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

[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)

 

[pixel]

Then I thought isn't this possible too that if we apply a very small force dF for displacement x...

x is not a displacement. dx is a displacement.

 

[AlphaLearner]

x is not a displacement. dx is a displacement.

'dx' is too small. 'x' says a significant value of displacement. I feel Mr Real gets closer but the second part of his explanation in #70 saying 'dF' force causing displacement 'x' is not a valid case.

 

[nasu]

No, x is a specific value of the position. $\Delta x$ is a finite displacement . And dx is an infinitesimal displacement. xdF will be the value of position times the change in force, while at this position. There is no displacement involved in this expression. If there is a change in force is not due to a change in position. This expression may be nonzero maybe only for a time-variable force.

 

[Mr Real]

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

So does x being a position instead of displacement makes x.dF not meaningful?

 

[jack action]

So does x being a position instead of displacement makes x.dF not meaningful?

I don't know if it is not meaningful, but it is not work for sure. Like someone already said, it defines a force that varies while staying at the same position. By definition, to have work, you need a displacement.

 

[Mr Real]

Can you provide a reference that says that x is not displacement in this case, its just position? Because if it's true then it gives the answer to this discussion.

 

[AlphaLearner]

Can you give a reference that says that x is not displacement in this case, its just position?

Even I need some kind of proof, how 'x' is position.

 

[AlphaLearner]

In coordinate geometry, 'x' is called position/coordinate only if it has displaced from origin and that magnitude is what displacement means.

 

[AlphaLearner]

Simply to say 'x' is magnitude of length between two points.

 

[sophiecentaur]

Simply to say 'x' is magnitude of length between two points.

You clearly want to find out about this and to get it 'right' in your head. In that case, you should use a better disciplined source for your information than a thread that starts as this one did. Nothing was defined and it was assumed that we all knew what the OP meant; we clearly didn't. The resulting chat has probably confused you more than you needed.
This hyper physics link puts the calculation in a well behaved way and for a force that varies with displacement. That is how work is defined. No other definition can be relied on to give the right answer for a general case. Frankly, I don't see any point in investigating other combinations of symbols on the off chance that it will deliver the same answer as the accepted method.
A good textbook is worth its weight in gold and it is a shame that people don't rely on them. Even when money is a problem, there are on line sources that will do the same job better than you can ever rely on a pleasant chat amongst people with a range of levels of knowledge.

 

[Mr Real]

Sophiecentaur, if you are online, can you please reply to post #48?
Thank you for the link, but I had already gone through that and that I have another question now ( thanks to that link). How is work a line/path integral? Because I had read that a path integral depends on the path followed but work doesn't?
And now I know that x.dF cannot be the work done. But my question isn't that, now it is: What am I doing wrong when I see x.dF as applying very small force dF for a finite displacement x?

 

[AlphaLearner]

This hyper physics link puts the calculation in a well behaved way and for a force that varies with displacement.

Thanks for reference, It was all as I studied. Wasn't able to put my thoughts clearly. No coverings/excuses, Seriously.

A good textbook is worth its weight in gold and it is a shame that people don't rely on them. Even when money is a problem, there are on line sources that will do the same job better than you can ever rely on a pleasant chat amongst people with a range of levels of knowledge.

I have learned the concept of work on my own referring my textbooks, made no less use of them. Nobody taught me. Even this is the first time seeing a question like this and made my try onto it putting my thoughts so even I get better understanding.
I realized that people are not getting what I mean to say. I quitted from this discussion regarding it after Dale's warning and once again learning integration so I see where I am going wrong in expressing.

 

[jack action]

Can you provide a reference that says that x is not displacement in this case, its just position? Because if it's true then it gives the answer to this discussion.

It is the usual way of defining $x$ in a cartesian coordinate system. In the case of a one-dimension system, the distance between 2 points - i.e. a displacement - is $\sqrt{\left(x_2 - x_1\right)^2}$ or simply $x_2 - x_1$.

But it could be defined as you said, i.e. a distance from some reference origin. Although, if that was true, it would only work with a straight path (unless $x$ would be defined as the path traveled from the origin) and if you wanted to compare $xdF$ with $Fdx$, then what would be the meaning of $dx$? A small displacement beginning at the origin? It wouldn't make sense.

 

[Mr Real]

It is the usual way of defining $x$ in a cartesian coordinate system.

But can you provide a reference that explicitly says that in this case, i.e. in the case we are considering (work ); by x we mean a fixed position(or coordinate)?

 

[Mr Real]

It is the usual way of defining $x$ in a cartesian coordinate system. In the case of a one-dimension system, the distance between 2 points - i.e. a displacement - is $\sqrt{\left(x_2 - x_1\right)^2}$ or simply $x_2 - x_1$.

