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How do you determine if a force field is conservative?

  1. Feb 1, 2012 #1
    I'm doing an undergraduate ODE course and the question gives us an equation.

    F(x,y) = (2y, 2x)

    Then it asks "determine whether the force field F is conservative".0
    How do we interpret (2y, 2x)? Is 2y, the x-component and 2x the y-component? Also what does it mean by conservative forcefield? And how do we determine that?
     
  2. jcsd
  3. Feb 1, 2012 #2

    HallsofIvy

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    Yes, the notations, (u, v), <u, v>, and ui+ vj all indicate the same two dimensional vector.

    A vector function u(x,y)i+ v(x,y)j is "conservative" (That's actually a physics term. I prefer the mathematics term, "exact differential") if and only if there exist a scalar function F(x,y) such that [itex]\nabla F= u(x,y)i+ v(x,y)j[/itex]. One can show that u(x,y)i+ v(x,y)j is conservative (or an exact differential) if and only if
    [tex]\frac{\partial v}{\partial x}= \frac{\partial u}{\partial y}[/tex]
     
  4. Feb 1, 2012 #3

    jambaugh

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    The field is conservative if it is the gradient of a scalar field.
    That means: [itex] F(x,y) = ( \partial u /\partial x , \partial u / \partial y )[/itex]

    The field you gave is indeed x and y components. you could also write it in terms of basis vectors: [itex] F = 2y\mathbf{i} + 2x\mathbf{j}[/itex].

    To determine if it is conservative you can find the actual function u such that [itex]F=\nabla u[/itex].
    But a sufficiency test is to take the "cross partial derivatives".
    Since [itex]\partial^2 u / \partial x \partial y[/itex] does not depend on the order in which you take partial derivatives you can take the partial w.r.t. y of the x component of F and the partial w.r.t. x of the y component of F and if F is conservative these will be equal (and equal to the above cross partial).

    To find [itex] u [/itex] such that [itex] F = \nabla u[/itex] you should integrate the first component with respect to x (treating y as a constant) and remember that the arbitrary constant (w.r.t. x) will be an arbitrary function of y. Likewise integrate the 2nd component of F w.r.t. y. Both integrals must be equal so match up the arbitrary "constant" terms.

    Example: Consider the constant field [itex] F = 2\mathbf{i} + 3\mathbf{j}[/itex].
    [tex] u = \int 2 dx = \int 3 dy[/tex]
    [tex]u = \int 2 dx = 2x + C_1(y)[/tex]
    [tex]u = \int 3 dy = 3y + C_2(x)[/tex]
    [tex] 2x + C_1(y) = 3y + C_2(x)[/tex]
    [tex] C_1(y) = 3y + C, \text{ and } C_2(x)=2x+C[/tex]
    Set C = 0 (or 42 or pi or whatever) since you don't care which u you get as long as it works.
    [tex] F = \nabla (2x+3y)[/tex]
    and so is conservative!
     
  5. Feb 1, 2012 #4
    Thanks that makes sense.
     
  6. Feb 3, 2012 #5

    LCKurtz

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    It is much better after integrating with respect to x and getting something like$$
    u(x,y) = f(x,y) + C(y)$$ to differentiate that with respect to y and set it equal to the y component of F. The problem is that after integrating the y component of F, "matching up" isn't well defined and can lead to wrong answers.
     
  7. Feb 3, 2012 #6

    jambaugh

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    Whether it is "much better" may be a matter of the specifics of the problem. I doubt it is substantially better, the method is intrinsically the same (you'll have to integrate twice).
    I'll agree it may be better and will be "as good as".

    As to "matching up" leading to wrong answers, I do not see how.

    Once you integrate each part you have an equation:
    [tex]f(x,y)+C(y) = g(x,y)+D(x)[/tex]
    Given a solution, a C and D function for which you have equality satisfied, you will have a function whose gradient IS the original field. A solution is a solution.
     
  8. Feb 4, 2012 #7

    LCKurtz

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    I don't disagree with anything you have said here. The problem I have encountered is when students get a bit too informal with it. All too many times I have seen students, when faced with finding a potential function ##\phi## for a vector field like $$
    \vec V = \langle y^2+2e^{2x},2xy -\sin y\rangle$$ they, in stead of correctly "matching up" the two results, do the following:

    ##\phi_x = y^2+2e^{2x}## so ##\phi = xy^2 +e^{2x}##
    ##\phi_y = 2xy-\sin y## so ##\phi = xy^2+\cos y##.
    They now look at those two results and build an answer by taking "anything new" but not repeating any common terms, so they don't write the ##xy^2## twice and give a final answer
    ##\phi(x,y) = xy^2 +e^{2x}+\cos y##
    which is correct. And building the answer this way frequently, I would even say almost always, gives them the correct answer in spite of the fact that they have ignored the "constants" of integration. I used to give them an example where, when they did that, they got the wrong answer because the common part of the two answers didn't look the same since they differed from each other by a constant, which wasn't obvious to them. So they put it in twice when building their final version of ##\phi##.

    I would go so far as to state that, if for no other reason, the method I have suggested (which is standard in any calculus book I have seen) is never more complicated than yours and frequently easier.
     
  9. Feb 7, 2012 #8

    jambaugh

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    I see your point and do not disagree. I further do not see that much difference hence my quibble. As I see it, in presenting the initial exposition "what one is trying to do" it is better expressed as I presented it as this maintains the symmetry of the initial problem. Then in a an exposition of "how to get there" your procedure is the most systematic means.

    Armed with both the student should see both the best starting point (choosing which term to integrate first) and the best path (the procedure you outlined).
     
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