Understanding Work Done and Integration: Exploring Small Work and Electric Flux

[AlphaLearner]

I was agreeing with your point that if we consider x.dF then it wouldn't work for a constant force and that goes a step ahead in proving why it can't be considered ( But I need more explanation for the case when the force is variable). Thanks for making it a little more clearer.

Yes, dF is clearly saying that force is not constant. But did you see that 'x' alone? Says that displacement is constant. Work done is clearly zero if that 'x' does not vary in any way with respect to time.

 

[Mr Real]

Yes, dF is clearly saying that force is not constant. But did you see that 'x' alone? Says that displacement is constant. Work done is clearly zero if that 'x' does not vary in any way with respect to time.

But x.dF doesn't say that x is constant just like F.dx doesn't say that F is constant, it only says that F is constant for a small displacement dx.

 

[AlphaLearner]

dF says the reading of Force at that moment in the process only and its going to change in the next moment to some other value. That's why you added that small 'd' in front of that 'F'.

 

[AlphaLearner]

Just 'F' denote that the force was same and unchanged at all moments throughout the process.

 

[Dale]

But x.dF doesn't say that x is constant

Correct. But it does say that F is not constant*, and often F is constant.

*or at least it says that if F is constant then the integrated quantity is 0

 

[AlphaLearner]

'dx' says displacement was 'x' only at that moment and the very next moment it just changed (Increased) in this case.

 

[Mr Real]

Just 'F' denote that the force was same and unchanged at all moments throughout the process.

If we are considering F.dx it means that F was constant during the displacement dx, it can vary during the whole process. Similarly x.dF means x was constant for force dF and
it can vary for the whole process.

 

[Dale]

That isn't the problem. The problems are that dF is just weird to begin with, the quantity is zero if F is constant, and that there is nothing physically interesting or relevant about x.dF or its integral.

I don't know what more can be said.

 

[AlphaLearner]

Mr real, you are facing problem understanding mathematical expressions leading you to misconceptions in physics. I suggest you to have a look at this book "Calculus made easy" giving a clear idea of what 'dx' means and think about how it is applied in physics. You will find solution for this answer too.

 

[AlphaLearner]

You can find a PDF of it at Archives.org

 

[Mr Real]

Mr real, you are facing problem understanding mathematical expressions leading you to misconceptions in physics. I suggest you to have a look at this book "Calculus made easy" giving a clear idea of what 'dx' means and think about how it is applied in physics. You will find solution for this answer too.

You said that x.dF means that x is constant which it doesn't have have to be that's what I explained to you. Then you said what dx means which i am already pretty clear with. Please read the questions carefully before answering them.

 

[AlphaLearner]

For now, I say that in physics, all processes are studied by dividing them into smaller moments (aka. instantaneous) moments. Similarly, in a process F.dx gives work done only at a moment in that process. Not total work done throughout that process. For total work done throughout process, formula is just F.X. 'F' denote force applied (consistently throughout the process) and X says total displacement (distance traveled) throughout the process.

 

[Mr Real]

I know now that x.dF is not meaningful when F is constant and I have thought of a reason why x.dF is not meaningful at all. I think that this reasoning might be a little flawed but I hope it's correct. Please correct me if I'm wrong. Here goes: F.dx is meaningful as it is possible to get dx displacement by applying force for an infinitesimal time but it would require infinite time to produce a displacement x from dF force, which is not possible.

 

[Mr Real]

Since there is no 'd' in front of 'x' it clearly states, 'x' is constant. Similarly no 'd' in front of 'F' says 'F' is constant.

But no one ever asked you regarding the meaning of dF or dx. Your comments are off-topic so please stop posting replies that are not relevant to the discussion.

 

[AlphaLearner]

Well, I wasn't trying to make you understand this thing through physics or general way but through mathematics. That why sound like "Off the topic". What you reasoned above does not even need to be that complicated. You can simply say, body stood rock solid for even instantaneous force so no work is done.

 

[Dale]

Since there is no 'd' in front of 'x' it clearly states, 'x' is constant. Similarly no 'd' in front of 'F' says 'F' is constant.

This is not correct. For instance, for a spring F=-kx. So F is not constant, but the integral of F.dx is well defined.

 

[Mr Real]

Can i have someone's reply (who knows what he's talking about) for post #48( especially sophiecentaur or Dale)?

 

[AlphaLearner]

B

This is not correct. For instance, for a spring F=-kx. So F is not constant, but the integral of F.dx is well defined.

But... F=-kx is an 'equation' expressing how force vary with change in displacement proportionally. F.dx is just an expression and when this small part F.dx is made to join with other remaining parts (integrating), even remaining parts should also be having same 'F' but different 'x'. Its like grouping up different members 'x' under same team 'F'.

 

[sophiecentaur]

Differentiating or integrating in presence of a constant (multiplied or divided) by function will not give zero.

It's a definite integral between equal limits. That gives zero.

 

[AlphaLearner]

It's a definite integral between equal limits. That gives zero.

Prove me then differential or integral of a function like f(x) = 2x will give a zero.

 

[Mr Real]

It's a definite integral between equal limits. That gives zero.

(I didn't know how to approach you so I'm replying to this post to get your attention). Please can you reply to post #48.

 

[sophiecentaur]

(I didn't know how to approach you so I'm replying to this post to get your attention). Please can you reply to post #49.

I have no idea what #49 is about.

 

[Dale]

Please correct me if I'm wrong. Here goes: F.dx is meaningful as it is possible to get dx displacement by applying force for an infinitesimal time but it would require infinite time to produce a displacement x from dF force, which is not possible.

