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What's the vector theta?

  1. Jun 18, 2015 #1
    If the vector r is (x,y), so, what is the vector θ? BY THE WAY is (y,-x) ?
     
    Last edited: Jun 18, 2015
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  3. Jun 18, 2015 #2

    HallsofIvy

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    I have no idea what you mean by "the vector [itex]\theta[/itex]". Could you please explain that? Where did you see a reference to a "vector [itex]\theta[/itex]"?
     
  4. Jun 18, 2015 #3

    Fredrik

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    The question "is (y,-x)?" doesn't make sense either.
     
  5. Jun 18, 2015 #4

    HallsofIvy

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    I didn't notice the "(y, -x)"! If a vector is given by r= (x, y) then its length is |r|= [itex]\sqrt{x^2+ y^2}[/itex] and the angle it makes with the x-axis, it that is what you mean by "[itex]\theta[/itex]", is given by [itex]arctan(y/x)[/itex] as long as x is not 0, [itex]\pi/2[/itex] if x= 0 and y is positive, [itex]3\pi/2[/itex] if x= 0 and y is negative.

    Given a vector (x, y), the vector (y, -x) is the result of rotating (x, y) through an angle of [itex]pi/2[/itex] radians.
     
  6. Jun 18, 2015 #5

    WWGD

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    The angle ##\theta## depends on your frame of reference : the positive x-axis does not have to represent the angle ## 0 ## , it can represent anything as long as the choices are made consistently, i.e., the angle with the negative x-axis must be ##\pi ## larger than the choice on the positive x-axis, as is done, e.g., with branches of the Complex logarithm.
     
  7. Jun 18, 2015 #6

    HallsofIvy

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    Well, that depends on exactly what Bruno Tolentino means by "the vector [itex]\theta[/itex]". I asked that earlier and he still hasn't answered.
     
  8. Jun 18, 2015 #7

    WWGD

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    Another possible interpretation is that ##\theta(x,y)=(-y,x)## is a _vector field_ , assigning to each point/tangent space at ##(x,y)##, the
    vector ##(-y,x)##.
     
  9. Jun 18, 2015 #8
    The ideia of vector θ come from following: if the vector dr is the tangent vector to parametric curve and the o vector dn is the normal vector:

    333.png

    And if the UNIT vector r^ is normal to UNIT vector θ^:

    70d796839d040d2b0fa8bcfc6a21df62.png
    0087580bf7a31a2e9556f337f6f14145.png

    So: the vector dn = dθ and therefore θ = (-y,x)!?

    EDIT: but confront with the following: if θ = (-y,x), so θ = (- r sin(θ), r cos(θ)) = r (- sin(θ), cos(θ)) = r θ^

    Is known that the vector r = r r^

    But, is correct to affirm that: θ = r θ^?

    The vector θ wouldn't: θ = θ θ^
     
    Last edited: Jun 18, 2015
  10. Jun 19, 2015 #9

    Fredrik

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    In this context (polar coordinates), there's no standard definition of the notation θ. It's definitely non-standard to use that notation for the vector ##r\hat{\theta}##.

    I think I see what you were thinking now: Since ##\hat{\mathbf r}## is a normalized version of ##\mathbf r##, it makes sense to ask if there's a vector that you can normalize to get ##\hat\theta##. There is, but it's not denoted by θ.

    The vectors ##\hat{\mathbf r}## and ##\hat\theta## are defined as what you get when you normalize ##\frac{\partial\mathbf r}{\partial r}## and ##\frac{\partial\mathbf r}{\partial\theta}##.
    \begin{align*}
    &\mathbf r=(x,y)=(r\cos\theta,r\sin\theta)\\
    &\hat{\mathbf r}=\frac{\frac{\partial\mathbf r}{\partial r}}{\left|\frac{\partial\mathbf r}{\partial r}\right|} =\frac{(\cos\theta,\sin\theta)}{1} =\frac{\mathbf r}{r}=\frac{(x,y)}{\sqrt{x^2+y^2}}\\
    &\hat{\theta} =\frac{\frac{\partial\mathbf r}{\partial\theta}}{\left|\frac{\partial\mathbf r}{\partial\theta}\right|} =\frac{(-r\sin\theta,r\cos\theta)}{r} =\frac{(-y,x)}{\sqrt{x^2+y^2}}
    \end{align*}
     
    Last edited: Jun 19, 2015
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