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My calculus manual suggests a gradient field is just a special case of a vector field. That implies that there are vector fields that there are not gradient fields. The gradient field is composted of a vector and each $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$ component (using 3 dimensions) is multiplied by a scalar that is a partial derivative. Is this because the scalar may be of the form that there is no antiderivative of? Perhaps of a function that can not be differentiated? Maybe a vector field that has sharp turns but I can't come up with any that I can't find a derivative for. Does anybody have any so I can get some intuition of the problem?

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    $\begingroup$ Consider a "rotational" vector field, for example $(-y,x)$ in 2D. Suppose it was a gradient of a scalar field; apply the gradient theorem on a circle around the origin and arrive at a contradiction. $\endgroup$ – user856 Aug 15 '17 at 14:32
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    $\begingroup$ See also Helmholtz decomposition theorem. Roughly speaking any vector field in $\mathbb{R}^3$ can be decomposed as ${\bf V} = \nabla \phi + \nabla\times {\bf A}$ for some scalar field $\phi$ and vector field ${\bf A}$. $\endgroup$ – Winther Aug 15 '17 at 15:31
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A vector field is just a function $\mathbf{F}:\mathbb{R}^n\to\mathbb{R}^n$. (Or the domain can be some subset $D$ of $\mathbb{R}^n$, not all of it, of course. To avoid making this remark repetitively, I'll just stick with $\mathbb{R}^n$, as an example). From an intuitive/geometrical point of view, it is best visualized as having a vector attached to (or starting from) each point of the domain $D$. Given a multivariable function $f:\mathbb{R}^n\to\mathbb{R}$, its gradient $\nabla f$ is precisely that kind of a function, as it acts from $\mathbb{R}^n$ to $\mathbb{R}^n$, so it is a vector field.

What the book says is that not all vector fields can be obtained in this way. Intuitively, it should make sense: the fact that the components of $\nabla f$ are derived from the same original function $f$ should result in some relationships among them. If you put random functions as components, most likely they wouldn't be related in that special way. It's like siblings: being the descendants of the same parents they have some things in common, something that random people wouldn't.

In more mathematical terms, one of such special relationships is the condition of equality of mixed partial derivatives. Say, for a function of two variables $f:\mathbb{R}^2\to\mathbb{R}$, we must have $\displaystyle \frac{\partial^2f}{\partial y\partial x}=\frac{\partial^2f}{\partial x\partial y}$, assuming these derivatives are continuous. Therefore, if $\displaystyle \mathbf{F}=\nabla f=\left\langle\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}\right\rangle=\langle M,N\rangle$ is its gradient, then we have to have $$\color{blue}{\frac{\partial M}{\partial y}}=\frac{\partial^2f}{\partial y\partial x}=\frac{\partial^2f}{\partial x\partial y}=\color{blue}{\frac{\partial N}{\partial x}}.$$

So being the gradient is very special! This condition allows you to construct vector fields that are not gradients. Pretty much anything you would put into a vector field at random almost surely wouldn't be a gradient. For example, $\mathbf{F}(x,y)=\langle 2x+3y,4x+5y\rangle$ isn't a gradient, because $$\frac{\partial M}{\partial y}=3\neq4=\frac{\partial N}{\partial x}.$$

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  • $\begingroup$ That is amazing....but from an intuitive geometrically visual point of view what does the geometry of the second partials being equal do to the vector field! After all when you map that vector field that happens to be a gradient you draw the First derivative vectors , correct??? the second derivative vectors are the change of slope of the First derivative vectors, correct ? so if the order does not matter and you get the same change what does this mean visually and geometrically...maybe you can tell me? $\endgroup$ – Sedumjoy Aug 15 '17 at 20:41
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    $\begingroup$ @Sedumjoy : The gradient is irrotational. For an example of something that looks like a rotational gradient, see Escher's Ascending and Descending. Note that there is no way for that staircase to exist as it appears to. (Going around the circuit once, you descend always. You must be lower when you return to your starting point. But this is absurd, the function whose gradient you are taking only has one value at that point.) $\endgroup$ – Eric Towers Aug 15 '17 at 21:34
  • $\begingroup$ Why we must have $$\frac{\partial^2f}{\partial y\partial x}=\frac{\partial^2f}{\partial x\partial y}$$? $\endgroup$ – Ivan Gonzalez May 25 at 22:56
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    $\begingroup$ @IvanGonzalez: See Clairaut's Theorem. $\endgroup$ – zipirovich May 26 at 1:04
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If you have a gradient field $(X=\partial_xf,Y=\partial_yf,Z=\partial_zf)$,

by using the relation ${\partial^2f\over{\partial x\partial y}}={{\partial^2f}\over{\partial y\partial x}}$ you obtain $\partial_yX=\partial_xY$. This is not true for every vector field.

For example $(x,x^2y^2,z)$ is not a gradient vector field.

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    $\begingroup$ And working out the converse so that you can decide when a given field is a gradient field is important in physics and mathematics. $\endgroup$ – Ethan Bolker Aug 15 '17 at 15:06
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General vector fields can generate flows and circulations. A gradient fields and only gradient fields (under some additional regularities) always generate circulations that amount to zero. Try to compute the circulation of a gradient field in a plane around an elementary rectangle centered at any point whose sides are parallel to the coordinate axes. You will get an elementary circulation that is zero simply from the fact that mixed second order partial derivatives of the scalar field does not depend on the order of derivation. Notice that the density of such elementary circulations (that is, the curl of the gradient field in every point) is also null.

So gradient fields and only gradient fields (under additional regularities) have curl identically equals to zero.

You can also see that there are fields whose flows (and elementary flow density in every point, that is their divergence) always amount to zero.

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  • $\begingroup$ You wouldn't happen to know of any two or three dimensional pictures of graphs that show a gradient field that you can share with me do you....I mean so I can compare that to a vector field that goes in a circle? $\endgroup$ – Sedumjoy Aug 15 '17 at 21:02
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    $\begingroup$ @Sedumjoy: A Google image search turns up plenty of examples. Although a few of those are examples of vector fields that are specifically not gradient fields, so do click through to the actual web pages and read the image descriptions there. $\endgroup$ – Ilmari Karonen Aug 15 '17 at 22:30
  • $\begingroup$ @llmari Karonen Your Google...brings up some fine examples. . Thank you $\endgroup$ – Sedumjoy Aug 16 '17 at 3:49

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