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

[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.

 

[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.