Fermat's Principle with Calculus of Variations

Homework Statement

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.]

Homework Equations

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

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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

1 person
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

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

Orodruin
Staff Emeritus
Homework Helper
Gold Member
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.

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

Orodruin
Staff Emeritus
Homework Helper
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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.

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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?

Yes, the z is set z=0.

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

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

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