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Laplacian of 1/r Explodes at Origin

  1. Sep 7, 2013 #1
    Ok, there are a couple of other threads about this, but they don't seem to answer my question.

    If I take the double derivative of 1/r, I'll get 2/r^3, but if I take the laplacian, I get something different. Why?

    Namely:

    [itex]\frac{d}{dr}\frac{d}{dr}(\frac{1}{r}) = \frac{d}{dr} (\frac{-1}{r^{2}}) = \frac{(-1)(-2)}{r^{3}} = \frac{2}{r^{3}}[/itex]

    However:
    [itex]∇^{2}(\frac{1}{r}) = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{1}{r})] = 0 [/itex] or [itex]- \frac{δ(r)}{r^{2}}[/itex] or [itex]-4\piδ(r)[/itex]
    Which of these answers is correct depends on who you ask - and the situation you are in. I am not interested in this debate. My question is not which one of these answers is correct; my question is more general.


    Why they are different at all. Why is it easily differentiable in one case and not the other? What basic calculus definition did I forget or never learn?
     
  2. jcsd
  3. Sep 7, 2013 #2
    Also, I'm new to this forum. I have some level of uncertainty how to tag this.
     
  4. Sep 7, 2013 #3

    D H

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    The Laplacian in three dimensional space using spherical coordinates is not given by ##\nabla^2 f = \frac {d^2f}{dr^2}##. It is instead given by ##\nabla^2 f = \frac 1{r^2} \frac{\partial}{\partial r}\left(r^2\frac {\partial f}{\partial r}\right) +## terms involving ##\frac {\partial}{\partial \theta}## and ##\frac {\partial}{\partial \phi}##. So of course you are going to see a discrepancy between ##\frac {df^2}{dr^2}## and ##\nabla^2 f##, even for a function that depends only on radial distance.

    Why is the Laplacian in spherical coordinates expressed that way? The Laplacian is defined in terms of Cartesian coordinates. That more complex expression for spherical coordinates is what is needed to arrive at a result consistent with the definition of the Laplacian.

    With regard to ##\nabla^2 \frac 1 r##, what do you do want to with point masses, point charges? Declare them illegal? Of course not. That three dimensional Dirac delta is what is needed to allow for point objects with a 1/r potential.
     
    Last edited: Sep 7, 2013
  5. Sep 7, 2013 #4
    Sorry, I guess I need to be more clear with my example.

    I am assuming:

    [itex]\frac{1}{r^{2}}\frac{d}{dr}r^{2}\frac{d}{dr}(\frac{e^{ar}}{r}) = \frac{1}{r^{2}}\frac{d}{dr} r^{2}(\frac{-e^{ar}}{r^{2}}+\frac{ae^{ar}}{r}) = \frac{1}{r^{2}}\frac{d}{dr} (-e^{ar}+are^{ar}) = \frac{1}{r^{2}} (-ae^{ar}+a^{2}re^{ar}+ae^{ar}) = \frac{a^{2}e^{ar}}{r} [/itex]

    Am I wrong about that?

    And then:
    [itex]∇^{2}(\frac{e^{ar}}{r}) =\frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{e^{ar}}{r})] = [/itex] something different.

    One solution I've seen is
    [itex]∇^{2}(\frac{e^{ar}}{r}) =\frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{e^{ar}}{r})] =\frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}(\frac{ae^{ar}}{r}+e^{ar}\frac{∂}{∂r}\frac{1}{r})]=\frac{1}{r^{2}}\frac{∂}{∂r}(are^{ar}+r^{2}e^{ar}\frac{∂}{∂r} \frac{1}{r})=[/itex]
    [itex]\frac{1}{r^{2}}(a^{2}re^{ar}+ae^{ar}+\frac{∂}{∂r}r^{2}e^{ar}\frac{∂}{∂r} \frac{1}{r})=
    \frac{a^{2}e^{ar}}{r}+\frac{ae^{ar}}{r^{2}}+\frac{1}{r^{2}}\frac{∂}{∂r}r^{2}e^{ar}\frac{∂}{∂r} \frac{1}{r}=\frac{a^{2}e^{ar}}{r}+\frac{ae^{ar}}{r^{2}}-4\pi∂(r)[/itex]

    From a credible source. What am I not understanding?
     
  6. Sep 7, 2013 #5
    I hope I typed that in right
     
  7. Sep 7, 2013 #6

    D H

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    That looks correct for r>0. But what about r=0?

    The term ##\frac{ae^{ar}}{r^2}## looks completely bogus. It does not agree with your obvious solution for r>0. To be pedantically correct, that final term should be the three dimensional delta distribution ##-4\pi\delta^3(\vec r)##.
     
