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Why dxdy=-dydx?

  1. Aug 6, 2010 #1
    why we postulate dxdy=-dydx

    not dxdy=dydx?

    i am now studying exterior differential forms and i always feel that i cannot appreciate the germ of the theory
     
  2. jcsd
  3. Aug 6, 2010 #2
    Because it naturally introduces minus signs in places where you would otherwise need to specify them explicitly.

    The most important operation on differential forms is exterior differentiation, denoted d. With the minus signs occurring naturally, you get the nice property that d(d(differential form)) = 0 for all differential forms. That has all sorts of nice algebraic consequences, and you can use algebraic techniques to extract topological information from your analytical objects.
     
  4. Aug 6, 2010 #3
    dx^dy = - dy^dx captures orientation. Minus signs for orientations show up all over the place: Changing the order of two columns of a square matrix changes the sign of its determinant. Changing the order of two vectors in a cross product changes the sign of the resulting vector. The definition of curvature relies on orientation, as do various constructions in topology.
     
  5. Aug 7, 2010 #4
    ok I didn't undertstand answers to this good question
     
  6. Aug 7, 2010 #5

    Mute

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    The dx dy notation is a shorthand notation for dx ^ dy. Given a tensor product [itex]dx \otimes dy[/itex], the anti-symmetrized product is

    [tex]dx \wedge dy \equiv dx \otimes dy - dy \otimes dx[/tex]
    which is clearly antisymmetric under interchanging x and y.

    Since [itex]dx \wedge dy[/itex] appears more than the bare tensor product or a symmetrized version, we just write dx dy for simplicity.
     
  7. Aug 9, 2010 #6
    but suppose we want to calculate the area of a triangle

    i think it is okay to change the order of dx and dy

    the area is surely a positive number
     
  8. Aug 9, 2010 #7
    Yes, but we are not solely interested in area; the full richness of differential geometry requires the inclusion of the study of signed areas, volumes, and n-volumes in order to capture orientation of surfaces, volumes, and n-dimensional objects. That is to say, we want to be able to tell when we're dealing with a right-handed glove versus a left-handed glove. They have the same volume, but there is no rigid transformation in 3-dimensional Euclidean space between the two, irrespective of coordinates. This gives us another coordinate-independent tool to use in analysis of geometry.
    Especially interesting is the existence of non-orientable objects; objects that have properties that would otherwise be inaccessible to analysis by considering only positive areas.
     
    Last edited: Aug 9, 2010
  9. Aug 9, 2010 #8
    You have already seen signed areas in at least two different contexts:

    1) Calculus: one interpretation of an integral is the "area under the curve." But it's the signed area. The integral of sin(x) over [0,2 Pi] is 0, not 4. If you computed integrals with absolute areas instead, you wouldn't get nice properties such as [itex]\int f +g = \int f + \int g [/itex].

    2) Linear Algebra: the determinant of a 2x2 matrix with columns v1 and v2 computes the signed area of the parallelogram spanned by v1 and v2. (This works in all dimensions, but you need to define what you mean by a parallelogram.)
     
  10. Aug 9, 2010 #9
    I don't know if this is what you are confused about, but I would like to point out that the symbol dx^dy as a differential form has a very different meaning from the symbol dx dy in

    [tex]\int f(x,y) \, dx dy.[/tex]

    Don't get them confused.

    The differentials in an integral are just dummy symbols. They indicate which variables you are integrating over. You can write dx dy = dy dx because it doesn't matter if you first integrate with respect to x or with respect to y.

    Differential forms are not just a mental crutch to help you remember if you are integrating with respect to x or a or t. They are functions that take a tangent vector to a manifold at a point and spit out a scaler. The wedge operation dx^dy is an operation defined on those functions, which satisfies dx^dy = -dy^dx.
     
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