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Don't understand critical part of derivation in textbook.

  1. Dec 2, 2012 #1
    I've been working my way through Intro to Electrodynamics (Griffiths), and in Chapter 3, one of the derivations comes out to

    ∫sin(n[itex]\pi[/itex]y/a) sin(n'[itex]\pi[/itex]y/a) dy ={ 0 if n'[itex]\neq[/itex]n
    a/2 if n'=n

    where the function is integrated from 0 to a.

    I assume there is some logical interpretation that allows me to reduce the answer down to just those two cases, but for the life of me I cannot figure out how. He skips over it as if it were trivial. Maybe it is supposed to be, but I didn't have any issues with all of the PDE's, so I find it frustrating that this is causing such a gap in my understanding.

    If it is of any help, that answer I got is :

    (an*sin([itex]\pi[/itex]m)cos([itex]\pi[/itex]n)-am*cos([itex]\pi[/itex]m)sin([itex]\pi[/itex]n)
    /([itex]\pi[/itex](m^2-n^2))


    edit: if anyone has the book, it is possible that I'm just no seeing a previous relation that simplifies this, so it may help to look
     
    Last edited: Dec 2, 2012
  2. jcsd
  3. Dec 2, 2012 #2


    $$\sin a\sin b=\frac{1}{2}\left(\cos(a-b)-\cos(a+b)\right)\Longrightarrow\int_0^a\sin\frac{n\pi y}{a}\sin\frac{n'\pi y}{a}\,dy\Longrightarrow\frac{1}{2}\int_0^a\left(\cos\left[(n-n')\frac{\pi y}{a}\right]-cos\left[(n+n')\frac{\pi y}{a}\right]\right) dy=$$

    $$=\left.\frac{1}{2}\frac{a}{(n-n')\pi}\sin\left[(n-n')\frac{\pi y}{a}\right]\right|_0^a-\left.\frac{1}{2}\frac{a}{(n+n')\pi}\sin\left[(n+n')\frac{\pi y}{a}\right]\right|_0^a=0\,\,\,,\,\text{if}\,\,\,n\neq n'$$

    since we get to evaluate above the function sine in integer multiples of [itex]\,\pi\,[/itex].

    OTOH, if [itex]\,n=n'\,[/itex] , then we get:

    $$\int_0^a\sin^2\frac{n\pi y}{a}dy=\left.\frac{a}{2\pi n}\left(\frac{n\pi y}{a}-\sin \frac{n\pi y}{a}\cos\frac{n\pi y}{a}\right)\right|_0^a=\frac{a}{2\pi n}(n\pi)=\frac{a}{2}$$
     
  4. Dec 2, 2012 #3
    Ah, I see. I didn't realize that you were supposed to redo the entire integral after substituting for n' = n. I thought you were supposed to be able to infer the answer from the result of the first calculation. Thank you, the first 2/3 of your work looks very similar to mine.

    kind of feel like a dufus, but I don't think I'd have ever though of that solution
     
  5. Dec 2, 2012 #4
    It would have been really helpful if you had provided the full reference to Griffiths.

    However I always cringe when posters bring this book up it seems to generate more than its share of queries.

    I assume you are trying to proceed from equations 3.30 / 3.31 page 130.

    Here is an extended derivation.

    Apply the boundary condition V=1 at x=0.

    [tex]1 = {C_1}\sin \frac{{\pi y}}{a} + {C_2}\sin \frac{{2\pi y}}{a} + {C_3}\sin \frac{{3\pi y}}{a}.........[/tex]

    This is a standard Fourier sine series which may be treated as follows

    Consider the Fourier sine expansion of f(by)

    [tex]f(by) = {a_1}\sin by + {a_2}\sin 2by + {a_3}\sin 3by...... + {a_n}\sin nby[/tex]

    Where the an are given by

    [tex]{a_{{n_n}}} = \frac{2}{\pi }\int\limits_0^\pi {f(by)\sin n(by)} d(by)[/tex]

    In this series if f(by) = 1 then

    [tex]{a_n} = \left[ {\begin{array}{*{20}{c}}
    {\frac{4}{{n\pi }},\;n\;odd} \\
    {0,\;n\;even} \\
    \end{array}} \right][/tex]

    and

    [tex]f(by) = 1 = \frac{4}{\pi }\left( {\sin by + \frac{1}{3}\sin 3by + \frac{1}{5}\sin 5by.....} \right)[/tex]

    Comparing this with the original series above leads to


    [tex]{C_1} = \frac{4}{\pi },\quad {C_2} = 0,\quad {C_3} = \frac{4}{{3\pi }},\quad {C_4} = 0,\quad {C_5} = \frac{4}{{5\pi }}[/tex]

    Does this help?
     
    Last edited: Dec 2, 2012
  6. Dec 2, 2012 #5
    It does help, though this is later in the derivation. Sorry I didn't include the full reference.

    The part that was giving me grief has been dealt with. I appreciate the extra info on the Fourier bit though. I still need to get more comfortable with that.
     
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