# I Derivation of work done

1. May 16, 2017

### Mr Real

My question is:why is small work done,dW taken as F.dx instead of x.dF because integrating the latter should also give the same result(taking x out of the integral and integrating dF gives x (F) ).Visualising this integration graphically seems to suggest the same.
Similar is the case with quantities like electric flux (=E.dA),why isn't it dE.A ?
p.s.There are similar questions here but answers to them didn't help.

2. May 16, 2017

### Staff: Mentor

I am not even sure what dF would mean in this context. In most cases I can think of, F is a function of x, but x isn't a function of F.

3. May 17, 2017

### Mr Real

Okay I got what you mean but we can make the force dependent on distance (i.e. a function of distance) by varying it with increasing distance (e.g. force is given by the relation:F=3x,where x is distance )??
And can't all of it be seen in the graph? dF can be seen as an infinitesimal portion of the graph, so we get a very small area under the curve and summing up all these areas still gives the total area(which I think,will be the total work done..)?

4. May 17, 2017

### Staff: Mentor

Sure. Since F can be written F(x) then F(x).dx makes sense. In contrast x(F).dF doesn't make sense precisely because x often cannot be written as x(F), such as when F is constant.

No, that doesn't work. Suppose for the sake of argument we have some invertable force profile such that we can write both F(x) and x(F). Then F(x)dx is an infinitesimal element of the area between the curve and the x axis while x(F)dF is an infinitesimal element of the area between the curve and the F axis. They are opposite parts of a rectangle, not the same area.

5. May 17, 2017

### Mr Real

About getting different areas under the curve,that is a part of my doubt too.Why do we only consider the area that the curve makes with the displacement axis (x-axis here ) and not the area it makes with the force axis (y-axis here)?
But isn't displacement always a function of force(because it changes with the force applied)? And if it isn't always true, then what happens when we consider the case when it is?Then, wouldn't x.dF would be meaningful?

6. May 17, 2017

### pixel

Work is defined in a way so that the work done is equal to the change in kinetic energy. That's true for dW = F dx. It's not true for dW = x dF.

7. May 17, 2017

### Staff: Mentor

Not in general, no. If you have F=3x then you can indeed invert that and obtain x=F/3, both of which are valid functions. But if you have a constant force, F=3, then you cannot invert it to solve for x. Remember the mathematical definition of a function. A function has one output value for each of its input values, but in the case of a constant force, if you input the force then you have multiple output positions. So position is not a function of force.

That other area isn't useful for analyzing physics as far as I know.

8. May 17, 2017

### Mr Real

Yeah, but if we consider the case when displacement Is a function of force; then wouldn't the product dF.x be meaningful and then it could be integrated?
That's what I don't understand. Why do we only take the area made with displacement axis and not the one with force axis?

9. May 17, 2017

### Staff: Mentor

Yes. Your example of F=3x is one such case.

Because that is the only one that has proven to be useful. As @pixel mentioned, it is useful because it is equal to the change in KE.

The other one is certainly not equal to the change in KE, and as far as I know it isn't equal to any other interesting quantity either.

10. May 17, 2017

### Staff: Mentor

The integral of Fdx is not the same as the integral of xdF. What made you think that they are the same. I think a little review of calculus would be helpful here.

11. May 17, 2017

### Mr Real

So even in this case can't the work done be given by the integration of dF.x?
Is there some convention regarding this or can you provide a link where it's clearly mentioned (or even the history regarding this derivation). I want this as I have seen a similar question on Quora by a user named suraj tiwari ( I don't know how to provide the link) and there different reasons were given, therefore I want the right reason.

Last edited: May 17, 2017
12. May 17, 2017

### Mr Real

But on integrating x.dF, x being constant comes out of the integral and integrating dF gives F; now dot product is commutative so F.x=x.F . I don't know where I am doing the mistake because thinking about it and visualising it graphically seems to suggest the same.

13. May 17, 2017

### Staff: Mentor

The only way that Fdx=xdF is if F = kx, where k is a constant.

14. May 17, 2017

### Mr Real

But on integrating x.dF , x being constant comes out of the integral and integrating dF gives force so that we get x.F; Now dot product being commutative F.x=x.F so according to me it should also give work done. I don't know what i'm doing wrongly because thinking about it or visualising it graphically seem to suggest the same.

15. May 17, 2017

### Staff: Mentor

I am sure there will be some cases where, by coincidence, it happens to work. But the relationship does not hold in general. I don't know which cases would coincidentally work and which would not. I would guess that F=3x would work but F=3x+1 would not.

Often there are multiple valid reasons for the same thing, so I wouldn't get worked up by this. Every valid reason is "right".

I definitely do not have a reference that clearly states that x.dF is not useful (I cannot imagine why an author would explicitly make that statement, it would be a distraction that would confuse the majority of readers and any good editor would remove it). I can provide many references that show that F.dx is useful.

I would recommend working it out. Try F=3, F=3x, F=3x+1, and F=3x^2. Integrate all of those from x=0 to x=2. Invert all of them, adjust your limits of integration, and integrate. Report the results, I am particularly curious about the F=3x+1 case.

Last edited: May 17, 2017
16. May 17, 2017

### Staff: Mentor

If x is constant (at say x1), then the integral of Fdx is not Fx. It is equal to x1(F2-F1), where F1 and F2 are the initial and final values of F.

17. May 17, 2017

### Staff: Mentor

I also think that this is correct.

18. May 17, 2017

### Mr Real

I think I got some of it. Please correct me if i'm wrong. In post #12 I had said that we can take x out of the integral as it is constant; this, in fact is not true if we assume displacement constant we would get a straight line parallel to the force axis and then work done(area under the curve will be zero). Therefore the 2 integrals will be different.

19. May 17, 2017

### Mr Real

But at the beginner level I think it's imperative to state the reason that why is F.dx considered and not x.dF as most(if not all) beginners will probably get confused about it. I cannot imagine that there did not arise this issue when the formula was first derived or even that is an explanation about this confusion in authors' books.

20. May 17, 2017

### Mr Real

Now, as a result of this discussion i know that integral of x.dF has been proven not to give the net work done. But some parts of my original question still stands, why doesn't it give the work done instead of F.dx or is not even a recognised quantity even though it seems to be closely connected with work? And also similar quantities like electric flux ( why is it equal to integral of E.dA and not that of dE.A ?). Is the experimental result the only evidence? Or was it decided as the convention for some unknown reason?
(I know, now it seems as though i'm asking things that are not known but I can't think of a possible reason of it not being discussed anywhere!)