Finding the Basis Vectors for a Coordinate System

  • #1
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Homework Statement:
Find the basis vectors for the given coordinate system for the arbitrary constant ##a##.
Relevant Equations:
The question is working with elliptic cylindrical coordinates. Refer to https://mathworld.wolfram.com/EllipticCylindricalCoordinates.html for more information.

Equations:
$$x = a \cosh{\mu} \cos{v}$$
$$y = a \sinh{\mu} \sin{v}$$
$$z=z$$
To my understanding, to get the basis vectors for a given coordinate system (in this case being the elliptic cylindrical coordinate system), I need to do something like shown below, right?

$$\hat{\mu}_x = \hat{\mu} \cdot \hat{x}$$
$$\hat{v}_z = \hat{v} \cdot \hat{z}$$

And do that for essentially 9 times in total. Despite that, though, how would I go about finding the actual resultant value? Also, perhaps this is more akin to a "semantic" question, what does it truly mean to find a basis vector for the arbitrary constant ##a##? Does it mean to find the basis vectors in terms of Cartesian basis vectors? Or perhaps something else entirely?

Ultimately, any help to assist me through the provided problem would be greatly appreciated. Thank you!
 

Answers and Replies

  • #2
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And do that for essentially 9 times in total. Despite that, though, how would I go about finding the actual resultant value? Also, perhaps this is more akin to a "semantic" question, what does it truly mean to find a basis vector for the arbitrary constant ##a##? Does it mean to find the basis vectors in terms of Cartesian basis vectors? Or perhaps something else entirely?

Ultimately, any help to assist me through the provided problem would be greatly appreciated. Thank you!

Hi,

I think the question is asking you to find a new set of basis vectors to describe the system. We have started in the cartesian basis vectors ## \hat i ##, ## \hat j ##, and ## \hat k ## and now we want to find ## \hat \mu ##, ## \hat v ##, and ## \hat z ##. One way to look at the overview of this problem is to consider that in cartesian coordinates we have ## d \vec r = dx \hat i + dy \hat j + dz \hat k ## and we want to find a way to write it in the following form: $$ d \vec r = h_{\mu} d\mu \hat \mu + h_v dv \hat v + h_z dz \hat z $$
where the h coefficients are called the metric coefficients. We need those new basis vectors to allow us to work in this coordinate system. I think the wording about ## a ## just means that is is a constant and not a function of either ## \mu ##, ## v ##, or ## z ##

Below is a brief overview (I would definitely recommend reading more around the topic so you understand where the formulae come from):

It turns out that the basis vectors are given by:
$$ \hat \mu = \frac{1}{h_{\mu}} \left( \frac{\partial x}{\partial \mu} \hat i + \frac{\partial y}{\partial \mu} \hat j + \frac{\partial z}{\partial \mu} \hat k \right) $$

where $$ h_{\mu} = \sqrt{ \left( \frac{\partial x}{\partial \mu} \right)^2 + \left( \frac{\partial y}{\partial \mu} \right)^2 + \left( \frac{\partial z}{\partial \mu} \right)^2} $$

and equivalent expressions can be obtained for the other two variables ## v ## and ## z ##. By inspection, you might realize that one of these three new variables is likely to be the same as its cartesian counterpart, but you can confirm that with the maths.

One place to refer to for further reading is: https://en.wikipedia.org/wiki/Curvilinear_coordinates

I hope that makes some sense. If not, I will be happy to clarify
 
  • #3
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@Master1022 , thank you so much for the incredibly helpful explanation!

While I did some minor rearrangements to what you have shown and explained above, below is the equation I began to use to find the basis vectors.

$$\hat{\mu} = h_{\mu} \bigg( \frac{\partial \mu}{\partial x} \hat{i} + \frac{\partial \mu}{\partial y} \hat{j} + \frac{\partial \mu}{\partial z} \hat{k} \bigg)$$

The same equation (with minor adjustments) can be written for both ##\hat{v}## and ##\hat{z}## as well.

From there, I was thankfully able to find for the scale factor or metric coefficient ##h_{\mu} = a \sqrt{\sinh^2 \mu + \sin^2 v}##.

The same answer also applies to ##h_v## too. ##h_z##, on the other hand, remains ##1##.