Well, that sounds like a reasonable heuristic, but mathematically neither F nor dx has any restrictions related to time. So I am not sure that it is exactly on point.

The reason that x.dF is not physically meaningful is simply that it is not related to any other physical quantity of interest. It is not related to work, it isn't independently conserved, it is not the outcome of any measuring device. I think there is nothing mathematically wrong with it (despite peculiarities like evaluating to 0 for constant force), but it doesn't compute a quantity that we want to know.

 

[Mr Real]

I have no idea what #49 is about.

Sorry I meant post #48.

 

[Dale]

Prove me then differential or integral of a function like f(x) = 2x will give a zero.

$$\int_1^1 2x \, dx = 0$$

 

[Mr Real]

Well, that sounds like a reasonable heuristic, but mathematically neither F nor dx has any restrictions related to time

I meant that x.dF is not physically possible at all as we can only get a finite displacement from an infinitesimal force if we apply it for an infinite time. So maybe this is also a plausible argument for x.dF not having anything to do with work.

 

[sophiecentaur]

Prove me then differential or integral of a function like f(x) = 2x will give a zero.

Yes, of course the Indefinite Integral (the inverse of the derivative) may not be zero but the Definite integral would be appropriate when calculating Work Done. That involves subtracting the indefinite integral,, evaluated for one value of F from the Integral, evaluated for another value of F. When the two values of F are the same, the definite integral will be zero.
Perhaps you need to find out what definite integrals are all about. If you are working from a calculus textbook than you need to look for the chapter on definite integrals or you may find this link useful. You will need to follow it all through.

 

[AlphaLearner]

Thank you Dale and sophiecentaur for making me understand definite Integrals, I kept indefinite Integrals in mind and asked it .You did not understand post #48 because you thought F.dx was work done for entire process, I thought that F.dx is just a piece of total part F.X! And that's why we find a need to integrate F.dx to get total work done in the process, at least between two points.

 

[jack action]

Here is a graphical representation of $F=x^2$ (black line):

The blue area represents the integral:
$$\int_4^8 Fdx$$
The orange area represents the integral:
$$\int_{16}^{64} xdF$$

This should help you visualize how different can be $Fdx$ and $xdF$.

Why nobody uses $xdF$? Because $x$ is a position that is dependent on how you define the frame of reference, so it is rather useless. But the difference between two positions ($dx$) is the same no matter what is the frame of reference.

For example, say you want to know the distance between 2 points (imagine 2 cities on a map). One point has the position (0) and the other one (10). The distance is is 10 - 0 = 10. But if we change the system of reference where the origin is somewhere else, the first point may hold the position (32) and the other one (42). The distance is still 10 (= 42 - 32).

In such a case evaluating something based on the position would be meaningless as it would change depending on your frame of reference.

But there are cases where you can have meaning for both integrals. In thermodynamics, on a Pressure-Volume diagram, you can evaluate $Vdp$ or $pdV$ to evaluate the work done. The $pdV$ form is called boundary work and relates to a closed system, like when a fluid is enclosed within a chamber that can vary in volume (piston-cylinder arrangement), but the mass inside stay constant. The $Vdp$ form is called isentropic shaft work and relates to open systems where fluid comes in and out like with a turbine. In that case the physical volume of the turbine doesn't change, but the output pressure is different from the input pressure. $Vdp$ considers the $pdV$ part, but also the work needed to maintain the flow within the system.

 

[Mr Real]

Yes, of course the Indefinite Integral (the inverse of the derivative) may not be zero but the Definite integral would be appropriate when calculating Work Done. That involves subtracting the indefinite integral,, evaluated for one value of F from the Integral, evaluated for another value of F. When the two values of F are the same, the definite integral will be zero

(Again, replying to this to get your attention)Please can you reply to post #48? (Sorry, I mistakenly said #49 last time)

 

[Dale]

So maybe this is also a plausible argument for x.dF not having anything to do with work.

Work is defined as the integral of F.dx. Since x.dF is not equal to F.dx it is proven that it is not equal to work. You don't need a plausible argument, you have a definite mathematical proof.

 

[jack action]

Post #48:

F.dx is meaningful as it is possible to get dx displacement by applying force for an infinitesimal time but it would require infinite time to produce a displacement x from dF force, which is not possible.

2 things are wrong.

First, time has nothing to do with this. Nobody assumes that the displacement is the consequence of the force. If a force is moving (i.e. there is a displacement) then, by definition, there is work.

Second, x is not a displacement, it is a position. dx is a displacement, i.e. the difference between 2 positions.

 

[Nidum]

Can this be looked at by considering the idea of duality ?

A physical problem can be formulated as a basic mathematical model or as the dual of the basic mathematical model .

Integral F. dx has meaning in the basic model and integral x.dF has meaning in the dual ?

 

[Dale]

Can this be looked at by considering the idea of duality ?

A physical problem can be formulated as a basic mathematical model or as the dual of the basic mathematical model .

Integral F. dx has meaning in the basic model and integral x.dF has meaning in the dual ?

Do you have a reference which discusses x.dF and its physical meaning in the dual model you are talking about?

 

[Mr Real]

I'll just say why I had the doubt about this in the first place, maybe answers to this will help in clearing up my confusion.
Okay, so when i first read that small work done is F.dx , I was able to see it as if we apply a force F for a small displacement dx then it would give the work done for this displacement (small work done) and then if we keep on applying force F (which can be constant or variable) for x displacement, then we would get the work for x displacement (total work done). Then I thought isn't this possible too that if we apply a very small force dF for displacement x, this would also be small work done and then if we continue applying force till it reaches the value F, we would get the total work done as F.x (this x would be different from the x for dF force)