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

[AlphaLearner]

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

 

[AlphaLearner]

In your case, 'x' is a constant and 'F' is the varying part and the value you want to integrate right?

 

[Dale]

I was talking about the formula x.dF that if we integrate this for a constant force then the integral would be zero. I was not talking about F.dx.

I can write down an infinite number of different quantities. Like mpxhg or mpxhg^2 or mpxhg^3 etc. Just because I can write it down doesn't mean that it is physically interesting. In order for it to be interesting I would need to show that it has some useful relationship to other quantities, or some special properties that make it otherwise useful. I certainly wouldn't expect any textbooks to spend any time discussing why mpxhg is not useful.

x.dF is not physically useful or interesting

 

[Mr Real]

I really have no idea what point you are trying to make. Random maths doesn't necessarily have any relation to Science. There are a million maths arguments that will tell you that 0=0 or 1=2 but they have no meaning.

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.

 

[AlphaLearner]

And if you say dF is constant yes, integrating it will make it a 0. But I feel it not a physics question you were asking, only the values in your question correspond here.

 

[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.

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