Question on transformation of coordinates.

[yungman]

This is usually in the first chapter of the EM books and I never see this in any of my calculus, vector calculus, PDE books even though it is really a math question.

This is about transforming from rectangular coordinates to spherical coordinates of a cross product of two position vectors A and B. I found that if I perform the cross product in rectangle coordinates and then transform to spherical coordinates, everything worked. But if I first transform A and B to spherical coordinates and then perform the cross product, the answer is ALWAYS ZERO if they are position vectors. This is because a position vector only contain radius component in spherical coordinates. Please explain to me why I get two different answer in two different method.

Let

$$\vec A = \hat x A_x +\hat y A_y + \hat z A_z \;\hbox { and }\; \vec B = \hat x B_x +\hat y B_y + \hat z B_z$$

1) Perform cross product in rectangular coordinates and then transform to spherical coordinates:

as you can find in all calculus books, unless the angle between the two vector is zero, the result is not going to be zero as long as both are non zero vector. And it would be non zero after transformation.



2) Transform to spherical coordinates first then perform the cross product.

$$\vec A = \hat x A_x +\hat y A_y + \hat z A_z \;\Rightarrow\; |\vec A| = \sqrt{A_x^2 +A_y^2 + A_z^2} \;\hbox { and }\; A_x = |\vec A| sin\theta_A cos\phi_A,\; A_y=|\vec A| sin\theta_A sin\phi_A$$

When transform to spherical coordiantes:

$$\vec A_R = A_x sin\theta_A cos \phi_A + A_y sin\theta_A sin \phi_A + A_z cos\theta_A = |\vec A|(sin^2\theta_A cos^2\phi_A+sin^2\theta_A sin^2\phi_A + cos^2\phi_A) \;=\; |\vec A|$$

$$\vec A_{\theta} = A_x cos\theta_A cos \phi_A + A_y cos\theta_A sin \phi_A - A_z sin\theta_A = |\vec A|( cos\theta_A sin\theta_A cos^2\phi_A+cos\theta_A sin\theta_A sin^2\phi_A - cos\theta_A sin\theta_A) \;=\; 0$$

$$A_{\phi_A} = -A_x sin\phi_A+A_y cos\phi_A = |\vec A|(- sin\theta_A cos\phi_A sin\phi_A + sin\theta_A cos\phi_A sin\phi_A) \;=\;0$$

As you can see, only $A_R$ is non zero

$$\Rightarrow\; \vec A = \hat R A_R$$

This is also true for B. If we perform the cross product AX in spherical coordinates:


$$\vec A X \vec B \;=\;\left|\begin{array}{ccc}\hat x & \hat y & \hat z \\A_R & 0 & 0\\ B_R & 0 & 0\end{array}\right|\;=\; 0$$

As you see, I get two different answer with two different method. What is the reason?

 

[vanhees71]

I do not understand your calculation with method (2).

You always have to perform the calculation in one fixed basis. So you have to find out first, according to which basis are you supposed to calculate the cross product in case (2)?

 

[bapowell]

A general vector in spherical coordinates will have non-vanishing $\hat{\phi}$ and $\hat{\theta}$ components. Not sure why you are only retaining the $\hat{r}$ component.

 

[yungman]

A general vector in spherical coordinates will have non-vanishing $\hat{\phi}$ and $\hat{\theta}$ components. Not sure why you are only retaining the $\hat{r}$ component.

This is a position vector, where the tail always at the origin and the arrow head at the point specified. This is not a vector field, this is a very very important point. For a position vector, the components are a constant respect to the coordinates, it can be a function of time.

For vector field in EM cases, they have all 3 components.

 

[yungman]

I do not understand your calculation with method (2).

You always have to perform the calculation in one fixed basis. So you have to find out first, according to which basis are you supposed to calculate the cross product in case (2)?

As I explained in the other post, this is a position vector, that is the reason I can have the angle figure out. This derivation cannot be used in any vector field where the components are function of the coordinates. That is the tricky part of it.

A position vector can only contain the radius component. That is the whole point of this post is that I cannot make them equal for the position vector and none of the book specified this.

 

[bapowell]

This is a position vector, where the tail always at the origin and the arrow head at the point specified. This is not a vector field, this is a very very important point. For a position vector, the components are a constant respect to the coordinates, it can be a function of time.

