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I was wondering how exactly to convert a vector in cartesian coordinates, to one in cylindrical coordinates. Given

A $= 5x/(x^2+y^2) \hat i + 5y/(x^2+y^2) \hat j + z \hat k$

how would I convert A in terms of r, theta, and z?

Sorry in advance for the awkwardness in the math script.

Aiva
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  • This question gives you the reverse question. – Jessica B Oct 10 '14 at 19:07
  • In case you're wondering, to get from azarel's answer (which is a Cartesian representation in terms of cylindrical variables) to Yves's answer (which may or may not have been what you wanted), you just need to convert the unit Cartesian vectors as well. The identity (which you should prove yourself if you can) is $\hat i = \cos(\theta)\hat u_\rho - \sin(\theta)\hat u_\theta$, $\hat j = \sin(\theta)\hat u_\rho + \cos(\theta)\hat u_\theta$, and of course $\hat k = \hat u_z$, where I guess Yves chose $u$ for "unit length". –  Oct 10 '14 at 21:36

2 Answers2

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In cylindrical coordinates, $A=\frac5\rho\vec{u_\rho}+z\vec{u_z}$. There is no tangential component ($\vec{u_\theta}$).

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Recall that $x=r\cos\theta,\ y=r\sin\theta$ and $z=z$. Substituting we get $$A=\frac{5r\cos\theta}{r^2}\vec{i}+\frac{5r\sin\theta}{r^2}\vec{j}+z\vec{k}=\frac{5\cos\theta}{r}\vec{i}+\frac{5\sin\theta}{r}\vec{j}+z\vec{k}$$

azarel
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  • One notes that the result as given is not in terms of cylindrical basis vectors. That is a separate, additional computation (though it is quite common). – Muphrid Oct 10 '14 at 18:56