1. Oct 8, 2018

### random_soldier

If the radius unit vector is giving us some direction in spherical coordinates, why do we need the angle vectors or vice versa?

2. Oct 8, 2018

### kuruman

Think about it. In three dimensions you need three quantities to define a point in space. They could be {x,y,z} or {r,θ,φ}. If I were to tell you that a fly is at position $1.5~\hat r$ meters from the corner of a room (the origin), where would you look to find it?

3. Oct 8, 2018

### random_soldier

I mean I suppose in that case I would have to tilt my head by a certain θ and a certain Φ to see the fly but then my question is why the need to have r as a vector? I thought the whole point in spherical coordinates as opposed to cartesian was the fact that r was just the distance from the center and to actually arrive at the required point meant offsetting by a certain θ and a certain Φ.

4. Oct 8, 2018

### kuruman

Because it makes the mathematical description in some cases easier to formulate and manipulate. Take the example of a point charge $Q$ at the origin. The electric field due to this charge can be written in spherical coordinated as$$\vec E(\vec r)=\frac{Q}{4 \pi \epsilon_0}\frac{\vec r}{r^3}.$$The same field in Cartesian coordinates is$$\vec E(x,y,z)=\frac{Q}{4 \pi \epsilon_0}\frac{x~\hat x + y~\hat y+z~\hat z}{(x^2+y^2+z^2)^{3/2}}.$$The top expression is easier to visualize as a radial field and often easier to work with if one has to take dot or cross products with other vectors or use vector calculus.

5. Oct 8, 2018

### random_soldier

Pardon my curtness but I think I am now just confused about what exactly the radius unit vector is giving the direction of. May I please know?

6. Oct 9, 2018

### kuruman

In the example of the electric field that I gave you, imagine a line from the origin where charge $Q$ is located to point $P$ where you want to find the field. The unit vector $\hat r$ is along this line and points from the origin to point $P$. Its direction of course depends on where $P$ is. You can specify that direction in the Cartesian representation and write $$\hat r=\frac{x~\hat x + y~\hat y+z~\hat z}{(x^2+y^2+z^2)^{1/2}}.$$You tell me where point $P$ is, i.e. you give me $x$, $y$, and $z$ and I will be able to draw $\hat r$. In terms of the standard spherical angles $\theta$ and $\phi$ that give an alternate way to find point $P$, you have $\hat r=\sin\theta \cos\phi~\hat x+\sin \theta \sin \phi~\hat y+\cos\theta ~\hat z$. You give me the angles and I will be able to draw $\hat r$.

7. Oct 9, 2018

### Orodruin

Staff Emeritus
By definition you will never need any other basis vector in spherical coordinates when you express the position vector or any other vectors pointing radially. However, this is not the only possible vector. For example, your fly may be moving non-radially, which will give it a velocity vector with non-radial components. Same thing with the velocity vector of a planet orbiting a star. Since the orbit is typically almost circular, the radial component is small and the tangential component large.

Last edited: Oct 9, 2018
8. Oct 9, 2018

### FactChecker

Can you clarify what you mean? 'r' is not a vector. It is a real number indicating magnitude and has no direction. The vector with a direction is $\vec r$, whose direction can be defined using a unit direction vector $\vec u = \vec r / \mid \vec r \mid$. The unit vector direction $\vec u$ can also be described using angles off of a coordinate system.