For vector field in EM cases, they have all 3 components.

I think you're confused. By retaining only the $\hat{r}$ component, all you know about the vector is its magnitude. But presumably this vector points in a specific direction -- this is given by the point $(\phi,\theta)$; you have apparently lost this data.

 

[yungman]

I think you're confused. By retaining only the $\hat{r}$ component, all you know about the vector is its magnitude. But presumably this vector points in a specific direction -- this is given by the point $(\phi,\theta)$; you have apparently lost this data.

Not at all confused. In fact I spent days on this. Two of the EM book actually gave example of transforming a position vector from rect co to spherical co and only the radius component remain.

Try do the transformation of $$\vec A =\hat x 4 + \hat y 5 + \hat z 6$$

Remember a position vector ALWAYS originate at the origin and point to a single point in space. The components cannot be a function of the coordinates. Unless my understanding of "position vector" is totally wrong, all the components are constant respecting to coordinates. This is how it works in calculus. Don't confuse the position vector with vector field, all the transformation work perfect for vector field.

 

[bapowell]

So then you are claiming that all vectors in spherical coordinates have only $\hat{r}$ components? Here:

$${\boldmath A} = (r,\phi,\theta) = (1,2,3)$$

There's a vector, with base positioned at the origin that has nonzero $\phi$ and $\theta$ components.

 

[yungman]

So then you are claiming that all vectors in spherical coordinates have only $\hat{r}$ components? Here:

$${\boldmath A} = (r,\phi,\theta) = (1,2,3)$$

There's a vector, with base positioned at the origin that has nonzero $\phi$ and $\theta$ components.

No, a vector field has all the components in spherical coordinates. I don't think your example can be from the origin. Remember spherical coordinates are at a point in space respect to the origin. Then you set up the coordinates according to the position vector at that point.

To make it more simple, if you take two constant vector and do the transformation to spherical coordinates and then do the cross products and you'll see.

Say let

$$\vec A =\hat x 4 + \hat y 5 + \hat z 6$$


$$\vec B = \hat x 6 + \hat y 7 + \hat z 8$$


Transform both A and B to spherical co and do the cross product and you'll see.

What I am trying to drive at is There is something missing. Either I am so out of it, or the transformation don't work in constant ( or position vector). Chances are I am wrong, but I have spent quite a bit of time and I don't see it. I want to find out.

Thanks

Alan

 

[bapowell]

I don't think your example can be from the origin.

Sure it can. Start at the origin. Walk out 1 pace along x. Move 2 units in the azimuthal direction. Move $\pi/2 - 3$ paces in the polar direction:

What I am trying to drive at is There is something missing. Either I am so out of it, or the transformation don't work in constant ( or position vector). Chances are I am wrong, but I have spent quite a bit of time and I don't see it. I want to find out.

I understand your frustration and I didn't mean to be short by calling you confused. There is nothing special about position vectors. They are just like any other old vector. Try this:

$$(x, y, z) \rightarrow (r,\phi,\theta) = (\sqrt{x^2 + y^2 + z^2},\tan^{-1}(y/x),\cos^{-1}(z/r))$$

These are the components of your vector in spherical coordinates.

 

[RedX]

I think you're confused about a lot of things, but a good start would be that the basis vectors depend on coordinate.

So for example $$\vec{A}=|A|e_r(r_A,\theta_A,\phi_A)$$

and $$\vec{B}=|B|e_r(r_B,\theta_B,\phi_B)$$

In fact, you can simply this by noting that there is no dependence on r.

So $$e_r(\theta_A,\phi_A) \times e_r(\theta_B,\phi_B) \neq 0$$

 

[yungman]

Sure it can. Start at the origin. Walk out 1 pace along x. Move 2 units in the azimuthal direction. Move $\pi/2 - 3$ paces in the polar direction:

I understand your frustration and I didn't mean to be short by calling you confused. There is nothing special about position vectors. They are just like any other old vector. Try this:

$$(x, y, z) \rightarrow (r,\phi,\theta) = (\sqrt{x^2 + y^2 + z^2},\tan^{-1}(y/x),\cos^{-1}(z/r))$$

These are the components of your vector in spherical coordinates.

