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Math Help - Excerpt from 3-D Schrodinger Eqaution...I'm stuck...

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Excerpt from 3-D Schrodinger Eqaution...I'm stuck...

    I'm solving the Three Dimensional Schrodinger Equation, and there is a little part that I'm stuck on...

    I'm starting with this DE:



    and I have to show that if I change the independent variable from theta to z, using the substitution



    and the identity



    I should show that the DE has the form:



    My problem is going from:




    I'm probably overlooking something pretty minor, but I'd appreciate any help!

    --Chris
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    I'm solving the Three Dimensional Schrodinger Equation, and there is a little part that I'm stuck on...

    I'm starting with this DE:



    and I have to show that if I change the independent variable from theta to z, using the substitution



    and the identity



    I should show that the DE has the form:



    My problem is going from:




    I'm probably overlooking something pretty minor, but I'd appreciate any help!

    --Chris
    Which part are you having problems with? I got it, and we know that you are a thousand times better at DE's than me, so I must be overlooking something to. The 1-z^2 is obvious right? and the rest...so which part?
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Which part are you having problems with? I got it, and we know that you are a thousand times better at DE's than me, so I must be overlooking something to. The 1-z^2 is obvious right? and the rest...so which part?
    Well...*ahem*...



    --Chris
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  4. #4
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    Quote Originally Posted by Chris L T521 View Post
    Well...*ahem*...



    --Chris
    Instead of differentiating a second time wrt theta, differentiate wrt z and use the chain rule. Then you have d/dz [....] times dz/dtheta = d/dz [....] times (- sin theta).

    And sin^2 theta = 1 - cos^2 theta = (1 - z^2) .....
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    Quote Originally Posted by Chris L T521 View Post
    My problem is going from:

    No wonder you are confused. I am confused. Typical physics "math" garbage what does "change variable from cos@ to z" is even supposed to mean? It is stuff like this made me spend much time translating physics "math" into formal understanding. Anyway, I guess I know what you have to do - whether it actually makes any sense I live it to you to decide, because it makes no sense to me (eventhough I have experience in trying to understand what the problem is asking).

    Open up the derivative,
    (-1/sin@)*[cos@*(dT/d@)+sin@*(d^2T/d@^2)] ----- [1]
    Now,
    dT/d@ = (dT/dz)*(dz/d@)=(dT/dz)*(-sin @)
    And,
    d^2T/d@^2 = d/d@(dT/d@)=d/d@(- dT/dz*sin @)=
    -d/d@(dT/dz)*sin@ - dT/dz*cos @= d^2T/dz^2 * sin @ - dT/dz*cos@
    Go back to [1] and substitute (and divide out),
    -cos@*dT/dz+d^2T/dz^2*sin@ - dT/dz*cos@
    Use the fact that z=cos @ to get,
    d^2T/dz^2 *sin@ - dT/dz*(1+z)
    And that is supposed to give the thing you are trying to show.

    I probably made a mistake somewhere but it is late and I cannot check it now. See if that mess above makes any sense?
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Instead of differentiating a second time wrt theta, differentiate wrt z and use the chain rule. Then you have d/dz [....] times dz/dtheta = d/dz [....] times (- sin theta).

    And sin^2 theta = 1 - cos^2 theta = (1 - z^2) .....
    Quote Originally Posted by ThePerfectHacker View Post
    No wonder you are confused. I am confused. Typical physics "math" garbage what does "change variable from cos@ to z" is even supposed to mean? It is stuff like this made me spend much time translating physics "math" into formal understanding. Anyway, I guess I know what you have to do - whether it actually makes any sense I live it to you to decide, because it makes no sense to me (eventhough I have experience in trying to understand what the problem is asking).

    Open up the derivative,
    (-1/sin@)*[cos@*(dT/d@)+sin@*(d^2T/d@^2)] ----- [1]
    Now,
    dT/d@ = (dT/dz)*(dz/d@)=(dT/dz)*(-sin @)
    And,
    d^2T/d@^2 = d/d@(dT/d@)=d/d@(- dT/dz*sin @)=
    -d/d@(dT/dz)*sin@ - dT/dz*cos @= d^2T/dz^2 * sin @ - dT/dz*cos@
    Go back to [1] and substitute (and divide out),
    -cos@*dT/dz+d^2T/dz^2*sin@ - dT/dz*cos@
    Use the fact that z=cos @ to get,
    d^2T/dz^2 *sin@ - dT/dz*(1+z)
    And that is supposed to give the thing you are trying to show.

    I probably made a mistake somewhere but it is late and I cannot check it now. See if that mess above makes any sense?
    It finally clicked. Thanks TPH and Mr. Fantastic!

    --Chris
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