• Support PF! Buy your school textbooks, materials and every day products Here!

Light in an optical fiber - variational calculus

  • Thread starter physman55
  • Start date
  • #1
13
0

Homework Statement



e5fkb5.jpg



Homework Equations



[tex]\frac{\delta{F}}{\delta{y}} - \frac{d}{dx}\frac{\delta{F}}{\delta{y'}} = 0[/tex]


The Attempt at a Solution



I'm having trouble setting this one up. If I let the functional be

[tex]F(x,x',y) = n(y)\sqrt{1+(x')^2}[/tex]

Applying the LE equation I obtain:

[tex]\frac{d}{dy}\frac{x'}{\sqrt{1+(x')^2}}n_0(1-\frac{y^2}{a^2})=0[/tex]

And this is where I stopped (I haven't used either of the hints). Any hints?
 

Answers and Replies

  • #2
13
0
Any ideas please? If p/a is small then I could write n(y)=n_0; but then I'm minimizing path length... so the answer is a straight line. Obviously this problem couldn't be that easy.
 
  • #3
213
8
it tells you y should be the principal variable so your functional should read

[tex]F(y,y')[/tex]= [tex]n(y)[/tex][tex]\sqrt{1+y'^{2}}[/tex] and then calculate the EL equations

by my calculations you should get

[tex]\frac{\partial}{\partial x}[/tex][tex](\frac{y'}{\sqrt{1+y'^{2}}})[/tex]= [tex]-2\frac{y}{a^{2}}\sqrt{1+y'^{2}}[/tex]

disregarding the (y/a)^2 term
 
Last edited:
  • #4
13
0
By "y" being the principal value my prof means that "y" is the independent variable.

For the last line, why did you ignore n(y) when you took the partial with respect to y' (on the left)?
 
  • #5
213
8
yeah Y is the degree of freedom and the Lagrangian are always expressed in term of the degrees of freedom L(q,q') and this makes more sense since you are asked to find y(x)

I didn't ignore the n(y) its just that y is comparible to [tex]\rho[/tex] and since [tex]\frac{\rho}{a}[/tex] is small n(y) = n0
 
  • #6
13
0
Ok thanks for the explanation, that makes sense. But shouldn't the d/dx be a total derivative and not a partial; and secondly; how the hell do you solve that ODE?
 
  • #7
213
8
i'm not sure about the entire thing but since dy is comparable to rho you can deduce that y' is approximate to alpha and thus all the y'^2 terms are negligible and you get the resulting ODE

[tex]y''=\frac{-2y}{a^{2}}[/tex]

which has solution

[tex]Acos(\frac{\sqrt{2}x}{a}) +Bsin(\frac{\sqrt{2}x}{a})[/tex] which kind of makes sense since you expect it to bounce off the walls in a periodic fashion
 

Related Threads on Light in an optical fiber - variational calculus

Replies
3
Views
10K
  • Last Post
Replies
1
Views
2K
Replies
0
Views
2K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
20
Views
3K
  • Last Post
Replies
0
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
Top