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Action integral

  • Thread starter JohanL
  • Start date
158
0
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?
 
1,222
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]
 

dextercioby

Science Advisor
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Extremizes...

Daniel.
 
158
0
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!
 

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