How Do Changes in Curvilinear Coordinates Affect Unit Vectors?

In summary, the author is trying to find a way to compute the partial derivative of e_i with respect to q_j. They found the above equation, but can't seem to prove how to do it.
  • #1
Swapnil
459
6
Given an orthogonal curvilinear coordinate system [tex](q_{1},q_{2},q_{3})[/tex] with standard orthonormal basis vectors [tex](\hat{e}_{1},\hat{e}_{2},\hat{e}_{3})[/tex], how would you prove the following?:

[tex]\frac{\partial \hat{e}_{i}}{\partial q_{j}}= \hat{e}_{j}\frac{1}{h_{i}}\frac{\partial h_{j}}{\partial q_{i}}\qquad \forall i \neq j[/tex]

where
[tex]h_{i}= \Big|\frac{\partial\vec{r}}{\partial q_{i}}\Big| = \sqrt{{\Big(\frac{\partial x}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial y}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial z}{\partial q_{i}}\Big)}^{2}}[/tex]

and
[tex]\hat{e}_{i}= \frac{1}{h_{i}}\frac{\partial \vec{r}}{\partial q_{i}}[/tex]

where
[tex]\vec{r}= x(q_{1},q_{2},q_{3})\hat{x}+y(q_{1},q_{2},q_{3})\hat{y}+z(q_{1},q_{2},q_{3})\hat{z}[/tex]


I don't know what am I missing, it probably involves some clever manipulation of the partial derivatives but I just can't figure it out.
 
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  • #2
Huh?

I take it you have a line element like
[tex] ds^2 = (A \, du)^2 + (B \, dv)^2 + (C \, dw)^2[/tex]
where [itex]A,B,C[/itex] are functions of the coordinates [itex]u,v,w[/itex]. Then you can take the obvious frame field
[tex] \vec{e}_1 = \frac{1}{A} \, \partial_u, \; \;
\vec{e}_2 = \frac{1}{B} \, \partial_v, \; \;
\vec{e}_3 = \frac{1}{C} \, \partial_w [/tex]
But then you seem to introduce without comment some coordinate transform to another chart. From the names of the coordinates, I suspect this might even be a Cartesian chart, and if so, your manifold must be flat, presumably [itex]E^3[/itex]. I can't help suspecting that either this is a problem in a textbook (on vector calculus?), which you misunderstood, or else you for some reason you didn't try very hard to explain what you are trying to do. It's always a good idea to say what book a problem comes from, at the very least.

Since I have no idea what you are up to, I'll just make a general suggestion: whenever you get confused, if all else fails go back to basics and recall that a vector field is a first order linear homogeneous operator on functions. So you can compute things like [itex]\partial_v \left( \vec{e}_1 f \right)[/itex].
 
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  • #3
Chris Hillman said:
I take it you have a line element like
[tex] ds^2 = (A \, du)^2 + (B \, dv)^2 + (C \, dw)^2[/tex]
where [itex]A,B,C[/itex] are functions of the coordinates [itex]u,v,w[/itex]. Then you can take the obvious frame field
[tex] \vec{e}_1 = \frac{1}{A} \, \partial_u, \; \;
\vec{e}_2 = \frac{1}{B} \, \partial_v, \; \;
\vec{e}_3 = \frac{1}{C} \, \partial_w [/tex]
Yeah, I guess this is one way to put it.
Chris Hillman said:
But then you seem to introduce without comment some coordinate transform to another chart. From the names of the coordinates, I suspect this might even be a Cartesian chart, and if so, your manifold must be flat, presumably [itex]E^3[/itex]. I can't help suspecting that either this is a problem in a textbook (on vector calculus?), which you misunderstood, or else you for some reason you didn't try very hard to explain what you are trying to do. It's always a good idea to say what book a problem comes from, at the very least.
I just found the above identity on some vector calculus pdf online and was trying to prove it using my current knowledge of vector calculus.

Chris Hillman said:
Since I have no idea what you are up to, I'll just make a general suggestion: whenever you get confused, if all else fails go back to basics and recall that a vector field is a first order linear homogeneous operator on functions. So you can compute things like [itex]\partial_v \left( \vec{e}_1 f \right)[/itex].
Yeah, but how?
How would you compute
[tex]\partial_v (\vec{e}_1)[/tex]?

The above identity I found says that it should be
[tex]\partial_v (\vec{e}_1) = \vec{e}_2\frac{{\partial}_u(B)}{A}[/tex]

but I can't prove how?
 
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  • #4
Come on!? Has anyone even attempted to answer my question (succesully or unsucessfully)?
 
