# [Physics] For what reason is the difference of potential energy $\Delta U=-W$ equal to the *opposite* of the work done

conservative-fieldconventionsnewtonian-mechanicspotential energywork

In my classical mechanics physics textbook (a translation of the Walker-Halliday-Resnick Fundamentals of Physics) the difference of potential energy is defined as

$$\Delta U = -W \qquad (1)$$

I have done extensive research (taking me 5+ hours) and I claim to have a reasonable understanding of this model. In particular, I understand that if we throw a solid object in a straight upward direction then the work (i.e., the quantity of kinetic energy conveyed or subtracted from a body) exerted by the Earth's gravitational force is negative because they act on opposite directions: $W = \vec{F} \cdot \vec{d} = F \cdot cos(\phi) \cdot d$, where $cos(\phi) = -1$ due to $\phi$, the angle between the movement and the gravitational force, being $180°$.

However, I couldn't find anywhere an explanation for this. I was demonstrated that for a conservative force $\vec{F}$ doing work along a path $ab$, $W_{ab} = -W_{ba}$, and I also know that we can always associate a potential energy to a conservative force. But I'm still missing a link, and not knowing how the negative work of a force relates to its potential energy gives me brain fog.

Can you please provide an explanation, or an appropriate proof, for $(1)$? Please note that my physics knowledge only extends up to what is taught in university-level Physics I and Physics II courses.

The potential is defined as a function $U$ such that the conservative force $\vec F$ that we are studying is given by the gradient $\vec F = -\nabla U.$ Since you probably have not seen vector calculus yet, let me be very careful to write this out as the components, $$F_x = -\frac{\partial U}{\partial x},\\ F_y = -\frac{\partial U}{\partial y},\\ F_z = -\frac{\partial U}{\partial z}.$$These "partial derivatives" are evaluated as normal derivatives treating the other variables as constant, so for example the potential $U=k~x^2~y + p~z^2$ would generate $F_x = -2k~x~y, F_y = -k~x^2, F_z = -2p~z.$
Partial derivatives are the natural way to understand calculus on a function of many variables. In single-variable calculus, you were trying to approximate a curve with a tangent line; in this multi-variable calculus we are trying to approximate surfaces with planes. In particular, if you can make this approximation then it means that a function can be expanded around a point, $$f(x + \delta x, y+\delta y, z+\delta z)\approx f(x, y, z) + \frac{\partial f}{\partial x}~\delta x+ \frac{\partial f}{\partial y}~\delta y+ \frac{\partial f}{\partial z}~\delta z.$$Here by $\delta q$ I just mean "a little change in $q$", whatever $q$ is. One could also move this term $f(x,y,z)$ to the left-hand side and refer to that difference as $\delta f$, if you'd like. This understanding of expanding out a multivariable function will be important.
The power exerted by a force on a particle is the dot product of that particle's velocity with the force, and the work done over the path is the time integral of power exerted by that force. It is common to denote the position of the particle as a vector $\vec r(t)$ with components $r_{x,y,z}(t)$ and then this is: $$P(t) = F_x~\frac{dr_x}{dt} + F_y~\frac{dr_y}{dt} + F_z~\frac{dr_z}{dt}\\ W = \int_{t_0}^{t_1}dt~P(t).$$ The combination of these two definitions is what you seem to be asking about: but it is not very complicated at all. Combine the two and then stare for a second at the following: $$P(t)~\delta t = -\frac{\partial U}{\partial x} ~\frac{dr_x}{dt}~\delta t -\frac{\partial U}{\partial y} ~\frac{dr_y}{dt}~\delta t -\frac{\partial U}{\partial z} ~\frac{dr_z}{dt}~\delta t.$$What should now stand out to you is that this is very much like the above expression for $\delta f$ above, if we defined $\delta x = \frac{dr_x}{dt}~\delta t$ and so on for $\delta y, \delta z.$ And those are very natural definitions, as $r_x$ represents an $x$-component of position and if we take this time-derivative we get a component of velocity, and multiplying against a short time $\delta t$ we get a small change in this $x$-component due to the particle's current motion.
There is a more formal way to do this and it is to invoke the chain rule, which says that when we apply some function $U(x, y, z)$ to these time-varying components $x = r_x(t)$ and so forth, we find that:$$\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) = \frac{\partial U}{\partial x}~\frac{dr_x}{dt} + \frac{\partial U}{\partial y}~\frac{dr_y}{dt} + \frac{\partial U}{\partial z}~\frac{dr_z}{dt}.$$ Therefore what we have found above is simply, $$P(t) = - \frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big).$$The work is the time integral of power, but integrals perfectly undo derivatives, and therefore when we do this definite integral we get from the fundamental theorem of calculus, \begin{align} W &= -\int_{t_0}^{t_1} dt~\frac{d}{dt} \Big(U\big(r_x(t),~r_y(t),~r_z(t)\big)\Big) \\ &= -\Big(U\big(r_x(t_1),~r_y(t_1),~r_z(t_1)\big) - U\big(r_x(t_0),~r_y(t_0),~r_z(t_0)\big)\Big)\\ &=-\Delta U.\end{align} That's really all there is to it: for conservative forces $\vec F = -\nabla U$ the power $\vec F \cdot \frac{d\vec r}{dt}$ is immediately seen to be a chain rule expression $\frac{d}{dt} U(\vec r) = \nabla U\cdot \frac{d\vec r}{dt}$, which identifies it as a total time derivative, and therefore the work, which is just the time integral of the power, must be the overall change in the quantity: in this case the quantity is $-U$ and so $W = \Delta(-U) = -\Delta U.$