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Convert a spherical vector into cylindrical coordinates

  • Thread starter Mulz
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  • #1
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Homework Statement


Convert the vector given in spherical coordinates to cylindrical coordinates:

[tex] \vec{F}(r,\theta,\varphi) = \frac{F_{0}}{arsin\theta}\bigg{[}(a^2 + arsin\theta cos\varphi)(sin\theta \hat{r} + cos\theta \hat{\theta}) - (a^2 + arsin\theta sin\varphi - r^2 sin^2\theta) \hat{\varphi} \bigg] [/tex]

Homework Equations



[tex] \theta [/tex] is elevation:

[tex] \rho = rcos\theta [/tex]
[tex] \varphi = \varphi [/tex]
[tex] z = rsin\theta [/tex]

The Attempt at a Solution



I attempted to solve the problem by simple inserting all of the conversions into the spherical vector field. The problem was I keep on getting a bunch of r left that I don't know what to do. This is what I got by simply inserting the conversion above, I got them from https://en.wikipedia.org/wiki/Cylindrical_coordinate_system.

[tex] \vec{F}(\rho,\varphi,z) = \frac{F_{0}}{az}\bigg{[}(a^2 \frac{z}{r} + az\frac{z}{r}cos\varphi)\hat{\varphi} + (a^2 \frac{\rho}{r} + az\frac{\rho^2}{r^2})\hat{\varphi} - (a^2 + azsin\varphi - z^2)\hat{z}\bigg{]} [/tex]

Which I think is wrong since I have a lot of r's left. Can anyone make the conversion? I have tried for days.
 
Last edited:

Answers and Replies

  • #2
PeroK
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What about using:

##r^2 = x^2 + y^2 + z^2 = \rho^2 + z^2##
 
  • #3
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What about using:

##r^2 = x^2 + y^2 + z^2 = \rho^2 + z^2##
After when I inserted the new coordinates or before, in the original expression?
 
  • #4
PeroK
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After when I inserted the new coordinates or before, in the original expression?
I thought you just needed to replace the remaining ##r## coordinates in terms of the relevant cylindrical coordinates. I haven't checked anything else, by the way.
 
  • #5
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I thought you just needed to replace the remaining ##r## coordinates in terms of the relevant cylindrical coordinates. I haven't checked anything else, by the way.
If I put it in the first expression I would simply rewrite ##r## in terms of cylindrical coordinates I think. Not sure how it goes for ##\theta## and ##\phi##.
 
  • #6
PeroK
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Why not just replace ##r## in this expression and then you're done?

[tex] \vec{F}(\rho,\varphi,z) = \frac{F_{0}}{az}\bigg{[}(a^2 \frac{z}{r} + az\frac{z}{r}cos\varphi)\hat{\varphi} + (a^2 \frac{\rho}{r} + az\frac{\rho^2}{r^2})\hat{\varphi} - (a^2 + azsin\varphi - z^2)\hat{z}\bigg{]} [/tex]

Which I think is wrong since I have a lot of r's left. Can anyone make the conversion? I have tried for days.
PS I guess that first unit vector should be ##\hat{\rho}##.
 
  • #7
kuruman
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It's all trig and algebra. You can do the conversion without too much struggle if first you draw a picture of an arbitrary point in space and on it draw the unit vectors associated with both the spherical and the cylindrical coordinates. The orthogonal ##\hat z## and ##\hat \rho## for cylindrical coordinates lie in the same plane as the orthogonal ##\hat r## and ##\hat \theta## for spherical coordinates. Knowing that the spherical angle ##\theta## is formed by ##\hat z## and ##\hat r##, you can easily find the coefficients ##a_{ij}## in the transformation equations ##\hat r=a_{11}~ \hat \rho +a_{12} ~\hat z## and ##\hat \theta=a_{21}~ \hat \rho +a_{22} ~\hat z.## They all involve sines and cosines of ##\theta##. This will allow you to substitute the unit vectors in the spherical expression. Note that ##\hat \varphi## is the same in both spherical and cylindrical.

The second step is to eliminate the spherical ##r## and ##\theta##. No need to do anything with ##\varphi##. Draw another drawing. Note that ##r=\sqrt{\rho^2+z^2}##, ##\cos\theta=z/\sqrt{\rho^2+z^2}## and ##\sin\theta =\rho/\sqrt{\rho^2+z^2}##. Substitute all of these in the given expression and simplify.

The final step is to pull it all together and cast it in the form ##\vec F(\rho,\varphi,z)=F_{\rho}~\hat \rho+F_{\varphi}~\hat \varphi+F_z~\hat z.##

Good luck.
 
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  • #8
Ray Vickson
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Homework Statement


Convert the vector given in spherical coordinates to cylindrical coordinates:

[tex] \vec{F}(r,\theta,\varphi) = \frac{F_{0}}{arsin\theta}\bigg{[}(a^2 + arsin\theta cos\varphi)(sin\theta \hat{r} + cos\theta \hat{\theta}) - (a^2 + arsin\theta sin\varphi - r^2 sin^2\theta) \hat{\varphi} \bigg] [/tex]

Homework Equations



[tex] \theta [/tex] is elevation:

[tex] \rho = rcos\theta [/tex]
[tex] \varphi = \varphi [/tex]
[tex] z = rsin\theta [/tex]

The Attempt at a Solution



I attempted to solve the problem by simple inserting all of the conversions into the spherical vector field. The problem was I keep on getting a bunch of r left that I don't know what to do. This is what I got by simply inserting the conversion above, I got them from https://en.wikipedia.org/wiki/Cylindrical_coordinate_system.

[tex] \vec{F}(\rho,\varphi,z) = \frac{F_{0}}{az}\bigg{[}(a^2 \frac{z}{r} + az\frac{z}{r}cos\varphi)\hat{\varphi} + (a^2 \frac{\rho}{r} + az\frac{\rho^2}{r^2})\hat{\varphi} - (a^2 + azsin\varphi - z^2)\hat{z}\bigg{]} [/tex]

Which I think is wrong since I have a lot of r's left. Can anyone make the conversion? I have tried for days.
Your equations are confusing and somewhat ambiguous: It is not clear if your ##arsin \theta## mean ##ar \sin \theta## or is a mis-spelling of ##\arcsin \theta## (although taking the arcsine of something called ##\theta## would be very unusual). Better: in LaTeX, type "\sin" instead of "sin"; the difference is between ##sin \theta## and ##\sin \theta##, and the second is much more readable. (The same applies to the other trig functions and their inverses, to "log" or "ln", to "max", "min", "lim" and many others.)

It is also not clear which of the two conventions you are using for spherical polar coordinates. One convention (perhaps used mostly in physics) is that ##\theta## is the polar angle (like latitude) and ##\phi## is the longitude; the other convention (used a lot by mathematicians) swaps the meaning of ##\theta## and ##\phi.##
 
  • #9
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Isn't ##sin\theta \hat{r} + cos\theta \hat{\theta}=\hat{\rho}##? This would mean that the vector F is parallel to the x-y plane.
 
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  • #10
kuruman
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Isn't ##sin\theta \hat{r} + cos\theta \hat{\theta}=\hat{\rho}##? This would mean that the vector F is parallel to the x-y plane.
That is correct. It simplifies to that. Also, ##r\sin\theta=\rho## and those two realizations are all there is to this problem. The vector field is 2-dimensional and independent of ##z##.
 

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