- #1

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Is it just ∇

^{-1}with the vector hat?- Thread starter ainster31
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- #1

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Is it just ∇^{-1} with the vector hat?

- #2

SteamKing

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- #3

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I mean like so:

$$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F } \\ \\ where\quad \phi \quad is\quad the\quad scalar\quad potential\quad function$$

- #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.I mean like so:

$$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F } \\ \\ where\quad \phi \quad is\quad the\quad scalar\quad potential\quad function$$

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.

- #5

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Here is the usage in the following context: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.

Also, the nabla has to have the arrow or else the equation doesn't make sense because a scalar cannot equal a vector.

- #6

Office_Shredder

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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

- #7

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You can find ##\phi## given ##\vec{F}## as follows. Pick a point ##\vec{x}_0## and define ##\phi(\vec{x})## via the line integral$$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F }$$

[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."

- #8

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Also, if we take a look at the OP's picture:The del symbol can be interpreted as a vector of partial derivative operators

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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