Derivative of a constant scalar field at a point

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Summary:

The Derivative of a scalar field at a point is normally described as the "cotangent vector of the field at that point". Can we define the derivative of the field at a point, for a constant scalar field, as the "linear density" of the field at the point?
Wikipedia defines the derivative of a scalar field, at a point, as the cotangent vector of the field at that point.

In particular;

The gradient is closely related to the derivative, but it is not itself a derivative: the value of the gradient at a point is a tangent vector – a vector at each point; while the value of the derivative at a point is a cotangent vector .

Now if a scalar fields value varies with x according to F = f(x) then the derivative is clearly f` (x) which may be different from zero (depending in the form of f(x)).

If the field has a constant value, over x, e.g Field = 4x. Can we say that the derivative = 4 and represents the "linear density" of the Field along the x axis ? In such a case the Gradient will be a vector (having magnitude = 4) being coincident with the x-axis. The Directional derivative of the field, at that point, will only exist along the x-axis (except as we may define values for the Field along the Y and Z axis) and will also have, in this case, a constant value of 4 in the direction of the x-axis.

I am only accustomed to thinking of the above in the case where f(x) is a function whose derivative is not a constant value.

Hope this question makes sense ? If I am correct then the answer to the above is trivial but will clarify things in my own mind.
 

Answers and Replies

  • #2
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Consider ##x \longmapsto x^3##.

Then the ##3x_0^2## represents the slope of the tangent at ##x_0##.

But at the same time it is the linear function ##\nabla_{x_0}## in
\begin{align*}f(x_0+v)&=(x_0+v)^3\\&=x_0^3+3x_0^2v+3x_0v^2+v^3\\
&=(x_0)^3 + 3x_0^2\cdot v+v^2(3x_0+v)\\&=f(x_0)+ \nabla_{x_0} \cdot v + r(v^2)
\end{align*}
which is "multiply the direction by ##3x_0^2##": ##v\longmapsto 3x_0\cdot v##.

So one perspective is the evaluated value at the point to construct the tangent, and the other is the linear function ## \nabla\, : \, x_0 \longmapsto L(3x_0^2)## where ## \nabla## maps a point to a linear function, namely the multiplication ## L(3x_0^2)\, : \,v \longmapsto 3x_0^2\cdot v##.

One is a vector: ##3x_0^2\cdot \vec{v}## and one a covector, a function: ##x\longmapsto L(3x^2)##.

Maybe it's better explained here:
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/
 
  • #3
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Hi Fresh-42

Thank you for your answer.

I know that is is very easy to confuse the concepts of, Gradient, derivative and Directional Derivative. That is not the issue here. I believe I have these concepts separated in my mind.

My question is, for example, if we had a scalar quantity which is spread in space having spherical symmetry (for example). The derivative of the field, at any point, gives the rate of change of the field at that point. If you take the dot product of this along vector X you get the directional derivative of the field along this vector. Usually we might have a field which changes along X meaning we would have a different value for the directional derivative at any particular X. Now what if the field has constant value all along X? Then surely the Directional derivative along X is zero. Unless, when moving along X, we were "adding up" how much of the Field we passed through (maybe its a constant electric field and we are moving a test charge along X). In this example we would accumulate Work moving the charge along X ant the sum of E.dx gives the total Work done. Hence we could think of the derivative of the field, along X, which is a constant, as the "density of work done moving a test charge along X" ?

I think ?
 

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