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Three strange quantities

  1. May 25, 2012 #1

    ShayanJ

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    Consider the velocity vector [itex] v^i=\dot{x}^i \hat{e}_i [/itex].
    We know that in the plane polar coordinates,it becomes [itex] \vec{v}=\dot{r}\hat{r}+r\dot{\theta}\hat{\theta} [/itex]
    We also know that it is a contravariant vector so if we transform it covariantly from cartesian coordinates to plane polar coordinates,we should
    get the second equation above from the first.

    [itex]
    v^1_{polar}=\frac{\partial r}{\partial x} \dot{x} + \frac{\partial r}{\partial y} \dot{y} = \frac{x}{\sqrt{x^2+y^2}} \dot{x}+
    \frac{y}{\sqrt{x^2+y^2}}\dot{y}=\cos{\theta} ( \dot{r} \cos{\theta}-r \dot{\theta} \sin{\theta}) + \sin{\theta} ( \dot{r} \sin{\theta} +
    r \dot{\theta} \cos{\theta}) = \dot{r} \cos^2{\theta} -r \dot{\theta} \sin{\theta} \cos{\theta} + \dot{r} \sin^2{\theta} + r \dot{\theta} \sin{\theta}\cos{\theta}=\dot{r}
    [/itex]

    Well,this has no problem

    [itex]
    v^2_{polar}=\frac{\partial \theta}{\partial x}\dot{x}+\frac{\partial \theta}{\partial y} \dot{y}=-\frac{y}{x^2+y^2} \dot{x}+\frac{x}{x^2+y^2} \dot{y}=
    -\frac{\sin{\theta}}{r} ( \dot{r} \cos{\theta}-r \dot{\theta} \sin{\theta})+\frac{\cos{\theta}}{r}( \dot{r} \sin{\theta} + r \dot{\theta} \cos{\theta})=
    -\frac{\dot{r}}{r} \sin{\theta} \cos{\theta} + \dot{\theta} \sin^2{\theta} + \frac{\dot{r}}{r} \cos{\theta} \sin{\theta} + \dot{\theta} \cos^2{\theta}=\dot{\theta}
    [/itex]

    You see?!It lacks r !

    It seeme that velocity does not transform contravariantly under transformation from cartesian coordinates to plane polar coordinates!
    I don't mean it transforms covariantly because that one gives sth completely different.
    So looks like velocity transforms not contravariantly NOR covariantly!

    Now consider the displacement vector [itex] \vec{r}=x \hat{i} + y \hat{j} [/itex]
    We know that its a contravariant vector,and we have seen the proof so I don't bother writing it.
    But maybe someone bored,wants to have some fun and tries to transform it covariantly to plane polar coordinates(Well,that was not my reason)

    [itex]
    r^1_{polar}=\frac{\partial x}{\partial r} x + \frac{\partial y}{\partial r} y = \cos{\theta} x + \sin{\theta} y = r \cos^2{\theta} + r \sin^2{\theta} =r
    [/itex]

    [itex]
    r^2_{polar}=\frac{\partial x}{\partial \theta} x + \frac{\partial y}{\partial \theta} y = -r \sin{\theta} x + r \cos{\theta} y=-r^2 \sin{\theta} \cos{\theta} + r^2 \cos{\theta}\sin{\theta}=0
    [/itex]

    You see,it gives the same vector as the contravariant transformation does.Looks like displacement vector transforms contravariantly AND covariantly under transformation from
    cartesian to plane polar coordinates.

    Sorry to make it too long,but also consider[itex] \phi=xy [/itex]
    It seems to be a scalar meaning if we rotate the coordinate system,It remains invariant and we get [itex] \phi'=x'y' [/itex] and [itex] \phi'=\phi [/itex]

    [itex]
    \phi'=x'y'=(x \cos{\psi} - y \sin{\psi})(x \sin{\psi} + y \cos{\psi} )=x^2 \sin{\psi} \cos{\psi} +xy \cos^2{\psi} - xy \sin^2{\psi} -y^2 \sin{\psi} \cos{\psi}=
    (x^2-y^2)\sin{\psi}\cos{\psi}+xy(\cos^2{\psi}-\sin^2{\psi})
    [/itex]

    So it seems [itex] \phi'=\phi [/itex] only if [itex] \psi=k \pi [/itex] which suggests that [itex] \phi [/itex] is not a scalar.
    So what is it?
    Also I should say that [itex]\vec{\nabla} \phi [/itex] behaves like what I said about velocity vector!

    Thanks
     
    Last edited: May 25, 2012
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  3. May 26, 2012 #2

    haruspex

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    Shouldn't that be
    [itex]
    v^2_{polar}=\frac{r\partial \theta}{\partial x}\dot{x}+\frac{r\partial \theta}{\partial y} \dot{y}[/itex]
    ?
     
  4. May 27, 2012 #3

    ShayanJ

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    As I remember the transformation law for contravariant vectors is as below:
    [itex]
    A'^i=\frac{\partial x'^i}{\partial x^j} A^j
    [/itex]
    I see no reason for putting that r!
     
  5. May 27, 2012 #4

    haruspex

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    Dimensionally, v is L/T, yes? Without the r, the RHS has dimension 1/T.
     
  6. May 27, 2012 #5

    ShayanJ

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    I see,your right!
    But what about the thing I said about [itex] \phi=xy [/itex]?
    Thanks
     
  7. May 27, 2012 #6

    haruspex

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    I couldn't understand why you thought xy ought to be invariant under coordinate rotation.
     
  8. May 28, 2012 #7

    ShayanJ

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    Isn't xy,a scalar?
    If it is,it should be invariant under rotations!
     
  9. May 28, 2012 #8

    haruspex

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    I think you're confusing two very different usages of the term scalar.
    xy is a scalar mathematically, but not in the sense of a scalar physical attribute (such as distance between two points, mass, charge, value of a scalar field at a point...)
     
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