# Curvilinear Unit Vectors

1. Apr 13, 2007

### 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.

Last edited: Apr 13, 2007
2. Apr 14, 2007

### 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
$$\vec{e}_1 = \frac{1}{A} \, \partial_u, \; \; \vec{e}_2 = \frac{1}{B} \, \partial_v, \; \; \vec{e}_3 = \frac{1}{C} \, \partial_w$$
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 $\partial_v \left( \vec{e}_1 f \right)$.

Last edited: Apr 14, 2007
3. Apr 15, 2007

### Swapnil

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

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.

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?

Last edited: Apr 15, 2007
4. Apr 17, 2007

### Swapnil

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

5. Apr 21, 2007

### Swapnil

Can't anyone help me out?

6. Apr 21, 2007

### pmb_phy

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. Apr 22, 2007

### pmb_phy

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

Pete

Last edited: Apr 22, 2007
8. Apr 22, 2007

### pmb_phy

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. Apr 22, 2007

### Swapnil

Last edited: Apr 22, 2007
10. Apr 23, 2007

### Doodle Bob

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.