How Do You Convert Cartesian Vectors to Cylindrical Coordinates?

[roam]

Homework Statement

I am trying to convert the following vector at (1, 1, 0) to cylindrical polar coordinates, and show that in both forms it has the same direction and magnitude:

$4xy\hat{x}+2x^2\hat{y}+3z^2\hat{z}$

Homework Equations

$\rho^2=x^2+y^2$

$tan \phi = \frac{y}{x}$

$z=z$

The Attempt at a Solution

[/B]
So for the Cartesian we have:

$|V|=\sqrt{(4(1)(1))^2+(2(1)(1))^2+(3(0))^2} = \sqrt{20}$

But how can we find the direction? I know how to find the direction angle using trigonometry in the 2D case, but how does one do this in the 3D situation?

Now to convert to cylindrical I am using the above relationships we get:

At that point we have x=4(1)(1)=4, y=2(1)=2, and z=3(0)=0. So

$\rho=\sqrt{20}$

$z=0$

$\phi= tan^{-1} (1/2)$ ($\phi$ being the azimuth angle)

Do we use degrees here?

Now how do we find the magnitude in this polar form? Unfortunately my lecture notes does not explain this.

Any explanation or link is greatly appreciated.

 

[HallsofIvy]

You cannot give a direction in three dimensions as a single number. One thing you can do is give the "direction cosines". They are the cosines of the angles the vector makes with each of the coordinate axes in turn. And cosine is "near side divided by hypotenuse". The "hypotenuse" is, of course, the length of the vector and the "near side" is the component in the direction of the given axis. In other words, the three "direction cosines" for the vectors <a, b, c> are $\frac{a}{\sqrt{a^2+ b^2+ c^2}}$, $\frac{b}{\sqrt{a^2+ b^2+ c^2}}$, and $\frac{c}{\sqrt{a^2+ b^2+ c^2}}$.
In other words, the "direction cosines" of a vector are precisely the components of a unit vector in that direction.

Unless you are dealing with a problem that specifically gives angles in polar form, you should always use radians, not degrees. A for "what is the magnitude", whether you are writing a vector in Cartesian or Polar form, it is still the same vector and it length (magnitude) does not change.

 

[vela]

Are you sure of what you're being asked to do? My interpretation of the problem is that you're being asked to express the vector in the form $A_\rho\hat{\rho} + A_\phi\hat{\phi}+A_z\hat{z}$ where $\hat{\rho}$, $\hat{\phi}$, and $\hat{z}$ are the cylindrical-coordinate unit vectors at the point (1, 1, 0).

 

[roam]

You cannot give a direction in three dimensions as a single number. One thing you can do is give the "direction cosines". They are the cosines of the angles the vector makes with each of the coordinate axes in turn. And cosine is "near side divided by hypotenuse". The "hypotenuse" is, of course, the length of the vector and the "near side" is the component in the direction of the given axis. In other words, the three "direction cosines" for the vectors <a, b, c> are $\frac{a}{\sqrt{a^2+ b^2+ c^2}}$, $\frac{b}{\sqrt{a^2+ b^2+ c^2}}$, and $\frac{c}{\sqrt{a^2+ b^2+ c^2}}$.
In other words, the "direction cosines" of a vector are precisely the components of a unit vector in that direction.

Unless you are dealing with a problem that specifically gives angles in polar form, you should always use radians, not degrees. A for "what is the magnitude", whether you are writing a vector in Cartesian or Polar form, it is still the same vector and it length (magnitude) does not change.

Thank you so much.

So for the Cartesian case, the three direction cosines would be:

$cos a = \frac{A_x}{\sqrt{(A_x)^2+ (A_y)^2+ (A_z)^2}} = \frac{4}{\sqrt{20}} \implies a = 0.46 \ rad$

Here a is the angle the unit vector $\hat{x}$ makes with x-axis.

Similarly, $cos b = \frac{2}{\sqrt{20}} \implies b = 1.107 \ rad$, and $c=0 \ rad$, with respect to y and z-axes respectively.

Is this the correct idea?

I think I need to apply this "direction cosines" method to the cylindrical polar form (the point of the exercise, I believe, is to show that in both coordinate systems the direction is the same).

 

[roam]

Are you sure of what you're being asked to do? My interpretation of the problem is that you're being asked to express the vector in the form $A_\rho\hat{\rho} + A_\phi\hat{\phi}+A_z\hat{z}$ where $\hat{\rho}$, $\hat{\phi}$, and $\hat{z}$ are the cylindrical-coordinate unit vectors at the point (1, 1, 0).

Thank you for your response. The question only asks to find the magnitude and direction in Cartesian, then convert it to cylindrical form and show that it still has the same direction and magnitude as the other counterpart.

