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What is a conservative vector field?

  1. Dec 19, 2013 #1
    I see how our line integral is a method for calculating work along a path by taking infinitesimally small 'slices' of our dot product of Force over our curve (distance). No problem here.

    Next we look to see if our field is conservative and if so then we know that regardless of the path the amount of work done will be equal so long as both particles have the same starting and subsequent ending points. How wonderful, but what does it mean?

    Assuming a 2D case I could see how if the force is equal and in one direction that this would seem to work out but I am seeing problems that have some rather complex Forces present.

    The best I can figure is that a conservative field is one that regardless of this 'complexity' it is static and unchanging while in a non conservative field the Force is changing at each point as the particle moves along it's path (as time lapses). Are these assumptions valid or is this the wrong way to looking at these problems?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited: Dec 19, 2013
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  3. Dec 19, 2013 #2

    HallsofIvy

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    The term "conservative" is really a physics term. It refers to a force field in which the force necessary to move in one direction is exactly the same as the force pushing in the opposite direction. We can then define a "potential energy" such that the force is the negative of the derivative of that energy function.

    Mathematically, we say that a vector function, [itex]\vec{f}(x,y,z)[/itex], is an "exact derivative" if there exist a function F(x, y, z) such that [itex]\vec{f}= \nabla F[/itex]. An important property of an "exact derivative" is that the integral from one point to another is independent of the path followed from the first point to the second. From that it follows that the intergral around any closed path is 0.

    Physically, that means that the work done in moving against or with the force field from one point to another, in a "conservative force field", is independent of the path and the work done in moving around a closed path, starting and ending at the same point, is 0.
     
  4. Dec 19, 2013 #3

    LCKurtz

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    The force varies along paths for either conservative or non-conservative forces. That isn't the difference. Think of lifting an object to a higher position. The work of lifting it will be reflected in its higher potential energy, and it will return that work if it is lowered. This is in the ideal case with maybe lifting it with frictionless pulleys. Now suppose the object is lifted the same height again but this time by being dragged up an inclined plane with friction. It will have the same new potential energy as before but you will have done more work getting it there to counteract the friction. Ditto letting it slide back down the plane. You don't get back the work wasted by friction. Similarly, if you drag it along a long gently sloping path you may waste even more energy to friction so the path now makes a difference.
     
  5. Dec 19, 2013 #4
    Okay so conservative fields are just like in PHY101 with the car rolling down the hill (frictionless, no drag, etc) and how energy is conserved regardless of the path taken as long as the vertical distance is equal.

    If we have paths where there is say wind blowing, mud on the slope, or whatever that puts a different Force on the particle for different paths so then it is non conservative.

    The mathematical approach for determining if a field is conservative takes the derivative with respect to 'y' along the 'x' component of our Force vector and 'x' with respect to the 'y' component and if equal we say we have a conservative field. Here we are looking at jerk on the particle(s) yes?
     
  6. Dec 19, 2013 #5
    Very good. So basically if the total energy in the system (which is just the particle) changes dependent on path then it is non conservative and if the energy remains the same regardless it is conservative. We can have multiple forces acting on our 'system' but they have to be equal regardless of the path.

    We have several examples where we have some fairly complex Forces dependent on position that are still conservative although I am still thinking about the 'test' for conservatism.
     
  7. Dec 20, 2013 #6

    vela

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    No, the jerk is the third derivative of displacement with respect to time, which is the derivative of the acceleration with respect to time.

    The work ##dW## done on a particle by a force ##\vec{f} = f_x\hat{i}+f_y\hat{j}## over an infinitesimal displacement ##d\vec{r} = dx\,\hat{i}+dy\,\hat{j}## is
    $$dW = \vec{f}\cdot d\vec{r} = f_x\,dx + f_y\,dy.$$ As you noted, the force is conservative when
    $$\frac{\partial f_x}{\partial y} = \frac{\partial f_y}{\partial x}.$$ In light of what Halls said above about ##f## being an exact derivative, does this look familiar?
     
  8. Dec 20, 2013 #7
    Makes sense, this is just the dot product of our Force components dotted with the derivative of the components of our position vectors to get our derivative of Work of which we take the integral to get the total Work done, very good.

    This next part is still a bit fuzzy,

    It is this partial that was the reference I was making about 'jerk' in my post. Is that not an appropriate term when taking the derivative of a Force? Is it because we are only looking at the components derivative with respect to the opposing axes and is also dependent on the result of taking this derivative? As in if we get a constant then no jerk but if we get a variable then there is?
     
  9. Dec 20, 2013 #8

    vela

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    No, the jerk is the derivative of acceleration with respect to time.
     
  10. Dec 20, 2013 #9
    I was thinking about that but the particle does not instantaneously move from one point to the next so I was assuming this implies time passes with change in position, (similar to what we see in dynamics)
     
  11. Dec 20, 2013 #10

    vela

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    Using that logic, the acceleration of a particle is given by dv/dx as well as dv/dt because x changes with time.
     
  12. Dec 20, 2013 #11
    That was the idea yes, is it not okay to view this system in this manner?
     
  13. Dec 20, 2013 #12

    vela

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    No, it isn't. dv/dx doesn't even have the right units.
     
  14. Dec 20, 2013 #13
    I am sorry, I do not believe I am conveying myself properly.

    I am not saying switching one for the other in the formulas, I am simply 'looking' at the derivatives of the partials as 'jerk' if assuming time passes with position. I am not trying to replace one with another in the math, just trying to take a closer 'look' using more familiar terms. Or are we supposed to be assuming no time passes with change in position in these problems?

    I am going to spend some time with my books on this one, maybe I am missing something more fundamental that is causing confusion. I appreciate your patience.
     
  15. Dec 20, 2013 #14
    Got it!

    Okay, while I was busy being side-tracked by 'jerk' I just noticed how the 'conservative test' works when re-reading Anton.

    This comes down to our original dot product and how it essentially works out to,

    ∫ydx + xdy

    If ydx is = to xdy then the dot product is zero which means so is our integral hence our 'conservative field test'! It is so simple!

    Now I can go back to jerk, thanks for all the help everyone.
     
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