# Action integral

1. Apr 19, 2005

### JohanL

If you have an action integral

$$\int_{A}^{B} \sqrt{F\mathbf{(r)}} dr$$

and

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

the minimum of the action integral is equivalent to

$$\frac{d}{dt}\frac{dG}{\dot{z}}-\frac{dG}{z}=0$$

where
$$G=\sqrt{F}$$

or am i doing this in a completeley wrong way?

2. Apr 19, 2005

### Crosson

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):

$$\Int F[y , y' , x ] dx$$

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

$$\frac {d}{dx} (\frac {\partial F}{\partial y'}) - \frac{\partial F}{\partial y} = 0$$

3. Apr 19, 2005

### dextercioby

Extremizes...

Daniel.

4. Apr 20, 2005

### JohanL

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

$$\epsilon(\vec{r})$$

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

$$\int_{A}^{B} \sqrt{\epsilon(\vec{r})} |d\vec{r}|$$

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

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

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)

$$\frac{d}{dt}\frac{d(\sqrt{\epsilon(\vec{r})} )}{\dot{z}}-\frac{d(\sqrt{\epsilon(\vec{r})} )}{z}=0$$

and with this you get

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

and this must be wrong!