I think for the unit vectors I need to use the following conversion matrix:

$\begin{pmatrix} cos \phi & sin \phi & 0\\ -sin \phi & cos \phi & 0\\ 0 & 0 & 1 \end{pmatrix}$

Since at the point (1, 1, 0), $x=4, \ y =2 \implies \phi = \tan^{-1} (y/x)= 0.46 \ rad$

So here's what I did:

$\begin{pmatrix} \hat{\rho} \\ \hat{\phi} \\ \hat{z} \end{pmatrix} = \begin{pmatrix} cos (0.46) & sin (0.46) & 0\\ -sin (0.46) & cos (0.46) & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \hat{x} \\ \hat{y} \\ \hat{z} \end{pmatrix}$

This yields the following:

$\hat{\rho} = 0.89 \hat{x} + 0.44 \hat{y}$

$\hat{\phi} = 0.89 \hat{x} - 0.44 \hat{y}$

$\hat{z}=\hat{z}$

Is this what we need?

And to transform x, y, z, do I need to use these relations:

$x = \rho \ cos \phi \implies 4xy = \rho (0.89)$

$y = \rho \ sin \phi \implies 2x^2 = \rho (0.44)$

$z=z =3z^2$

I'm confused here, I'm not sure how to work out $A_\rho, A_\phi, A_z$ in $A_\rho\hat{\rho} + A_\phi\hat{\phi}+A_z\hat{z}$.

 

[roam]

Finally, I've managed to get the correct cylindrical form. But I've encountered another problem. Here is what I did:

$\begin{pmatrix} \hat{\rho} \\ \hat{\phi} \\ \hat{z} \end{pmatrix}\begin{pmatrix} cos \phi & sin \phi & 0\\ -sin \phi & cos \phi & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 4xy \\ 2x^2 \\ 3z^2 \end{pmatrix}$

Using the relations in my first post above we get:

$A_\rho = 4 \rho^2 cos^2 \phi sin \phi + 2 \rho^2 cos^2 \phi sin \phi$

$A_\phi = -4 \rho^2 cos \phi sin^2\phi + 2 \rho^2 cos^3 \phi$

$A_z = 3z^2$

I've been told that this is the correct cylindrical polar form. I then substituted $\rho=\sqrt{20}, \ \phi=0.463, z=0$ (as found above) into this to find the magnitude of the Cylindrical vector field at (1, 1, 0). Here is what I did:

$|A|= \sqrt{A_\rho^2 +A_\phi^2 +A_z^2}= \sqrt{42.9^2+14.37^2+0^2} = 45.24$

But this answer is clearly wrong, because for the Cartesian case we had $|A|=\sqrt{20}$. So why is this wrong, and how can I get the correct magnitude? (I need to confirm that in both cases the magnitudes equal)

 

[roam]

Shouldn't the magnitude be the same in both coordinate systems?

 

[SammyS]

Shouldn't the magnitude be the same in both coordinate systems?

Yes, it should.

You are confusing components of the vector, A_ρ, A_φ, and A_z with the direction of the cylindrical-coordinate unit vectors at (x, y, z) = (1, 1, 0).

 

[roam]

Yes, it should.

You are confusing components of the vector, A_ρ, A_φ, and A_z with the direction of the cylindrical-coordinate unit vectors at (x, y, z) = (1, 1, 0).

Yes, thank you.

So:

$\begin{pmatrix} A_\rho \\ A_\phi \\ A_z \end{pmatrix}\begin{pmatrix} cos \phi & sin \phi & 0\\ -sin \phi & cos \phi & 0\\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 4xy \\ 2x^2 \\ 3z^2 \end{pmatrix}$

$4xy cos \phi + 2x^2 sin \phi$
$-4xy sin \phi + 2x^2 cos \phi$
##3z^2 \end{pmatrix}##

Using the relationships $x=\rho cos \phi, \ y= \rho sin \phi, \ z=z$ we get:

$A_\rho = 6 \rho^2 cps^2 \phi sin \phi$
$A_\phi = -4\rho^2 cos \phi sin^2 \phi + 2 \rho^2 cos^3 \phi$
$A_z =3z^2$

Substituting values for (1, 1, 0) we get the correct answer of $||A||=\sqrt{20}$. Correct answer!

But now now I still have trouble finding the direction of the cylindrical form (to confirm it has the same direction as the Cartesian one)...

 

[roam]

In the Cartesian case we had with respect to x-axis: $A(1,1,0)=4(1)(1) \hat{x} + 2(1)^2 \hat{y} + 3 (0)^2 \hat{z} = 4 \hat{x} + 2 \hat{y}$

And the angle with respect to x-axis was $\phi= tan^{-1} (A_y/A_x)=0.46 \ rad$

I even worked out the "direction cosines", as suggested by HallsofIvy.

In cylindrical form we have $A(1,1,0)=(4 cos (0.46)+2 sin(0.46)) \hat{\rho} + (2 cos (0.46) - 4 sin (0.46))\hat{\phi}$

$=4.4721 \hat{\rho} + 0.0163 \hat{\phi}$

How can I use this to show that it has the same direction as the Cartesian counterpart?

