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Orthogonality of Legendre Polynomials from Jackson

  1. Oct 4, 2011 #1
    Hello all!

    I am trying to work through and understand the derivation of the Legendre Polynomials from Jackson's Classical electrodynamics. I have reached a part that I cannot get through however. Jackson starts with the following orthogonality statement and jumps (as it seems) in his proof:

    Equation 3.17 states:

    [itex]\int P_{l'}[\frac{d}{dx} ([1-x^{2}]\frac{dP_{l}}{dx})+l(l+1)P_{l}(x)]dx=0[/itex]

    He mentions integration by parts on the "first term" but I don't see how he gets to

    Equation 3.18:

    [itex]\int [(x^{2}-1)\frac{dP_{l}}{dx} \frac{dP_{l'}}{dx} +l(l+1)(P_{l'}(x)P_{l}(x))]dx=0[/itex]

    I don't see this. Could someone please explain or give a hint to the intermediate step here? I'm afraid I just do not see it. Thanks.
  2. jcsd
  3. Oct 4, 2011 #2


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    You understand integration by parts, yes? To wit,

    [tex] \int udv = uv-\int vdu [/tex]

    So by the chain rule we can also say

    [tex] \int u \frac{dv}{dx}dx = uv-\int v \frac{du}{dx}dx [/tex]

    So it is easy to see that [itex]u=P_{\ell'}[/itex] and [itex]v=\left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right][/itex]. In this manner we arrive at

    [tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \left. P_{\ell'} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \right|^{x=1}_{x=-1} - \int_{-1}^1 \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]

    The first term works out to be zero obviously and thus we say that,

    [tex] \int_{-1}^1 P_{\ell'} \frac{d}{dx} \left[ \left(1-x^2\right)\frac{dP_\ell}{dx}\right] dx = \int_{-1}^1 \left[ \left(x^2-1\right)\frac{dP_\ell}{dx}\right] \frac{d P_{\ell'}}{dx} dx [/tex]
  4. Oct 4, 2011 #3
    Of course I understand Integration by parts, and in fact I just barely started it myself so this whole conversation is null. Thank you for your time, but I just had an aha! moment :) Sorry about that.
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