With that said, though, I am having a really hard time finding for ##\frac{\partial \mu}{\partial x}## and ##\frac{\partial v}{\partial x}## or the likes simply because I do not know how ##\mu## is defined in terms of Cartesian coordinates.

I searched on the web and found the below equations:

$$\frac{x^2}{a^2 \cosh^2 \mu} + \frac{y^2}{a^2 \sinh^2 \mu} = \cos^2 v + \sin^2 v = 1$$

and

$$\frac{x^2}{a^2 \cos^2 v} - \frac{y^2}{a^2 \sin^2 v} = \cosh^2 \mu - \sinh^2 \mu = 1$$

Source: https://en.wikipedia.org/wiki/Elliptic_cylindrical_coordinates

However, none of these equations solely define ##\mu## in terms of Cartesian coordinates.

If I was working with a different coordinate system (e.g. ##r = \sqrt{x^2 + y^2 + z^2}##), I could easily perform the below equation.

$$\frac{\partial r }{\partial x} \hat{i} = \frac{\partial \big(\sqrt{x^2 + y^2 + z^2} \big)}{\partial x} \hat{i} = \frac{x}{r} \hat{i}$$

However, I am able to perform the above calculation as I perfectly know what ##r## translates to in terms of Cartesian coordinates. For ##\mu## (and ##v##), though, I am just not sure how to proceed. Therefore, any additional help and clarification would be really appreciated! Once again, thank you for all your help thus far!
 
  • #4
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Hi,

@Master1022 , thank you so much for the incredibly helpful explanation!

While I did some minor rearrangements to what you have shown and explained above, below is the equation I began to use to find the basis vectors.

$$\hat{\mu} = h_{\mu} \bigg( \frac{\partial \mu}{\partial x} \hat{i} + \frac{\partial \mu}{\partial y} \hat{j} + \frac{\partial \mu}{\partial z} \hat{k} \bigg)$$

The same equation (with minor adjustments) can be written for both ##\hat{v}## and ##\hat{z}## as well.
I think there are different ways of doing coordinate transformations based on the form of the expressions provided. We can be provided with either:
1. OLD variables in terms of NEW variables
2. NEW variables in terms of OLD variables

The processes differ slightly between each one. We are working with scenario (1) here as we have expressions for ## x ##, ## y ##, and ## z ## in terms of the three new variables. In all honesty, I haven't that expression for the basis vector that you have written above so I will need to look into it. However, I would work from using the old variables in terms of the new ones (i.e. I wouldn't spend time trying to find expressions for the new ones in terms of the old ones).

From there, I was thankfully able to find for the scale factor or metric coefficient ##h_{\mu} = a \sqrt{\sinh^2 \mu + \sin^2 v}##.

The same answer also applies to ##h_v## too. ##h_z##, on the other hand, remains ##1##.
Agreed

With that said, though, I am having a really hard time finding for ##\frac{\partial \mu}{\partial x}## and ##\frac{\partial v}{\partial x}## or the likes simply because I do not know how ##\mu## is defined in terms of Cartesian coordinates.
No easy way has become become apparent to me. As mentioned above, I would work with the formulae that I provided as the derivatives can all be found immediately. In theory, this method that you are doing should work, but it seems like will require much more effort though.

If I was working with a different coordinate system (e.g. ##r = \sqrt{x^2 + y^2 + z^2}##), I could easily perform the below equation.

$$\frac{\partial r }{\partial x} \hat{i} = \frac{\partial \big(\sqrt{x^2 + y^2 + z^2} \big)}{\partial x} \hat{i} = \frac{x}{r} \hat{i}$$

However, I am able to perform the above calculation as I perfectly know what ##r## translates to in terms of Cartesian coordinates. For ##\mu## (and ##v##), though, I am just not sure how to proceed. Therefore, any additional help and clarification would be really appreciated! Once again, thank you for all your help thus far!
As you have written, in the cylindrical polar scenario we are given the NEW variables in terms of OLD variables. Therefore, it makes sense to use the method that you did.

However, given that we are working with OLD in terms of NEW, I would use a different method accordingly (I think it will save lots of work).

I would try the method that I provided above and see if that gets you to the required answer. Considering you found the metric coefficients correctly, you have already calculated all 9 derivatives and it is just a case of putting them into the vectors.

I will see whether I can find any good literature online to refer you too (although I often struggle to find good online notes for this topic).

