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Conservative vector field or not?

  1. Apr 24, 2007 #1
    To show that a vector field F=(P,Q,R) is conservative, is it enough to show that DP/DY = DQ/DX?
  2. jcsd
  3. Apr 24, 2007 #2

    I believe you have to show that the curl of your vector field is the zero vector (this is really equating the mixed partials as you'll see when doing it). Moreover, the domain of the vector field must be simply connected. A simple proof of this is by using Stoke's theorem I believe.
    Last edited: Apr 24, 2007
  4. Apr 24, 2007 #3
    I mean only showing that a certain vector field in R3 is conservative.
  5. Apr 24, 2007 #4

    matt grime

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    What is the definition of conservative (and just so we're clear: I know what it is, I'm not asking for my benefit)?
  6. Apr 24, 2007 #5
    A vector field Fvec is conservative if it has a potential function f, so that grad f = Fvec

    My book says that in R2, you only have to check that DP/Dy=DQ/Dx. However, it doesn't say anything aboud conservative vector fields in R3.

    Is it enough that DP/Dy=DQ/Dx here as well?
  7. Apr 24, 2007 #6

    matt grime

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    Clearly not. R^3 has x,y, and z. You can't just ignore the z. Where did the R go to? I suggest you operate with a nicer definition of conservative, for R^3, such as its curl is 0.
    Last edited: Apr 24, 2007
  8. Apr 24, 2007 #7
    We haven't learnt about conservative fields in R^3 yet. Only in R^2, and then it's sufficent that DP/dy=DQ/Dy.

    But the curl-definition is better.
  9. Apr 24, 2007 #8


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    You can try figuring out the original scalar potential based on what your P,Q and R are:

    If grad(Phi(x,y,z)) = F = (P, Q, R) then:

    P = dPhi/dx
    Q = dPhi/dy
    R = dPhi/dz

    Integrate P with respect to x, Q with respect to y and R with respect to z (remember to add "constant" functions that are unknown but functions of the other two variables... so when integrating P w.r.t. x, you have to add G(y,z) to your answer).

    It's likely you can eyeball the answer and figure out what the potential is without actually solving methodically the final part

    Obviously this isn't as clean as taking the curl, but it's possibly the method you're intended to use
  10. Feb 21, 2010 #9
    for R3
    you need to check all
    x: d/dy , d/dz
    y: d/dx, d/dz
    z: d/dx, d/dy
  11. Feb 22, 2010 #10


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    To say that <P(x,y), Q(x,y), R(x,y)> is a conservative vector field (that's really physics terminology- I prefer saying that "P(x,y)dx+ Q(x,y)dy+ R(x,y)dz is an exact differential) means that there exist some function F(x,y,z) such that
    [tex]dF= \frac{\partial F}{\partial x}dx+ \frac{\partial F}{\partial y}dy+ \frac{\partial F}{\partial z}dz= P(x,y)dx+ Q(x,y)dy+ R(x,y)dz[/tex]

    IF that is true, then look at the mixed second derivatives:
    [tex]\frac{\partial^2F}{\partial x\partial y}= \frac{\partial Q}{\partial x}[/tex]
    must be equal to
    [tex]\frac{\partial^2 F}{\partial y\partial x}= \frac{\partial P}{\partial y}[/tex]

    [tex]\frac{\partial^2 F}{\partial z\partial x}= \frac{\partial P}{\partial z}[/tex]
    must be equal to
    [tex]\frac{\partial^2 F}{\partial x\partial z}= \frac{\partial R}{\partial x}[/tex]

    [tex]\frac{\partial^2 F}{\partial y\partial z}= \frac{\partial R}{\partial y}[/tex]
    must be equal to
    [tex]\frac{\partial^2 F}{\partial z\partial y}= \frac{\partial Q}{\partial z}[/tex].

    To be sure the field is "conservative" you must check all three of those equations.
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