• My assumption here is that in the definition of elementary work :

$dW = F ds$

symbol $d$ represents a differential.

  • But a differential implies a function :

$dy =_{df} d[f(x)] = f'(x) \Delta x = f'(x)dx$.

  • This is why, in order to better understand the notion of " elementary work" ( and in general, " elementary quantty" , such as " elementary displacement, etc. ), I'd like to express elementary work as the differential of a function.

  • Is this possible?

  • It seems to me that the function should have distance ( " s") as independent variable. Is it the case?

My question is not much about work than about the notation that (I believe) is used to define other physical quantities.

In University Physics ( vol.1) I find this ( with $dW$ standing for "increment of work", and $d\vec r$ standing for infinitesimal displacement):

$$dW= \vec F . d\vec r$$

(Source: text page 328 of https://d3bxy9euw4e147.cloudfront.net/oscms-prodcms/media/documents/UniversityPhysicsVolume1-LR.pdf)

(In the Wikipedia article, I find the same thing with "$\delta W$" instead of "$d W$", so $ \delta W = Fds$).

Should $dW$ (or $\delta W $) be understood as a differential (defined as: $dy = d[f(x)] = f'(x) dx$)?

I mean, is it correct to read the equation as a definition of the differential of the work function, say $W(s) =$ work done for a distance $s$?

Can the equation $d W= Fds$ be read as:

$$d(W(s)) = Fds~?$$

I guess I am wrong since, being given the general definition of a differential (i.e. $dy=d[f(x)] = f'(x)dx$), my last line would mean:

$$ d[W(s)] = W'(s)ds= F ds$$

implying that force $F$ is the derivative of work (which I don't think is true).

In brief (1) I don't understand the "small change" / "infinitesimal change" interpretation of $d$ (2) and the "differential" interpretation does not seem to work either.

So, please, what does $d$ rigorously mean? (Sorry, my style certainly lacks clarity, due to the confusion I'm in regarding this $d$-notation).

PS: another interpretation is that $dW$ is a derivative of the work function (since work is defined as the integral of $dW$). But in that case, why not use the ordinary derivative notation?

  • 1
    $\begingroup$ Related: physics.stackexchange.com/q/153791 $\endgroup$ Apr 12 at 13:18
  • $\begingroup$ @Mathphys meister. - I've read the question you are referring to; but my question is much more elementary. $\endgroup$ Apr 12 at 13:21
  • $\begingroup$ Personally, I did not understand the answer of the question you point to. $\endgroup$ Apr 12 at 13:22

The $\delta$ indicates in general that the work depends on the particular path taken for doing the integration. In case of exact differential, it doesn't matter the path because integral depends only on the initial and final points. If it's possible to write the force like a gradient of a scalar function, then it's $dW$. For example if the force is constant or depends upon the coordinates. Take a look to conservative field. If it's not the case, the path chosen in integration change the results. This can happen for example in thermodynamics.


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