Hope that is of some help
 
Last edited:
  • #5
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@Master1022 , thank you so much for the clarifications here! It made a lot of sense and it most certainly made the calculation process much easier.

I will see whether I can find any good literature online to refer you too (although I often struggle to find good online notes for this topic).

I highly agree with this statement you made here. Despite searching everywhere on the net, I couldn't quite find any site that taught how to derive the basis vectors (for different coordinate systems). I wonder why ...

That aside, though, I did apply what I have learned here and calculated for the basis vectors. Do the solutions below look right to you?

$$\hat{u} = \frac{1}{a \sqrt{\sinh^2 {u} + \sin^2 {v}}} (a \cos{v} \sinh{u} \; \hat{i} + a \cosh{u} \sin{v} \; \hat{j})$$

$$\hat{v} = \frac{1}{a \sqrt{\sinh^2 {u} + \sin^2 {v}}} (-a \cosh{u} \sin{v} \; \hat{i} + a \sinh{u} \cos{v} \; \hat{j})$$

$$\hat{z} = 1 \hat{z}$$

Once again, thank you for all your kind assistance thus far!
 
  • #6
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Happy to be of some help!

That aside, though, I did apply what I have learned here and calculated for the basis vectors. Do the solutions below look right to you?

$$\hat{u} = \frac{1}{a \sqrt{\sinh^2 {u} + \sin^2 {v}}} (a \cos{v} \sinh{u} \; \hat{i} + a \cosh{u} \sin{v} \; \hat{j})$$

$$\hat{v} = \frac{1}{a \sqrt{\sinh^2 {u} + \sin^2 {v}}} (-a \cosh{u} \sin{v} \; \hat{i} + a \sinh{u} \cos{v} \; \hat{j})$$

$$\hat{z} = 1 \hat{z}$$

These look correct to me (although I am definitely not an expert!). One other thing that seems promising is that the basis vectors are mutually orthogonal.

I highly agree with this statement you made here. Despite searching everywhere on the net, I couldn't quite find any site that taught how to derive the basis vectors (for different coordinate systems). I wonder why ...

Yes, I really don't know why. I haven't been able to find any simple proofs for the basis vector formula on the internet and I feel the proof may help you see where these metric coefficients and terms come from. Just so you have an overview (I don't know if I can type the whole thing right now as it is somewhat intensive), this is what the method basically entails:

Method for forming u, v, w basis vectors from cartesian basis vectors?
1. Start with the fact that you have a position vector and a dr vector

$$ \vec r = x(u, v, w) \hat i + y(u, v, w) \hat j + z(u, v, w) \hat k $$
$$ \vec r = dx \hat i + dy \hat j + dz \hat k $$
and we want to turn the position vector into: ## d \vec r = h_u du \hat u + h_v dv \hat v + h_w dw \hat w ##

2. Now find expressions for dx, dy, dz. For example, dx will look like:
$$ dx = \frac{\partial x}{\partial u} du + \frac{\partial x}{\partial v} dv + \frac{\partial x}{\partial w} dw $$
You can find similar expressions for dy and dz

3. Substitute in to the equation for ## d \vec r ## and group terms based on du, dv, and dw.
For example, the 'du' group will look like:
$$ d \vec r = \left( \frac{\partial x}{\partial u} \hat i + \frac{\partial y}{\partial u} \hat j + \frac{\partial z}{\partial u} \hat k \right)du + ... $$
then there will be two other portions for dv and dw

4. Compare terms
if you compare this to the target expression for ## d \vec r ##, then you can see that
$$ h_u du \hat u = \left( \frac{\partial x}{\partial u} \hat i + \frac{\partial y}{\partial u} \hat j + \frac{\partial z}{\partial u} \hat k \right)du $$
and equivalent expressions for the other two variables. Given that ## \vec u ## is a unit vector, you can just normalize the vector and get expressions for the vector ## \hat u ## and the metric coefficient ## h_u ##.

Hope that makes some sort of sense. At least that is one way you can derive the expressions for yourself from scratch. Happy to provide clarification if needed.
 
  • #7
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This is great! Thank you so much for the concrete explanation! I found it incredibly helpful and now I officially know how the expressions for basis vectors are derived. If I do end up having additional questions on the topic, I will be sure to ask it here. In the meantime, I will be studying more on the topic!
 

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