Explanation for $M{\ddot{r}}=-\nabla \phi$

Question

Could someone please explain this equation $$M\bf {\ddot{r}}=-\nabla \phi$$ Where $\bf r$ is a position vector and $\phi$ is the potential function. Could someone brief explain the potential function and tell me why we've got minus sign before the nabla operator?

Answer 1

The minus sign is only there by convention. You could replace $\phi$ with $-\phi$ and the minus sign would go away. Note that $\nabla \phi$ points in the direction of steepest ascent for $\phi$, whereas $-\nabla \phi$ points in the direction of steepest descent. Perhaps it seems nice for the force on an object to be pointing in a direction of descent for $\phi$, as if the object is trying to move to a location where $\phi$ is as small as possible.

It's simply a mathematical fact that if a vector field $F$ is conservative, then there exists a scalar-valued function $\phi$ such that $-\nabla \phi = F$. If $F$ is a force field that some object is moving in, then we also have $F = Ma$, which yields your equation \begin{equation*} M {\ddot r} = -\nabla \phi. \end{equation*}

Assuming that a vector field $F$ is conservative, how do we show mathematically the existence of $\phi$? You can do it by first selecting some reference point $x_0$ arbitrarily, then defining \begin{equation} \phi(x) = -\int_{x_0}^x F \cdot dr. \end{equation} This line integral is taken along any path connecting $x_0$ to $x$. You get the same answer no matter which path you take, because $F$ is conservative.

Then note that \begin{align} \phi(x + \Delta x) - \phi(x) &= -\int_x^{x + \Delta x} F(r) \cdot dr \\ &\approx -\int_x^{x + \Delta x} F(x) \cdot dr \\ &= \langle -F(x), \Delta x \rangle. \end{align} The approximation is good when $\Delta x$ is small. Comparing this with the equation \begin{equation} \phi(x + \Delta x) - \phi(x) \approx \langle \nabla \phi(x), \Delta x \rangle \end{equation} shows that $\nabla \phi(x) = - F(x)$, or in other words \begin{equation*} -\nabla \phi(x) = F(x). \end{equation*}

Answer 2

It's total convention. The idea is that one can caluclate, for any path:

$$\Delta\left(\frac{1}{2}mv^{2}\right) = \int_{\rm path}{\vec F}\cdot d{\vec s}$$

One can then split the left hand side into a conservative bit and a nonconservative bit. (with the definition of conservative being ${\vec \nabla}\times{\vec F} = 0$). Then, since we know that any curlless vector $V$ can be expressed as ${\vec \nabla}f$ for some function $f$, we choose $f = -\phi$, and then the above expression becomes

$$\begin{align}\Delta {\rm KE} &= -\Delta \phi + \int_{\rm path}{\vec F_{\rm noncon}}\cdot d{\vec s}\\ \Delta {\rm KE} + \Delta \phi &= \int_{\rm path}{\vec F_{\rm noncon}}\cdot d{\vec s} \end{align}$$

So, we can identify $\phi$ with the potential energy, and we see that ${\vec F} = -{\vec \nabla}\phi$

We choose the minus sign just to make the energy conservation equation look prettier, really.