Why is work done expressed as F.dx instead of x.dF?

[jack action]

If you want x to vary then you need to state what it varies with thus:- x(F)dF

But it still wouldn't define work as x(F) is not a displacement but just a definition of a fixed position x with respect to a force F. The dx have meaning that x(F) doesn't have.

 

[Mr Real]

If you want x to vary then you need to state what it varies with thus:- x(F)dF.

Yes, that's correct but what I said there is that x.dF doesn't mean that x is constant, because it can vary (see I didn't say there that x.dF definitely means that x is variable, rather I pointed out to the user that he was wrongly assuming that x.dF means x is constant, the point is: x can vary)

At this stage I have a feeling that you just don't want to be wrong, rather than use this thread as a learning experience.

Actually, no I have no problem in being wrong, I just want a satisfactory answer, that's why I've spent so much time posting messages here, replying to them, pondering over other people's replies,etc.

 

[Dale]

I pointed out to the user that he was wrongly assuming that x.dF means x is constant, the point is: x can vary

I agree. The integrand is not assumed to be constant. Even when doing numerical integration it is uncommon to approximate it as piecewise constant.

I just want a satisfactory answer

You have had several correct answers. Have you been satisfied with them? If not, why not; shouldn't a correct answer be satisfactory?

 

[Mr Real]

You have had several correct answers. Have you been satisfied with them? If not, why not; shouldn't a correct answer be satisfactory?

Yes, I have got several good answers and they have partially answered my questions, but not completely.

 

[sophiecentaur]

, I just want a satisfactory answer,

You have been given the perfectly satisfactory answer that your arbitrary bit of maths can't be assumed to have a physical interpretation. It's just not satisfactory for you.

 

[Dale]

Yes, I have got several good answers and they have partially answered my questions, but not completely.

I don't see any question of yours which has been partially answered.

1) You wanted to know why the integral of x.dF did not also give work just like F.dx. The complete answer was that it's a calculus error to switch the quantities like that. The area under a curve is not the same as the area to the left of the curve. So whatever x.dF might represent it is not generally work.

2. You wanted to know why textbooks don't spend time discussing this. The complete answer is that there are an infinite number of mistakes you could make, and this is a very rare one. Authors focus primarily on the correct ways, not one of the infinite incorrect ways. When they do address mistakes, it is only the common ones.

3. You wanted to know why the integral of x.dF is not useful in its own right. The complete answer is twofold, first, it is just an odd expression. It is not common to express x as a function of F, and it is often not even possible to do so, and the quantity dF is a weird quantity without physical use elsewhere. Second, there is no known relationship of that quantity to other known physical quantities of interest, nor does it posses known properties that make it a primary quantity of interest.

What could possibly be considered incomplete about this discussion? Not only has this community correctly and completely answered your question, we have been exceptionally patient with you. This discussion should have ended on the first page as soon as it was pointed out that it is a mathematical mistake to switch the terms that way. So your answer has been correct, complete, and exceptionally supportive.

It is highly frustrating to have such thoughtful and helpful replies from the community regarded as insufficient.

At this point I am going to close this thread. I only hope that in the future you can recognize when you have received such high quality responses, and be a little more encouraging to the community that provided them.