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Homework Help: Expression a vector in different basis

  1. Mar 13, 2010 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider the vector [tex]\vec A[/tex] whose origin is [tex]\vec r[/tex].
    1)Express the vector [tex]\vec A[/tex] in a basis of Cartesian coordinates, cylindrical and spherical ones.
    2)Repeat part 1) if the origin is [tex]\vec r[/tex] + [tex]\vec r_0[/tex].

    2. Relevant equations
    None given.


    3. The attempt at a solution
    1)In Cartesian coordinates, [tex]\vec A=(r_x+a_x,r_y+a_y, r_z+a_z)[/tex]. If we change the origin as in part 2), then [tex]\vec A=(r_x+r_{0x}+a_x,r_y+r_{0y}+a_y, r_z+r_{0z}+a_z)[/tex].

    For cylindrical coordinates, I've sketched a 3 dimensional graphic but I don't see how I can express [tex]\vec A[/tex]. Life would be easier if I could use some linear algebra maybe. I just don't know how to start and if it's a good idea.
    Thanks for any suggestion.
     
  2. jcsd
  3. Mar 13, 2010 #2
    Going from cylindrical coordinates to cartesian coordinates, we can use the transformations [tex] x = r \cos(\theta) [/tex], [tex] y = r \sin(\theta) [/tex], and, of course, [tex] z = z_{\text{cyl}}[/tex].

    Now, going from cartesian to cylindrical isn't as obvious, but is still simple. For example, the radius, r, is simply [tex] r = \sqrt{x^2 +y^2} [/tex].

    This might help: http://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Cartesian_coordinates"
     
    Last edited by a moderator: Apr 24, 2017
  4. Mar 13, 2010 #3

    fluidistic

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    I think I'm stupid, I already knew this. I don't know why I didn't think about applying it. I'm going to try this tomorrow (it's too late). So what I've done so far is ok?
     
    Last edited by a moderator: Apr 24, 2017
  5. Mar 13, 2010 #4

    vela

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    No, you've misunderstood the question. The vector A is assigned to the point r. The problem is asking you to express A in terms of the coordinate unit vectors at that point, and when it's parallel transported it to the point r+r0.

    Remember that the direction of most of the unit vectors, like [itex]\hat{r}[/itex], depends on what point you're talking about.
     
  6. Mar 14, 2010 #5

    fluidistic

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    I'm not sure I'm understanding. I've sketched the point r and from it I drew [tex]\vec A[/tex]. I mean, I took the origin of [tex]\vec A[/tex] being r.
    At point r, the unit vectors in Cartesian coordinates are the same as in the origin 0. Or not? I mean [tex]\hat i[/tex], [tex]\hat j[/tex] and [tex]\hat k[/tex]. So [tex]\vec A[/tex] would be [tex]a_x \hat i + a_y \hat j + a_z \hat k[/tex]. Hmm I don't think so...
    Also, I don't understand when you talked about "when it's parallel transported it to the point...".

    In fact I don't understand what they ask me to do. It doesn't look so hard, but I don't know what to do.
     
  7. Mar 14, 2010 #6

    vela

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    That's right, actually.
    I just mean the vector A is moved to the new point without changing its magnitude and direction.
    It's probably easiest to explain using an example. I'll just use R2 to keep it simple. Let A=(1,0). Now consider the points P=(1,1) and Q=(2,0) in the xy plane; in polar coordinates, you'd have rP=sqrt(2), θP=π/4 for P and rQ=2, θQ=0 for Q. Therefore, at P, you get:

    [tex]\hat{r}_P=(\cos\theta_P,\sin\theta_P)=\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)[/tex]

    [tex]\hat{\theta}_P=(-\cos\theta_P,\sin\theta_P)=\left(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)[/tex]

    [tex]\vec{A}=1\hat{i}=\frac{1}{\sqrt{2}}\hat{r}_P-\frac{1}{\sqrt{2}}\hat{\theta}_P[/tex]

    and at Q, you get:

    [tex]\hat{r}_Q=(\cos\theta_Q,\sin\theta_Q)=(1,0)[/tex]

    [tex]\hat{\theta}_Q=(-\cos\theta_Q,\sin\theta_Q)=(0,1)[/tex]

    [tex]\vec{A}=1\hat{i}=1\hat{r}_Q[/tex]
     
  8. Mar 14, 2010 #7

    fluidistic

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    Thanks, this was really useful!

    I'm having a problem. So in Cartesian coordinates, we've said that [tex]a_x \hat i + a_y \hat j + a_z \hat k[/tex]. Now I want to do cylindrical ones.
    I've sketched the vector r. Considering only the x-y plane, I have that [tex]\hat r =(\cos \varphi, \sin \varphi)[/tex] and [tex]\hat \phi=(-\cos \varphi, \sin \varphi)[/tex]. Now I must write [tex]\vec A[/tex] as a linear combination of [tex]\hat r[/tex] and [tex]\hat \varphi[/tex].
    I already know that [tex]a_z=a_z \hat k[/tex]. It's clear that [tex]a_x[/tex] can be written as a (linear ?) combination of only [tex]\hat r[/tex] and [tex]\hat \varphi[/tex]. To me it's clear it will be something of the form [tex]B\hat r - B\hat \phi[/tex]. Is B worth [tex]\frac{a_x}{2 \cos \varphi}[/tex]?
    If so, then I'd get [tex]\vec A=\left ( \frac{a_x}{2 \cos \varphi} \hat r-\frac{a_x}{2 \cos \varphi} \hat \varphi, \frac{a_y}{2\sin \varphi} \hat \varphi +\frac{a_y}{2 \sin \varphi} \hat r, a_z \hat k \right )[/tex]. I'd appreciate a feedback. I'm not confident at all.
     
  9. Mar 14, 2010 #8

    vela

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    Looks good except for the way you wrote some stuff down.
    You're mixing up the notation. If you include the unit vectors explicitly, you should write it as a sum:

    [tex]\vec A=\left(\frac{a_x}{2 \cos \varphi} \hat r-\frac{a_x}{2 \cos \varphi} \hat \varphi\right)+\left(\frac{a_y}{2\sin \varphi} \hat \varphi +\frac{a_y}{2 \sin \varphi} \hat r\right)+ a_z \hat k[/tex]

    After combining terms, you get

    [tex]\vec A=\left(\frac{a_x}{2 \cos \varphi} + \frac{a_y}{2 \sin \varphi}\right) \hat r+\left(-\frac{a_x}{2 \cos \varphi}+\frac{a_y}{2\sin \varphi}\right) \hat \varphi + a_z \hat k[/tex]

    Now if you write it as an ordered triplet [itex](r,\varphi,z)[/itex], you'd say:

    [tex]\vec A=\left(\frac{a_x}{2 \cos \varphi} + \frac{a_y}{2 \sin \varphi},-\frac{a_x}{2 \cos \varphi}+\frac{a_y}{2\sin \varphi}, a_z \right)[/tex]
     
  10. Mar 14, 2010 #9

    fluidistic

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    Ah yes, you're right. I can't believe I got it right. I'm going to tackle the rest. Thanks a lot for all.
     
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