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I was studying a Spherical Coordinate System. And I kinda stuck in process where it's coordinate is represent in term of vector notation . Any point $P$ in standard vector space can be represented by three unit vectors $\mathbf i,\ \mathbf j$ and $\mathbf k$.

But in case of Spherical Coord. Sys. there are different basis vector $\theta,\ r$ and $ρ$. As we know magnitude of $\mathbf{i, \ j, \ k}$ are unit or $1$. How about magnitude $\rho$. Is it unit? But $ρ$ is an angle. How it can be unit (as unit is for distance)? Is it related to polar coordinate system?

Can anyone please explain me how does it all happen and how does same point can be represented by both Cartesian and spherical vector coordinate system?

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  • $\begingroup$ It depends whether you follow the vector calculus convention (then all of the basis vectors will be unit length as is standard) or if you follow the differential geometry convention, where basis vectors are partial derivative operators, which are not necessarily unit length. $\rho$ will be unit length in both systems, while the angular ones will not. Also $\rho$ is not an angle. $\endgroup$ Commented Aug 8, 2020 at 10:10
  • $\begingroup$ In vector calculus. and i mean to say whether or not ϕ is unit lenght not ρ. I've edited the question. $\endgroup$ Commented Aug 8, 2020 at 10:19
  • $\begingroup$ Then they will all be unit vectors. The length is not the issue. Being a $\phi$ unit vector means pointing in the $\phi$ direction, it has nothing to do with curvilinear distance/length. $\endgroup$ Commented Aug 8, 2020 at 10:21
  • $\begingroup$ But basis vector θ is used to represent the orientation. As in case of 2i+3j it means unit vector i is scaled by 2 and j by 3 and they are added. Then do basis vector θ also scale like i? $\endgroup$ Commented Aug 8, 2020 at 10:25
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    $\begingroup$ No coordinate system's vectors line up with the grid coordinates like Cartesian does if that is what you are asking. In fact, this is a defining property of Cartesian coordinates which makes it special since it is the only coordinate system where the basis vectors are not functions of the coordinates, but rather are constants. $\endgroup$ Commented Aug 8, 2020 at 10:32

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What do you think the basis vectors might be in a system of spherical co-ordinates ? The vector from the origin to $r=1, \theta=0, \phi=0$ is simple enough. But what about the vector from the origin to $r=0, \theta=1, \phi=0$ ? Or to $r=0, \theta=0, \phi=1$ ?

The problem with spherical co-ordinates (or any system of non-Cartesian co-ordinates) is that the direction from a point $P$ along which one co-ordinate changes and the other two remain constant depends on the location of $P$. In Cartesian co-ordinates the directions and lengths of the basis vectors are independent of location. But you are trying to generalise this property of Cartesian co-ordinates to other co-ordinate systems where it just does not hold.

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  • $\begingroup$ Cartesian co-ordinates the directions and lengths of the basis vectors are independent of location. Can you please elaborate this? $\endgroup$ Commented Aug 8, 2020 at 10:54
  • $\begingroup$ @NervousHero In Cartesian co-ordinates the $\vec i$ vector, which points in the direction of increasing $x$ co-ordinate, is always parallel to the $x$ axis. So its direction is independent of location. In spherical co-ordinates the $\vec r$ vector, which points in the direction of increasing $r$ co-ordinate, always points away from the origin - so its direction depends on location. $\endgroup$ Commented Aug 8, 2020 at 11:15
  • $\begingroup$ @NervousHero See math.stackexchange.com/questions/1854909/… which has a very comprehensive answer to your question. $\endgroup$ Commented Aug 8, 2020 at 11:18

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