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Assertion Consider the motion of a particle of mass $m$ under the influence of a potential $V(x)$ in 1-dimension. The force on that particle is given by $\textbf{F}(x)=-\frac{dV}{dx}\hat{\textbf{x}}$. From this, it is clear that the $\textbf{F}(x)$ is always directed downhill at any point on $x$-axis i.e., towards a local minimum of the potential. This is true independent of the form of the potential $V(x)$.

Question In 3-D, $\textbf{F}(x,y,z)=-\boldsymbol{\nabla} V(x,y,z)$. What can we say in this case about the direction of the force at any point $(x,y,z)$? Again I'm looking for the most general conclusion that can be drawn for an arbitrary V(x,y,z) but reasonably well-behaved.

SRS
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  • What is the form of $V(x,y,z)$? – Riccardo.Alestra Jul 18 '17 at 10:02
  • My question is for arbitrary $V(x,y,z)$ what is the most general conclusion. Note that my assertion in 1-D is quite general and true, independent of the exact form of $V(x)$. – SRS Jul 18 '17 at 10:04
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    The force is in the direction of steepest descent, this need not be in the direction of a (local/global) minimum. However, if you follow the force infinitesimally, then you will end up at a local minimum. – Peter Jul 18 '17 at 10:05
  • @Peter Is your comment about 1d or 3d? – SRS Jul 18 '17 at 10:10
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    About any dimension greater than one. – Peter Jul 18 '17 at 10:16
  • Can you give the proof? – SRS Jul 18 '17 at 10:19
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    @Peter "if you follow the force infinitesimally, then you will end up at a local minimum" is true unless, of course, no local minimum exists, or in other similar situations (e.g. V(x,y,z) = x). Compactness of domain is sufficient. – AnonymousCoward Jul 18 '17 at 13:11
  • This is a good point, the phycisict in me assumes that potentials are bounded below, which might explain this oversight – Peter Jul 18 '17 at 13:15

2 Answers2

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Let $V: \mathbb{R}^3 \rightarrow \mathbb{R}$ be a potential function.

The force is now $F = -\nabla V$, that is, it is minus the gradient of $V$. The gradient is the direction of steepest ascent. Hence, minus the gradient is the direction of steepest descent.

Now there are two claims:

  • Following the force (or minus the gradient) infinitesimally will give you a local minimum, provided that the potential is bounded below$^1$.
  • The gradient need not point in the direction of a global or local minimum.

For the second point I invite you to think about a hilly landscape, here we can define a function $h: \mathbb{R}^2 \rightarrow \mathbb{R}$ to be the height of the current position. Now, a river will flow along the direction of steepest descent (that is $-\nabla h$), but if you are standing on the bank of a river and looking downstream there is no guarantee that you are looking in the direction of the lake where the river ends up. Indeed the river may meander back and forth, and may even stream in the opposite direction before doubling back on itself!

The first point is the justification for the method of gradient descent for finding minima/maxima of functions.

A proof would presumably use the gradient theorem, but I don't have time to write anything up right now.

(1): A local minimum could still be located "at infinity", if this bothers you, you could ask for the domain of $V$ to be compact.

Peter
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$$F_x=-\dfrac{\partial V(x,y,z)}{\partial(x)}$$ $$F_y=-\dfrac{\partial V(x,y,z)}{\partial(y)}$$ $$F_z=-\dfrac{\partial V(x,y,z)}{\partial(z)}$$ So the total force is: $F=F_x+F_y+F_z$ Obviously this is a vector sum. As you said the $V(x,y,z)$ must be a reasonably well behaved potential. In that case the direction of the force is directed in a direction orthogonal to the tangent plane of $V(x,y,z)$ in any point $P(x,y,z)$

Riccardo.Alestra
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    How does it answer my question? What you've done is really expanded the relation I wrote $\textbf{F}=-\nabla V$. – SRS Jul 18 '17 at 10:16
  • You ask about the direction of the force: this is a direct consequence of the $F$ obtained by the gradient of $V(x,y,z)$ If you are able to calculate $F$ you know also the direction of the force in a proper reference frame. – Riccardo.Alestra Jul 18 '17 at 10:21
  • So towards which point is the force directed at any point (x,y,z) in space? – SRS Jul 18 '17 at 10:23
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    The force is orthogonal in every point to the tangent plane of $V(x,y,z)$ – Riccardo.Alestra Jul 18 '17 at 10:27
  • Well...This makes sense. But let me think what that means (and physically). And I'll get back to you. – SRS Jul 18 '17 at 10:29
  • This direction is given by the unit vector $\hat{\textbf{n}}=-\frac{\boldsymbol{\nabla} V}{|\boldsymbol{\nabla} V|}$. Do I get it right? – SRS Jul 18 '17 at 10:34
  • It's right. Obviously as you said the $V(x,y,z)$ must be a reasonably well behaved potential – Riccardo.Alestra Jul 18 '17 at 10:37
  • If you add your comments in the answer or modify the answer a bit I can accept it. – SRS Jul 18 '17 at 10:42
  • Well, shouldn't the total force be $\vec F=(F_x,F_y,F_z)$? – Frenzy Li Jul 18 '17 at 11:22
  • @FrenzyLi: Yes it is. In my answer I considered $F$ as the vectorial sum. The components of $F$ are as you wrote: $(F_x,F_y.F_z)$ and with this components you can calculate the cosine direction of the force respect to a reference frame in which you have the $V(x,y,z)$. – Riccardo.Alestra Jul 18 '17 at 11:24