# Defining potential energy in Taylor's Classical Mechanics

I'm trying to understand this sentence in introducing potential energy in John Taylor's book:

If all forces on an object are conservative, then can define a quantity called potential energy, $$U (\mathbf{r}),$$ with the property that $$(\text{total mechanical energy}) = E = KE + PE = T + U(\mathbf{r})$$ is constant.

I'm trying to understand why this is important/interesting. My mathematical instinct is that it has something to do with the fact that $$U$$ is only a function of $$\mathbf{r}$$, because otherwise, I could always define a function $$U := T$$ for which $$E$$ above is constant. But it seems like there is something special in the fact that $$U$$ only depends on $$\mathbf{r}$$, whereas $$T$$ depends on time (I think)?

Can someone explain this initial setup without getting into further topics like force gradients, etc...?

Here is a simple answer. The work done by a force is $$W = \int_{}^{} \vec F \cdot d \vec r$$, where $$\vec F$$ is the force and $$\vec r$$ is the path. The work done by a force causes a change in kinetic energy, $$T$$; specifically, $$W = \Delta T$$. For a conservative force, the work done is independent of the path and a potential energy can be defined that describes the negative of the work done by the force: $$W = - \Delta PE$$, where $$\Delta PE$$ is the change in potential energy. So $$\Delta T + \Delta PE = 0$$. One reason for using potential energy is that it is can simplify a calculation. For example, the work for the force of gravity near the earth is described by the potential difference $$mg \Delta h$$ where $$m$$ is mass, $$g$$ is the acceleration of gravity, and $$\Delta h$$ is difference in elevation. No matter how tortuous the path, the work by gravity is accounted for simply considering $$mg \Delta h$$, instead of having to evaluate $$W = \int_{}^{} \vec F \cdot d \vec r$$.

(For a conservative force $$\nabla \times \vec F = 0$$ and $$\vec F = - \nabla U$$ where U is potential energy. for example, see Symon, Mechanics.)

Let's see if we can answer your question without formalism, but it is going ot be a somewhat long trip so... stay with me!

KI is the so called kinetic energy, which depends on the speed of your object. Assume your object changes speed (i.e. accelerates or decelerates). This means that there is a force acting on it (because of Newton's law). Suppose the object was at the beginning at point $$A$$ and is at the end at point $$B$$ (of course, in general $$A$$ and $$B$$ are position vectors, but we will denote them simply by "position $$A$$" and "position $$B$$").

Now, there are two options:

a) the force depends on the path the object takes from $$A$$ to $$B$$. Imagine your object is moving on a high friction surface: of course the longer the path from $$A$$ to $$B$$, the more the object will slow down because of friction. This is also why when you walk from home to the supermarket you try to take the shortest path, because you know what the longer the path, the more you will be tired when you arrive. This kinds of forces are called dissipative because the energy you loose is (from the ob#ject's point of view) gone forever. If when you arrive at the supermarket you are extremely tired already, there is no way you are going to be able to go back home! The importance of this will be more clear when we consider the other case:

b) the force does not depend on the path you take. This is the case of gravity in absence of friction or of the electromagnetic field. Now, let's call $$U_{AB}$$ the energy you spend in going from $$A$$ to $$B$$. We can define it univocally, because it does not the depend on the way we got there! However, it still depends on both the starting point $$A$$ and the arrival point $$B$$, so we use a small trick: we define another position $$0$$, which we use as a reference position, and define the energy $$U(A)$$ at position $$A$$ as the energy one spends to go from the reference position $$0$$ to position $$A$$. At the same time, the energy $$U(B)$$ is defined as the energy we spend, again starting from position $$0$$, to get to position $$B$$.

Why is this useful? Because now, as you noted, we have a function $$U$$ which is only dependent on the position ($$A$$, $$B$$ in our case, but r in your example) and on the reference position $$0$$. This is possible because the exact path we take is not relevant.

Can we now get rid of the reference position? Sort of, because if we are interested in the energy we spend in going from $$A$$ to $$B$$, which we called $$U_{AB}$$, we can do the following: start in $$0$$, go to $$A$$ (spending the energy $$U(A)$$) and then go to $$B$$ (spending the extra energy $$U_{AB}$$). But, because we started at $$0$$ and ended up in $$B$$ and because the path is irrelevant, the total energy we spent must have been $$U(B)$$ which means $$U(B)=U(A)+U_{AB}$$ and from this we get $$U_{AB}=U(B)-U(A)$$

This shows that if we have a function $$U$$ which is only dependent on the position r we can always compute the energy we spend to go between two points as the difference between the energy of the final position minus the energy of the starting position.

So yes, the fact that $$U$$ only depends on the position is fundamental because it allows us to compute energy difference. It also means that

$$U_{AB}=-U_{BA}$$

which means that all the energy we lost in going from $$A$$ to $$B$$ can be recovered if we go back from $$B$$ to $$A$$.

An example: start with a ball at 1m of height, let it fall and assume it bounces back without loosing any energy. Then the ball will climb back at exactly 1m! As the ball goes from 1m to 0, it looses energy (in this case, gravitational energy) but it gains speed (kinetic energy), as it bounces back it will gain all the energy it lost before but loose kinetic energy.

From Newton's equations, you can get indeed that the sum of KI and the energy you "spend" during motion are constant (conservation of energy: you can convert kinetic energy into other kinds of energies and vice versa). If your force is conservative, the energy you spend in doing something is not lost foreveer, but stored somewhere else, ready to be given back.

Anothe example: you spend some energy to compress a spring, but then you release it and all the energy comes back at you!

And yes, this is all a consequence of $$U$$ being only a constant of the position r and independent of the path taken (the two things are interconnected).

Summing up: Kinetic energy+energy lost = constant if U=U(r) then the "energy lost" can be converted back into kinetic energy.

For a non conservative force, after you convert your kinetic energy into energy lost, you can not take it back, so that the sum is constant, but energy cannot go back into kinetic energy. For a conservative force, not only the sum is constant, but it can change between the two kinds of energy as $$KI+U(\bf{r})=c$$ means that the KI you have at any position is $$KI(\bf{r})=c-U(\bf{r})$$ i.e. also the kinetic energy is now a function of the distance. If you go back in position, you recover the lost energy!

Final example, a bit absurd: you attach a spring to your door and the other end you attach it to yourself and then you walk to the supermarket, dilating the spring. You will spend a lot of energy in dilating it, but whatever path you take, even if you arrive at the supermarket, you just have to let go and the spring will "bounce" you back home, giving you back all the energy you lost ;)

P.S. BOTH the potential and the kinetic energy depend on time, but their sum is constant and always the same over time. In one case the transformation between different kinds of energy is reversible, in the other not.

U(r) is not an arbitrary definition. Neglecting sign conventions, it is the work done to move an object from an initial to a final position in the field of the conservative force. By virtue of the force being conservative, $$\Delta U$$ is independent of the path. Work done is associated with a change in energy. If an object is moved from one rest position to another rest position, there is no change in its kinetic energy but there is a change in its potential energy.

U(r) is useful since the total energy T + U(r) is constant. Thus, dropping an object from a given height above the earth's surface, we can find its final velocity when it strikes the ground just by knowing the change in its potential energy.