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Homework Help: Spherical coordinates

  1. Mar 26, 2016 #1
    1. The problem statement, all variables and given/known data
    transform the following vectors to spherical coordinates at the points given

    10ax at P (x = -3 , y = 2, z=4)

    2. Relevant equations
    x y z can be chage into x = rsinθcosφ , y=rsinθsinφ , z=cosθ

    3. The attempt at a solution
    ax vector can be expressed ar,aθ,aφ so, I can change x , y, z through 2.

    therefore we have to find θ, φ, r

    also, we can know the sinθcosφ = x/r sinθsinφ = y/r z=cosθ

    10ax = 10ax ar + 10ax aθ + 10ax aφ

    = 10sinθcosφ + 10cosθcosφ - 10sinφ

    >>>> I can fill out innerproduct between x and r but how to solve the others,

    Is it right answer? I want to find more correctly one (it make lots of number, because find each variables through lots of calculation)

    >>>> I want to more objective soultion !

    >>>>> want to know how to chage between sehperical coordinates and cartesian coordinates

    Thank you for your attention for me
  2. jcsd
  3. Mar 26, 2016 #2

    Simon Bridge

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    You have three equations and three unknowns.
    It is better to use geometry though ... ferinstance "r" is the magnitude of the vector ##\vec r##, which is given by ##r^2=x^2+y^2+z^2##
    Try sketching the vector.

    I notice that "10ax" is not a vector though.
  4. Mar 27, 2016 #3

    Simon Bridge

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    Of course you could always just look up the transformation from cartesian to spherical ...
  5. Mar 27, 2016 #4

    Ray Vickson

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    When you write ##10ax## do you really mean ##10 a \vec{r} = 10 a \langle x,y,z \rangle?## One of these is a vector and the other is not.
  6. Mar 28, 2016 #5

    Simon Bridge

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    Hmmm ... looking at the later notation: OP may be using "a" to indicate the unit vector ... see ar aφ etc later on, as in
    So 10ax would mean ##\vec r = 10\hat a_x = (10,0,0)## cartesian ... which is very easy to put in spherical coordinates.
    OTOH: that does not fit so well with the rest of the problem statement: the vector does not depend on position for example.

    OP has been back since I replied and "liked" the reply ... presumably got what was needed.
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