THat is not how the spherical coordinates are defined. What you are doing is actually still in the rectangular coordinate as in all the calculus book for the multiple variable only, these are just the polar form representation of the vector in rectanglar coordinates. The spherical coordinates are totally different. If you look at books like Griffiths or Cheng or Ulaby, spherical coordinates is established at a point defined by a position vector and the radius vector is the R/B] direction and they define the $\theta, \phi$ direction that is orthogonal to the radius vector direction at that given point. It is totally different thing. That is the reason I find it so hard to get info. Only the PDE text touched the surface of it in the 2D poisson's equation. Look at my post #1. I show the forumla of the transformation, these are povened and published by all books on how you get the magnitude of each components in spherical coordinates. If you just look at any of your EM books and you will follow what I am doing.

As I said, if you just do the conversion using my example of A and B with the formulas given in post #1, you'll see. I don't even know how many time I double check. I am really looking for some major things that I am too blind to see! I can tell you, the formulas work for any VECTOR FIELDS. Because the vector fields is defined at the point in space. Then you have all three components. It is a very very distinct difference.

I don't mean to sound rude also, I spent a lot of time, I actual posted in other math forums and got no answer, been looking at all the definitions and formulas and I just cannot get the answer. This in not even supposed to be a difficult question as it is in the first chapter of all 5 EM books I have.

Thanks

Alan

 

[yungman]

I think you're confused about a lot of things, but a good start would be that the basis vectors depend on coordinate.

So for example $$\vec{A}=|A|e_r(r_A,\theta_A,\phi_A)$$

and $$\vec{B}=|B|e_r(r_B,\theta_B,\phi_B)$$

In fact, you can simply this by noting that there is no dependence on r.

So $$e_r(\theta_A,\phi_A) \times e_r(\theta_B,\phi_B) \neq 0$$

Did you try doing the transformation of A and B and then do the cross product as given in post #9 using the formulas given in post #1? I showed every single step why I come up with the question. Can you please do that first? This is very very specific. I check very carefully before I posted.

 

[RedX]

Did you try doing the transformation of A and B and then do the cross product as given in post #9 using the formulas given in post #1? I showed every single step why I come up with the question. Can you please do that first? This is very very specific. I check very carefully before I posted.

vanhees71 had it right.

You are assuming A and B are in the same direction when you write them as: (A,0,0) and (B,0,0) and use the determinant thing as a cross-product. But they're not in the same direction in general.

$$\hat{e}_r=\frac{\vec{r}}{|r|}$$ where $$\vec{r}$$ is the position.

$$\frac{\vec{r}_A}{|r_A|} \neq \frac{\vec{r}_B}{|r_B|}$$ since A and B are pointing in different directions.

Using unit vectors so as to not be confusing:

$$\hat{r}_A \neq \hat{r}_B$$

 

[yungman]

I have a copy of the picture of spherical coordinate axis:

It is not 3D polar coordinates where you take the magnitude and swing an angle $\phi$ from the x-axis and then swing $\theta$. It is defined AT a point by a position vector.

You see the point P? That is the point pointed by a position vector. Then you set the spherical coordinates at that point as shown in red color.

 

[yungman]

vanhees71 had it right.

You are assuming A and B are in the same direction when you write them as: (A,0,0) and (B,0,0) and use the determinant thing as a cross-product. But they're not in the same direction in general.

$$\hat{e}_r=\frac{\vec{r}}{|r|}$$ where $$\vec{r}$$ is the position.

$$\frac{\vec{r}_A}{|r_A|} \neq \frac{\vec{r}_B}{|r_B|}$$ since A and B are pointing in different directions.

Using unit vectors so as to not be confusing:

$$\hat{r}_A \neq \hat{r}_B$$

I don't assume anything. I did strictly according to the formulas given in the transformation. Please verify my formulas in post #1 with your books and transform one of the vector example I gave and see what you think. I actually hope you can prove my formulas are wrong but please give me the reason why they are wrong.

 

[RedX]

Maybe it's clearer if I write it like this:

It is true that $$\vec{A}=|A|e_{rA}$$ and $$\vec{B}=|B|e_{rB}$$.

