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Line integral of work done

  1. May 16, 2012 #1
    In the line integral of work done by conservative forces, should we care about the direction of force and differential displacement if we choose different coordinate system? The formula of potential energy comes out to be different if we choose different coordinate system.
     
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  3. May 16, 2012 #2

    Andrew Mason

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    The potential energy at a certain position is, of course, relative to some other position so the co-ordinate system does not matter. For example, the potential energy of a body at point A relative to point B is equal to the work done ON the body in moving from point B to point A.

    What makes you think that this depends on the co-ordinate system?

    [itex]U_{A→B} = \int_B^A \vec{F}\cdot d\vec{s} = \int_B^A|F||ds|cos\theta[/itex] where [itex]\theta[/itex] is the angle between the force and the incremental displacement vectors. So direction of the force is important. Ultimately, however, the only thing that matters in a conservative field is position, not path.

    AM
     
  4. May 16, 2012 #3
    However, in calculating gravitational potential energy of system of mass M and m, if we place our origin at the initial (or final) position of the particle of mass m,(which we are free to do so, because only difference in potential matters), we get one of the term as infinity in the formula{-GMm(1/r2-1/r1)}.
     
  5. May 17, 2012 #4

    Andrew Mason

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    This not a matter of changing co-ordinates! Coordinates have nothing to do with it. You are using distances or displacements. r1 and r2 are the distances from the centre of the mass.

    Besides, one never measures potential energy relative to the centre of a mass. This formula only works for point masses or for positions outside the surface of the masses. The force at the centre of a mass is 0.

    AM
     
  6. May 17, 2012 #5
    Yes, but so does the displacement. The dot product between the force and the displacement vector is unchanged by switching coordinate systems, even if the two of those do individually change in the new coordinate system.
     
  7. May 18, 2012 #6

    Andrew Mason

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    The distances r1 and r2 are the distances of the centre of mass of m from the centre of mass of the mass M. That does not depend on the choice of origin.

    Neither the force nor displacement depend on the co-ordinate system. The displacement of the centre of mass of m from the centre of mass of M determines the force. This displacement is always determined by the displacement vector of the centre of mass of m from the origin MINUS the displacement of the centre of mass of M from the origin. The incremental change of displacement of m from position 1 to position 2 is always the displacement from the origin at 2 MINUS the displacement from the origin at 1.

    AM
     
    Last edited: May 18, 2012
  8. May 18, 2012 #7
    Force is a vector quantity. The vector's components do in fact change depending on the coordinate system. What you are saying (and what is true) is that the magnitude of the force (which is a scalar quantity) does not change from coordinate system to coordinate system. The same is true for the displacement. The force and displacement will always have the same magnitude, and the angle between them will always be the same, but nevertheless the force and displacement are different in different coordinate systems.
     
  9. May 19, 2012 #8

    Andrew Mason

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    How are you defining direction?

    If the force vectors in different systems were different vectors but had the same magnitude, they would have to differ in direction. The direction is always the same for the force. It is in the radial direction between the centres of mass of m and M. It cannot be in any other direction! The displacement between point 1 and point 2 is always the same direction regardless of the co-ordinates. It is an arrow with the tail on the point 1 and head on point 2.

    AM
     
  10. May 19, 2012 #9
    I didn't use the word direction, and I'm not disputing that the force vector (or displacement vector) always points in a direction between the two bodies. But a given vector that points between the two bodies will have different components if you (for example) rotate your coordinate system by 30 degrees.
     
  11. May 19, 2012 #10

    Andrew Mason

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    Yes it will. But it is the same vector. So the force and displacement vectors do not change when you change co-ordinate systems. This appears to be the concern expressed in the OP.

    AM
     
  12. May 19, 2012 #11

    Ken G

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    We should agree that vectors and scalars are both independent of coordinates (indeed, that is pretty much the reason for using them in physics, their objective character). I think we have a simple confusion here between the difference between a vector, and its components in some coordinates. Andrew Mason is saying that the force and displacements vectors (pictured as arrows if you like) are independent of coordinates, and Steely Dan is saying that the vector components depend on coordinates. Those are both true, and both approaches end up with the same work because the dot products come out the same if you regard the two vectors being dotted as being coordinate independent (which they are), or if you regard their components as being coordinate dependent but in such a way as to keep the dot products the same.

    Note the advantage in thinking in Andrew Mason's terms-- the invariance of the dot product does not have to seem like magic if you recognize that the vectors are themselves invariant, so will generate invariant scalars when you dot them. A lot of places try to teach us that a vector is a list of numbers, so the vector is its components, but that is not correct-- the vector has a life of its own that transcends the components, that's the value in picturing vectors as arrows instead of lists of components.
     
    Last edited: May 19, 2012
  13. May 19, 2012 #12

    Ken G

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    I'm afraid that is not at all the point of a vector. Pick up any book or website devoted to the use of vectors in physics, you will find that vectors are considered coordinate invariant, and that's why they appear in theories like special relativity. Your key will be understanding the difference between the vector itself, and its components in some coordinate system. I can write the general expression like this:
    |v> = X|x> + Y|y> = X'|x'> + Y'|y'> = |v>
    where things inside | > are vectors, things that look like |x> are "basis vectors" and things that look like X are "components" (and equal the dot product between |v> and |x>, often written <x|v>). You will note that |v> is just what it is-- it is entirely independent of the choices for |x> or |x'>. That's why |v> appears at both ends of the chain I wrote. The transformation rules between the |x> and |x'> and the X and X' must be such that the above equality holds, i.e., that |v> is the same entity regardless of coordinate choice.
     
    Last edited: May 19, 2012
  14. May 19, 2012 #13

    Ken G

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    Two words: "4 vector."
    Vectors, tensors, and scalars all have the property of being coordinate invariant, and that's why special relativity is is built from all three. Indeed, the fundamental statement of relativity is that physical laws must be expressible in terms of these three types of elements, and indeed I think you'll find, if you do a bit more reading on tensors, that vectors and scalars are actually subclasses of tensors.

    As for finding a source that considers vectors to be coordinate invariant, I can't possibly imagine where I would look for one that doesn't. Of course I've seen countless ones that do, I've even taught vector-based physics from such sources. Why are you hijacking this thread, and where on Earth is this idea coming from that scalars and tensors are coordinate invariant but vectors aren't? I can only imagine that you invented the whole idea, because all you have mentioned to support your idea is vector components, and everyone knows that vector components are coordinate dependent, since vector components are inner products between vectors and some chosen set of basis vectors.
     
    Last edited: May 19, 2012
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