How Do Changes in Curvilinear Coordinates Affect Unit Vectors?

[Swapnil]

Given an orthogonal curvilinear coordinate system (q_{1},q_{2},q_{3}) with standard orthonormal basis vectors (\hat{e}_{1},\hat{e}_{2},\hat{e}_{3}), how would you prove the following?:

\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

where
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}}

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

where
\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}


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.

 

[Chris Hillman]

Huh?

I take it you have a line element like
ds^2 = (A \, du)^2 + (B \, dv)^2 + (C \, dw)^2
where A,B,C are functions of the coordinates u,v,w. Then you can take the obvious frame field
\vec{e}_1 = \frac{1}{A} \, \partial_u, \; \; <br /> \vec{e}_2 = \frac{1}{B} \, \partial_v, \; \; <br /> \vec{e}_3 = \frac{1}{C} \, \partial_w
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 E^3. 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 \partial_v \left( \vec{e}_1 f \right).

 

[Swapnil]

I take it you have a line element like
ds^2 = (A \, du)^2 + (B \, dv)^2 + (C \, dw)^2
where A,B,C are functions of the coordinates u,v,w. Then you can take the obvious frame field
\vec{e}_1 = \frac{1}{A} \, \partial_u, \; \; <br /> \vec{e}_2 = \frac{1}{B} \, \partial_v, \; \; <br /> \vec{e}_3 = \frac{1}{C} \, \partial_w

Yeah, I guess this is one way to put it.

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 E^3. 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.

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 \partial_v \left( \vec{e}_1 f \right).

Yeah, but how?
How would you compute
\partial_v (\vec{e}_1)?

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

but I can't prove how?

 

[Swapnil]

Come on!? Has anyone even attempted to answer my question (succesully or unsucessfully)?

 

[Swapnil]

Can't anyone help me out?

 

[pmb_phy]

Given an orthogonal curvilinear coordinate system (q_{1},q_{2},q_{3}) with standard orthonormal basis vectors (\hat{e}_{1},\hat{e}_{2},\hat{e}_{3}), how would you prove the following?:

\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

where
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}}

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

where
\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}


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

 

[pmb_phy]

Sorry Swapnil. I tried to work out that proof but to no avail. Sorry.

Pete

 

[pmb_phy]

\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

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

 

[Swapnil]

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:

 

[Doodle Bob]

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.