Unit Vectors and Spherical Coordinates

In summary: Calculating these partials is the same procedure as you've done before in the Cartesian case; you take the derivative of each component with respect to the corresponding variable, holding the other two constant. So for example, to get the partial with respect to r, you would hold theta and phi constant and take the derivative of the first component of the vector, rsin(theta)cos(phi), with respect to r. In summary, the conversation is about finding the unit vector \hat r in spherical coordinates. The process involves taking the derivatives of the vector r with respect to each coordinate variable and then dividing the resulting vector by its length. This is the same procedure as finding the unit vector in Cartesian coordinates.
  • #1
jesuslovesu
198
0

Homework Statement


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

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

Normally if I were going to find the unit vector I would just say the unit vector u hat = u/|u|
 
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  • #2
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
 
  • #3
jesuslovesu said:

Homework Statement


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

I want to find [tex]\hat r[/tex] and I understand everything except:
Why is [tex]\hat r = \frac{\frac{d\mathbf{r}}{dr} }{|\frac{d\mathbf{r}}{dr}|} [/tex] (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).
 

1. What are unit vectors?

Unit vectors are vectors that have a magnitude of 1 and are used to indicate direction in a particular coordinate system. They are typically denoted by the symbols 𝑚, 𝑛, and 𝑜 in the x, y, and z directions, respectively.

2. How are unit vectors used in spherical coordinates?

In spherical coordinates, unit vectors are used to represent direction in three-dimensional space. The unit vector in the radial direction, 𝑚, points from the origin to a specific point and has a magnitude of 1. The unit vector in the polar direction, 𝑛, points in the direction of increasing polar angle and has a magnitude of 1. The unit vector in the azimuthal direction, 𝑜, points in the direction of increasing azimuthal angle and also has a magnitude of 1.

3. How do you convert from Cartesian coordinates to spherical coordinates?

To convert from Cartesian coordinates (x, y, z) to spherical coordinates (r, 𝜃, 𝛼), first calculate the magnitude of the vector using the Pythagorean theorem: r = √(x2 + y2 + z2). Then, use the inverse tangent function to find the polar angle (𝜃): 𝜃 = tan-1(y/x). Finally, use the inverse cosine function to find the azimuthal angle (𝛼): 𝛼 = cos-1(z/r).

4. What are the advantages of using spherical coordinates?

Spherical coordinates are useful for solving problems involving symmetry or objects with spherical symmetry. They are also particularly convenient for describing phenomena that occur in three-dimensional space, such as fluid dynamics and celestial mechanics. Additionally, spherical coordinates can make certain calculations easier, such as calculating distances and angles between points in three-dimensional space.

5. Can spherical coordinates be used to describe motion in a three-dimensional space?

Yes, spherical coordinates can be used to describe the position, velocity, and acceleration of an object in three-dimensional space. The radial, polar, and azimuthal unit vectors can be used to represent the direction of motion in each respective direction. However, Cartesian coordinates are often preferred for describing motion in three-dimensional space, as they are more intuitive and easier to use in many cases.

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