So it would be wrong to write $$\vec{A}=|A|e_{rA}+(\mbox{something }\times) e_{\theta A}$$

since it's just $$\vec{A}=|A|e_{rA}$$.

However, $$e_{rB}$$ is *not* $$e_{rA}$$. However, you can write

$$e_{rB}=\lambda_1 e_{rA} + \lambda_2 e_{\theta A}+\lambda_3 e_{\phi A}$$

So when you take the cross product it is no longer $$\vec A X \vec B \;=\;\left|\begin{array}{ccc}\hat x & \hat y & \hat z \\A_R & 0 & 0\\ B_R & 0 & 0\end{array}\right|\;=\; 0$$

but the last row has all 3 components. This is because we are expressing the vector B in the basis describing the vector A.

You always have to perform the calculation in one fixed basis.

 

[yungman]

This is the formulas from the book on the transformation:

 

[D H]

You always have to perform the calculation in one fixed basis.

That is worth repeating because failing to do this is the source of all of the OP's confusion.

So I'll repeat it.

You always have to perform the calculation in one fixed basis.

Yungman, while it is true that both of your position vectors are purely radial ($$\vec A = A\,\hat r_A$$ and $$\vec B = B\,\hat r_B$$) the unit vectors $$\hat r_A$$ and $$\hat r_B$$ are different unit vectors. You are implicitly assuming that $$\hat r_A$$ and $$\hat r_B$$ are the same unit vectors and hence their cross product is zero. That is not the case here. $$\hat r_A \times \hat r_B$$ is not the zero vector.

 

[yungman]

Maybe it's clearer if I write it like this:

It is true that $$\vec{A}=|A|e_{rA}$$ and $$\vec{B}=|B|e_{rB}$$.

So it would be wrong to write $$\vec{A}=|A|e_{rA}+(\mbox{something }\times) e_{\theta A}$$

since it's just $$\vec{A}=|A|e_{rA}$$.

However, $$e_{rB}$$ is *not* $$e_{rA}$$. However, you can write

$$e_{rB}=\lambda_1 e_{rA} + \lambda_2 e_{\theta A}+\lambda_3 e_{\phi A}$$

So when you take the cross product it is no longer


$$\vec A X \vec B \;=\;\left|\begin{array}{ccc}\hat x & \hat y & \hat z \\A_R & 0 & 0\\ B_R & 0 & 0\end{array}\right|\;=\; 0$$

but the last row has all 3 components. This is because we are expressing the vector B in the basis describing the vector A.

You always have to perform the calculation in one fixed basis.

A and B are specified both a position vector and are totally independent to each other. Take a look at the copy of the book by Ulaby, which is consistant with Griffiths and Cheng.

 

[yungman]

That is worth repeating because failing to do this is the source of all of the OP's confusion.

So I'll repeat it.

You always have to perform the calculation in one fixed basis.

Yungman, while it is true that both of your position vectors are purely radial ($$\vec A = A\,\hat r_A$$ and $$\vec B = B\,\hat r_B$$) the unit vectors $$\hat r_A$$ and $$\hat r_B$$ are different unit vectors. You are implicitly assuming that $$\hat r_A$$ and $$\hat r_B$$ are the same unit vectors and hence their cross product is zero. That is not the case here. $$\hat r_A \times \hat r_B$$ is not the zero vector.

Thanks, this is the first statement that make sense. So the question is how do you perform the cross product of two position vectors? I have been question about how you take the angle into account. None of the book even mention this.

What I presented works perfect for vector fields. The problem is none of the book even hint about what you said. Ulaby actually shown an example that the transformation of rect to cylindrical and the $\hat {\phi}$ term is zero.

Also the table of transformation is consistance on all the books, so are they missing the point?

I looked very carefully on how to perform the cross product in spherical coordinates, and to the best of my knowledge, it is like what I wrote, which has no angle information of the vector. And I can tell you I search through all the EM books already.

 

[yungman]

That is worth repeating because failing to do this is the source of all of the OP's confusion.

So I'll repeat it.

You always have to perform the calculation in one fixed basis.

