# I Convert from rectangular to Spherical Coordinates

1. Oct 5, 2017

### Philosophaie

How do you convert this to Spherical Components?

Spherical Convention = (radial, azimuthal, polar)

$\vec r = |\vec r| * \cos{(\theta)} * \sin{(\phi)} * \hat x +|\vec r| * \sin{(\theta)} * \sin{(\phi)} * \hat y +|\vec r| * \cos{(\phi)} * \hat z$

Is this correct?

$\vec r =\sqrt{(x^2 + y^2 + z^2)} * \hat r + \arctan{(\frac{y}{x})} * \hat \theta + \arccos{(\frac{z}{r})} * \hat \phi$

2. Oct 5, 2017

### Orodruin

Staff Emeritus

3. Oct 5, 2017

### Philosophaie

Cartisian Coordinates
$\vec r = x * \hat x + y * \hat y + z * \hat z$
$x = |\vec r| \cos{\theta} \sin{\phi}$
$y = |\vec r| \sin{\theta} \sin{\phi}$
$z = |\vec r| \cos{\phi}$
Spherical Coordinates
$r = \sqrt{(x^2 + y^2 + z^2)}$
$\theta = \arctan{(\frac{y}{x})}$
$\phi = \arccos{(\frac{z}{r})}$
$\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi$
or is it
$\vec r = r \hat r + \frac{\theta}{r\sin{\phi}} \hat \theta +\frac{\phi}{r} \hat \phi$

4. Oct 5, 2017

### Orodruin

Staff Emeritus
Neither of those is correct. You are even adding vectors with different physical dimension ... I suggest you express the Cartesian basis vectors in terms of the spherical basis vectors and start from the Cartesian expression for the position vector.

5. Oct 5, 2017

### Philosophaie

What would be the expression in terms of Spherical Coordinates
$\vec r = r(x,y,z) \hat r + \theta(x,y,z) \hat \theta + \phi(x,y,z) \hat \phi$
$r(x,y,z)=?$
$\theta(x,y,z)=?$
$\phi(x,y,z)=?$

6. Oct 5, 2017

### phyzguy

Why don't you try drawing it out? As you said, in Cartesian Coordinates the position vector is $\vec r = x * \hat x + y * \hat y + z * \hat z$. When you draw this out, in what direction is it pointing in spherical coordinates? How much of it is pointing along the r, θ, and φ unit vectors?

7. Oct 5, 2017

### Philosophaie

If I draw it out the Spherical Coordinates are:
$r = \sqrt{(x^2 + y^2 + z^2)}$
$\theta = \arctan{(\frac{y}{x})}$
$\phi = \arccos{(\frac{z}{r})}$
$\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi$
$\hat \theta$ and $\hat \phi$ are along the arcs.

8. Oct 5, 2017

### phyzguy

Right. So if I decompose the Cartesian position vector into the spherical coordinate unit vectors, how much of the Cartesian position vector points along these arcs?

9. Oct 5, 2017

### Philosophaie

Is correct then?

10. Oct 5, 2017

### Philosophaie

$r = \sqrt{(x^2 + y^2 + z^2)}$
$\theta = \arctan{(\frac{y}{x})}$
$\phi = \arccos{(\frac{z}{r})}$
$\vec r = r \hat r + \theta \hat \theta + \phi \hat \phi$
is correct then?

11. Oct 5, 2017

### phyzguy

No. You didn't answer my question.

12. Oct 5, 2017

### Philosophaie

$\arctan{(\frac{y}{x})}$ points in the $\hat \theta$ direction.
$\arccos{(\frac{z}{r})}$ points in the $\hat \phi$ direction.

13. Oct 5, 2017

### Orodruin

Staff Emeritus
A scalar cannot "point" in any direction. You were asked to decompose the Cartesian basis vectors into the spherical ones in post #4 and to make the drawing with the position vector relative to a point in spherical coordinates twice in posts #6 and #8. I suggest that you follow either or both of these suggestions.

14. Oct 5, 2017

### Philosophaie

Uncertain how to convert x,y,z to $(\hat r, \hat \theta, \hat \theta)$. Can you give a hint?

15. Oct 5, 2017

### Orodruin

Staff Emeritus
Are you familiar with how to express the spherical basis vectors in Cartesian coordinates?

16. Oct 5, 2017

### Philosophaie

Look above. Post #1.

17. Oct 5, 2017

### Orodruin

Staff Emeritus
You have not done so above.

18. Oct 5, 2017

### phyzguy

Have you drawn out the vectors yet? If so, can you post a picture?

19. Oct 5, 2017

### Staff: Mentor

$$\vec{r}=r\hat{r}$$

20. Oct 6, 2017

### Philosophaie

Attached is a drawing. r is in a 3d box or x,y,z. On the xy plane starting on x-axis is $\theta$ starting on the arc is $\hat \theta$ as a unit vector. On the z-xy plane directly down from r to the xy plane. Starting from the z-axis to r to the xy plane is $\phi$ starting on the arc $\hat \phi$ as a unit vector.

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Last edited: Oct 6, 2017