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Infinitesimal volume element in different coordinate system

  1. Jun 27, 2011 #1
    I've already post this, but i've done it in the wrong section!!

    So here I go again..

    I've a doubt on the way the infinitesimal volume element transfoms when performing a coordinate transformation from [itex]x^j[/itex] to [itex]x^{j'}[/itex]
    It should change according to [tex]dx^1dx^2...dx^n=\frac{\partial (x^1,x^2...x^n)}{\partial (x^{1'}x^{2'}...x^{n'})}dx^{1'}dx^{2'}...dx^{n'}[/tex]where [itex]\frac{\partial (x^1,x^2...x^n)}{\partial (x^{1'}x^{2'}...x^{n'})}[/itex] is the Jacobian of the transformation.


    So i tried to do this in a concrete example: the transformation between cartesian [itex]x,y[/itex] to polar [itex]r,\theta[/itex] coordinates.
    The jacobian of this transformation is [itex]r[/itex] and so, according to what i've written above[tex]dxdy=rdrd\theta[/tex]but 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] i get to [tex]dV=r(cos\theta dx+sin\theta dy)(-\frac{1}{r}sin\theta dx+\frac{1}{r}cos\theta dy)=(-sin\theta cos\theta dx^2+sin\theta cos\theta dy^2+cos^2\theta dxdy-sin^2\theta dxdy)[/tex]and this is not equal to [itex]dxdy[/itex], the volume element in cartesian coordinate, as it should be!!

    Where am I mistaking?

    Thanks!
     
  2. jcsd
  3. Jun 27, 2011 #2
    Are you sure that [itex]\mathrm{d}r=\cos\theta\;\mathrm{d}x+\sin\theta\; \mathrm{d}y[/itex]? It seems to me that
    [tex]\mathrm{d}r=\sqrt{\mathrm{d}x^2+\mathrm{d}y^2}.[/tex]
    Also, the formula for the angle also seems to be wrong. Please explain your formulas.
     
  4. Jun 27, 2011 #3

    HallsofIvy

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    [itex]x= r cos(\theta)[/itex], [itex]y= r sin(\theta)[/itex]
    so that [itex]dx= cos(\theta)dr- r sin(\theta)d\theta[/itex] and
    [tex]dy= sin(\theta)dr+ r cos(\theta)d\theta[/tex]

    The differential of area is
    [tex]dxdy= (cos(\theta)dr- r sin(\theta)dy)\times(sin(\theta)dr+ r cos(\theta)d\theta)[/tex]
    where the "[itex]\times[/itex] is the cross product:
    [tex]\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ cos(\theta)dr & -r sin(\theta)d\theta) & 0 \\ sin(\theta)dr & r cos(\theta) d\theta & 0 \end{array}\right|= (r cos^2(\theta)+ rsin^2(\theta))drd\theta\vec{k}= r drd\theta\vec{k}[/tex]
     
  5. Jun 27, 2011 #4
    My [itex]dr[/itex] and [itex]d\theta[/itex] are the differentials of[tex]r=\sqrt{x^2+y^2}[/tex]and[tex]\theta=arctg\frac{y}{x}[/tex]

    Isn't this the right way to perform the calculation?
     
  6. Jun 27, 2011 #5

    Hurkyl

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    Two important properties of differential forms:
    [tex]dx \, dx = 0[/tex]
    [tex]dx \, dy = -dy \, dx[/tex]​
    Try doing your original calculation again, now that you know that the product of differential forms don't commute. (In particular, be careful when you expand the product so as not to reorder things!)

    EDIT: I guess I really out to also point out that
    [tex]f \, dx = dx \, f[/tex]​
    where f is real-valued. (or complex-valued)
     
    Last edited: Jun 27, 2011
  7. Jun 27, 2011 #6
    I knew that there was some sort of commutation rule for infinitesimal!!

    The [itex]dx^{n}=0[/itex] with [itex]n>1[/itex] propriety is ok (they re higher order terms in a taylor expansion of the function, so I guess they're trascurable by definition); but where does the anticommutation rule come from??
     
  8. Jun 27, 2011 #7

    Hurkyl

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    Geometrically, it deals with orientation.

    Algebraically, it's the same rule. :smile: Let [itex]z = x+y[/itex], and simplify this expression in two different ways:
    [tex](dz)^2 - (dx)^2 - (dy)^2[/tex]​
     
  9. Jun 27, 2011 #8

    Hurkyl

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    Geometrically, it deals with orientation.

    Algebraically, it's the same rule. :smile: Let [itex]z = x+y[/itex]. Then:
    [tex](dz)^2 = (dx + dy)^2[/tex]​
    and so....
     
  10. Jun 27, 2011 #9
    So becaouse [itex](dz)^2[/itex] must be zero, we have [itex]dxdy=-dydx[/itex] !!

    Thanks a lot for your help guys!!!!!!!
     
  11. Jun 27, 2011 #10

    HallsofIvy

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    Since it is dxdy you want to replace, it make more sense to me to calculate dx and dy in terms of [itex]dr[/itex] and [itex]d\theta[/itex].
     
  12. Jun 27, 2011 #11
    Isn't [itex]\mathrm{d}y\mathrm{d}x[/itex] equal to 0? (I haven't really learnt the rules but it seems more intuitive that way).
     
  13. Jun 28, 2011 #12

    HallsofIvy

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    No, it isn't. It is, however, a "second order differential" so that while "dx" and "dy" separately are differentials of length, "dxdy" is a differential of area.
     
  14. Jun 28, 2011 #13
    You're absolutely right HallsofIvy; but I wanted to check the rule by doing the backwards calculation!!
     
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