Yungman, while it is true that both of your position vectors are purely radial ($$\vec A = A\,\hat r_A$$ and $$\vec B = B\,\hat r_B$$) the unit vectors $$\hat r_A$$ and $$\hat r_B$$ are different unit vectors. You are implicitly assuming that $$\hat r_A$$ and $$\hat r_B$$ are the same unit vectors and hence their cross product is zero. That is not the case here. $$\hat r_A \times \hat r_B$$ is not the zero vector.

The more I think, the more it make sense, this explain why it work on vector fields because when you talk vector field, it is relative to one single point, so the two vector field share the same coordinates so their $\hat R, \hat {\theta}, \hat {\phi}$ are the same for both vector.

But being that, why the books never mention this point and actually show an example that a constant vector transform to spherical coordinates and show only the radial component exist? Why non of the books said anything about not able to use the transformation as is and need more translations. I am talking about well known books like Chengs "Field and Wave" that is use in many universities and Ulaby "Engineering Electromagnetics" that used in SJSU.

Do you have links that talk about this specifically in spherical and cylindrical coordinates only. I try not the get into culver coordinates that itself is a big topic include non orthogonal stuff. It should be quite simple to include the unit vectors.

 

[vanhees71]

You should write out the expansion of your vectors with respect to the bases used. You have to use one fixed set of basis vectors at each point of space for both vectors, $$\vec{A}$$ and $$\vec{B}$$. You can not change the basis and just mix two column vectors that refer to different bases with each other. As you see at your example this leads to wrong results.

As I said already before, I do not know what you are supposed to do. Can you post the exact wording of your question as given on the problem sheet?

 

[yungman]

Maybe it's clearer if I write it like this:

It is true that $$\vec{A}=|A|e_{rA}$$ and $$\vec{B}=|B|e_{rB}$$.

So it would be wrong to write $$\vec{A}=|A|e_{rA}+(\mbox{something }\times) e_{\theta A}$$

since it's just $$\vec{A}=|A|e_{rA}$$.

However, $$e_{rB}$$ is *not* $$e_{rA}$$. However, you can write

$$e_{rB}=\lambda_1 e_{rA} + \lambda_2 e_{\theta A}+\lambda_3 e_{\phi A}$$

So when you take the cross product it is no longer


$$\vec A X \vec B \;=\;\left|\begin{array}{ccc}\hat x & \hat y & \hat z \\A_R & 0 & 0\\ B_R & 0 & 0\end{array}\right|\;=\; 0$$

but the last row has all 3 components. This is because we are expressing the vector B in the basis describing the vector A.

You always have to perform the calculation in one fixed basis.

I think I get what you mean. So in order to do the cross product in spherical co, we need to set up the spherical co axis. So in this case, we use A as the reference and do the transformation and will have the radial component only.

Then we set up the spherical coordinate system at the origin using the direction of A as the radial direction and obtain the $\hat {\theta}, \hat {\phi}$ as I draw here:

In the drawing, I show the spherical coordinates are set up in referenced to A. You can see the $\hat {\theta}$ has x=A_x, y=A_y. and z is set up so $\hat {\theta}$ is orthogonal to R. $\hat {\phi}$ is in xy plane orthogonal to both R and $\hat {\theta}$.

Then we have to translate B into the spherical coordinates that we set up reference to A. With this, B will have all three components in the spherical coordinates.

Am I getting it.

I want to verify in the vector field situation, it work because the spherical coordinates are set up by the position vector pointed at a specific point, so A and B just use the coordinates and both will have three components.

Am I getting this now?

Thanks

Alan

 

[vanhees71]

Yes, I think now you got the concept right.

 

[yungman]

Yes, I think now you got the concept right.

Thanks

I am a self studier, don't have an instructor with me. Maybe if I had taken the class, the instructor might have taught this. I am surprised none of the EM books emphasis on this fact that you have to be very careful with how you set up the coordinates. It almost as if they are just trying to drop a little hint and run!

 

[vanhees71]

Which books on E+M are you studying? Usually these books do not give a detailed study of vector and tensor calculus, which is known to be a somewhat tricky subject in the first semesters of physics studies. In my opinion, the best books on classical physics ever written, although somewhat outdated nowadays, are Sommerfeld's Lectures on Theoretical Physics. In these old days hydrodynamics was still part of the physics curriculum, and thus in the corresponding volume 2 of this book he gives a quite detailed introduction to vector and tensor analysis, which I can only strongly recommend first, before you move on with electromagnetism, which is also very well represented in volume 3 (and 4 for the optics part). These books are available in English too.

