MHB General orthogonal scale factor identity

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The discussion focuses on deriving expressions related to the orthogonal scale factor identity in generalized curvilinear coordinates. The initial attempt establishes that the scale factor \( h_1 \) can be expressed as the square root of the sum of the squares of the partial derivatives of the coordinates. The second part explores the relationship between the derivatives of the unit vectors and the scale factors, confirming that the calculations are valid. The third part seeks guidance on deriving a specific expression involving the derivative of the unit vector \( \hat{q}_1 \) with respect to \( q_1 \), indicating a struggle to find the correct approach. Overall, the discussion emphasizes the importance of confirming the validity of the derived expressions and seeking hints for further derivations.
ognik
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Please be patient as I struggle with latex here ...
Part 1 of the problem says to start with:
$ \frac{\partial\bar{r}}{\partial{q}_{1}} ={h}_{1} \hat{q}_{1} $ and then to find an expression for $ {h}_{1} $ that agrees with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $

My attempt is:
$ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$
but $ \hat{q}_{1}.\hat{q}_{1}=1$
Then $ \left({h}_{1}\right)^{\!{2}} =\left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}$
Now $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)+\left(\frac{\partial{z}}{\partial{q}_{1}}\right) $
so $ \left(\frac{\partial\bar{r}}{\partial{q}_{1}}\right)^{\!{2}}= \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}} $
Now $ {g}_{ii} $ is defined = $ {h}_{i}^{\!{2}}$ and comparing with $ {g}_{ij}=\sum_{l} \frac{\partial{x}_{l}}{\partial{q}_{i}}\frac{\partial{x}_{l}}{\partial{q}_{j}} $ above...
Therefore $ {h}_{1}= \sqrt{ \left(\frac{\partial{x}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{y}}{\partial{q}_{1}}\right)^{\!{2}}+\left(\frac{\partial{z}}{\partial{q}_{1}}\right)^{\!{2}}} $ QED - but have I done anything illegal here?
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Part 2 is a derivation:
Again starting with $ \hat{q}_{1} = \frac{1}{h}_{1} \frac{\partial\bar{r}}{\partial{q}_{1}}$
Then $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}}\frac{\partial{}^{2}\bar{r}}{\partial{q}_{1}\partial{q}_{2}} = \frac{1}{{h}_{1}}\frac{\partial}{\partial{q}_{1}}\left(\frac{\partial\bar{r}}{\partial{q}_{2}}\right)^{\!{}} $ (${h}_{1}$ constant w.r.t. 2)
but $ \frac{\partial\bar{r}}{\partial{q}_{2}}={h}_{2} \hat{q}_{2} $
so $ \frac{\partial\hat{q}_{1}}{\partial{q}_{2}}=\frac{1}{{h}_{1}} \frac{\partial\left({h}_{2}\hat{q}_{2}\right)}{\partial{q}_{1}}= \hat{q}_{2}\frac{1}{{h}_{1}}\frac{\partial{h}_{2}}{\partial{q}_{1}}$
Which is QED - but again I have this uncertain feeling so would appreciate confirmation there I have done nothing untoward
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Part 3 has me so far, I would appreciate a hint ...
Derive $ \frac{\partial\hat{q}_{1}}{\partial{q}_{1}}= -\sum_{j\ne{1}}^{} \hat{q}_{2}\frac{1}{{h}_{2}}\frac{\partial{h}_{1}}{\partial{q}_{2}}$
I have tried a few things without success ... probably there is a trick I haven't encountered before?
 
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In addition: I have assumed perhaps incautiously, that people would recognize the subject here - it is around generalised curvilinear coordinates, with the expression for $ {g}_{ij}$ representing the mixed coordinates, but ${h}_{i}$ excludes the mixed coordinates, IE it is for an orthogonal basis. I feel fairly confident of my working for the first 2 parts, but would like confirmation that I haven't done anything that isn't justifiable. For the 3rd part, I have spent ages trying different things - I thought I was onto it when I differentiated the starting eqn w.r.t. BOTH i and j, but still could not get a term in $ \frac{\partial{\bar{q}}_{i}}{\partial{q}_{i}} $. So I'd also really appreciate a tip as to what approach might product that term, thanks.
 
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