1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why elementary work is not an exact differential?

  1. Dec 14, 2011 #1
    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. jcsd
  3. Dec 14, 2011 #2
    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.
  4. Dec 14, 2011 #3
    I'm not talking about U(potential energy function), I'm asking about W.
    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!
  5. Dec 14, 2011 #4

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    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).
  6. Dec 14, 2011 #5
    Thanks, it was helpful!
  7. Dec 14, 2011 #6
    Whenever you write df it implies that the differential operator is applied to the function f, so if you write work as an exact differentail dW implicitly you are saying force can be written as the derivative of some function of spatial coordinates.
  8. Dec 15, 2011 #7
    Yeah! right!
    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.

  9. Dec 15, 2011 #8
    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.
  10. Dec 15, 2011 #9
    If force is a derivative of some function of spatial coordinates, this function is called the potential energy, and the force is conservative. Non-conservative forces are certainly not derivatives of any function.
  11. Dec 17, 2011 #10
    well... thanks,
    I concluded the same!!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook