Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Action integral

  1. Apr 19, 2005 #1
    If you have an action integral

    [tex]
    \int_{A}^{B} \sqrt{F\mathbf{(r)}} dr
    [/tex]

    and

    F=a-bz^2 , b>0, a-bd^2>0

    the minimum of the action integral is equivalent to

    [tex]
    \frac{d}{dt}\frac{dG}{\dot{z}}-\frac{dG}{z}=0
    [/tex]

    where
    [tex]
    G=\sqrt{F}
    [/tex]

    or am i doing this in a completeley wrong way?
     
  2. jcsd
  3. Apr 19, 2005 #2
    You jump around a lot with your notation. See if this is clear:

    If you have an Integral of this form(where y' is the derivative of y with respect to x):

    [tex] \Int F[y , y' , x ] dx [/tex]

    Then the function y which minimizes the integral is the solution to this differential equation:

    [tex] \frac {d}{dx} (\frac {\partial F}{\partial y'}) - \frac{\partial F}{\partial y} = 0 [/tex]
     
  4. Apr 19, 2005 #3

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Extremizes...

    Daniel.
     
  5. Apr 20, 2005 #4
    It was those equations i was trying to use...but probably in a wrong way.
    To explain my notation i give you the problem:

    A light ray's path in a medium with variating dielectricity constant

    [tex]
    \epsilon(\vec{r})
    [/tex]

    between the two points A and B, minimizes, according to Fermat's principle, the action integral

    [tex]\int_{A}^{B} \sqrt{\epsilon(\vec{r})} |d\vec{r}| [/tex]

    A plane piece of glass with thickness 2d has a dielectricity function which, when the piece is in |z|<=d in a cartesian coordinatesystem, can be written

    [tex]
    \epsilon(\vec{r})=a-bz^2 ,b>0, a-bd^2>0

    [/tex]

    Calculate the general form for the light ray's path in the piece of glass.

    Solution:

    The minimum of the action integral is equivalent to
    (But this is probably wrong)

    [tex]\frac{d}{dt}\frac{d(\sqrt{\epsilon(\vec{r})} )}{\dot{z}}-\frac{d(\sqrt{\epsilon(\vec{r})} )}{z}=0[/tex]

    and with this you get

    -1/2(a-bz^2)^(-1/2) * 2bz = 0

    and this must be wrong!
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook