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Volume element in different coordinate system

  1. Jun 26, 2011 #1
    Very simple question:

    Let [itex]x^0,x^1,...,x^n[/itex] be some fixed coordinate system, so that the infinitesimal volume element is [itex]dV=dx^0dx^1...dx^n[/itex].
    Then any change to a new (primed) coordinate system [itex]x^{0'},x^{1'},...,x^{n'}[/itex] transforms the volume to [tex]dV=\frac{\partial (x^0,x^1,...,x^n)}{\partial (x^{0'},x^{1'},...,x^{n'})}dx^{0'}dx^{1'}...dx^{n'}[/tex] where [itex]\frac{\partial (x^0,x^1,...,x^n)}{\partial (x^{0'},x^{1'},...,x^{n'})}[/itex] is the determinat of the jacobian of the transformation.

    So let's try to do this in a concrete example: the transformation from cartesian [itex]x,y[/itex] to polar [itex]r,\theta[/itex] coordinates.
    The Jacobian is simply [itex]r[/itex] and so i get to [itex]dV=dxdy=rdrd\theta[/itex].

    Doing the math i get [tex]dV^{pol.}=r(\cos\theta dx+\sin\theta dy)(-\frac{1}{r}\sin\theta dx+\frac{1}{r}\cos\theta dy)[/tex]since [itex]dr=\cos\theta dx+\sin\theta dy[/itex] and [itex]d\theta=-\frac{1}{r}\sin\theta dx+\frac{1}{r}\cos\theta dy[/itex].
    But then:[tex]dV^{pol.}=-\cos\theta\sin\theta dx^2+\cos^2\theta dxdy-\sin^2\theta dxdy + \cos\theta\sin\theta dy^2[/tex] and is clear that [itex]dV^{cart.}\neq dV^{pol.}[/itex] !!!
    Where is my mistake?? Thanks for your help guys!!
     
    Last edited: Jun 26, 2011
  2. jcsd
  3. Jun 27, 2011 #2
    You can't just multiply differentials.

    Try using the formal rules

    dx dx = dy dy = 0

    and

    dx dy = -dy dx
     
  4. Jun 27, 2011 #3

    HallsofIvy

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    [tex]x= r cos(\theta)[/tex] [tex]y= r sin(\theta)[/tex]

    [tex]dx= cos(\theta)dr- r sin(\theta)d\theta[/tex]

    [tex]dy= sin(\theta)dr+ r cos(\theta)d\theta[/tex]

    But when you multiply differentials, you have to use "skew product" Petr Mugver mentions.

    A good way of thinking about it is as vectors
    [tex]dx= cos(\theta)dr\vec{i}- r sin(\theta)d\theta\vec{j}[/tex]
    [tex]dy= sin(\theta)dr\vec{i}+ r sin(\theta)d\theta\vec{j}[/tex]

    Now the product is the cross product of the two "vectors".

    [tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ cos(\theta) & - rsin(\theta) & 0 \\ sin(\theta) & rcos(\theta) & 0 \end{array}\right|= (rcos^2(\theta)+ rsin^2(\theta))drd\theta\vec{k}= r drd\theta \vec{k}[/tex]
     
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