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I Spherical coordinates via a rotation matrix

  1. May 31, 2016 #1
    First, I'd like to say I apologize if my formatting is off! I am trying to figure out how to do all of this on here, so please bear with me!

    So I was watching this video on spherical coordinates via a rotation matrix:

    and in the end, he gets:

    x = \rho * sin(\theta) * sin(\phi)
    y = \rho* cos(\theta) * sin(\phi)
    z = \rho cos(\theta)

    Clearly, this doesn't look like the equations we normally get for spherical coordinates, which are:

    x = \rho* cos(\theta) * sin(\phi)
    y = \rho * sin(\theta) * sin(\phi)
    z = \rho * cos(\theta)

    What I don't understand is he says that they are not the same as some of the spherical coordinates as we've seen, but selecting measures is an arbitrary choice. Additionally, most resources I find that get the answer I'm more used to.

    However, it seems to me that if someone was using these equations, they would get completely the wrong stuff. So is there a way to get our usual equations using this method? Or what's going on here that makes this ok? I don't see how we should get a completely different result. I really like this method because it makes a lot of sense to me, but I don't like how it gets what I consider a completely wrong result.

    From what I can tell, the issue seems to come from the rotation table/matrix from the a to n coordinate frame which is:

    R = [cos(\theta), sin(\theta), 0
    sin(\theta), cos(\theta), 0
    0, 0, 1]

    However, I can't seem to think of a way that this is inherently wrong given the geometry, nor how we would get the usual equations from this.

    Can someone please help/elaborate/explain this situation to me?
  2. jcsd
  3. May 31, 2016 #2


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    Gold Member

    This is not the usual spherical coordinates. The correct usual ones are:
    ## x=\rho \sin\theta \cos\phi \\ y=\rho \sin \theta \sin\phi \\ z=\rho \cos\theta ##
    You don't get the wrong stuff. Its just a convention. Conventions are just arbitrary choices that don't change the results but of course you need to be consistent about them. You can't use one definition of spherical coordinates for some part of the problem and another definition for another part. This is what makes you think this convention gives the wrong stuff. As long as you're consistent, it doesn't matter what definition you use. So you're free to use this definition of spherical coordinates but you need to be careful that anywhere you see spherical coordinates, you should make sure you can make it consistent with this definition you're using.
    There is nothing wrong with his method. And the reason he gets different results than usual is not his method! His method can be used to find out the definition for the usual convention of spherical coordinates too. What makes he get different results, is just the names he assigns to the angles and the axis he chose from which he measures ## \theta ##. Rename ## \phi \to \theta ## and ## \theta \to \phi ## and measure the new named ## \phi ## from the x axis instead of the y axis that he uses. Then do the calculations as he explains and you'll get the usual definition of the spherical coordinates.
  4. May 31, 2016 #3
    I can share one bad experience. I once had 2 points out of 10 deducted from a physics qualifying exam by solving a problem (correctly) but using the "latitude" angle from the equatorial plane rather than the polar angle downward from the z-axis. It turned out I was returning to graduate school from work in an area where we used an Earth-based coordinate system, so that latitude was more natural. In addition, I had to take the exam in January, after starting in September, without the first year grad courses, so I did not have much chance to see how the physics texts define the angles. (I was placed in 2nd year graduate courses).

    By the way, it gets worse. Some rotation matrices use the x-y-x convention for Euler angles, rather than the x-y-z convention or the z-x-z convention. If this does not make sense to you right now, forget about it.

    Try to do what your grader expects, but if you differ, you can usually do the problem OK if you are consistent. If you get a different answer from the book, you can usually square your answer with the book if you are consistent.
  5. May 31, 2016 #4

    Thank you so much! This explained it clearly! I went back with your suggestions and rederived the equations and got the same result that we would consider the usual definition using his method. I truly appreciate your time!
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