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Thread: simple doubt : vector field in cylindrical coordinates

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    simple doubt : vector field in cylindrical coordinates

    Hello Mathematicians! I have the answer to a very simple problem, but I dont understand it.
    Following vector field is given:

    G(x,y,z) = (x - yz, y+xz, z)

    I am asked to write the representation Ĝ(r, fi, z) of the given field in cylindrical coordinates (where fi is the angle).

    So here´s what I wrote:

    Ĝ(r, fi, z)= ( r( cos(fi)-zsin(fi) ) , r( sin(fi) + zcos(fi) ) , z ) I just substituted x= rcos(fi) and y=rsin(fi)

    But this is the answer that comes in the answers paper:

    Ĝ(r, fi, z) = (r , zr, z) Doesnt make any sense to me... Could it be a mistake in the solutions paper?

    Could someone explain why the answer is so, and why what I wrote is wrong?
    Thanks!!
    Last edited by Phyba; Jul 29th 2019 at 03:08 PM.
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    Forum Admin topsquark's Avatar
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    Re: simple doubt : vector field in cylindrical coordinates

    Quote Originally Posted by Phyba View Post
    Hello Mathematicians! I have the answer to a very simple problem, but I dont understand it.
    Following vector field is given:

    G(x,y,z) = (x - yz, y+xz, z)

    I am asked to write the representation Ĝ(r, fi, z) of the given field in cylindrical coordinates (where fi is the angle).

    So here´s what I wrote:

    Ĝ(r, fi, z)= ( r( cos(fi)-zsin(fi) ) , r( sin(fi) + zcos(fi) ) , z ) I just substituted x= rcos(fi) and y=rsin(fi)

    But this is the answer that comes in the answers paper:

    Ĝ(r, fi, z) = (r , zr, z) Doesnt make any sense to me... Could it be a mistake in the solutions paper?

    Could someone explain why the answer is so, and why what I wrote is wrong?
    Thanks!!
    You transformation is wrong. You have G(x, y, z) = (x - yz, y + xz, z). Here the "x" coordinate is actually x - yz, etc. So...
    $\displaystyle r = \sqrt{ (x - yz)^2 + (y + xz)^2}$

    etc.

    -Dan
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    Re: simple doubt : vector field in cylindrical coordinates

    Thank you Dan for your answer! So clearly I am not understanding the principle of transforming coordinates at all...
    I'm very confused , but trying to understand how one carries out the transformation...
    So you wrote r in term of the "x" and "y" coordinates
    simple doubt : vector field in cylindrical coordinates-capture.png

    Now, how is the first funcion of the field Ĝ1 = r ? How is simple doubt : vector field in cylindrical coordinates-untitled.pngequivalent to (x - yz)^2 ....
    And how is Ĝ2 = rz ?

    Could you please explain it briefly or provide a link with an explanation that will allow me to understand how it works?

    Regards,

    Phyba
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    Forum Admin topsquark's Avatar
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    Re: simple doubt : vector field in cylindrical coordinates

    Given a vector in Cartesian coordinates $\displaystyle (a_x, a_y, a_z)$ then we have
    $\displaystyle r = \sqrt{a_x ^2 + a_y ^2}$
    $\displaystyle \phi = tan^{-1} \left ( \dfrac{a_y}{a_x} \right )$
    $\displaystyle z = z$

    In this instance you have, in Cartesian coordinates for G, $\displaystyle g_x = x - yz$, $\displaystyle g_y = y + xz$, and $\displaystyle g_z = z$. The tranformation above gives G in terms of cylindrical coordinates as a function of x, y, z.

    -Dan

    Addendum: Perhaps I should mention that x, y, and z are not the Cartesian components of G. The notation is, IMHO, somewhat confusing.
    Last edited by topsquark; Jul 30th 2019 at 03:17 PM.
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    Re: simple doubt : vector field in cylindrical coordinates

    Ok, so I get what (r, ϕ ,z )are written in terms of (x,y,z)...
    Now, how do we get to ĝ1 = r , ĝ2 = rz ?

    ĝ3 = z does makes sense to me on its own, as there is nothing to transform in this coordinate
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    Forum Admin topsquark's Avatar
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    Re: simple doubt : vector field in cylindrical coordinates

    Okay. Sorry, I misread what you were asking. I have no idea how to get to their answer. My guess is a typo, but that's one heck of a typo. Is it possible that this is the solution to another problem?

    I would have to say that the answer key is wrong.

    -Dan
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    Re: simple doubt : vector field in cylindrical coordinates

    What's going on is that you are neglecting to restate the field in the new unit vectors

    The field is given as

    $G = (x-yz)\hat{x} + (y+x z) \hat{y} + z \hat{z}$

    we want the field components as

    $G = G_r \hat{r} + G_\theta \hat{\theta} + G_z \hat{z}$

    we do this by dotting the cartesian unit vectors with the cylindrical unit vectors

    $G_r = G_x \hat{x}\cdot \hat{r} + G_y \hat{y}\cdot \hat{r}$
    $G_r = G_x \cos(\theta) + G_y \sin(\theta)$

    $G_\theta = G_x \hat{x} \cdot \hat{\theta} + G_y \hat{y}\cdot \hat{\theta}$
    $G_\theta = -G_x \sin(\theta) + G_y \cos(\theta)$

    $G_r = r(\cos(\theta)- z\sin(\theta))\cos(\theta) + r(\sin(\theta)+z\cos(\theta))\sin(\theta) = r$

    $G_\theta =-r(\cos(\theta)- z\sin(\theta))\sin(\theta) + r(\sin(\theta)+z\cos(\theta))\cos(\theta) = rz$

    I leave it to you to confirm the algebra.
    Last edited by romsek; Jul 30th 2019 at 09:11 PM.
    Thanks from topsquark and Phyba
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    Re: simple doubt : vector field in cylindrical coordinates

    Ahhhhh... I forgot about transforming the unit vectors.

    Good catch!

    -Dan
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    Re: simple doubt : vector field in cylindrical coordinates

    Ah, I see now! I forgot that one must take the unit vectors into account. Thank you so much, now I know how it works.
    I upload here the solution worked out by myself in case somebody might find it useful in the future.

    To carry out the dot-product of unit vectors, I used the cartesian2cylindrical-transformation matrix that I found in the following link (I didnt know myself how the cylindrical unit vectors look like) :
    https://en.wikipedia.org/wiki/Vector...al_coordinates

    Thank you so much!simple doubt : vector field in cylindrical coordinates-vector-field-cyl.jpg
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