Physics 351 — Monday, February 2, 2015

I HW #3 due Friday.

I You read Chapter 6 (calculus of variations) for today.

I Remember quiz Weds: basically one HW1 problem, ∼ 15 min.

I On HW3 XC1, I should add, “Use substitutionsin(φ/2) = Au, where A = sin(Φ/2).”

Fermat: light travels along “path of least time”

Snell’s law works for any number of discrete interfaces between niand ni+1. But what happens if n(x) is a continuous function of x?Then the time of flight for path y(x) is an integral:

t =∫ xf

x=xi

dsv(x)

=1c

∫ xf

x=xi

ds n(x) =1c

∫ xf

xi

dxn(x)√

1 + (y′)2 = T [y]

For example, if n(x) = ax then we want to minimize the integral

T [y] =a

c

∫ xf

xi

dxx√

1 + (y′)2

i.e. we want to figure out the path y(x) that makes this integral assmall as possible.

Static equilibrium occurs when potential energy U is a minimumw.r.t. possible movements of the system.

In ordinary calculus, we have a “function” g(x) that maps somenumber x into some other number g(x). When we minimize afunction, we seek some number x0 for which a function g(x) is(locally) flat, i.e.

0 =[

dgdx

]x=x0

= g′(x0)

The result is a point x0 at which

limε→0

g(x0 + ε)− g(x0)ε

= 0

In calculus of variations, we have a “functional” I[y] that mapssome function y(x) into a (scalar) number:

I[y] =∫ xf

xi

dx f(x, y, y′)

For example, I[y] may be the total path length (f =√

1 + (y′)2)

I[y] =∫ xf

xi

dx√

1 + (y′)2

We want to find the function y(x) that “minimizes” this integral.

Near the local minimum x0 of an ordinary function,

g(x0 + ε)− g(x0) = g′(x0) ε+O(ε2) = 0 +O(ε2)

In other words, when we change x by ε, the change in the functiong(x) has no first-order dependence on ε.

I[y] =∫ xf

xi

dx f(x, y, y′)

Let’s try varying y(x) and insist that “small” variations (∝ ε) iny(x) will cause no change in I to first order in ε.

Consider a “nearby” path Y (x, ε) = y(x) + εη(x).

But we insist that Y (x) coincide with y(x) at the endpoints, i.e.η(xi) = η(xf ) = 0.

I[Y ] =∫ xf

xi

dx f(x, Y, Y ′)

This I[Y ] is a function of ε, so we’ll write I(ε).

I(ε) = I(ε = 0) + ε

[dIdε

]ε=0

+O(ε2)

where I(ε = 0) is just I[y], i.e. where Y (x) = y(x). We want thered term to vanish when Y (x) = y(x), i.e. we want I(ε) to haveno first-order dependence on ε along the path for which ε = 0.

0 =[

dIdε

]ε=0

=[

ddε

∫ xf

xi

dx f(x, Y, Y ′)]ε=0

=∫ xf

xi

dx[∂f

∂Y

∂Y

∂ε+

∂f

∂Y ′∂Y ′

∂ε

]ε=0

(∗∗)

Annoyingly, ∂Y ′/∂ε = η′(x), while ∂Y/∂ε = η(x). We want towork on the second term to eliminate the η′(x) dependence, sothat both terms are multiplied by η(x), not η′(x). Notice that

ddx

(∂f

∂Y ′∂Y

∂ε

)=(

ddx

∂f

∂Y ′

)∂Y

∂ε+

∂f

∂Y ′

(d

dx∂Y

∂ε

)

ddx

(∂f

∂Y ′∂Y

∂ε

)=(

ddx

∂f

∂Y ′

)∂Y

∂ε+

∂f

∂Y ′

(d

dx∂Y

∂ε

)=(

ddx

∂f

∂Y ′

)∂Y

∂ε+

∂f

∂Y ′

(∂

∂ε

dYdx

)=(

ddx

∂f

∂Y ′

)∂Y

∂ε+

∂f

∂Y ′

(∂Y ′

∂ε

)Now integrating both sides,∫ xf

xi

dxd

dx

(∂f

∂Y ′∂Y

∂ε

)=[∂f

∂Y ′∂Y

∂ε

]xf

xi

=

=∫ xf

xi

dx(

ddx

∂f

∂Y ′

)∂Y

∂ε+∫ xf

xi

dx∂f

∂Y ′∂Y ′

∂ε

Then solving for the last term, we get∫ xf

xi

dx∂f

∂Y ′∂Y ′

∂ε=[∂f

∂Y ′∂Y

∂ε

]xf

xi

−∫ xf

xi

dx(

ddx

∂f

∂Y ′

)∂Y

∂ε

The boundary term is zero because ∂Y/∂ε = η(x), and byconstruction η(xi) = η(xf ) = 0.

[∂f

∂Y ′∂Y

∂ε

]xf

xi

=[∂f

∂Y ′η(x)

]xf

xi

=[∂f

∂Y ′

]η(xf )−

[∂f

∂Y ′

]η(xi) = 0

because η(xi) = η(xf ) = 0 by construction. So then∫ xf

xi

dx∂f

∂Y ′∂Y ′

∂ε= −

∫ xf

xi

dx(

ddx

∂f

∂Y ′

)∂Y

∂ε

Now we plug this into the (**) equation above (which = 0):∫ xf

xi

dx[∂f

∂Y

∂Y

∂ε+

∂f

∂Y ′∂Y ′

∂ε

]ε=0

=∫ xf

xi

dx[∂f

∂Y

∂Y

∂ε−(

ddx

∂f

∂Y ′

)∂Y

∂ε

]ε=0

Now remember that ∂Y/∂ε = η(x).

0 =∫ xf

xi

dx[η(x)

(∂f

∂Y− d

dx∂f

∂Y ′

)]ε=0

And that η(x) was an arbitrary function, as long as η(x) = 0 atthe two endpoints. Imagine η(x) zero everywhere except for asmall blip at x = x∗, where xi < x∗ < xf .

Imagine η(x) zero everywhere except for a blip at x = x∗. Then[∂f

∂Y− d

dx∂f

∂Y ′

]ε=0

= 0

at x = x∗. But x∗ is an arbitrary point along xi < x∗ < xf . So forthe entire path corresponding to ε = 0, we must have

∂f

∂y− d

dx∂f

∂y′= 0

where I let ε = 0 and used Y (x, ε) = y(x) + εη(x), so Y (x) = y(x)at ε = 0. So a necessary condition for the function y(x) to“extremize” the functional

I[y] =∫ xf

xi

dx f(x, y, y′)

is that everywhere along the path y(x), we have

∂f

∂y=

ddx

∂f

∂y′

(We didn’t get to this part — continue from hereWednesday.)

Suppose we want to find the path y(x) from xi to xf thatminimizes the total arc length

L[y] =∫

ds =∫ xf

xi

dx√

1 + (y′)2

Use the Euler-Lagrange equation

∂f

∂y=

ddx

∂f

∂y′

to find the equation for such a path. Try it!

Example of using Mathematica for last week’s XC online (PDF andNB) in “files” area:

positron.hep.upenn.edu/p351/files/0202_nick_hw2ec.pdf

positron.hep.upenn.edu/p351/files/0202_nick_hw2ec.nb

Physics 351 — Monday, February 2, 2015

I HW #3 due Friday.

I Remember quiz Weds: basically one HW1 problem, ∼ 15 min.You can bring one sheet of paper, handwritten, one or bothsides.

I On HW3 XC1, I should add, “Use substitutionsin(φ/2) = Au, where A = sin(Φ/2).”