spherical coordinates

Discussion in 'Math' started by emt, Oct 30, 2015.

  1. emt

    Thread Starter New Member

    Oct 27, 2015
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    how can analysis this form graph :
    from where is this! ar= sinθ aρ +cosθ az ?

    and
    aθ= cosθ aρ - cos(90-θ) az ?

    and
    aΦ= -sinθ aρ + cosθ az ?

    cb5fc23c12b4ddca90beafc6e5d25d36a802e6a432098e204605633ae6a2f34a.jpg
     
  2. Papabravo

    Expert

    Feb 24, 2006
    10,138
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    First of all there are a great many symbols there without much in the way of explanatory text. I will assume that you are familiar with Cartesian coordinates and the unit vectors for the Cartesian system. The next thing you have to work through is the transformation of a Cartesian triple (x,y,z) into a spherical triple (r,θ,φ). Can you write out this process?

    You might find the following paper helpful.
    http://www.physics.purdue.edu/~jones105/phys310/coordinates.pdf
     
    Last edited: Oct 30, 2015
  3. emt

    Thread Starter New Member

    Oct 27, 2015
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    0
    6a63523995afc1bb96799f332893963a5436033541400c01b393043b5696c636.jpg
    and
    Aρ = cosθ ax +sinθ ay

    I want know from where came this

    ar= sinθ aρ +cosθ az ?

    and
    aθ= cosθ aρ - cos(90-θ) az ?

    and
    aΦ= -sinθ aρ + cosθ az ?
     
  4. Papabravo

    Expert

    Feb 24, 2006
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    Just like the Cartesian unit vectors form a mutually perpendicular basis, we seek a mutually perpendicular basis for spherical coordinates as well. ar is the unit vector in the radial direction. When θ is equal to zero ar = az as it should. When θ is 90° the ar = aρ which is again how it ought to be. aρ is then seen to be a unit vector in the plane parallel to the xy plane. It can be seen to have unit magnitude since ∀ θ∈[0°, ... , 90°]

    cos^2 (\theta) + sin^2(\theta)=1

    Can you work out the geometry to the remaining unit vectors?
     
    emt likes this.
  5. emt

    Thread Starter New Member

    Oct 27, 2015
    12
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    no please help me
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    So aθ is a unit vector that must be perpendicular to ar, whic I discussed above. When θ=0°, we know that ar = az, the unit vector in the z direction, and aθ is equal to aρ. aρ is in the plane which is parallel to the xy plane which intesects the origin at z = 0. When θ = 90°, aθ no points in the negative z direction. At all points in between, θ=0°, and θ=90°, it has a magnitude of one and is perpendicular to ar.

    After giving you two examples you should be able to construct the argument for aφ. Can you do it?
     
  7. emt

    Thread Starter New Member

    Oct 27, 2015
    12
    0
    sorry I didn't understand :confused:
     
  8. Papabravo

    Expert

    Feb 24, 2006
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    You're looking to understand the formulas given for the orthogonal basis vectors for spherical coordinates. I gave you a heuristic description of how the formula for ar works and I gave you a description of how the formula for aθ. Using those two descriptions, can you tell me how the formula for aφ works. I'm only asking you to tell me which direction aφ points in when θ=0° and 90°. If you still don't understand what to do I suggest you drop Advanced Calculus in favor of something else.
     
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