What are the advantages of $\frac{dp}{dt}$ over $\frac{dp}{ds}$ as the definition of force?

Question

Suppose we want to have a formula which measures how hard we push or pull. Now, momentum is a measure of quantity of motion. And, our external push is something which changes it, so our formula must have something to do with change in momentum.

Now, if we push a box continuously with large effort, then it's momentum changes by a large amount in a short time interval. And, if we push the same box with a feeble effort, then to produce the same change in momentum, we have to push it for a longer time interval. So, the formula $\frac{dp}{dt}$ should make sense as a measure of force.

But, we can also think of the problem like this: If we push a box with large effort even through a small distance, then its momentum changes by large amount. But if we push the same box with a feeble effort continuously, then we've to push it through a larger distance to produce the same change in momentum. So, $\frac{dp}{ds}$ can also be a measure of how hard we push or pull.

Then, what are the advantages of choosing $\frac{dp}{dt}$ as force?

Answer 1

The reason for defining force (like many other quantities) in the way that we do, is that the laws of physics take simple mathematical forms if we use these quantities. In the case of force, if we use $F=\frac{dp}{dt}$, then Hooke's law, Newton's third law, Newton's law of gravitation, the Lorentz force law (and no doubt many other laws) take simple forms – which they would not do if we used your proposed alternative.

Answer 2

If $F$ were to be $\displaystyle \frac{dP}{dx}$, then:

$$F = \frac{dP}{dx}$$

$$=m\frac{dv}{dx}$$

$$= m\frac{a}{v}$$

That is pretty inconvenient to write (if $v$ goes to zero, $F$ goes to infinity).

The consequences are:

$$W = Fv.dx$$

This is inconvenient and ugly too.

Most importantly, force is easier to measure as $ma$ than $m\frac{a}{v}$. Well, $m\frac{a}{v}$ has no real physical meaning as far as I know. It would be of no use.

If we did define force to be that way, we would have eventually define a new quantity called [insert word] which would be $\frac{dP}{dt} = ma$ because it is more useful.

Answer 3

The physical quantity we call force can be introduced consistently in a number of different ways. If you want to define force as a measure of how hard something is being pushed or pulled, then the statement $F=\frac {dp} {dt}$ becomes an inescapable law, rather than a definition we are free to pick or choose.

To see this, we have to work out how to objectively measure the strength of a push or pull (in a way that matches our intuition about what a push or pull feels like). The usual way is to measure the deformation of a standard object which is elastic enough to return to its original shape after each experiment. The simplest example is a spring, for which the deformation (and hence the force) can be measured by the extent of its compression or extension. (You don't have to use a spring, but let's use one in our thought experiment.)

Now imagine we have a series of different objects on a frictionless surface. By applying our spring we push (or pull) each object, being careful to maintain the same compression or extension in each case, so that we can confidently argue that the same force has been applied to each. If we measure the resulting motion each object experiences whilst undergoing this force, what do we find?

We find that all objects, whether big/small or light/heavy, have the same value of $\frac {dp} {dt}$!

Why did I bold all that? Because that is an empirical fact about the world that wasn't true by choice, but rather by experiment. Namely, the force (as we've chosen to define it: as a measure of distortion) on an object uniquely determines the rate of change of momentum the recipient of the force experiences.

By calibrating the spring appropriately, we get the following law of nature: $F=\frac {dp} {dt}$.

The point I'm trying to bring home is that we have no choice in this result. When we set up an objective quantitative definition of force that matches the idea of 'extent of push/pull' it turns out that this quantity uniquely determines the rate of change of momentum and that's all there is to it!

Note: I phrased something carefully at the start of this answer: "If you want to define force as a measure of how hard something is being pushed or pulled...". I did that because you can introduce force by a mathematical definition if you want to, and then the answer to your question is a little different. If you choose to define force by $F=\frac {dp} {dx}$, for example, you'll find that the quantity does not in any way match your intuitive feel for force as a measure of push or pull.

In particular, when you push or pull two different objects with exactly the same push or pull (as measured roughly by your intuitive experience) they will, in general, have very different values of $\frac {dp} {dx}$, and this is a sure sign that you have defined force incorrectly. Thus, even if this quantity turns out to be useful in physics, and therefore deserving of a name, we wouldn't call it force.

Answer 4

Well it works in your example because you limit yourself to the case when the box is not moving so much to start with. Now take the case where the box is moving at considerable speed. $\frac{dp}{ds}$ will not be representing the effort you are making since it is biased by speed $v$ ($ds$ is very big). It can help to write the relationship between the two: $$F=\frac{dp}{dt} = \frac{ds}{dt}\cdot \frac{dp}{ds} = v \cdot \frac{dp}{ds}$$ You can see there than the speed is essential to understand how a change in position actually corresponds to a change in momentum.

Now I also totally agree with Phillip Wood who talks about the simplicity of physical laws. The current definition of force makes it elegantly consistent with the laws of Newtonian dynamics and other parameters like momentum, speed, acceleration, work, energy and so on.