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My Mind going to blow! Why Conservative field Path independ? by TMSxPh

  1. Apr 5, 2014 #1

    Kin

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    Searching for 'Conservative fields intuition' comes up with a thread pretty high up on the list with this question. Which was closed years ago.

    "

    Hi

    I'm now reading about Vector fields, everything is clear and intuitive for me as curl divergence ..ect , except one simple thing that I'm straggling with for the last 4 days! I searched Internet and a lot of math & physics books but in vain :

    Why Conservative field is Path independent? or why work between two point will be path independent, by definition conservative field is the one generated out of gradient of some scalar, and purely mathematically it's very straight forward and clear, but it's not intuitive for me at all, I can't find any geometric explanation for that even so it seems to be very neat , Beautiful and simple result...

    Any I ideas? thx in advance.

    "
     
  2. jcsd
  3. Apr 5, 2014 #2

    Kin

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    Can you say that conservative fields have no circulation, so every path between two points can be collapsed into a straight line between two points?

    Or every circulation containing the straight line has the same value, so all the other paths must be the same as that line?
     
    Last edited: Apr 5, 2014
  4. Apr 5, 2014 #3

    jtbell

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    Staff: Mentor

    Try a concrete example. Let the scalar field h(x,y) be the height (elevation) above sea level, of a point on the ground, whose location is specified by x and y (think of gridlines on a map). h(x,y) is larger at the top of a hill, and smaller down in a valley. The gradient of this field is a vector ##\vec S = \vec \nabla h##, which we might call the "slope field". At a given point, the direction of ##\vec S## is the (horizontal) direction of the steepest uphill slope, and its magnitude is the value of that steepest slope.

    If we walk a short distance (a single step, perhaps) in some (horizontal) direction, specified by the vector ##d \vec r##, we gain or lose a little bit of height. We can calculate that change in height as ##dh = \vec S \cdot d \vec r##. If we take a lot of steps along some path, we find the total change in height by adding up the little changes, that is, we integrate:
    $$\Delta h = \int {\vec S \cdot d \vec r}$$
    If we come back to our starting point, we obviously have to come back to our starting elevation, no matter which path we take. So
    $$0 = \oint {\vec S \cdot d \vec r}$$
    independently of the path. We say that ##\vec S## is therefore a conservative field.
     
  5. Apr 5, 2014 #4

    Nugatory

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    An example that may make it a bit more intuitive:

    Gravity is a conservative field, and the scalar potential corresponds to the height relative to some arbitrary zero (such as sea level).

    If you travel from one point to another, your net elevation change will always be equal to the difference in the elevation of the two points. If you choose a roundabout path that crosses a high mountain path you'll have to climb the mountain, increasing your elevation, but then you have to descend the other side of the mountain; the elevation changes up and down cancel. Likewise, gravity is doing positive work on you as you climb the mountain, but it's also doing negative work on you when you descend the other side so you end up with zero net work from going up the mountain and down again.

    No matter what path you choose, the net elevation change will be just the difference between the elevation of the starting point and the elevation of the ending point. That's also the change in gravitational potential and the net work done as you traverse a path between the two points.
     
  6. Apr 5, 2014 #5

    BiGyElLoWhAt

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    The mathmatical answer for this comes from Maxwell's definition of a conservative field : del x f = 0
     
  7. Apr 5, 2014 #6

    AlephZero

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    Suppose the work was NOT independent of the path.

    If you move along some path from point ##A## to point ##B##, you do some work ##W_{AB}##. If you move back to A along a different path, you do some work ##W_{BA}##.

    If ##W_{AB} \ne -W_{BA}##, then when you move around a closed loop from A to B and back, either you have to do some work on the system, or the system does some work on you. Either way, the amount of energy in the system changes. There are plenty of real-world systems where this happens. A simple example is sliding an object on a horizontal table, and doing work against friction. You have to do work to slide the object, and that work gets converted into heat energy.

    "Conservative" just means the energy in the system does NOT change when you move around a closed loop. An example would be compressing a spring and releasing it again. As you compress the spring you store energy in it. When it returns to its original length, the energy is released. OK, for a real-world spring a small amount of energy is always lost, but the small energy loss can often be ignored in a simple math model of the spring.
     
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