{t}-kz+\delta)) \cdot (-\hat x \omega E_{x_0}\sin(\omega {t} -kz)-\hat y \omega E_{y_0}\sin(\omega{t}-kz+\delta)) = 0$$

    $$- \omega E_{x_0}^2 \sin(\omega {t} -kz) \cos(\omega {t} -kz) - \omega E_{y_0}^2 \sin(\omega{t}-kz+\delta) \cos(\omega{t}-kz+\delta) = 0$$

    Since ##\sin(2x) = 2 \sin x \cos x##, we get:

    $$E_{x_0}^2 \sin(2(\omega {t} -kz)) + E_{y_0}^2 \sin(2(\omega{t}-kz)+2\delta)) = 0$$

    You can further simplify this with more trig formulas, yielding a solution for t, and after that, the 4 extreme points you are looking for.
     
  4. Jan 19, 2013 #3
    Thank you for the help. Now it make sense. It's been a while I worked with vectors and their tangents. The moment you mentioned this, now I remember.

    Thanks again

    Alan
     
  5. Jan 19, 2013 #4
    I want to verify the rest of the steps:

    For z=0

    [tex]E^2_{x0}\sin(2\omega t) + E^2_{y0}\sin(2\omega t+2\delta)=0\;\Rightarrow \sin(2\omega t)\;=\;-\frac{E^2_{y0}}{E^2_{x0}}\sin(2\omega t +2\delta)[/tex]

    [tex]\sin(A+B)=\sin A \cos B\;+\;\cos A \sin B\;\Rightarrow[/tex]
    [tex]\frac{E^2_{y0}}{E^2_{x0}}\;=\;-\frac{\sin 2\omega t}{\sin 2\omega t \cos 2 \delta + \cos 2\omega t \sin 2\delta}\;=\; -\frac{\tan 2\omega t}{\tan 2\omega t \cos 2 \delta + \sin 2\delta}[/tex]
    [tex]\Rightarrow\; \tan 2\omega t \;=\;-\frac{E^2_{y0}}{E^2_{x0}} tan 2\omega t \cos 2\delta \;-\;\frac{E^2_{y0}}{E^2_{x0}} \sin 2\delta\;\Rightarrow\; \tan 2\omega t (1+\frac{E^2_{y0}}{E^2_{x0}} \cos 2\delta)\;=\; -\frac{E^2_{y0}}{E^2_{x0}} \sin 2\delta[/tex]
    [tex]\Rightarrow \tan 2\omega t\;=\; -\frac { \sin 2 \delta}{\frac{E^2_{x0}}{E^2_{y0}}+\cos 2\delta}[/tex]
    Therefore:
    [tex]\omega t\;=\;\frac {1}{2} \tan^-1 \left(-\frac {\sin 2 \delta}{\frac{E^2_{x0}}{E^2_{y0}}+\cos 2\delta}\right)[/tex]

    Since [itex]\omega t [/itex] is in radian, so I just measure the angle from x-axis. I should get two angles.
     
  6. Jan 20, 2013 #5

    I like Serena

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    Looks good. ;)

    Due to the use of ##\tan##, you will get one angle between ##- \frac \pi 4## and ##+ \frac \pi 4##.
    From periodicity, the other 3 angles are a multiple of ##\frac \pi 2## from there.

    If you fill in your solutions in ##\vec E## and take their lengths, you'll find the half long axis and the half short axis.
     
  7. Jan 20, 2013 #6
    Thanks

    Alan
     
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