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## 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