Curvilinear coordinates question

1. Dec 28, 2009

mnb96

Hello,
given a system of curvilinear coordinates $x_i=x_i(u_1,\ldots,u_n)$; $u_i=u_i(x_1,\ldots,x_n)$ and considering the position vector $\mathbf{r}=x_1\mathbf{e}_1+\ldots+x_n\mathbf{e}_n$ there is the well-known identity that defines the reciprocal frame:

$$\frac{\partial \mathbf{r}}{\partial u_i }\cdot \nabla u_j = \delta^i_j$$

I tried to verify it by myself but I cannot see where is the mistake:

$$\frac{\partial \mathbf{r}}{\partial u_i }\cdot \nabla u_j=$$

$$=(\frac{\partial x_1}{\partial u_i }\mathbf{e}_1+ \ldots + \frac{\partial x_n}{\partial u_i }\mathbf{e}_n )\cdot (\frac{\partial u_j}{\partial x_1 }\mathbf{e}_1+ \ldots + \frac{\partial u_j}{\partial x_n }\mathbf{e}_n ) =$$

$$=n\frac{\partial u_j}{\partial u_i} =$$

$$=n\delta^i_j$$

Why am I getting that wrong multiplication by n ?

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(Funny) EDIT: since there is a well-known proof in every book which correctly shows that the aforementioned inner-product equals $$\delta^i_j$$, if no-one manages to find my mistake, our world will have an "amazing" proof that 1=2=...=n for any integer n :)

Last edited: Dec 29, 2009
2. Dec 29, 2009

Astronuc

Staff Emeritus
Why? $$n\frac{\partial u_j}{\partial u_i}$$

where does one get n?

What's with the c in $$\frac{\partial \mathbf{r}}{\partial u_i }\cdot \nabla u_jc=$$ ?

3. Dec 29, 2009

mnb96

ops...ignore that 'c': it was a typo and I removed it.
Back to the main question, let's assume i=j, we have:

$$(\frac{\partial x_1}{\partial u_i }\mathbf{e}_1+ \ldots + \frac{\partial x_n}{\partial u_i }\mathbf{e}_n )\cdot (\frac{\partial u_j}{\partial x_1 }\mathbf{e}_1+ \ldots + \frac{\partial u_j}{\partial x_n }\mathbf{e}_n ) =$$

$$=\frac{\partial x_1}{\partial u_i }\frac{\partial u_i}{\partial x_1} + \frac{\partial x_2}{\partial u_i }\frac{\partial u_i}{\partial x_2 }+\ldots+\frac{\partial x_n}{\partial u_i} \frac{\partial u_i}{\partial x_n }=$$

$$=\frac{\partial x_1}{\partial x_1}+\ldots+\frac{\partial x_n}{\partial x_n}=$$

$$=1+1+1+\ldots+1=$$

$$=n$$

I know for sure there is a very trivial mistake I should be ashamed of, but I would like to realize where and why it is.

4. Dec 29, 2009

JSuarez

You have a problem when you apply the chain rule. When you expand:

$$(\frac{\partial x_1}{\partial u_i }\mathbf{e}_1+ \ldots + \frac{\partial x_n}{\partial u_i }\mathbf{e}_n )\cdot (\frac{\partial u_j}{\partial x_1 }\mathbf{e}_1+ \ldots + \frac{\partial u_j}{\partial x_n }\mathbf{e}_n )$$

You do get:

$$\frac{\partial x_1}{\partial u_i }\frac{\partial u_i}{\partial x_1} + \frac{\partial x_2}{\partial u_i }\frac{\partial u_i}{\partial x_2 }+\ldots+\frac{\partial x_n}{\partial u_i} \frac{\partial u_i}{\partial x_n }$$

But:

$$\frac{\partial x_j}{\partial u_i}\frac{\partial u_i}{\partial x_j}\neq\frac{\partial x_j}{\partial x_j}$$

Because you have:

$$u_i\left(x_1\left(u_1,...,u_n\right),...,x_n\left(u_1,...,u_n\right)\right)=u_i$$

Then:

$$\frac{\partial u_i}{\partial u_j} = \sum^{n}_{k=1}\frac{\partial u_i}{\partial x_k}\frac{\partial x_k}{\partial u_j} = \delta^{i}_{j}$$

But that sum, when $$i=j$$, is exactly what you obtain when you expand the inner product, and there's no $$n$$. Sometimes, the Chain rule is tricky.

Last edited: Dec 29, 2009
5. Dec 30, 2009

mnb96

Thanks a lot JSuarez!
actually once you realize that the situation we had was like [tex]u_i\left(x_1\left(u_1,...,u_n\right),...,x_n\left( u_1,...,u_n\right)\right)=u_i[/itex] everything makes perfectly sense.

In this case the lesson to learn for the novice (me) is probably that when making calculations with partial derivatives, one should always have very clear in mind the actual function he/she is manipulating. Naively relying on notation apparently leads to wrong conclusions.