Need Help With Change of Variables from Cartesian to Spherical Coordinates

Aug 2016
3
0
Texas
Hello there!

Cutting straight to the chase, I need help with the derivation best described in this video here @ 7:48: https://youtu.be/AjDSLq-Pzcs?t=7m48s

By this point in the video, Adam Beatty has already established:
x=r*cos(Φ)*sin(θ) , y=r*sin(Φ)*sin(θ) , z=r*cos(θ)

It also makes sense to describe a general cartesian vector as:
A = Ax i + Ay j +Az k

And to then transfer the same logic to spherical and say:
A = Ar r + A
θ θ + AΦΦ

This makes the somewhat logical progression that we can just sub-in our x,y,z values into this general spherical vector formula to get

A =
r*cos(Φ)*sin(θ) i + r*sin(Φ)*sin(θ) j + r*cos(θ) k

He describes getting Ar, A
θ, and AΦ as simply just doing the partial derivative of our A function by the respective variable (r, θ , Φ)

And this is what gives us the "End Result" that I can only ever seem to find.

Ar = cos(Φ)*sin(θ) i + sin(Φ)*sin(θ) j + cos(θ) k
Aθ = r*cos(Φ)*cos(θ) i + r*sin(Φ)*cosθ) j - r*sin(θ) k
AΦ = -r*sin(Φ) i + r*cos(Φ) j

My issue with this is: "How"?
Multiple comments point out that there appear to be the missing [
sin(θ)]'s for the AΦ term which one would expect to still be there, but he never quite responded. In other words, it should look like this, no?:

AΦ = -r*sin(Φ)*sin(θ) i + r*cos(Φ)*sin(θ) j


______________________________________________________

All other tutorials on the matter kind of just go into the Jacobian and spit out the ultimate final answer (saying that you need r2sin(θ)drdθdΦ when doing the change of coordinates)

But still, I mange to consistently see those terms [
Ar, Aθ, and AΦ] in that layout. I never see such tutorials explain where those answers come from; this YouTube video was the closest it got to explaining (save the otherwise seemingly erroneous typo)

An additional resource which appears to kind of just conjure up the answer via Jacobian may be found here:
https://en.wikipedia.org/wiki/Vector_fields_in_cylindrical_and_spherical_coordinates#Spherical_coordinate_system

But I'm not the best when it comes to following abstract math when it's just delivered with such little context. And yet again, it still fails to have that sin(θ) which is my cause for all the confusion in the first place.


______________________________________________________________

Anyway, thank you in advance for whatever help you may provide.

-Joshua

I suppose I should also mention that the way Adam works with the spherical coordinates is that

r = Length of vector
Φ = Angle within the xy-plane (Azimuthal angle)
θ = Angle off of the Z-axis


(Also, terribly sorry if formatting is a bit choppy, evidently these Theta and Phi symbols throw off the text in some kind of way.)
 
Aug 2016
3
0
Texas
That's a rather nice PDF, and you're absolutely spot on that it shows the method I have found in the video.

But when all the PDF does is demonstrates how to get the r-hat unit vector, then says this

wolfram 3.png

wolfram 1.png
wolfram 2.png

Ultimately leaving me as thus

wolfram 4.png


I will admit, I am being lazy by trying to use a calculator to do this number crunching for me, but I've never been fond of juggling around various trig-identities to make something fit; and as far as I can tell, that would be the only way to get rid of the monstrosities pictured in the second Wolfram Alpha screenshot.
 

romsek

MHF Helper
Nov 2013
6,723
3,029
California
That's a rather nice PDF, and you're absolutely spot on that it shows the method I have found in the video.

But when all the PDF does is demonstrates how to get the r-hat unit vector, then says this

View attachment 36054

View attachment 36052
View attachment 36053

Ultimately leaving me as thus

View attachment 36055


I will admit, I am being lazy by trying to use a calculator to do this number crunching for me, but I've never been fond of juggling around various trig-identities to make something fit; and as far as I can tell, that would be the only way to get rid of the monstrosities pictured in the second Wolfram Alpha screenshot.
the square root term in the denominator simplifies to just $r \sin(\theta)$
 
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Aug 2016
3
0
Texas
the square root term in the denominator simplifies to just $r \sin(\theta)$
By golly you're right! (factoring out the [r^2 * sin(theta)^2] seems ridiculously simple now that I'm fresh to the problem. Not quite sure how I missed it; but regardless, thank you so very much for your assistance.