(P.S. I'm also not sure how the "direction cosines" apply to the cylindrical case)

 

[SammyS]

In the Cartesian case we had with respect to x-axis: $A(1,1,0)=4(1)(1) \hat{x} + 2(1)^2 \hat{y} + 3 (0)^2 \hat{z} = 4 \hat{x} + 2 \hat{y}$

And the angle with respect to x-axis was $\phi= tan^{-1} (A_y/A_x)=0.46 \ rad$

I even worked out the "direction cosines", as suggested by HallsofIvy.

In cylindrical form we have $A(1,1,0)=(4 cos (0.46)+2 sin(0.46)) \hat{\rho} + (2 cos (0.46) - 4 sin (0.46))\hat{\phi}$

$=4.4721 \hat{\rho} + 0.0163 \hat{\phi}$

How can I use this to show that it has the same direction as the Cartesian counterpart?

(P.S. I'm also not sure how the "direction cosines" apply to the cylindrical case)

You need to find the direction of each of $\hat{\rho}$ and $\hat{\phi}$ at the point (x,y,z)=(1,1,0).

Also, I'm pretty sure that to find the cylindrical components of A, the values of $\rho\text{ and } {\phi}$ you should use are based on the point (x,y,z)=(1,1,0) .

 

[roam]

You need to find the direction of each of $\hat{\rho}$ and $\hat{\phi}$ at the point (x,y,z)=(1,1,0).

Also, I'm pretty sure that to find the cylindrical components of A, the values of $\rho\text{ and } {\phi}$ you should use are based on the point (x,y,z)=(1,1,0) .

But how do I do that?

I mean, using (1,1,0) I worked out $\rho = \sqrt{x^2+y^2}=\sqrt{2}$ and $\phi=tan^{-1}(y/x)=0.46 \ rad$. But how do I find their direction?

I know that $\hat{\rho}= cos \phi \hat{x} + sin \phi \hat{y}$ and $\hat{\phi}= -sin \phi \hat{x} + cos \phi \hat{y}$ but I'm not sure if that helps...

 

[SammyS]

But how do I do that?

I mean, using (1,1,0) I worked out $\rho = \sqrt{x^2+y^2}=\sqrt{2}$ and $\phi=tan^{-1}(y/x)=0.46 \ rad$. But how do I find their direction?

I know that $\hat{\rho}= cos \phi \hat{x} + sin \phi \hat{y}$ and $\hat{\phi}= -sin \phi \hat{x} + cos \phi \hat{y}$ but I'm not sure if that helps...

OK.

I see you did use that angle. Of course, tan^-1(1) is better known as 45° = π/4 radians. (I didn't recognize 0.46 rad as being approx = π/4 rad .)

The z-component of vector A is zero, so verifying the angle can be simplified. Right?

What angle does A make with $\hat{\rho}$ and what angle does $\hat{\rho}$ make with the x-axis?

 

[roam]

OK.

I see you did use that angle. Of course, tan^-1(1) is better known as 45° = π/4 radians. (I didn't recognize 0.46 rad as being approx = π/4 rad .)

The z-component of vector A is zero, so verifying the angle can be simplified. Right?

What angle does A make with $\hat{\rho}$ and what angle does $\hat{\rho}$ make with the x-axis?

It was actually $tan^{-1}(1/2)=0.46 \ rad$.

When we have the expression for $\hat{\rho}$ and A (in cylindrical), how do we calculate the angle between the two vectors? (That's my question)

 

[SammyS]

It was actually $tan^{-1}(1/2)=0.46 \ rad$.

That's $\hat{\phi}$ for the x & y components of vector A so that's not π/4. (I didn't use a calculator, but should have recognized that.)

So, back to what I said in post #11. Use ${\phi}$ based on (1, 1, 0) .

 

[roam]

That's $\hat{\phi}$ for the x & y components of vector A so that's not π/4. (I didn't use a calculator, but should have recognized that.)

So, back to what I said in post #11. Use ${\phi}$ based on (1, 1, 0) .

Following your advice in post #11, at point (x,y,z)=(1,1,0) and by substituting $\phi=0.46$ we have:

$\hat{\phi}=-4 sin \phi + 2 cos \phi = 0.0163$

$\hat{\rho}=4 cos \phi + 2 sin \phi = 4.4721$

How does this establish that the vector now in its cylindrical form, has the same direction as the Cartesian counterpart?

 

[SammyS]

Following your advice in post #11, at point (x,y,z)=(1,1,0) and by substituting $\phi=0.46$ we have:

$\hat{\phi}=-4 sin \phi + 2 cos \phi = 0.0163$

$\hat{\rho}=4 cos \phi + 2 sin \phi = 4.4721$

How does this establish that the vector now in its cylindrical form, has the same direction as the Cartesian counterpart?

First of all, $\phi\ne 0.46\$ . At (1, 1, 0), $\phi=\pi/4\approx 0.785398\$ .

Beyond that, what you have above are not unit vectors.

Added in Edit:

Using the correct $\ \phi$ ,

$A_{\phi}=-4 \sin \phi + 2 \cos \phi$

$A_{\rho}=4 \cos \phi + 2 \sin \phi$