What is the notation of an inverse gradient?

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Discussion Overview

The discussion revolves around the notation and concept of an "inverse gradient," particularly in the context of vector fields and potential functions. Participants explore the mathematical implications and definitions associated with the gradient operator, as well as the appropriate notation for expressing these relationships.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants question the meaning of "inverse gradient," suggesting it could refer to either the multiplicative inverse of the gradient or a function that is the gradient of another function.
  • One participant presents an equation relating a vector field to a scalar potential function, proposing that the scalar potential can be expressed as the inverse of the gradient operator applied to the vector field.
  • Several participants assert that the gradient operator (∇) should not be treated as a vector and should not have an arrow over it, emphasizing that this notation is mathematically incorrect.
  • Another participant mentions that the notation for potential functions is typically expressed without an explicit inverse gradient notation, suggesting a more conventional approach to defining potential functions.
  • One participant describes a method for finding the potential function from a vector field using a line integral, which they refer to as an explicit notation for the "inverse gradient."
  • There is a discussion about the appropriateness of using arrows or bolding for vector notation, with some arguing that the del operator can be interpreted as a vector of partial derivative operators.
  • Another participant notes that there is no general form for the inverse of the gradient, indicating that only specific cases can be addressed, particularly due to the nature of indefinite integrals.

Areas of Agreement / Disagreement

Participants express differing views on the notation and definition of the inverse gradient, with no consensus reached on the correct approach or interpretation. The discussion remains unresolved regarding the appropriate mathematical representation and implications of the inverse gradient concept.

Contextual Notes

Participants highlight limitations in the clarity of notation and definitions, as well as the dependence on specific contexts for the application of the gradient operator and potential functions. There are unresolved questions about the mathematical validity of certain expressions and the conventions used in notation.

ainster31
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Is it just ∇-1 with the vector hat?
 
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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?
 
SteamKing said:
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?

I mean like so:

$$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F } \\ \\ where\quad \phi \quad is\quad the\quad scalar\quad potential\quad function$$
 
ainster31 said:
I mean like so:

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

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 \overrightarrow { F } =\overrightarrow { \nabla } \phi makes no sense mathematically.
 
SteamKing said:
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 \overrightarrow { F } =\overrightarrow { \nabla } \phi makes no sense mathematically.

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.
 
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 \phi be a potential function of F, such that \mathbf{F} = \nabla \phi"
 
ainster31 said:
$$\overrightarrow { F } =\overrightarrow { \nabla } \phi \\ \phi ={ \overrightarrow { \nabla } }^{ -1 }\overrightarrow { F }$$

You can find ##\phi## given ##\vec{F}## as follows. Pick a point ##\vec{x}_0## and define ##\phi(\vec{x})## via the line integral

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

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."
 
Those saying that the del operator shouldn't have an arrow are being picky. From the wikipedia article on Del:
The del symbol can be interpreted as a vector of partial derivative operators

Also, if we take a look at the OP's picture:
ainster31 said:
vKBVIpf.png

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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