What does this derivative notation mean in Goldstein's Classical Mech?

[zenterix]

Homework Statement

In the book "Classical Mechanics" by Goldstein, there is the following snippet

By a well-known theorem of vector analysis, a necessary and sufficient condition that the work $W_{12}$, be independent of the physical path taken by the particle is that $\vec{F}$ be the gradient of some scalar function of position

$$\vec{F}=-\nabla V(\vec{r})\tag{1.16}$$

where $V$ is called the potential, or potential energy. The existence of $V$ can be inferred intuitively by a simply argument. If $W_{12}$ is independent of the path of integration between the end points 1 and 2, it should be possible to express $W_{12}$ as the change in a quantity that depends only upon the positions of the endpoints. The quantity may be designated $-V$, so that for a differential path length we have the relation

Relevant Equations

$$\vec{F}\cdot d\vec{s}=-dV$$

or

$$F_s=-\frac{\partial V}{\partial s}$$

which is equivalent to 1.16.

My question is simply about the notation used here.

What does

$$F_s=-\frac{\partial V}{\partial s}$$

mean exactly?

 

[Orodruin]

Force component in the direction of the path = minus directional derivative of potential in the path direction.

 

[zenterix]

Ok, then this is how I have understood it using notation I am more familiar with.

We have a function $V(\vec{r})$.

Let $\hat{v}$ denote the unit velocity vector at $\vec{r}$.

Then, the directional derivative of $V$ at $\vec{r}$ in the direction of $\hat{v}$ is

$$V'(\vec{r};\hat{v})=\nabla V(\vec{r})\cdot\hat{v}=-\vec{F}\cdot \hat{v}=-F_s$$

$$F_s=-V'(\vec{r};\hat{v})=-\frac{\partial V}{\partial s}$$