  • #5
Can't anyone help me out?
 
  • #6
Swapnil said:
Given an orthogonal curvilinear coordinate system [tex](q_{1},q_{2},q_{3})[/tex] with standard orthonormal basis vectors [tex](\hat{e}_{1},\hat{e}_{2},\hat{e}_{3})[/tex], how would you prove the following?:

[tex]\frac{\partial \hat{e}_{i}}{\partial q_{j}}= \hat{e}_{j}\frac{1}{h_{i}}\frac{\partial h_{j}}{\partial q_{i}}\qquad \forall i \neq j[/tex]

where
[tex]h_{i}= \Big|\frac{\partial\vec{r}}{\partial q_{i}}\Big| = \sqrt{{\Big(\frac{\partial x}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial y}{\partial q_{i}}\Big)}^{2}+{\Big(\frac{\partial z}{\partial q_{i}}\Big)}^{2}}[/tex]

and
[tex]\hat{e}_{i}= \frac{1}{h_{i}}\frac{\partial \vec{r}}{\partial q_{i}}[/tex]

where
[tex]\vec{r}= x(q_{1},q_{2},q_{3})\hat{x}+y(q_{1},q_{2},q_{3})\hat{y}+z(q_{1},q_{2},q_{3})\hat{z}[/tex]


I don't know what am I missing, it probably involves some clever manipulation of the partial derivatives but I just can't figure it out.
I believe I see how to do this but it takes a lot of Latex which I'm rusty on. Try this: You have the definition of e_i etc. So what you need to do is to take the derivative of that expression with respect to q_j. Then work the calculus on the right hand side through until you get the right side of the equation your looking for. I'll try to beef up on my Latex and in the mean time you can give what I said a try, at least to see how far you can go. If you can't reach the conclusion that is required then we can work together to get your result. Sound good?

Pete
 
  • #7
Sorry Swapnil. I tried to work out that proof but to no avail. Sorry.

Pete
 
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  • #8
Swapnil said:
[tex]\frac{\partial \hat{e}_{i}}{\partial q_{j}}= \hat{e}_{j}\frac{1}{h_{i}}\frac{\partial h_{j}}{\partial q_{i}}\qquad \forall i \neq j[/tex]
Note: The indices on the right hand side do not conform to the Einstein summation conventions so make sure you don't attempt to use it or assume it from what is given. This, at least, can be gleaned from the question.

Pete
 
  • #9
Here's the exact place where I found that theorem:

www.math.wisc.edu/~milewski/321f04/fwnotescurvi2.pdf[/URL]

It does have some derivations. Maybe it can give you some new ideas Pete. :smile:
 
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  • #10
Swapnil said:
I don't know what am I missing, it probably involves some clever manipulation of the partial derivatives but I just can't figure it out.

what you are missing is the actual *reading* of the text from the top of page 2 to midway through page 3, wherein the author pretty much proves your desired statements.
 

FAQ: How Do Changes in Curvilinear Coordinates Affect Unit Vectors?

What are curvilinear unit vectors?

Curvilinear unit vectors are a set of three unit vectors that are used to describe the direction and orientation of a curve in three-dimensional space. They are typically denoted by er, eθ, and eφ and are used in spherical and cylindrical coordinate systems.

How are curvilinear unit vectors related to Cartesian unit vectors?

Curvilinear unit vectors are related to Cartesian unit vectors through a transformation matrix. This matrix allows for the conversion of coordinates and vectors between the two systems. Additionally, the direction and magnitude of the curvilinear unit vectors can be expressed in terms of Cartesian unit vectors.

What is the significance of unit vectors in curvilinear coordinate systems?

Unit vectors in curvilinear coordinate systems are essential for accurately describing the direction and orientation of a curve in three-dimensional space. They allow for the representation of complex curves and surfaces, making it easier to solve problems in fields such as physics, engineering, and mathematics.

How are the magnitudes of curvilinear unit vectors determined?

The magnitudes of curvilinear unit vectors vary depending on the coordinate system being used. In spherical coordinate systems, the magnitudes are equal to the radius of the sphere. In cylindrical coordinate systems, the magnitudes are equal to the radius of the cylinder. In both cases, the magnitudes are constant and do not change with position.

Can curvilinear unit vectors be used in other coordinate systems besides spherical and cylindrical?

Yes, curvilinear unit vectors can be used in other coordinate systems such as parabolic, ellipsoidal, and toroidal. These coordinate systems are used to describe more complex shapes and surfaces and require different unit vectors to accurately represent them. However, the underlying principles and equations for determining the unit vectors remain the same as in spherical and cylindrical coordinate systems.

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