Work, power and energy questions.
Why does small work done mean:
$$dw=f.ds$$
and why not:
$$dw=df.ds$$
and why not:
$$dw=s.df \ \ ?$$
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4$\begingroup$ In the expression $s.df$ and $f.ds$ first state clearly what physical quantities you are referring to: be very clear on what you mean by $s$ and what you mean by $df$ for example. Be clear on the difference between $s$ and $ds$ and the difference between $f$ and $df$. Apply your ideas to a specific physical example. This will help you ask the question more clearly, and you may find you have gone a long way towards answering it for yourself, or at least to make a start. $\endgroup$– Andrew SteaneCommented Oct 5, 2022 at 11:31
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2$\begingroup$ by definition of work $$W=\int dw=\int fds$$ $\endgroup$– Aslan MonahovCommented Oct 5, 2022 at 12:58
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2$\begingroup$ It's good to question why certain definitions are made (see terrytao.wordpress.com/career-advice/…). It often boils down to: a quantity is given a name when we can say something useful about it. Work is defined the way it is in part because with that definition we can prove the work energy theorem. You may also want to think more carefully about your proposed definition(s) involving $df$. Can you think of a procedure you could use to calculate a quantity defined in that way? $\endgroup$– AndrewCommented Oct 5, 2022 at 13:35
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3$\begingroup$ This is a question of a school kid, so it should have an answer according to its level and not closed. It is just a missunderstanding of what ds or df really means. $\endgroup$– trulaCommented Oct 5, 2022 at 14:13
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$\begingroup$ I'd also add that dw = df x ds would not be an appropriate definition as the right hand side is a higher order differential (not sure of the precise mathematical term) than the left side. $\endgroup$– Not_EinsteinCommented Oct 5, 2022 at 16:12
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2 Answers
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I think that it is better to start with power than with work.
Recall that momentum, $\vec p$, is conserved and that force, $\vec F$, is the rate of change of momentum. Also, recall that impulse, $\vec I$, is the change in momentum. So we can write $$\vec F = \frac{d\vec p}{dt} = m\vec a$$$$ \vec I = \Delta \vec p = \int \vec F \ dt $$
Similarly energy, $E$, is conserved and power, $P$, is the rate of change of energy. Also similarly work, $W$, is the change in energy. So just like above we have $$P = \frac{dE}{dt} = \vec F \cdot \vec v$$$$ W=\Delta E = \int P \ dt $$
Now, the rest is easy $$W = \int P \ dt = \int \vec F \cdot \vec v \ dt = \int \vec F \cdot \frac{d\vec s}{dt} \ dt = \int \vec F \cdot d\vec s$$$$ dW=\vec F \cdot d\vec s $$
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The reason is really that defining $dW = F\, ds$ leads to a useful law (the work-energy theorem), and the other options don't.
Let's think a little bit more about what all these quantities mean. To take the differential of something, you need a variation. In this case, the setup is that we have some object moving along a given path $x(t)$. The notation $dx$ means taking two times very close to each other, and calculating the difference in position between these two times. Similarly, $dF$ is the difference in the value of the force applied to the particle at these two times.
To calculate the integral $\int F\, dx$ we start by dividing the path into a bunch of little intervals. At each point between two intervals we calculate the force felt by the object, multiplying it by the difference in position between one point and the next, adding all these products together, and taking the limit as the size of the intervals goes to zero. The useful identity $d^2 x/dt^2 = \dot{x}\, d\dot{x}/dx$, together with Newton's law, lets us express this integral as the variation in kinetic energy.
Similarly, the integral $\int x\, dF$ involves multiplying the position of each point with the difference in force between two points, and adding. And at this point, some red flags start popping up. Don't get me wrong: the integral is perfectly well-defined, there's nothing mathematically wrong with it. It's just that "force differential" is not really a quantity that shows up very often or is very useful: Newton's law just talks about the force itself. And having a naked $x$ inside an integral is weird, because the result then depends on where you put your coordinate system. Most important quantities don't depend on that. Finally, there's no equivalent mathematical trick that applies here: given a path and a force you can calculate the integral no problem, but you can't really relate it to anything else, the way you can relate work and kinetic energy.
And as for $dW = dF\, ds$, the issue is that when you take the limit where the interval size goes to zero, the product $dF\, ds$ goes to zero too fast, and you just get zero for the integral.
