Confusion with $F=-\nabla V$, $F$ conservative I am rather confused by the relationship $F=-\nabla V$.
If a pen drops from a height it loses potential energy so $\nabla V$ is negative. From the above equation this means that the gravitational force acting on the pen is positive, but isn't it  acting in the negative direction (downwards)?
 A: I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.
A: I think what is missing is the definitions of $F$ and $V$.
Consider first the definition of potential energy.
The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point.
That force which you exert on the mass in doing the work is equal and opposite to the force due to the gravitational field (let it be produced by the Earth) on the mass, $ F_{mE}$ as there is no change in kinetic energy of the mass.
So $\Delta V =  F_{my} \cdot \Delta  r = -  F_{mE} \cdot \Delta  r \Rightarrow \frac{\Delta V}{\Delta r} = \nabla V = - F_{mE}$
So this equation is telling you that as you lift a mass by applying an upward force on the mass and the potential energy of the mass increases the force due to the gravitation attraction of the Earth is downwards.
