Expression a vector in different basis

[fluidistic]

Homework Statement

Consider the vector \vec A whose origin is \vec r.
1)Express the vector \vec A in a basis of Cartesian coordinates, cylindrical and spherical ones.
2)Repeat part 1) if the origin is \vec r + \vec r_0.

Homework Equations

None given.

The Attempt at a Solution

1)In Cartesian coordinates, \vec A=(r_x+a_x,r_y+a_y, r_z+a_z). If we change the origin as in part 2), then \vec A=(r_x+r_{0x}+a_x,r_y+r_{0y}+a_y, r_z+r_{0z}+a_z).

For cylindrical coordinates, I've sketched a 3 dimensional graphic but I don't see how I can express \vec A. 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.

 

[hbweb500]

Going from cylindrical coordinates to cartesian coordinates, we can use the transformations x = r \cos(\theta), y = r \sin(\theta), and, of course, z = z_{\text{cyl}}.

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

This might help: http://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Cartesian_coordinates"

 

[fluidistic]

Going from cylindrical coordinates to cartesian coordinates, we can use the transformations x = r \cos(\theta), y = r \sin(\theta), and, of course, z = z_{\text{cyl}}.

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

This might help: http://en.wikipedia.org/wiki/Cylindrical_coordinate_system#Cartesian_coordinates"

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?

 

[vela]

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 \hat{r}, depends on what point you're talking about.

 

[fluidistic]

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 \hat{r}, depends on what point you're talking about.

I'm not sure I'm understanding. I've sketched the point r and from it I drew \vec A. I mean, I took the origin of \vec A being r.
At point r, the unit vectors in Cartesian coordinates are the same as in the origin 0. Or not? I mean \hat i, \hat j and \hat k. So \vec A would be a_x \hat i + a_y \hat j + a_z \hat k. 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.

 

[vela]

I'm not sure I'm understanding. I've sketched the point r and from it I drew \vec A. I mean, I took the origin of \vec A being r.
At point r, the unit vectors in Cartesian coordinates are the same as in the origin 0. Or not? I mean \hat i, \hat j and \hat k. So \vec A would be a_x \hat i + a_y \hat j + a_z \hat k. Hmm I don't think so...

That's right, actually.

Also, I don't understand when you talked about "when it's parallel transported it to the point...".

I just mean the vector A is moved to the new point without changing its magnitude and direction.

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.

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 r_P=sqrt(2), θ_P=π/4 for P and r_Q=2, θ_Q=0 for Q. Therefore, at P, you get:

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

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

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

and at Q, you get:

\hat{r}_Q=(\cos\theta_Q,\sin\theta_Q)=(1,0)

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

\vec{A}=1\hat{i}=1\hat{r}_Q

 

[fluidistic]

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 r_P=sqrt(2), θ_P=π/4 for P and r_Q=2, θ_Q=0 for Q. Therefore, at P, you get:

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

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

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

and at Q, you get:

\hat{r}_Q=(\cos\theta_Q,\sin\theta_Q)=(1,0)

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

\vec{A}=1\hat{i}=1\hat{r}_Q

Thanks, this was really useful!

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

 

[vela]

Looks good except for the way you wrote some stuff down.

If so, then I'd get \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 ). I'd appreciate a feedback. I'm not confident at all.

You're mixing up the notation. If you include the unit vectors explicitly, you should write it as a sum:

\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

After combining terms, you get

\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

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

\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)

 

[fluidistic]

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.