Vector Projection; Space Coordinate System

Why is it that we have to provide a reference point \(P(x, y, z)\) to convert a Cartesian vector to a Cylindrical vector? Also, how would you interpret the vector \(A_\rho a_\rho + A_\phi a_\phi + A_z a_z \), geometrically?

#Physics #Engineering #Math

Note by JohnDonnie Celestre
7 years, 9 months ago

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Comments

Cylindrical polar coordinates are defined in terms of an axis (the axis of the cylinder). The relationship between the fundamental Cartesian coordinate vectors ax,ay,aza_x,a_y,a_z (which are unit vectors pointing along the xx-, yy- and zz-axes) and the fundamental cylindrical polar coordinate vectors aρ,aϕ,aza_\rho,a_\phi,a_z varies depending on the point where these vectors are being considered. While, for example, axa_x always points in the same direction, the vector aρa_\rho points in different directions at different points in space.

To be specific, if ax=(1,0,0)a_x=(1,0,0), ay=(0,1,0)a_y = (0,1,0) and az=(0,0,1)a_z = (0,0,1). Suppose we are at the point PP with coordinates (ρcosϕ,ρsinϕ,z)(\rho\cos\phi,\rho\sin\phi,z).

  • aρa_\rho is the unit vector in the "direction of increasing ρ\rho", namely the unit vector which is perpendicular to the zz-axis and radially outwards, so aρ=(cosϕ,sinϕ,0)a_\rho = (\cos\phi,\sin\phi,0).

  • aϕa_\phi is the unit vector in the "direction of increasing ϕ\phi", namely the direction you would go if ϕ\phi increases, while keeping ρ\rho and zz constant. If you keep ρ\rho and zz constant, you rotate about the zz-axis, and so aϕ=(sinϕ,cosϕ,0)a_\phi = (-\sin\phi,\cos\phi,0).

  • aza_z is still the unit direction in the "direction of increasing zz", namely (0,0,1)(0,0,1).

If we are converting between the Cartesian and cylindrical polar systems, we are in the business of describing the same vector aa in two different ways, and so a  =  Axax+Ayay+Azaz  =  Aρaρ+Aϕaϕ+Azaz a \; = \; A_xa_x + A_ya_y + A_za_z \; = \; A_\rho a_\rho + A_\phi a_\phi + A_z a_z Taking scalar products: Aρ=aaρ  =  Axcosϕ+AysinϕAϕ=aaϕ  =  Axsinϕ+AycosϕAz=aaz  =  Az \begin{array}{rcl} A_\rho & = & a \cdot a_\rho \; = \; A_x \cos\phi + A_y \sin\phi \\ A_\phi & = & a \cdot a_\phi \; = \; -A_x \sin\phi + A_y \cos\phi \\ A_z & = & a \cdot a_z \; = \; A_z \end{array} The last equation is trivial, since the zz-coordinate is the same coordinate in both coordinate systems. These are precisely the equations that are being used in the example you provided.

Mark Hennings - 7 years, 9 months ago

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Thank you Sir Mark :))

JohnDonnie Celestre - 7 years, 9 months ago
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