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What is the notation of an inverse gradient?

  1. Nov 20, 2013 #1
    Is it just ∇-1 with the vector hat?
     
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  3. Nov 20, 2013 #2

    SteamKing

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    It's not clear what you mean by 'inverse gradient.' Are you talking about the multiplicative inverse of the gradient or do you have some function which is the gradient of a second, unknown function?
     
  4. Nov 20, 2013 #3
    I mean like so:

    $$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F } \\ \\ where\quad \phi \quad is\quad the\quad scalar\quad potential\quad function$$
     
  5. Nov 20, 2013 #4

    SteamKing

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    Strictly speaking, the operator ∇ is not a vector and it is never written with an arrow over the top.

    The following article shows its definition and how it is applied to scalar and vector functions:

    http://en.wikipedia.org/wiki/Del

    AFAIK, your equation [itex]\overrightarrow { F } =\overrightarrow { \nabla } \phi [/itex] makes no sense mathematically.
     
  6. Nov 20, 2013 #5
    Here is the usage in the following context:

    vKBVIpf.png

    Also, the nabla has to have the arrow or else the equation doesn't make sense because a scalar cannot equal a vector.
     
  7. Nov 20, 2013 #6

    Office_Shredder

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    As far as notation goes, nabla should not have an arrow. Notice that the source that you posted doesn't even have an arrow over the nabla.

    I have never seen any kind of notation to explicitly denote the protential function of a vector field... typically one would just write
    "Let [itex] \phi[/itex] be a potential function of F, such that [itex] \mathbf{F} = \nabla \phi [/itex]"
     
  8. Nov 21, 2013 #7
    You can find ##\phi## given ##\vec{F}## as follows. Pick a point ##\vec{x}_0## and define ##\phi(\vec{x})## via the line integral

    [tex]\phi(\vec{x}) = \int_{\vec{x}_0}^{\vec{x}} d\vec{z} \cdot F(\vec{z})[/tex]

    where it doesn't matter what path you integrate along between ##\vec{x}_0## and ##\vec{x}## if ##\vec{F}## really is the gradient of some potential. So this line integral gives an explicit notation for the "inverse gradient."
     
  9. Nov 21, 2013 #8
    Those saying that the del operator shouldn't have an arrow are being picky. From the wikipedia article on Del:
    Also, if we take a look at the OP's picture:
    The F does not have an arrow above it either, even though it is a vector field. It is bolded instead--textbooks usually do this instead of putting an arrow above a vector. It is hard to tell but the del might also be bolded too. Therefore it is hard to draw conclusions from that picture whether or not del should be drawn with an arrow.

    The wikipedia article on the Gradient Theorem should answer your question, OP, if it hasn't been answered already. There is no general form of the inverse of the gradient, however, only one for specific lines, since the indefinite integral (which would be a candidate for the general form) adds a constant.
     
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