# on differentials again

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In a certain book i read and i quote :

"As in mechanics this work is defined by the integral.

W = $\int F.dl$.

where F is the component of the force acting in the direction of the displacement dl.In a differential form this equation is written.

δW = F.dl.

where δW represents a differential quantity of work"

Is that correct??

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I have moved some off-topic posts that don't answer your question to this thread, where perhaps they can be discussed properly.

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δW = F.dl.

where δW represents a differential quantity of work"

Is that correct??

$\delta W = dl \cdot F$

is an inexact differential.

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$\delta W = dl \cdot F$

is an inexact differential.

F.dl is an exact differential . If you consider δW as an inexact differential (which you may) ,then we have an inexact differential equal to an exact differential.

And again my problem is how do we mathematically prove that :

if $W =\int F.dl$ then δW = F.dl

Edited by triclino
correction

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You cannot consider $\partial W$ an exact differential because it is not an exact differential. Do you know what the phrase "exact differential" means?

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Sorry it was a typo i meant to say that δW is an inexact differential.

I corrected my post

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F.dl is an exact differential .

No, it is not.

Suppose you need to bring a gas from pressure P0 and volume V0 to pressure P1 and volume V1. The amount of work done depends on the path taken through the P,V phase state. For example, consider these two paths:

1. Change the volume from V0 to V1 while maintaining a constant pressure P0, then change the pressure from P0 to P1 at constant volume V1.
2. Change the pressure from P0 to P1 at constant volume V0, then change the volume from V0 to V1 while maintaining a constant pressure P1.

Both paths achieve the desired goal. The work done is P0*(V1-V0) for the first path, P1*(V1-V0) for the second. Clearly these are not the same except for the trivial cases V1=V0 or P1=P0.

That the result of some integration depends not only on the initial and final states but also on the path taken from the initial state to the final state is one of the signature characteristics of an inexact differential.

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from mathematics we have :

W=$\int F.dl$ => $\frac{dw}{dl} = F$ => dw = $\frac{dw}{dl}.dl$ => dw = F.dl.

Isn't dw an exact differential and thus F.dl ,since dw/dl = F??

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No. You cannot do that.

I posted this earlier, but my post was mistakenly moved along many other posts in the cleanup of the hijack of this thread. A better way to express work is

$W = \int_C \vec F \cdot d\vec l$

Force $\vec F$ and the differential length $d\vec l$ are vectors. What you did is invalid. You are once again treating vectors as if they are scalars.

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Note that the curve is specified in the above.

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You do not seem to have read my original post carefully .We are not examining

$W = \int_C \vec F \cdot d\vec l$ ,but the scalar product :$\int Fdl$ as the OP points out.

But let me ask you a question:

Suppose we are given the differential f(x)dx. How do we know whether this differential is excact or inexact ??

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You do not seem to have read my original post carefully .We are not examining

$W = \int_C \vec F \cdot d\vec l$ ,but the scalar product :$\int Fdl$ as the OP points out.

From the OP, (emphasis mine):

In a certain book i read and i quote :

"As in mechanics this work is defined by the integral.

W = $\int F.dl$.

where F is the component of the force acting in the direction of the displacement dl."

In other words, $\int_C\vec F\cdot d\vec l$. We are talking about exactly the same thing here, triclino. I am being explicit in showing that the integral is a path integral.

But let me ask you a question:

Suppose we are given the differential f(x)dx. How do we know whether this differential is excact or inexact ??

A differential will always be exact if the underlying space truly is one dimensional (which is implicit in what you wrote). Inexact differentials only appear in higher dimensional spaces.

A differential $\partial F \equiv \sum_i f_i(x) dx_i$ is exact if

• There exists some differentiable function $\phi(x_1,...,x_n)$ such that $\nabla \phi(\vec x) = \vec f(\vec x)$, or
• The integral $\int_{\vec a}^{\vec b} \partial F$ is the same for all paths that start at point $\vec a$ and end at point $\vec b$.

There are many other tests for an exact (or total) differential.

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If f(x) = 1 if x is rational and f(x) =0 if x is irrational ,then

f(x)dx is the differential of which function??

Is that an exact or inexact?

Edited by triclino
correction

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If you mean something like

$f(x) = \Biggl\{\array{ll} 0 & x=0 \\ \pi & x\ne 0\endarray$

then yes, $f(x)dx$ is exact. It integrates to

$F(x) = \pi x + c$

Note that $\frac {dF}{dx} = f(x) \,\text{a.e.}$

Edited by D H

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Sorry i meant :

f(x)= 1 if x is rational and,

f(x) =0 if x is irrational .

I corrected my post

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This integrates to F(x)=x+c.

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Sorry again to be exact the function is defined as follows:

$f : [0,1]\Longrightarrow R$,where

f(x)= 1 ,if x is rational and

f(x) =0 ,if x is irrational

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Oops. I misread your post. I thought you said f(x) = 1 if x is irrational, 0 if x is rational. That function integrates to F(x)=x+c. That is the standard gotcha problem used to introduce Lebesgue integration.

You switched the values so that f(x)=1 if x is rational, 0 if x is irrational. This function is zero almost everywhere. It integrates to a constant.

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Oops. I misread your post. I thought you said f(x) = 1 if x is irrational, 0 if x is rational. That function integrates to F(x)=x+c. That is the standard gotcha problem used to introduce Lebesgue integration.

You switched the values so that f(x)=1 if x is rational, 0 if x is irrational. This function is zero almost everywhere. It integrates to a constant.

Which is then the function F(x) whose derivative is equal to our function f(x),

so that the differential f(x)dx is exact??

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Almost. The derivative of F(x) is equal to f(x) almost everywhere. Click on the 'almost everywhere' in either my previous post, or in your quote of my post. Or google the term "almost everywhere".

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