# How is the definition of work motivated?

For most dynamical variables in classical physics, I can understand how one may have decided to introduce them as a result of some "incompleteness" in Newton's laws of motion. For example:

1. One would like to say that applying a force to an object over time "gives" it something, so momentum must be introduced as \begin{align} \mathbf p = \int _0 ^t \mathbf F(t') dt', \text{ or as some quantity such that }\mathbf F = \frac{d\mathbf p}{dt}. \end{align}
2. It is useful to describe forces that result in changes in angular velocity $$\mathbf{ \omega = v \times r}$$, so that leads to the definition of angular momentum and torque.
3. For some forces it is convenient to say that they serve to make a system approach some stable state, no matter what their initial condition is. Take a spinning top, for example. No matter how you spin it, it will always approach higher "stability" by falling onto the table. The definition of potential energy in the equation $$\mathbf F = -\nabla U$$ emphasizes the idea of increasing "stability", or decreasing potential, when a force acts on something.

These examples all involved some kind of physically obvious observation (exerting a force gives an object something, forces sometimes rotate things, some forces make things more stable), to which the response was to model the idea using Newton's laws. One can even do something similar with kinetic energy, by using the definition of work, by saying that a change in kinetic energy is just an alternate form for the work done on some object.

But there's something about work which just does not feel motivated at all to me. The work done on an object moving along a path $$\gamma$$ is defined as \begin{align} W = \int _\gamma \mathbf F \cdot d\mathbf r, \end{align} but I simply do not see how this is a quantity which "needed" to exist. Of course it is a very useful quantity, but I'm still having some trouble understanding what it really means intuitively.

Internet definitions (and even textbook ones) don't help much; often either kinetic energy is randomly defined from which the definition of work follows, work is randomly defined from which the definition of kinetic energy follows, or (my least favourite) a circular definition is employed, where energy is defined to be the ability to do work, and work is defined as the energy transferred to an object.

So what is the real meaning of the term "work", before even discussing anything about energy?

(I should clarify that I do not necessarily want to know the historical motivation for the definition of work. I am aware that sometimes things just "turn up" in physics because of experimentation with numbers, etc. I suppose I'm wondering what the "constructivist" approach to the definition of work is.)

• Why do you find "applying a force to an object over time "gives" it something" to be intuitive, while "applying a force to an object over distance "gives" it something" non-intuitive? Jan 16, 2021 at 16:53
• Andrew This does seems to be the closest to what I was asking for in my question. I suppose the notion of work in that sense violates my intuitions on causality. With momentum, you can clearly see that the force is giving the particle momentum because it acted on the particle for some time. But with the idea of a force acting over a distance, it's like time has been regarded as "unimportant". The work done is only indirectly related to the time over which it happened, as it is now directly related to the distance. However, this brief comment has given me a lot to think about already. Jan 17, 2021 at 4:49
• That's a good observation. One thing to think about is that work is a means of transferring energy from one form to another. For instance, a compressed spring has a lot of potential energy. When released, the spring can do work on a block connected to it, converting the potential energy to kinetic energy of the block. From this point of view, maybe we can see why time is not so important. The rate at which the potential energy of the spring is converted to work depends on the process (for example, do we apply a counter force), but the total amount of energy available for work does not. Jan 17, 2021 at 5:25

The motivation for me is all that ancient tools to facilitate moving things as levers, pulleys or gears.

It is clear that something keeps constant in the trade off between displacement and force. That constant receives the name of work.

Work is the thing that makes kinetic energy Galilean invariant. Suppose you have a ball with mass $$m$$, on a spring ($$k$$), compressed a distance $$L$$.

That system has energy, in terms of kinetic ($$T_i$$) and potential ($$U_i$$):

$$E_i = T_i + U_i = 0 + \frac 1 2 k L^2$$

Then you release the spring, and the balls rolls of with velocity $$u$$, so that:

$$E_f = T_f + U_f = \frac 1 2 m u^2 + 0$$

with $$u^2 = L^2k/m$$ so that:

$$\Delta E = E_i-E_f = 0$$

(Note that you can write $$u = \nu L$$, where $$\nu$$ is the period of the spring/mass-oscillator).

In terms of forces, of course, the force $$kL$$ was applied through a distance $$L$$, so the work done by the spring was:

$$W = \int_0^Lkxdx = \frac 1 2 kL^2$$

Now do that in a frame moving at $$v$$:

$$E'_i = T'_i + U'_i = \frac 1 2 m v^2 + \frac 1 2 kL^2$$

and release the spring:

$$E'_f = T'_f + U'_f = \frac 1 2 m (v+u)^2 + 0$$ $$E'_f = \frac 1 2 mv^2 + \frac 1 2 kL^2 + mvu$$

Now:

$$\Delta E' = E'_i - E'_f = mvu$$

This may be distributing. It's clear why momentum conservation survives a Galilean boost (as it's linear in velocity), but energy appears problematic.

Where did $$mvu$$ come from?

It's work. Because the spring is moving, it is applying a force over a much greater distance than $$L$$. The force is applied over a distance of:

$$d = L + \frac 1 4 \frac v {\nu}$$

and the work done is the moving frame is

$$W = \frac 1 2 k L^2 + mvu$$

• The problem I have with this answer is that it assumes the formula for kinetic energy to be $\frac{1}{2} mv^2$, before introducing the notion of work. The question then becomes "Why is kinetic energy a quantity which should exist?", which to me seems much harder to justify, seeing as it involves a squared term and a factor of 1/2. Of course you could then justify its existence in terms of work, which is the circular definition I dislike. Jan 17, 2021 at 4:14
• @BayleeV $\frac 1 2 m v^2$ is Noether's theorem 101.
– JEB
Jan 17, 2021 at 14:54

The fundamental forces in newtonian mechanics, namely gravity and columb forces depend on distance instead of time, thus it becomes sensible to write $$F=mv\frac {dv}{dx}$$ instead of $$F= m\frac {dv}{dt}$$

The following images are from my personal notes, I couldn't copy and paste the text due to some issues Apologies for that.

• While the proof of the work-energy theorem is correct, it doesn't really address the central issue - Why do work/energy have the formulas they do, and what experiments motivated their introduction? Apr 10 at 19:05
• @catalogue_number if I am correct then you are asking why our intuitive notion for energy takes the form of the formulas the way it does, In other words, what is the relation between the mv^2/2 and the physical process causing the object to move? Apr 10 at 19:21
• Hmmm.. after rereading the whole thread, I see where you're coming from. If you assume $F = \frac{dp}{dt} = -\nabla U$, and further that $p=mv$, then you see the equivalence between work and energy. Apr 10 at 19:27