Explore the relationship between conservative force and energy conservation

This article’s idea is originally from the lecture notes of David Tong:

Consider a particle moving in three dimensional space . It is possible to have energy conservation even if the force depends on the velocity. Conversely, forces that depend only on the position do not necessarily conserve energy: we need an extra condition. That is conservative force.

There exists a conserved energy if the force can be written in the form

for some potential energy function . For components of the force be the form

The conserved energy is given by

For one-dimension case, we can just differentiate the energy with respect to time.

For energy conservation, means that therefore . The only option for this equation is , next shows two ways to prove this.

Work Done

If a force acts on a particle and succeeds in moving it from to along a trajectory , then the work done by the force is defined to be

alternate form

We can use the work done to solve this problem. Replacing to get

where

is the kinetic energy. So the total work done is equal to the change of kinetic energy. If we want to have a energy conservation, then the change in kinetic energy must be equal to some compensating change in potential energy. This is true if the work done depends only on the end points, and , meaning that there’s a potential function so the work done can be written by

In another form, the energy conservation can be written like this

How can we prove

for some function .

From

where the last equality follows from the chain rule. But now we have the integral of a total derivative, so

which depends only on the end points as promised.

From

As defined

Obviously if we want force , should differentiate this equation, it can be written as

Alternatively, you can omit the brackets entirely since the content fits on one line.

Since the line represents should be the straight line in the direction. This means that the line integral projects onto the component of the vector . Since we’re integrating this over a small segment of length , the integral gives and, after taking the limit , we have

or

This is our desired result .