The same symbol may be used in many areas to mean different things.
Now, dx is a part of x and both are displacements, only x is finite but dx is infinitesimal in magnitude. Even in integration that's what we do, if we integrate all dx terms, we get x; now if dx is displacement then it's integral would also be displacement, it cannot be any other quantity. Also, we know work, W= F.x , here x is displacement by the definition of work, as it is said to occur only when a force produces a displacement , (therefore we take into account the displacement, not the position); for example, if a body was at position x=2 and after applying a force F it reached x=7 m position; we wouldn't take 7 in our formula, we would take x=5(7-2), so x is displacement.☺

Regards
Mr R

 

[jack action]

The general mathematical definition of work is:
$$W = \int_a^b \vec{F}(r) \cdot d\vec{r}$$
This is known as a line integral of a vector field. From Wikipedia:

Applications
The line integral has many uses in physics. For example, the work done on a particle traveling on a curve C inside a force field represented as a vector field F is the line integral of F on C.

So it does represent the work done.

where · is the dot product and r: [a, b] → C is a bijective parametrization of the curve C such that r(a) and r(b) give the endpoints of C.

What is parametrization? Again, from Wikipedia:

Parametric equations are commonly used to express the coordinates of the points that make up a geometric object such as a curve or surface, in which case the equations are collectively called a parametric representation or parameterization (alternatively spelled as parametrization) of the object. For example, the equations

form a parametric representation of the unit circle, where t is the parameter.

So it should be clear that $\vec{r}$ is an expression of the coordinates of the path.

If $\vec{F}(r)$ is always parallel to $d\vec{r}$ then you can simplify to:
$$W = \int_a^b F(r)dr$$
If $F(r)$ is constant, you can further simplify to:
$$W = \int_a^b Fdr$$
And if the path from $a$ to $b$ is along a single dimension (say $x$), then you simplify to:
$$W = F(x_b - x_a) = F\Delta x$$
Which is the simplest form found in physics textbooks.

now if dx is displacement then it's integral would also be displacement, it cannot be any other quantity.

Here is a representation of $x$ and $dx$ (shown as $r$ and $dr$):

$dr$ is the displacement along the path, which is what we care about. It is a distance. This is the same vector no matter where is the origin of the reference frame.

$r$ is the vector between the origin and the vector $dr$. It is also a distance. This vector depends on where is the origin.

If $dr$ begins at the origin, then $r = 0$.

So when we say $\Delta x = x_2 - x_1$, we actually simplify the vector equation $\vec{\Delta x} = \vec{x_2} - \vec{x_1}$ which looks like $\vec{dr} = \vec{(r+dr)} - \vec{r}$ in the previous figure.

 

[sophiecentaur]

The same symbol may be used in many areas to mean different things.
Now, dx is a part of x and both are displacements, only x is finite but dx is infinitesimal in magnitude. Even in integration that's what we do, if we integrate all dx terms, we get x; now if dx is displacement then it's integral would also be displacement, it cannot be any other quantity. Also, we know work, W= F.x , here x is displacement by the definition of work, as it is said to occur only when a force produces a displacement , (therefore we take into account the displacement, not the position); for example, if a body was at position x=2 and after applying a force F it reached x=7 m position; we wouldn't take 7 in our formula, we would take x=5(7-2), so x is displacement.☺

Regards
Mr R

Why are you wasting time with this. The simple derivation of work down with force F over a distance x is Fx. We know that. When the Force is not constant, the work has to be calculated using a Definite Integral between limits x1 and x2 of F(X)dx. Any other way of arranging those symbols has nothing to do with Words as we define it. There is nothing more to me said about the matter but the thread is 86 posts long. How can this be?

 

[pixel]

There is nothing more to me said about the matter but the thread is 86 posts long. How can this be?

I agree. Enough already.

 

[Dale]

Can you provide a reference that says that x is not displacement in this case, its just position? Because if it's true then it gives the answer to this discussion.

Here is a reference that describes the difference between position and displacement.

https://www.khanacademy.org/science...lacement-velocity-time/a/what-is-displacement

 

[AlphaLearner]

Enough. Even I have got enough from this and x.dF will not give a value we want just like how k is taken as 1 in F = k.ma through observation when we solve F α Δmv/Δt similarly, x.dF would have not given actual values when observed practically.
I'm done with this.

 

[Mr Real]

Okay, I think further discussions will not prove to be much fruitful at this point. But this discussion helped in clearing up some of my doubts. So thanks to everyone for helping me.
And I just hope that if someone at some point of time reads all this and knows what we all have been missing, they'll just post or start a conversation
Thank you all!

