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Newtonian potential in Helmholtz decomposition

  1. Apr 24, 2009 #1
    I'm trying to find a divergenceless vector field based on its curl, and discovered that I could use a http://en.wikipedia.org/wiki/Helmholtz_decomposition" [Broken], and the article I found on this didn't make much sense to me.

    First, can someone confirm that the dimension referred to in the Newtonian potential article is the number of spacial dimensions in which the field exists? If not, what is it?

    Next, the exponent on what appears to be the absolute value of x for more than two dimensions can only be negative, which doesn't make sense to me in the context of the problem I am working on. Also, is that an absolute value, or some other notation?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 27, 2009 #2

    clem

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    The d is the spatial dimension, so for our world d=3.
    2-3=-1 which means that G~1/|x| for 3 dimensions.
    Why does that bother you.
    Having G is trivial, doing the convolution is the whole problem.
    You have found out that Wiki is not for learning a subject.
     
  4. Apr 27, 2009 #3
    I am aware of the limitations of learning advanced materials on the internet. However, until I start college next year, it is my only option.

    The reason the negative exponent bothered me is because (I thought) it would mean that the final field is increasing as distance increases, which does not make sense in the problem I am doing (I am using Maxwell's equations)

    The Helmholtz Decomposition article uses the same notation as the Newtonian Potential article uses for the kernel, as opposed to the convolution. Is this misleading? Do I need to do the convolution for the Helmholtz Decomposition?

    If I do have to do a convolution, how will I do this on a three dimensional vector field?
     
    Last edited: Apr 27, 2009
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