I am having troubles trying to understand why is work defined as it is.

So, I know how work is defined: $W = \vec{F}\cdot{}\vec{d}$ (F is the force, d the displacement) and I am okay with it. This, and the quantity $E_k = \frac{1}{2}mv^2$, are a mathematical object useful to describe the state of a system more easily. I also know the more general definition of $W=\int{\vec{F}\cdot{}d\vec{r}}$ where the integral is calculated along the path of the body. It's also known that the four fundamental interactions are distance-depending forces. Now, it is possible to imagine a scenario in wich we have two bodies, say $A$ and $B$, in one dimension for simplicity (a line) and one of this four forces, say $\vec{f}$ (and of course $-\vec{f}$), acts on them. In the definition of work, the quantity $W$ is calculated assuming the force being constant, so the distance between the two bodies must remain a constant too, $d$. For this to be true, the two bodies have the same displacement $\vec{r}$. So the work on $A$, $W_A$, is the opposite of $W_B$, according to the third of the Newton's principles. This makes sense to me, because at this point (if what I am saying is not wrong, and I am sure it is indeed wrong) energy is actually something that is conserved, in every physical scenario, because we could (just theoretically) break it down to the four fundamental interactions and that's it. I know I am probably doing something wrong, and if this is the case, I would like to ask a question: is it possible to explain the mathematical intuition behind the definition of work using just Newton's laws? I've studied classical mechanics only. Thanks.

  • $\begingroup$ 1. The definition of "work" is much older than our understanding of the fundamental forces, so I'm puzzled how you think the two could be related. 2. "It's also known that the four fundamental interactions are distance-depending forces." - at the very least for the weak "force", I have trouble determining in what way you think this "is known" (obligatory xkcd). 3. Possibly related: physics.stackexchange.com/q/277347/50583 $\endgroup$
    – ACuriousMind
    May 26, 2022 at 16:33
  • $\begingroup$ Through experiment we can determine that an object possesses a certain qauntity. When metal balls are dropped into sand, the penitration depth, $p \propto v^2$ one could theorise that the a qauntity called kinetic energy is therefore proportional to v^2, kinetic energy is a useful qauntity to define, that relates forces to distance (/velocity). I'm sure if you type in "Émilie du Châtelet kinetic energy" you will be able to piece together why kinetic energy is defined the way it is and the experiments that led to its conception. $\endgroup$ May 26, 2022 at 17:05
  • $\begingroup$ You can find various derivations for "Work" here: en.wikipedia.org/wiki/Work_(physics) If you just start from the definitions of Force, Acceleration, Velocity, and Position, you can derive a relationship which ends up relating work (position) to kinetic energy (velocity) for a given force. $\endgroup$
    – Alwin
    May 26, 2022 at 17:14
  • $\begingroup$ @ACuriousMind I am sorry, I didn't mean that we defined work as it is after knowing all of the fundamental forces; I tought this could be an explanation of why (at least in classical mechanics) this model works: to move a box (in the void) with a constant force over 1 meter I have to move exactly at the same velocity of the box over the same distance, or the forces between the particles of my hand and those of the box will decrease. Because of your 2nd point, however, my "idea" doesn't make any sense, and I thank you for the answer. $\endgroup$
    – Due.Berto
    May 26, 2022 at 18:02

1 Answer 1


The construct $W = \int F(r) dr$ is useful because the fundamental interactions depend only on distance (and not other variables like velocity or time), and because F(r) is a conservative vector field for the fundamental interactions (so that W is not depending on path). And a derivation is able to connect this quantity to kinetic energy, which is a very useful shortcut.

However, the conservation of energy in all scenarios depends on whether the Hamiltonian is time-dependent, and whether your scenario is accounting for everything.

If you imagine a time dependent potential, energy will not be conserved. Imagine you are a ball at the bottom of a $x^2$ well, but the well suddenly begins moving to the side. You will have more kinetic energy now, because you are moving, even though you had 0 kinetic energy before. And your potential energy will be greater than or equal to your potential energy before, since you were at the bottom of the well. So your energy will strictly increase.

Why did this happen? Well, the potential of your system went from being time-independent ($P = (x-x_0)^2$) to time-dependent (let $x_0 = vt$, then $P = (x-vt)^2$).

Now, you can account for this by considering the total energy of the system, ball, well, and energy source pushing the well combined. But it is important to keep in mind my main takeaway: when the Hamiltonian of your system is time-dependent, then energy is not necessarily conserved.

  • $\begingroup$ Thank you for your answer. I am sorry but I don't know exactly what an Hamiltonian is, I will look it up. $\endgroup$
    – Due.Berto
    May 26, 2022 at 17:09

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