  8. Sep 8, 2013 #7
    Hi !

    The results are different because they correspond to different configurations of the field 1/r

    First :
    The field depends of a variable in one direction only : it depends of x only for example, then x=r
    [itex]∇^{2}(\frac{1}{r}) =\frac{∂^2}{∂r^2}(\frac{1}{r}) = 0 [/itex]
    In 3-D, in this case r is NOT the distance relatively to a point O "origin", but is the orthogonal distance relatively to à plane (the plane Oy,Oz)

    Second :
    The field depends of two variables, i.e. in two directions : it depends of x and y for example, then r=sqrt(x²+y²)
    [itex]∇^{2}(\frac{1}{r}) = \frac{1}{r}\frac{∂}{∂r}[r\frac{∂}{∂r}(\frac{1}{r})] = 0 [/itex]
    In 3-D, in this case r is the orthogonal distance relatively to à line (the line Oz)

    Third :
    The field depends of three variables, i.e. in three directions : it depends of x, y and z, then r=sqrt(x²+y²+z²)
    [itex]∇^{2}(\frac{1}{r}) = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{1}{r})] = 0 [/itex]
    In this case r is the distance relatively to a point : the "origin" O.

    They are many other configurations of field, to which a different formula of Laplacian corresponds.
    Of course, the results are different. We cannot say what is "the good one" without knowing what is the configuration of field that we intend to model.
     
  9. Sep 10, 2013 #8
    Thank you JJacquelin,

    Why is this 0 instead of [itex]\frac{2}{r^{3}}[/itex]?
     
  10. Sep 10, 2013 #9
    because I thought that you where considering the Laplace equation, i.e. :
    Laplacian(1/r) = 0.
    If you are not considering the Laplace equation, but the Laplacian itself, just suppress all the =0 from my preceeding post.
     
    Last edited: Sep 10, 2013
  11. Sep 10, 2013 #10
    I really appreciate your help. Why is the result of the Laplacian different from that of a double derivative?
     
  12. Sep 10, 2013 #11
    Your question is a non-sens. You cannot say "THE Laplacian", because what Laplacian are you thinking of, among the many that could be. Don't you understand the three examples of different Laplacians that I show you, among the many possible ?
    Moreover, you implicitly compare some expressions made with doubles partial derivatives to a simple double derivative. This is only comparable in ONE dimentional field. The Laplacian corresponding to this case is equal to the double derivative, not diifferent as you write. Look at the first example given in my first post.
     
  13. Sep 10, 2013 #12

    vanhees71

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    The Laplace operator is defined as [itex]\vec{\nabla} \cdot \vec{\nabla}[/itex]. In spherical coordinates it reads
    [tex]\Delta \Phi=\frac{1}{r}\frac{\partial^2}{\partial r^2}(r \Phi) + \frac{1}{r^2 \sin \vartheta} \frac{\partial}{\partial \vartheta} \left (\sin \vartheta \frac{\partial \Phi}{\partial \vartheta} \right ) + \frac{1}{r^2 \sin^2 \vartheta} \frac{\partial^2 \Phi}{\partial \varphi^2}.[/tex]
    Applying this to
    [tex]\Phi(r)=\frac{1}{r},[/tex]
    gives indeed [itex]\Delta \Phi(r)=0[/itex] for [itex]r>0[/itex].
     
  14. Sep 13, 2013 #13
    Here is what I've found as my answer, but had trouble formulating a question:

    [itex]∇^{2}(\frac{1}{r}) = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{∂}{∂r}(\frac{1}{r})] = \frac{1}{r^{2}}\frac{∂}{∂r}[r^{2}\frac{-1}{r^{2}}] = \frac{1}{r^{2}}\frac{∂}{∂r}(-1) =0 [/itex]

    However,

    [itex]\int_V ∇^{2}(\frac{1}{r})dV =\int_V ∇\cdot(∇\frac{1}{r}dV) =\int_S (∇\frac{1}{r})\cdot da =\int_S (\frac{∂}{∂r}\frac{1}{r}\hat{r})\cdot da =\int_S \frac{-1}{r^{2}} r^{2}sin\theta~ d\theta d\phi =-4\pi[/itex]

    Everybody can agree that [itex] \int_V 0 ~dV \neq -4\pi [/itex]

    What exactly is the problem? for 1/r there is a problem for r = 0. As r → 0, 1/r → ∞.

    So we put a delta dirac in there, and make it:

    [itex]∇^{2}(\frac{1}{r}) = -4\pi \delta(r)[/itex]

    Ok, I now have the information required to carry on and finish homework and what not. However, one unsettling question remains. What rule of differentiation am I forgetting? Is there some rule against discontinuities, and 1/r has a discontinuity at r=0?
     
  15. Sep 13, 2013 #14

    D H

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    The rules you learned in introductory calculus don't work at discontinuities. Consider the Heaviside step function H(x): H(x)=0 for x<0, 1 for x>0. What's the derivative of this function? It's zero everywhere but x=0. At x=0, the rules you learned say that it's undefined. "Physics math" (I'm guilty of using physics math) defines the derivative of H(x) to be some magical "function" δ(x) that when integrated regenerates the step function. This δ(x) is not a function. It's a distribution, a generalization of the concept of a function.

    This concept can be made rigorous, but you'll need to learn measure theory and Lebesgue integration first. Teaching that is not something that can be done in a question and answer type of internet forum such as this. You'll have to take an upper level math class (or more) to learn these concepts, rigorously.

    Or you can just do what physicists do: Use non-rigorous physics math. Here's a non-rigorous way to look at the delta function. Create a sequence of everywhere differentiable functions that in the limit becomes equal to the step function. Now create another sequence that comprises the derivatives of the functions in the original sequence. Think of the delta function as the limit of the sequence of derivative functions.
     
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