 

[yungman]

Which books on E+M are you studying? Usually these books do not give a detailed study of vector and tensor calculus, which is known to be a somewhat tricky subject in the first semesters of physics studies. In my opinion, the best books on classical physics ever written, although somewhat outdated nowadays, are Sommerfeld's Lectures on Theoretical Physics. In these old days hydrodynamics was still part of the physics curriculum, and thus in the corresponding volume 2 of this book he gives a quite detailed introduction to vector and tensor analysis, which I can only strongly recommend first, before you move on with electromagnetism, which is also very well represented in volume 3 (and 4 for the optics part). These books are available in English too.

I studied Cheng's "Field and Wave Electromagnetics", Ulaby "Engineering Electromagnetics", Griffiths "Introduction to Electrodynamics", Schwarz "Electromagnetics for Engineers" and some on Popovic "Introductary Electromagnetics". Reference a little of the Kraus "Electromagnetics with Applications".

So this is more in the Tensor analysis? My highest Math studies is PDE, maybe this is beyond my knowledge. I am planning on studying Complex analysis in the future before getting into JD Jackson "Classical Electrodynamics".

This is my third go around, first gone through Ulaby's book which is used in San Jose State, it is the easiest. Then I studied Cheng's which is a lot harder, it concentrate a lot on the phasor, transmission lines, smith charts designs. Now the third time, I spent more time with Griffiths which is more in line with the physics side of EM. Griffiths is equivalent to Cheng's in physics as Cheng in Engineering. They have like 60% in common and the rest quite different.

 

[yungman]

OK, I am trying to work out the problem:

1) I transform A into spherical co. so $$\vec A = \hat R_A |\vec A|$$ with $${\theta_A}, \hat {\phi_A}$$.

2)Then we want to transform $$\vec B$$ to $$R_A, {\theta_A}, \hat {\phi_A}$$.

$$B_{R_A} = \vec B \cdot \hat{R_A} = B_x sin(\theta_A)cos(\phi_A) + B_y sin(\theta_A)sin(\phi_A) + B_z cos (\theta_A)$$

$$B_{\theta_A} = \vec B \cdot \hat{\theta_A} = B_x cos(\theta_A)cos(\phi_A) + B_y cos(\theta_A)sin(\phi_A) - B_z sin (\theta_A)$$

$$B_{\phi_A} = \vec B \cdot \hat{\phi_A} = -B_x sin(\phi_A) + B_y cos(\phi_A)$$



$$\vec A X \vec B = \left|\begin{array}{ccc} \hat x & \hat y & \hat z \\|\vec A| & 0 & 0\\ \left(\begin{array}{c} B_x sin(\theta_A)cos(\phi_A)\\ +B_y sin(\theta_A)sin(\phi_A) \\ +B_z cos (\theta_A) \end {array}\right) & \left(\begin{array}{c} B_x cos(\theta_A)cos(\phi_A)\\ +B_y cos(\theta_A)sin(\phi_A) \\ - B_z sin (\theta_A) \end{array}\right ) & \left(\begin{array}{c} -B_x sin(\phi_A) \\ +B_y cos(\phi_A) \end{array}\right) \end{array}\right | = |\vec A| \left [ \hat{\theta_A} \left(\begin{array}{c} B_x sin(\phi_A) \\ -B_y cos(\phi_A) \end{array}\right) + \hat{\phi_A}\left(\begin{array}{c} B_x cos(\theta_A)cos(\phi_A)\\ +B_y cos(\theta_A)sin(\phi_A) \\ - B_z sin (\theta_A) \end{array}\right ) \right]$$

Am I getting this right?

Thanks

Alan

 

[yungman]

I just use this to calculate with some real numbers and it work. I got the exact same answer doing cross product in rect and transform both position vectors into spherical co, do the cross product and then translate back to rect co and they MATCH!:rofl:

The only thing that is different:

1) If I perform cross product in rect co and transform to spherical co, you get the usual radial component only.

2) If I transform both to spherical co first and perform the cross product. I get zero on the radial component. Then I get both $\hat{\theta},\hat{\phi}$.