I'm learning about work in my dynamics class right now. We have defined the work on a particle due to the force field from point A to point B as the curve Integral over the force field from point A to B. From math I know that if a vector field has a potential, we only need to evaluate the potential at point B minus the potential at point A to get the result of the curve Integral. In the text that I'm reading, it's explained that if the integral over a force-field is path-independent, then the force field $F = -{\rm grad}(V)$, where $V$ is the potential. Why is it defined as the negative gradient? Doesn't one determine the potential from $F$ mathematically. Why do we impose the sign on the potential?
Forces – Why is a Conservative Force Defined as the Negative Gradient of a Potential?
conventionsforcespotential energy
Related Solutions
I think you just forgot that the $\int_A^B F\,dl$ is not a scalar expression. Rather it should be written in a form $\int_A^B \vec{F}\cdot d\vec{l}$. Then it comes to the sign of the scalar product: $$\vec{F}\cdot d\vec{l}=F\,dl\,\cos\theta$$ where the angle $\theta$ is taken between the vector $\vec{F}$ and the direction of the tangent to the integration path from $A$ to $B$. Then, in your first example,
$W_{A \to B}=\int_{r_A}^{r_B} F(r) dr = \int_{r_A}^{r_B} \left(-\frac{GMm}{r^2} \right)dr$
the path could go with any slope, but the gravity is always directed downwards, along the $r$ axis. That means, we can always take $(\pi-\theta)$ as the angle between the vector $d\vec{l}$ and the $r$ axis, that is $$dl\,\cos(\pi-\theta)=dr$$ but $\cos(\pi-\theta)=-\cos\theta$ and thus we have $$\vec{F}\cdot d\vec{l}=-F\,dr=-\frac{GMm}{r^2}dr$$
For your second example:
...we also should change the sign, because the gravitational force is always a force of attraction.
what the authors actually mean is that: the Coulomb's and Newton's forces have exactly the same expressions, but the sign conventions for them are different. The Newton's force is defined that if all the quantities ($M$, $m$ and $r$) are positive, then the vector of the force is directed towards the other body. But for the Coulomb's force, if all the quantities ($q_1$, $q_2$ and $r$) are positive, then the vector of the force is directed away from the other charge. That becomes manifest if we take the vector expressions for these forces: $$\vec{F}_N=-\frac{GMm\,\vec{r}}{r^3}\qquad\vec{F}_C=\frac{q_1q_2\,\vec{r}}{4\pi\epsilon_0\,r^3}$$ Now the different signs are clearly seen.
"...from point $A$ to point $B$..." - ...as I understand it - the work that I must do is always $U_B-U_A$. However the work that the force that is being created by the field do_es_ is always $U_A-U_B$, am I correct?
Yes this is correct.
The mnemonic rule is very simple: $U$ is like the height of the slope. When you go up, $U_B>U_A$, and it is you who does the work. But when you go down, $U_A>U_B$, and it is the field force who does the work.
The work done that we consider here is the work done by external force (or by us).
So if the potential energy at a point is high then it means that the work done by us against the conservative field (say, gravity) will be also be high.
For example, if we lift a body, it's potential energy increases with the height because we are doing work against the conservative force (i.e, gravitational force) and this work (+ve work) gets stored in the body as its potential energy. In other words, higher we lift the body higher will be its potential energy.
Energy conservation plays a very important role here.
Now, if the body comes down the work done by gravity will be positive. The body's potential energy will be converted into kinetic energy. The body is coming from a point of high potential energy to a point of lower potential energy, so this loss will be the gain in the kinetic energy (or the work done by gravity).
Therefore, the work done by the conservative force (gravity) will be equal to the loss in potential energy.
Since the work done by gravity is $+ve$ and there is a loss in potential energy $(\Delta U=-ve)$,
$W_{conservative}=-\Delta U$
Best Answer
We introduce a minus sign to equate the mathematical concept of a potential with the physical concept of potential energy.
Take the gravitational field, for example, which we approximate as being constant near the surface of Earth. The force field can then be described by $\vec{F}(x,y,z)=-mg\hat{e_z}$, taking the up/down direction to be the $z$ direction. The mathematical potential $V$ would be $V(x,y,z) = -mgz+\text{Constant}$ and would satisfy $\nabla V=\vec{F}$. This would correspond with decreasing in height increasing in potential energy which would make us have to redefine mechanical energy as $T-V$ in order to maintain conservation.
Instead of redefining mechanical energy, we introduce the minus sign $\vec{F} = -\nabla V$ which equates the physical notion of potential energy with the mathematical notion of the scalar potential.