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Use of spherical unit vectors

  1. Jan 31, 2013 #1
    I'm not sure that I understand the vectors [itex]\hat{r}[/itex], [itex]\hat{\theta}[/itex], and [itex]\hat{\phi}[/itex] in spherical coordinates correctly. I was looking through this link earlier.

    I understand that [itex]\hat{r}[/itex] always points radially outward from the origin. That seems to imply to me that any position in space could be specified by a multiple of [itex]\hat{r}[/itex] alone. But that seems odd to me. What is the use of [itex]\hat{\theta}[/itex] and [itex]\hat{\phi}[/itex] if any point can be specified by a multiple of [itex]\hat{r}[/itex] in this way? Does this mean that when we specify a point as a vector in spherical coordinates, we simply write the vector as (r,0,0) where r is the distance from the origin, no matter where it is?

    I get what they're saying in that link, but it makes me more confused in other ways.
  2. jcsd
  3. Jan 31, 2013 #2

    Simon Bridge

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    A multiple of ##\hat{r}## would only tell you how far away the point was from the origin. If I told you to walk 20 paces radially outwards to find the treasure, you'd ask me which direction I mean right?

    You still need the angles... and you need to know which direction the angles are in.

    eg. if I tell you to turn 30 degrees from where you are facing, then walk 20 paces radially outwards ... you would still need to know if I mean you t turn to the left or to the right. The angle unit vectors are a way of writing that in math.
  4. Feb 1, 2013 #3
    Another way to put it would simply be to say that [itex]\hat{r}[/itex] changes direction depending on where you are in [itex]\theta[/itex] and [itex]\phi[/itex]. Thus you need the information of all 3 because otherwise you don't know the exact nature of [itex]\hat{r}[/itex].
  5. Feb 1, 2013 #4
    I understand that you need to know [itex]\theta[/itex] and [itex]\phi[/itex] to determine [itex]\hat{r}[/itex]. What I'm saying is that, if you follow the logic in that link, the unit vectors [itex]\hat{\theta}[/itex] and [itex]\hat{\phi}[/itex] seem to have no use.

    If coordinates are simply specifying a linear combination of unit vectors, it seems that every point in space would simply be a multiple of [itex]\hat{r}[/itex], and its spherical coordinates would be (R[itex]\hat{r}[/itex]+0[itex]\hat{\theta}[/itex]+0[itex]\hat{\phi}[/itex]), or (R,0,0) where R is its distance from the origin. I understand that to determine [itex]\hat{r}[/itex] you need more information, but every point IS a multiple of [itex]\hat{r}[/itex].

    Notice in the link that they ask for the students to express each point as a linear combination of unit vectors, and they expect the answer to be 5[itex]\hat{r}[/itex] for every point. This makes sense, but, taken from a linear algebra point of view, that would make [itex]\hat{r}[/itex] a basis for 3-space. Which seems silly, and I want to know if I'm missing something.
    Last edited: Feb 1, 2013
  6. Feb 1, 2013 #5

    Simon Bridge

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    You are sort-of missing something - you are assuming a particular convention for direction of

    Lets say I tell you to find something at 30 degrees, 100m away... even if you know you are facing 0degrees, how do you know which direction to turn 30 degrees in? Do you turn left or do you turn right? That is the information encoded into the unit vector for angle as stated in earlier posts.
  7. Feb 2, 2013 #6
    Yeah, basically if you write <r, 0, 0> it looks particularly deceiving. With this notation you're used to assuming that each variable is in reference to a scalar multiple of the same vector that never changes. If you write this expression out you have <r, 0, 0> = r [itex]\hat{r}[/itex] + 0 [itex]\hat{\theta}[/itex] + 0 [itex]\hat{\phi}[/itex]. Now the [itex]\hat{r}[/itex] is staring you in the face, and as we have mentioned that vector is not always the same vector (and it usually isn't).

    I mean, basically your whole problem is that in curvilinear coordinates like spherical coordinates there is a different vector space you use (called the tangent space) at each particular coordinate position once you've decided upon a way to arithmetize the space. So basically if I told you that you are at (r, [itex]\theta[/itex], [itex]\phi[/itex]) you are then given a particular vector space to work with at that point. Thus by your method in which <r, 0, 0> can somehow point to all points in space is neglecting the fact that that came at the expense of using an infinite number of different vector spaces to do so. The notation is affecting you for these kinds of things because you are stealing references of a vector ([itex]\hat{r}[/itex]) from an infinite number of different vector spaces! heh, it's an interesting information laundering scheme..
  8. Feb 3, 2013 #7
    Exactly. You need the information from all three variables, but if you use the usual convention that coordinates are a linear combination of the unit vectors, you get silly results. And in the link I provided, they explicitly point out that <[itex]r,\theta,\phi[/itex]> is an incorrect answer, and expect the correct answer to be 5[itex]\hat{r}[/itex] for every point given. This seems silly and useless to me, but I'm wondering if it is technically correct? What is considered the proper way to give spherical coordinates by mathematicians? And if we don't actually give coordinates in terms of the vectors [itex]\hat{\theta}[/itex] and [itex]\hat{\phi}[/itex], then what is the use of these vectors?
  9. Feb 3, 2013 #8


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    Hi Opus_723! :smile:

    The paper you linked to is simply pointing out that vectors that start from the origin are always multiples of ##\hat{r}##.

    However, most vectors start from somewhere else. :wink:
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