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Violation of energy conservation?

  1. May 6, 2012 #1
    Hello,

    I have a question, which is kinda stupid probably, but I cannot solve it for myself. Hence the fancy title.

    Imagine you have a point mass m near a more massive object M. Let's say there is some other force supporting the point mass, so it can stand still at [itex]r = \vec{r_0}[/itex]. Point mass then has a potential [itex]V(\vec{r_0})[/itex].

    Now, imagine now that this supporting force changes and exerts a sudden movement of the particle along the lines of the force field. Once particle starts moving, the supporting force becomes equal in value and opposite in direction to the gravity, so it effectively annuls it. What I'm trying to say here is that you imagine a force which acts upon the point mass, and then this point mass, due to this force, goes from resting to some constant velocity in the direction of unit vector [itex]\hat{r}[/itex] Mathematically, I'm not sure how I would model this force, maybe something like this
    [itex]F = \delta(\vec{r}-\vec{r_0})\hat{r} - G(\vec{r})[/itex], where [itex]G(\vec{r})[/itex] is a gravitational force.

    Here is my reasoning, and hence my problem: as this point mass moves away from an object M, its potential energy grows. This is logical, because the bigger the distance [itex]r[/itex] is, the bigger the potential energy is. Also, if you look it the other way around, force F is doing this work of moving the particle, and that invested energy must go somewhere, and it goes into the kinetic energy of the point mass (constant value) and into increasing the potential energy of the particle.

    The problem arises in my head when I try to compute this increase of the energy by calculating the integral of the force over the distance traveled. First I look at the point mass, and I say to myself:
    1) look all the forces that act on the particle, and find the net force
    2) calculate the integral of that force over the trajectory.
    Well, there are only two forces acting upon the point mass - force F and gravitational force G. If I add them up, the Gs will disappear and the only thing that will remain is the Dirac impulse. If I integrate that over any interval, I will always have the same value, and hence the same energy difference.
    This, of course, doesn't make sense. The bigger the distance traveled, the bigger potential energy the particle has. Or, if I use the potential function, the bigger the [itex]\vec{r}[/itex], the bigger [itex]V(\vec{r})[/itex] is, and therefore, [itex]V(\vec{r})-V(\vec{r_0})[/itex] must increase as well.

    Another variant of this scenario, that completely flabbergasted me, was to imagine the particle starting to move from resting at [itex]\vec{r_0}[/itex], in the direction of the unit vector, and then stopping at [itex]\vec{r_1}[/itex]. Force who could make such movement would have two Dirac impulses, one positive and one negative - positive to make the point mass going, and negative to stop it.
    [itex]F = \delta(\vec{r}-\vec{r_0})\hat{r} - G(\vec{r}) - \delta(\vec{r}-\vec{r_1})\hat{r}[/itex]

    Again, if I look at the net force, it will only have these two impulses. After I integrate it, the work will be zero! So that will mean that I could move the point mass to any point away from [itex]\vec{r_0}[/itex] at the expense of zero energy. Of course, the amount of the energy I would have spent is [itex]m \cdot (V(\vec{r_1})-V(\vec{r_0}))[/itex].

    I was wondering if anyone could tell me where am I at fault here. I can't seem to find it myself.
    Thanks.
     
  2. jcsd
  3. May 6, 2012 #2

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    Two points:
    1. The change in gravitational potential energy is the opposite of the work done by the gravitational force. The only time the gravitational force does no work is if the force and the velocity vector are perpendicular, or in other words the particle is moving in a circle around the large mass.

    2. The impulse requred to stop the motion is not necessarily the same magnitude as the impulse that started it. The difference between them will of course be the change in potetial energy of the particle.
     
  4. May 6, 2012 #3

    Dale

    Staff: Mentor

    Since you are finding the net force you can use the Work-Energy Theorem (I think that is probably your intention). Note that the Work Energy Theorem describes the increase in KE which is due to the net force. It does not describe any changes in PE due to individual forces. So the only time that there is a non-zero net force is the delta force, and that is the only time that the KE changes, as expected.

    No, it makes perfect sense, that is exactly what the work energy theorem predicts.

    This is also correct, but since it is a potential energy and not KE you need to calculate it based on the individual force to which it pertains, not the net force.
     
  5. May 7, 2012 #4
    Hm, wow, it seems I was living in a lie the whole time.
    I thought that [itex]\Delta K = -\Delta U[/itex] was always true, no mater what forces are there, only the net force is considered.

    Thank you guys, it seems I got some thinking to do!
     
  6. May 7, 2012 #5

    jtbell

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    Staff: Mentor

    No, it applies only to conservative forces, that is, forces that can be derived from a potential via ##\vec F = - \vec \nabla V##.

    $$\Delta K = W_{total} = W_{conservative} + W_{nonconservative}$$
    $$\Delta K = -\Delta U + W_{nonconservative}$$
     
  7. May 7, 2012 #6

    Dale

    Staff: Mentor

    Adding to jtbell's comments, in this case, the impulsive force is non-conservative. It changes the KE without chainging the PE.

    That does not imply that energy is not conserved.
     
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