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Conservative forces

  1. Feb 27, 2015 #1
    When we say that conservative forces don't vary with time, we are talking about a specific position, right? Because if the position is allowed to vary with time, then the force varies with time.
    In general, the (net) force on a body may be written (in one dimension) as ##F = m\ddot{x} = mv \frac{dv}{dx}##
    In other words, we can express it as a function of position or velocity: ##F = f(t) = g(x)##
    For ##F## to be conservative, should ##f'(t) = 0##? Or should ##f'(t) = 0## only if ##x = a## where ##a## is a constant?
    By the way, I have deliberately chosen the one dimensional case since I have very little knowledge of vector calculus.
    Last edited: Feb 27, 2015
  2. jcsd
  3. Feb 27, 2015 #2


    Staff: Mentor

    No, that isn't quite right. You're mixing a position at some point in the force field with your movement within the force field.

    If you're in a gravitational field then field is constant at any given point. That is if you move around and then return to the point the force is still has the same magnitude and direction at that point.

    Think of it in terms of Newton's gravitation law: f(r) = G * m1 * m2 / r^2

  4. Feb 27, 2015 #3
    What's the difference? If a body moves in the field, then its position always corresponds to some point in the field, right?
  5. Feb 27, 2015 #4


    Staff: Mentor

    The point is though if you monitor a collection of points in the conservative force field the force at those points doesn't change over time. The fact that you're using some object which moves from point to point to measure the force is irrelevant. The force is still dependent only on the position.
  6. Feb 27, 2015 #5
    I understand that the force field itself doesn't depend on time. It only depends on position. But the force between two masses for example would depend on time if their separation depends on time, right?
  7. Feb 27, 2015 #6
    Also, what is wrong with this statement: "##\frac{d}{dt} \vec{F}(\vec{r}) = \vec{F}'(\vec{r}) \frac{d\vec{r}}{dt}## implies that if ##\vec{v} = 0## then the force is constant"? In other words: "the force on a particular mass in a gravitational field becomes time-dependent if the mass begins to change position relative to the other mass".
    Doesn't this show that if the position does not change with time, the force would be constant (since it only depends on position)?
    Last edited: Feb 27, 2015
  8. Feb 27, 2015 #7
    This discussion raised a somewhat unrelated question. How can any force written as a function of position give two different results for the same position (nonconservative forces)? Will this become clearer when I study vector calculus?
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