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Why elementary work is not an exact differential? 
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#1
Dec1411, 03:35 AM

P: 20

Why elementary work is defined as δW=Fdr?
My ques. is not on the definition; it is on why it cannot be dW=Fdr? 


#2
Dec1411, 03:44 AM

P: 23

It is only an exact differential int he case of a conservative force. Precisely, if there is a potential we write
dU = Fdr which implies that F = dU/dr. 


#3
Dec1411, 03:54 AM

P: 20

I know that in case of conservative/potential field δW=dU. Reference: Fundamental Laws of Mechanics, IE irodov from equation 3.1 to 3.49 wherever needed he used δA for elementary work, in general! 


#4
Dec1411, 04:15 AM

Mentor
P: 15,153

Why elementary work is not an exact differential?
Work is a line integral:
[tex]W=\oint \mathbf F \cdot d\mathbf r[/tex] It is path independent in the case of a conservative force, but in general the integral depends on the path. In other words, [itex]\mathbf F \cdot d\mathbf r[/itex] is not an exact differential by definition (an exact differential is path independent). 


#5
Dec1411, 05:28 AM

P: 20




#6
Dec1411, 03:27 PM

P: 23




#7
Dec1511, 03:13 AM

P: 20

Now, what does it further imply? Cannot force be derivative of a function of spacial coordinates? I think I have got it, but request you to elaborate so that I may confirm. Thanks! 


#8
Dec1511, 05:12 AM

P: 84

If work was an exact differential, for any two points a and b, you could write that the work to go from one point is F(b)  F(a), where F' is work. But this is most certainly not true, as this is saying work is a function of state, i.e. if you have a point, you'd have a work associated to it. This is false, as work is something you use to go from one state to another. It's pretty much like heat. Heat is also not a function of state and depends on the path.



#9
Dec1511, 08:42 AM

P: 23




#10
Dec1711, 11:28 PM

P: 20

I concluded the same!! 


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