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I'm trying to convert a parametrization of a curve $$\vec \alpha(x,y,z) = \alpha_1(x,y,z) \hat x + \alpha_2(x,y,z) \hat y + \alpha_3(x,y,z) \hat z$$ to $$\vec \alpha(r, \theta, \phi) = \alpha_r(r, \theta, \phi) \hat r + \alpha_\theta (r, \theta, \phi) \hat \theta + \alpha_\phi (r, \theta, \phi) \hat \phi $$.

To do that, let $$x = x(r,\theta, \phi) \\ y = y(r, \theta, \phi) \\ z = z(r, \theta, \phi)$$ be given, and define $$\hat u = \left(\frac{\partial x}{\partial r} \hat x + \frac{\partial y}{\partial r} \hat y + \frac{\partial z}{\partial r} \hat z \right) / a \\ \hat v = \left(\frac{\partial x}{\partial \theta} \hat x + \frac{\partial y}{\partial \theta} \hat y + \frac{\partial z}{\partial \theta} \hat z \right) / b \\ \hat w = \left(\frac{\partial x}{\partial \phi} \hat x + \frac{\partial y}{\partial \phi} \hat y + \frac{\partial z}{\partial \phi} \hat z \right) / c,$$ where $a,b,c$ are the norm of the given expression so that $\vec u, \vec v, \vec w$ are unit vectors.

Now, from this point, how can I write $\alpha$ in terms of $\hat u, \hat v, \hat w$ ?

Note that, as far as I figured, $\hat u$ corresponds to $\hat r$ in a sense, and similarly for the others.

Edit:

I'm particularly interested with the general derivation, so please don't throw the result at me only.

Edit 2:

$u, v, w$ are mutually orthogonal, and $r, \theta, \phi$ are spherical coordinates.

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  • $\begingroup$ This question is a bit hard to follow. I think that $\alpha_1$, $\alpha_2$, and $\alpha_3$ in the second equation are not identical to the expressions in the first. In that case maybe they should be called $\alpha_r$, $\alpha_{\theta}$, and $\alpha_{\phi}$. Given that it's a curve, I suppose all variables are functions of a single parameter $t$: $x=x(t)$, $y=y(t)$, etc. Are $r$, $\theta$, and $\phi$ the standard spherical coordinates or more general? If the latter and $\hat u\cdot\hat v\ne0$ you can see how a similar situation is handled via the reciprocal lattice. $\endgroup$ – user5713492 Mar 6 '18 at 8:59
  • $\begingroup$ @user5713492 since we know what $x,y,z$ are in terms of $r, \theta, \phi$, we can directly substitude into the equation of $\alpha_i$, so and it should be the same map, because we only changed the variables, but just representation is changed, so I did not think that I need to use another symbol for them. $\endgroup$ – onurcanbektas Mar 6 '18 at 9:21
  • $\begingroup$ @user5713492 Moreover, see my edit, please. $\endgroup$ – onurcanbektas Mar 6 '18 at 9:23
  • $\begingroup$ If $\hat r$ isn't simply $\hat u$, then how is $\hat r$ defined? If you could explain that then we could take $\hat r\cdot\vec\alpha$ and find its component in that direction which you call $\alpha_1$ and I would prefer to call $\alpha_r$. $\endgroup$ – user5713492 Mar 6 '18 at 9:29
  • $\begingroup$ @user5713492 You are right, see my edit please. $\endgroup$ – onurcanbektas Mar 6 '18 at 9:45

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