Warm regards
Mr R

 

[Dale]

someone at some point of time reads all this and knows what we all have been missing

I don't think we have been missing anything. You may not like the answer, but that doesn't mean that the answer is missing.

 

[Mr Real]

I don't think we have been missing anything. You may not like the answer, but that doesn't mean that the answer is missing.

It's not that I didn't like the answers. As I said some of my doubts were cleared (e.g. when someone if force is constant then x.dF is equal to zero which is not so for F.dx that proved why when force is constant x.dF cannot have anything to do with work). Like this I think there is a simple answer to the question: does x.dF mean anything or has anything to do with work? and that'll be pointed out to us in due time.

 

[Dale]

I think there is a simple answer to the original question too

See post 4 and post 64

I think there is a simple answer to the question: does x.dF mean anything or has anything to do with work?

The answer is indeed simple: "no"

 

[jack action]

does x.dF mean anything or has anything to do with work?

Let me show you what you are trying to do with a simpler example. Here is a prism with known side length $a$, $b$ and $c$:

Just like $Fdx$ represents work, $ab$ represents the area of one side of this prism. $a$ is a length, $b$ is a length and length times a length is an area.

Asking what is the meaning of $xdF$ is like asking what is the meaning of $ac$. $ac$ is meaningless in this case. Even if $a$ is a length and $c$ is a length, $ac$ is not an area.

So the answer to your question is $xdF$ means nothing, just like $ac$ means nothing. And I doubt someone will ever find meaning to them, even if it is mathematically possible to multiply both variables together. If it is meaningless, then it certainly doesn't have anything to do with work.

 

[Pedro Zanotta]

I´m not a physics, so, sorry for my comentary (if it turns out to be too naive). But there is a fundamental problem, in my opinion, with your doubt. As I understand it, you presume a fixed distance (x) and the force is varying, right? Then you integrate it to get the work done (in terms of dF). Ok? But if x is constant, there is no deslocation. So, although the force had been applied, there were no work done (as you had said elsewhere). I think this is a definition, physiclly speaking. Now, if force has been applied, then could work has been done? That depends: suppose you have an ideal elastic ribbon. You tied it to a stone and to a tree. Depending of the stone and the tree, the system could stand still. So, the ribbon is "tense" (there are forces) but the stone and the tree stay there. Now, if you have a machine (suppose a caterpillar) then you could try to displace the stone. But the stone is too heavy for the catterpillar. So, the motor works, but the stone endures. The motor has done work (it rotates, burn oil, etc) that could be measured. But, as the stone didn´t move, no mechanical work (in terms of the displacement of the stone) has been done.
All of this make sense?
My two coins
Pedro
(Sorry for my english)

 

[sophiecentaur]

@ Pedro Zanotta I think that the right message is in there somewhere Basic thing is that no work is done on a system if there is no movement, however great the force is. Caterpillar is too weak to make any change so no work. Machine is strong enough to get over the stiction of the stone and can do work.
Remember that this definition of work is 'pure' and takes no account of sweat and groaning whilst pushing at something that just won't move or in supporting a dead weight that stays in the same place. This is why scenarios such as the one you propose tend to get in the way of the actual Physics. Physics tries to reduce things to the very basics.

 

[Mr Real]

As I understand it, you presume a fixed distance (x) and the force is varying, right? Then you integrate it to get the work done (in terms of dF). Ok? But if x is constant, there is no deslocation.

Actually, no x.dF does not mean that x is considered fixed, it is constant only for the small force dF but it can vary for the total force F (just like in F.dx we assume F constant for small displacement dx but it can vary for the total displacement, x).

 

[jack action]

Actually, no x.dF does not mean that x is considered fixed, it is constant only for the small force dF but it can vary for the total force F (just like in F.dx we assume F constant for small displacement dx but it can vary for the total displacement, x).

$xdF$ means $x(F_2 - F_1)$ and $Fdx$ means $F(x_2 - x_1)$. Do you see that $dF$ is not a force, it is a force increment and $dx$ is not a position, it is a position increment?

$F$ is not equivalent to $dF$ and $x$ is not equivalent to $dx$.

 

[sophiecentaur]

Actually, no x.dF does not mean that x is considered fixed,

Yes it does. If you want x to vary then you need to state what it varies with thus:- x(F)dF
X can be any function of F that fits a physical situation. And that is where you are getting things wrong. Maths follows the Science. If it is a model of a Physical situation then you can progress from there. You cannot invent a mathematical expression and assume that is has any relevance at all to Science.
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.