# Orthogonal coordinate systems - scale factors

## Homework Statement

Start from the 'relevant equation' below and derive
$$(1) \frac{\partial{\bf{\hat{q}}_{i}}}{{\partial{q}}_{j}}={\hat{q}}_{i}\frac{1}{{h}_{i}} \frac{\partial{h}_{i}}{{\partial{q}}_{j}}, {i}\ne{j}$$
$$(2) \frac{\partial{\bf{\hat{q}}_{i}}}{{\partial{q}}_{i}}= -\sum {\hat{q}}_{j}\frac{1}{{h}_{j}} \frac{\partial{h}_{i}}{{\partial{q}}_{i}}, {j}\ne{i}$$

## Homework Equations

$$\frac{\partial{\bf{r}}}{{\partial{q}}_{i}}={h}_{i}{\hat{q}}_{i}$$

## The Attempt at a Solution

I have finished the first part, using $$\frac{\partial{\bf{\hat{r}}}}{{\partial{q}}_{j}}={h}_{j}\hat{q}_{j}$$
But I am not getting anywhere with part 2), especially with the summation ... a hint or 2 would be much appreciated please

Related Calculus and Beyond Homework Help News on Phys.org
I will put down my attempt for deriving the (2) part, although not very useful I suspect ...
I used the chain rule to get $$\frac{\partial{\hat{q}_{i}}}{\partial{q}_{i}} = \frac{\partial\hat{{q}_{i}}}{\partial{q}_{j}}.\frac{\partial{{q}_{j}}}{\partial{q}_{i}} = \hat{q}_{j}\frac{1}{{h}_{i}}.\frac{\partial{{h}_{j}}}{\partial{q}_{i}}.\frac{\partial{{q}_{j}}}{\partial{q}_{i}}$$
So I've got the LHS and some part of the RHS - but without any way I can to end up with a -∑ term

Mark44
Mentor
Any problem involving partial derivatives is definitely not precalculus. I have moved this thread to the calculus section.

Sorry for the hassle, of course you're right - I must have been too focused on the problem (been struggling with it for weeks and going round in the same circles ..... I'm studying by correspondence but in practice forums are the only useful support I have)

$$I\: had\: one\: of\: those\: random\: thoughts\:while\: abed\:,\: to\:use\:\hat{q}_{i}.\hat{q}_{j}=0$$
$$Differentiating\: w.r.t.\: {q}_{i} \:gives:$$
$$\hat{q}_{j}.\frac{\partial{{q}_{i}}}{\partial{q}_{i}} =-\hat{q}_{i}.\frac{\partial{\hat{q}_{j}}}{\partial{q}_{i}}$$
$$From\: the\: result\: of\: part(1)\: above\: I\: can\: use\: \frac{\partial{\hat{q}_{j}}}{\partial{q}_{i}}=\hat{q}_{i}. \frac{1}{{h}_{j}}\frac{\partial{{h}_{i}}}{\partial{q}_{j}}$$
$$Therefore\: \hat{q}_{j}\frac{\partial{\hat{q}_{i}}}{\partial{q}_{i}}=-\hat{q}_{i}.\hat{q}_{i} \frac{1}{{h}_{j}}\frac{\partial{{h}_{i}}}{\partial{q}_{j}}\:and\: \hat{q}_{i}.\hat{q}_{i}=1$$
This is closer than my other efforts, but it still leaves me with a ∑ missing from the RHS and a unit vector on the wrong side ....then my brain threatened to explode, so hopefully someone will see this and help me complete it.

Fredrik
Staff Emeritus
Gold Member
I had another look at it today. (The first look was after oqnik linked to this thread from another). I still haven't been able to obtain those results for ##\frac{\partial\hat{\mathbf q}_i}{\partial q_j}##, but I think I got a bit closer. If I can prove that ##\frac{\partial^2\mathbf r}{\partial q_i\partial q_j}## is a vector in the plane spanned by ##\{\mathbf q_i,\mathbf q_j\}##, then I will have proved the ##i\neq j## result at least.

You asked specifically about the summation in the result for ##\frac{\partial\hat{\mathbf q}_i}{\partial q_j}##. That's the only part that I understand completely. For all vectors ##\mathbf v##, we have
$$\mathbf v=\sum_{j=1}^3 (\mathbf v\cdot\hat{\mathbf q}_j)\hat{\mathbf q}_j.$$ When we consider the special case ##\mathbf v=\frac{\partial\hat{\mathbf q}_i}{\partial q_i}##, the ##j=i## term is going to be zero. (I was able to prove that). So we get
$$\frac{\partial\hat{\mathbf q}_i}{\partial q_i} =\sum_{j\neq i} \left(\frac{\partial\hat{\mathbf q}_i}{\partial q_i}\cdot\hat{\mathbf q}_j\right)\hat{\mathbf q}_j.$$
You said that you were able to obtain the result for ##\frac{\partial\hat{\mathbf q}_i}{\partial q_j}## when ##i\neq j##. How did you do that? I'm not quite there yet. I feel like I must be missing something that makes the problem easier.

Hi Frederik - I didn't trust my latex with rewriting it, so I scanned the page I had done on deriving (1) in my initial post - so my solution for (1) should be viewable at https://drive.google.com/file/d/0B_fp1L8s2RmZZTMyQ25IQTI4Um8/view?usp=sharing

Your equation was all I needed (I think), but I would be grateful if you'd also check my solution to part (2) above - posted at https://drive.google.com/file/d/0B_fp1L8s2RmZTkFTdEpjWVJkcnc/view?usp=sharing

I have to admit I don't fully understand what most of the expressions mean geometrically, I was just doing the arithmetic .... I believe the h's represent the coefficiants of a generalized curvilinear orthogonal transformation, making the hat{q sub i}s the basis, so if you could also breakdown the 2 equations I 'proved' into something I can understand (maybe a sketch?), that would be very helpful as well :-) Thanks again, Alan

Last edited:
Fredrik
Staff Emeritus
Gold Member
In the first attachment, you say without explanation that ##\frac{\partial\hat{\mathbf q}_i}{\partial q_j}## is in the ##\hat{\mathbf q}_j## direction. Have you been able to prove that?

I still haven't worked out the first part completely, but I will show you what I have. I will save myself some typing by defining
$$\mathbf X_{ij}=\frac{\partial^2\mathbf r}{\partial q_i\partial q_j}.$$ My understanding of ##h_i## and ##\hat{\mathbf q}_i## is that the latter is defined as the unit vector in the direction of ##\frac{\partial\mathbf r}{\partial q_i}##, and the former is defined by ##\frac{\partial\mathbf r}{\partial q_i}=h_i\hat{\mathbf q}_i##. This implies that ##h_i=\big|\frac{\partial\mathbf r}{\partial q_i}\big| =\sqrt{\big(\frac{\partial\mathbf r}{\partial q_i}\big)^2}##. It's not hard to show that this implies that
$$\frac{\partial h_i}{\partial q_j}=\hat{\mathbf q}_i\cdot\mathbf X_{ij}.$$ I will use that result in the calculation below. Now let ##i,j,k## have three different values in the set ##\{1,2,3\}##. We have
\begin{align}
&\frac{\partial\hat{\mathbf q}_i}{\partial q_j} =\frac{\partial}{\partial q_j}\left(\frac{1}{h_i}\frac{\partial\mathbf r}{\partial q_i}\right) =-\frac{1}{h_i^2}\frac{\partial h_i}{\partial q_j}\frac{\partial\mathbf r}{\partial q_i}+\frac{1}{h_i}\mathbf X_{ij} =\frac{1}{h_i}\left( -\big(\hat{\mathbf q}_i\cdot\mathbf X_{ij}\big)\hat{\mathbf q}_i+\mathbf X_{ij}\right)\\
&=\frac{1}{h_i} \big(\big(\hat{\mathbf q}_j\cdot\mathbf X_{ij}\big)\hat{\mathbf q}_j +\big(\hat{\mathbf q}_k\cdot\mathbf X_{ij}\big)\hat{\mathbf q}_k\big) = \frac{1}{h_i} \left(\frac{\partial h_j}{\partial q_i}\hat{\mathbf q_j}+\big(\underbrace{\hat{\mathbf q}_k\cdot\mathbf X_{ij}}_{=0?}\big)\hat{\mathbf q}_k\right)
\end{align}

I haven't examined your second attachment yet.

