# Proving Drag is a Non-Conservative Force

I wanted to show that a force such as drag is not a conservative force but cannot apply my method of approach. Suppose we took the situation with drag at high speeds, so $\vec D=c|\vec v|\vec v$ for some $c$.

Since work is defined as $\int_C \vec F \cdot d\vec s$ along a path $C$, if I could show that this integral differs depending on the path taken, then I would have shown that drag is non-conservative.

So the integral I would have is: $\int_C -(c|\vec v|\vec v)\cdot d\vec s$ and we can write $\vec v$ as $|v|\hat{e_t}$ which is a vector in the tangential direction to the path. Since $d\vec s$ is defined as $\hat{e_t}\cdot ds$ then the integral becomes $$\int_C -(c|\vec v|\vec v)\cdot d\vec s=\int_C -(c|\vec v|^2)ds$$

How would I proceed from here?

• The easiest way to prove that something depends of the path taken is to write down some different paths, $\mathbf{x}(t)$, evaluate the integral, and see that you get different answers. – By Symmetry May 5 '17 at 14:45
• By showing it's false for one case, that will suffice right? Is there a more elegant method / approach to this problem? – user258521 May 5 '17 at 14:47
• Yes, you are trying to prove that something is not always true, so it is good enough to provide one counter example. I would say this method is fairly elegant. Since you only need to find one counter example you are free to choose very simple paths. You could find more abstract methods, but whether that is more elegant is a matter of taste. – By Symmetry May 5 '17 at 14:51
• In order for a force to be conservative, it has to give zero net work when an object is moved over a closed path. That means that the force has to be capable of both giving energy to as well as absorbing energy from the object. Isn't it apparent that drag doesn't fit that description and therefore must be non-conservative? Drag will always take energy from an object but will never return any energy to it. – user93237 May 5 '17 at 17:17
• @user258521 - You don't need to wrap everything up in equations in order to present a convincing mathematical proof. In fact, there are many branches of mathematics which don't even use any recognizable equations. All you need is to provide convincing logic. – user93237 May 5 '17 at 20:00

• In other words, if $|v|=cte.$ then $\int_C -(c|\vec v|^2)ds=-cv^2s\neq 0$ – user126422 May 5 '17 at 23:41