# Fermat's Principle with Calculus of Variations

1. Jul 25, 2014

### kq6up

1. The problem statement, all variables and given/known data
This is problem 6.3 in Taylor’s Classical Mechanics. It is in context of the calculus of variations.
Consider a ray of light traveling in a vacuum from point P1to P2 by way of the point Q on a plane mirror, as in Figure 6.8. Show that Fermat's principle implies that, on the actual path followed, Q lies in the same vertical plane as P 1and P2 and obeys the law of reflection, that 0 1= 02. [Hints: Let the mirror lie in the xz plane, and let P 1lie on the y axis at (0, yi, 0) and P2 in the xy plane at (x2, y2, 0). Finally let Q = (x, 0, z). Calculate the time for the light to traverse the path P1QP2 and show that it is minimum when Q has z = 0 and satisfies the law of reflection.]

2. Relevant equations
$\frac { \partial f }{ \partial x } =\frac { d }{ dt } \frac { \partial f }{ \partial x\prime }$ This would be reiterated for y and z.

3. The attempt at a solution
$S=\frac { 1 }{ c } \int _{ t_{ 1 } }^{ t_{ 2 } }{ \sqrt { x\prime ^{ 2 }+y\prime ^{ 2 }+z\prime ^{ 2 } } dt } \quad \therefore \quad f=\sqrt { x\prime ^{ 2 }+y\prime ^{ 2 }+z\prime ^{ 2 } }$

Since, $\frac{ \partial f }{ \partial x } =0$, then $\frac { \partial f }{ \partial x\prime}=Const.$

That means $\frac { \partial f }{ \partial y\prime}=Const.$ and $\frac { \partial f }{ \partial z\prime}=Const.$ is true too.
Subbing in f yields: $Const.=\frac { 2 }{ c } \frac { x\prime }{ \sqrt { x\prime ^{ 2 }+y\prime ^{ 2 }+z\prime ^{ 2 } } }$
for y: $Const.=\frac { 2 }{ c } \frac { y\prime }{ \sqrt { x\prime ^{ 2 }+y\prime ^{ 2 }+z\prime ^{ 2 } } }$
and for z: $Const.=\frac { 2 }{ c } \frac { z\prime }{ \sqrt { x\prime ^{ 2 }+y\prime ^{ 2 }+z\prime ^{ 2 } } }$

These can be divided to eliminate f, yielding a system of linear equations.
For x and y: $x-0={ k }_{ xy }^{ 1 }(0-y_{ 1 })\quad and\quad x_{ 2 }-x={ k }_{ xy }^{ 2 }(y_{ 2 }-0)$

For x and z: $x-0={ k }_{ xz }^{ 1 }(z-0)\quad and\quad x_{ 2 }-x={ k }_{ xz }^{ 2 }(0-z)$

For y and z: $0-y_{ 1 }={ k }_{ yz }^{ 1 }(z-0)\quad and\quad y_{ 2 }-0={ k }_{ yz }^{ 2 }(0-z)$

I am having trouble finding a way to prove the law of reflection with system of equations. For I think I know that ${ k }_{ xy }^{ 1 }=-{ k }_{ xy }^{ 2 }$, but assuming that would be circular reasoning.

I am going to look at eliminating variables. I do expect ${ k }_{ xz }=0\quad and\quad { k }_{ yz }=0$, but I am not sure I have enough info for that assumption. I am going to go back to looking at it to see if I am indeed stuck. Have I done ok so far?
Thanks,
Chris

2. Jul 25, 2014

### Orodruin

Staff Emeritus
You know already that the variational principle gives straight lines when traveling through the vacuum. Assume a reflection point and minimize the optical path (locally) wrt the reflection point.

3. Jul 26, 2014

### kq6up

I have everything set up now, but I am in personally uncharted waters mathematically speaking. I am not sure how to post my drawings and rational. I did show a friend using Open Sankoré and Skype. However, I think it was over his head.

There is the possibility I am overthinking this.

Chris

4. Jul 27, 2014

### kq6up

Ok, I see that I am not supposed to solve this with Euler, but assume what has already proven. That is were dealing with straight lines here. The pythagorean theory applies, so: $L=\sqrt { x^{ 2 }+{ y }^{ 2 }_{ 1 }+z^{ 2 } } +\sqrt { (x_{ 2 }-x)^{ 2 }+{ y }^{ 2 }_{ 2 }+z^{ 2 } }$ for the total length of the light path.

I set the first partial ($L_x=0$ and $L_z=0$) to zero. This gives me to equations with two unknows. Theoretically I should be able to find a reflection point Q from these two equations that give me the shortest path. Does this sound fine? I have crunched through all of the math and it is a mess, and still no proof yet. I wanted to double check to make sure I am going after it correctly then recheck my algebra.

Thanks,
Chris

5. Jul 28, 2014

### Orodruin

Staff Emeritus
Yes. This is the approach that is suggested. Note that the minimization in the z-direction is quite simple and you hardly need the derivative to see that it occurs for z=0.

6. Jul 29, 2014

### kq6up

Ok, I have proven z=0. However taking the partial derivative of wrt x of L up there, yields: $\newcommand{\Bold}[1]{\mathbf{#1}}\left( x, x_{2}, y_{1}, y_{2}, z \right) \ {\mapsto} \ \frac{x - x_{2}}{\sqrt{{\left(x - x_{2}\right)}^{2} + y_{2}^{2} + z^{2}}} + \frac{x}{\sqrt{x^{2} + y_{1}^{2} + z^{2}}}$ I have played with this for quite a while, but the simple ratio solution that I expect when $\theta _1=\theta _2$ eludes me.

Chris

7. Jul 30, 2014

### Orodruin

Staff Emeritus
Well, start by putting z=0 just to get rid of it. Then equate the entire derivative to zero for the minimization. How are the angles related to x, x2, y1, and y2?

Edit: Hint: since z=0 everything is 2 dimensional now - try drawing the situation on a paper if you have problems.

8. Jul 30, 2014

### kq6up

Yes, the z is set z=0.

I expected the equation to simplify to x/y1=(x2-x)/y2 since it should be a simple ratio between the opposite and adjacent sides (that is if the incoming and outgoing theta are equal). However, when I set the above partial (wrt x) to zero, I get some nasty algebra that I was not able to solve by hand. Wolfram alpha has 4th degree polynomials under a radical for the relationship. If I don't solve it, and merely do an implicit plot in sage, I see plots of two different equations. Something is wrong.

Thanks,
Chris

9. Jul 30, 2014

### kq6up

Duh, I see it now. I used the tangent to construct a ratio equation, but had I been more careful at observing my derivative wrt x, I would have seen it is a ratio equation with the cosines instead. The denominator is the hypotenuse of either side, and the x and the x2-x is the adjacent. Setting those equal to $cos \theta$ is now have the identity $\theta _1=\theta _2$.

FINALLY QED!

Thanks,
Chris