Unit Vectors and Spherical Coordinates

Click For Summary
The discussion focuses on understanding the derivation of the unit vector \(\hat{r}\) in spherical coordinates. It clarifies that \(\hat{r}\) is defined as \(\hat{r} = \frac{\frac{d\mathbf{r}}{dr}}{|\frac{d\mathbf{r}}{dr}|}\), emphasizing the use of derivatives to obtain the directional vectors corresponding to changes in the coordinates. The participants suggest that visualizing the surfaces created by holding each coordinate constant can aid in understanding the basis vectors in spherical coordinates. Additionally, it is noted that the unit vector formula \(u/\|u\|\) is indeed being applied, but to the vectors derived from the partial derivatives of the position vector \(\mathbf{r}\). This approach helps clarify the relationship between the spherical coordinate system and its unit vectors.
jesuslovesu
Messages
185
Reaction score
0

Homework Statement


\mathbf{r} = rsin(\theta)cos(\phi) \hat x + rsin(\theta)sin(\phi) \hat x + r cos(\theta) \hat z
I am kind of following the description of the process given at http://mathworld.wolfram.com/SphericalCoordinates.html

I want to find \hat r and I understand everything except:
Why is \hat r = \frac{\frac{d\mathbf{r}}{dr} }{|\frac{d\mathbf{r}}{dr}|} (why the derivatives)?

Normally if I were going to find the unit vector I would just say the unit vector u hat = u/|u|
 
Last edited:
Physics news on Phys.org
I'm not exactly sure what you're asking, but i have a feeling you want to find the base vectors in spherical. To do so, it might be easier to do so graphically. To find the vectors you need to look at the surfaces that are created when you hold each of r, theta, and phi constant, individually. So, when you hold r constant your surfaces are spheres. to find a unit vector that is normal to the surface, you just use the fact that x = rsin(phi)cos(theta), y = rsin(phi)sin(theta), z = r cos(phi), and then divide the vector by its length. When you hold theta constant you get a sheet that hangs from the z axis. when you hold phi constant you get cones. THe easiest way to find all of them is to find 2 and then do a cross product (in the appropriate order so your sign is correct) on them to find the third. hope this helps some
 
jesuslovesu said:

Homework Statement


\mathbf{r} = rsin(\theta)cos(\phi) \hat x + rsin(\theta)sin(\phi) \hat x + r cos(\theta) \hat z
I am kind of following the description of the process given at http://mathworld.wolfram.com/SphericalCoordinates.html

I want to find \hat r and I understand everything except:
Why is \hat r = \frac{\frac{d\mathbf{r}}{dr} }{|\frac{d\mathbf{r}}{dr}|} (why the derivatives)?

Normally if I were going to find the unit vector I would just say the unit vector u hat = u/|u|

They ARE using u/|u|. But the u's they are applying that to are the vectors pointing in the coordinate directions, the partials dR/dr, dR/dtheta and dR/dphi (where R is the vector r).
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
33
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 9 ·
Replies
9
Views
1K
Replies
3
Views
1K
Replies
3
Views
2K
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K