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Laplace equation charge reflection

  1. Sep 16, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that

    If [itex]\phi[/itex](x,y,z) is a solution of Laplace's equation, show that
    [itex]\frac{1}{r}\phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )[/itex] is also a solution


    2. Relevant equations



    3. The attempt at a solution

    let [itex]\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )[/itex] is a solution.

    Then in the spherical coordinate,

    [itex]\psi=\frac{1}{r} \phi ( \frac{1}{r} , \theta , \varphi )[/itex]

    So input [itex]\psi [/itex] to the spherical laplace equation.

    [itex]\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\psi}{∂r}) = \frac {2}{r^4} \frac{∂\phi}{∂r} - \frac{1}{r^3} \frac{∂^2 \phi}{∂r^2} [/itex]

    The derivation with other angles, same to the original one, except 1/r times factor.

    But the r part for the original one is

    [itex]\frac{1}{r^2}\frac{∂}{∂r} (r^2 \frac{∂\phi}{∂r}) = \frac {2}{r} \frac{∂\phi}{∂r} + \frac{∂^2 \phi}{∂r^2} [/itex]

    What's wrong with me?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 17, 2012 #2

    gabbagabbahey

    User Avatar
    Homework Helper
    Gold Member

    Here you are assuming what you are trying to prove. That is not usually a good idea.

    Instead, just compute the Laplacian of [itex]\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )[/itex] and show that it is zero if [itex]\nabla^2 \phi(x,y,z)=0[/itex]

    No. What are [itex]x[/itex], [itex]y[/itex] and [itex]z[/itex] in terms of spherical coordinates [itex]r[/itex], [itex]\theta[/itex] and [itex]\varphi[/itex]? What does that make [itex]\frac{x}{r^2}[/itex], [itex]\frac{y}{r^2}[/itex], and [itex]\frac{z}{r^2}[/itex]?
     
  4. Sep 17, 2012 #3


    [itex]\psi= \frac{1}{r} \phi (\frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} )= \frac{1}{r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})[/itex]

    [itex] \frac{∂}{∂r} \phi(\frac{sin \theta cos\varphi}{r}, \frac{sin \theta sin\varphi}{r},\frac{cos \theta}{r})= -( \frac{ sin\theta cos\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}+\frac{ sin\theta sin\varphi}{r^2} \frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})}+\frac{ cos\theta}{r^2}\frac{∂\phi}{∂(\frac{sin \theta sin\varphi}{r})} ) [/itex]

    Then how can I calculate the terms like [itex] \frac{∂\phi}{∂(\frac{sin \theta cos\varphi}{r})}[/itex]?
     
  5. Sep 17, 2012 #4

    gabbagabbahey

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    Homework Helper
    Gold Member

    good.

    Just use a substitution like [itex]\bar{x} \equiv \frac{\sin \theta \cos\varphi}{r}[/itex], [itex]\bar{y} \equiv \frac{\sin \theta \sin\varphi}{r}[/itex], and [itex]\bar{z} \equiv \frac{\cos \theta}{r}[/itex] to make it easier to write:

    [tex]\frac{\partial \phi\left( \frac{x}{r^2} ,\frac{y}{r^2} , \frac{z}{r^2} \right)}{\partial (\frac{\sin \theta \cos\varphi}{r})} = \frac{ \partial \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial \bar{x}}[/tex]?

    In the end, you should get something like [itex]\nabla^2 \psi = \text{some factor} * \left[ \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{x}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{y}} + \frac{ \partial^2 \phi( \bar{x}, \bar{y}, \bar{z})}{ \partial^2 \bar{z}} \right][/itex]

    Aside: Changing the size on all your posts makes them difficult to read (for my eyes anyways), please stop it.
     
  6. Apr 6, 2013 #5
    I am stuck on this too

    Hi!

    I was just trying to solve this problem through this same approach (make the substitutions [itex]\overline{x}=\sin(\theta)\cos(\phi)/r[/itex], etc...). But it turns out to be *very* complicated. Do you have finished this one? Can you show me more explicit steps? Thanks a lot!!
     
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