A) Yes, briefly - $$\hat{q}_{i}.\hat{q}_{i}=1,\: differentiating\:w.r.t.\:{q}_{i}\: gives\: \hat{q}_{i} \frac{\partial \hat{q}_{i}}{\partial {q}_{i}} =0\: \: \therefore \:\frac{\partial\hat{q}_{i}}{\partial {q}_{i}}\:must\:be\:in\:same\:direction\:as\:\hat{q}_{i}$$
B) You don't say, but is there a flaw in my attempt at (1)?
C) Maybe this helps :
Going back to basics, qi are generalized curvilinear (and orthogonal) coordinates, so that (q1, q2, q3) could, for example, be (r, θ, ϕ).
What is hi? It is a coefficient (with units of length), such that dsi = hi dqi - keeping with the above spherical example:
$$h_{r}=\frac{\partial \vec{r}}{\partial r}=(sin{\theta} cos{\phi}, sin{\theta}sin{\phi},cos{\theta}$$

Fredrik
Staff Emeritus
Gold Member
A) Yes, briefly - $$\hat{q}_{i}.\hat{q}_{i}=1,\: differentiating\:w.r.t.\:{q}_{i}\: gives\: \hat{q}_{i} \frac{\partial \hat{q}_{i}}{\partial {q}_{i}} =0\: \: \therefore \:\frac{\partial\hat{q}_{i}}{\partial {q}_{i}}\:must\:be\:in\:same\:direction\:as\:\hat{q}_{i}$$
If the dot product of two non-zero vectors is zero, then they're not in the same direction. Instead, one of them is in the plane that's orthogonal to the other.

Edit: I see now that the calculation you did there is an essential part of the proof of the result for ##\frac{\partial\hat{\mathbf q}_i}{\partial q_i}##. It explains why we can drop the ##j=i## term from the sum right away, and then use that ##j\neq i## in the remaining terms.

B) You don't say, but is there a flaw in my attempt at (1)?
I don't see anything that's wrong, but the proof isn't complete until you have proved that ##\frac{\partial\hat{\mathbf q}_i}{\partial q_j}## is in the direction of ##\hat{\mathbf q}_j##.

My attempt is no better, since I wasn't able to prove that ##\hat{\mathbf q}_k\cdot\mathbf X_{ij}=0##.

C) Maybe this helps :
Going back to basics, qi are generalized curvilinear (and orthogonal) coordinates, so that (q1, q2, q3) could, for example, be (r, θ, ϕ).
What is hi? It is a coefficient (with units of length), such that dsi = hi dqi - keeping with the above spherical example:
$$h_{r}=\frac{\partial \vec{r}}{\partial r}=(sin{\theta} cos{\phi}, sin{\theta}sin{\phi},cos{\theta}$$
##h_r## is the scalar you get when you take the dot product of that vector with itself, and then take the square root of the result.

I like this way of looking at it: We have ##dr_i=\sum_j \frac{\partial r_i}{\partial q_j}dq_j## and therefore
$$ds^2=\sum_i dr_i^2 =\sum_i\sum_j\sum_k \frac{\partial r_i}{\partial q_j}\frac{\partial r_i}{\partial q_k}dq_j dq_k =\sum_j\sum_k \frac{\partial\mathbf r}{\partial q_j}\cdot\frac{\partial\mathbf r}{\partial q_k}dq_j dq_k.$$ When we're dealing with an orthogonal coordinate system, that dot product is zero when ##j\neq k##, so we can simplify the above to
$$\sum_j\left(\frac{\partial\mathbf r}{\partial q_j}\right)^2 dq_j^2.$$ The ##h_i## are supposed to satisfy
$$ds^2=\sum_j h_j^2 dq_j^2,$$ so we have ##h_i^2=\big(\frac{\partial\mathbf r}{\partial q_j}\big)^2##, where the "square" on the right-hand side is a dot product.

Last edited:
Fredrik
Staff Emeritus