Relating 2nd order partial derivatives in a coordinate transformation.

[phsopher]

Homework Statement

Could some mathematically minded person please check my calculation as I am a bit suspicious of it (I'm a physicist myself). This isn't homework so feel free to reveal anything you have in mind.

Suppose I have two functions $$\phi(t)$$ and $$\chi(t)$$ and the potential V which is a function of these two. Suppose I introduce new variables $$\sigma$$ and $$s$$ such that

$$d\sigma = \cos\theta d\phi + \sin\theta e^{b(\phi)} d\chi$$ (1)
$$ds = \cos\theta e^{b(\phi)} d\chi - \sin\theta d\phi$$ (2)

where

$$\cos\theta = \frac{\dot{\phi}}{\sqrt{\dot{\phi}^2 + e^{2b(\phi)}\dot{\chi}^2}}$$ (3)
$$\sin\theta = \frac{e^{b(\phi)}\dot{\chi}}{\sqrt{\dot{\phi}^2 + e^{2b(\phi)}\dot{\chi}^2}}$$ (4)

where the overdot represents the derivative wrt t.

Denote partial derivatives as follows: $$A_x \equiv \frac{\partial A}{\partial x}$$.

I need to find second partial derivatives of V wrt to new variables in terms of second partial derivatives wrt old variables (i.e. $$V_{\sigma\sigma}, V_{\sigma s}, V_{ss}$$ in terms of $$V_{\phi\phi},V_{\phi\chi}$$ and $$V_{\chi\chi}$$).

Homework Equations

The Attempt at a Solution

The way I went about this is as follows (using as an example $$V_{\sigma\sigma}$$):

$$V = V_{\phi}d\phi + V_{\chi}d\chi$$ solving $$d\phi$$ and $$d\chi$$ from (1) and (2)
$$\Rightarrow V = V_{\phi}(\cos\theta d\sigma - \sin\theta ds) + V_{\chi}e^{-b(\phi)}(\sin\theta d\sigma + \cos\theta ds)$$
$$\Rightarrow V = (V_{\phi}\cos\theta + V_{\chi}e^{-b(\phi)}\sin\theta) d\sigma + (-V_{\phi}\sin\theta + V_{\chi}e^{-b(\phi)}\cos\theta) ds$$
$$\Rightarrow V_{\sigma} = V_{\phi}\cos\theta + V_{\chi}e^{-b(\phi)}\sin\theta$$ (5)

This seems right so far. Now taking $$V_{\sigma}$$ as the new function and repeating the exact same steps I get (just replacing $$V$$ with $$V_{\sigma}$$ in the above result):

$$V_{\sigma\sigma} = V_{\sigma\phi}\cos\theta + V_{\sigma\chi}e^{-b(\phi)}\sin\theta$$ and taking the apropriate derivatives of (5)
$$\Rightarrow V_{\sigma\sigma} = V_{\phi\phi}\cos^2\theta + 2 V_{\phi\chi}e^{-b(\phi)}\sin\theta\cos\theta + V_{\chi\chi}e^{-2b(\phi)}\sin^2\theta - b_{\phi}V_{\chi}e^{-b(\phi)}\sin\theta\cos\theta$$

What makes me suspicious is the last term which arises because of the $$\phi$$ dependence of b. For example it makes mixed derivatives non-symmetric i.e. $$V_{\sigma s} \neq V_{s \sigma}$$. Could that be right? I'm not 100 % sure of my method of arriving at the result so it would be great if someone with a firmer understanding of the mathematics involved could check this. Thanks.

 

[HallsofIvy]

Use the chain rule.

If $dV= V_{\phi}d\phi+ V_\chi d\chi$ and you are replacing $d\phi$ by $d\phi= cos(\theta)d\sigma- sin(\theta)ds$ and $d\chi= sin(\theta)d\sigma+ cos(\theta)ds$ (essentially, just a rotation), then
$$V_\phi= \frac{\partial V}{\partial \phi}= \frac{\partial V}{\partial \sigma}\frac{\partial \sigma}{\partial \phi}+ \frac{\partial V}{\partial s}\frac{\partial s}{\partial \phi}$$

Since $d\sigma= cos(\theta)d\phi- sin(\theta)d\chi$
$$\frac{\partial \sigma}{\partial \phi}= cos(\theta)$$
and since $ds= sin(\theta)d\phi+ cos(\theta)d\chi$
[tex]\frac{\partial s}{\partial \phi}= sin(\theta)[/itex]<br /> <br /> That is, <br /> [tex]V_\phi= \frac{\partial V}{\partial \phi}= cos(\theta)V_\sigma+ sin(\theta)V_s[/itex]<br /> <br /> To find $V_{\phi\phi}$, for example, do that again:<br /> $$V_{\phi\phi}= \frac{\partial}{\partial \phi}\left(cos(\theta)V_\sigma+ sin(\theta)V_s\right)$$<br /> $$= cos(\theta)\left(cos(\theta)V_\sigma+ sin(\theta)V_s\right)_\sigma+ sin(\theta)\left(cos(\theta)V_\sigma+ sin(\theta)V_s\right)_s$$[/tex][/tex]

 

[phsopher]

That's pretty much what I did as I explained in the previous post. Except in the other direction since I need to find $V_{\sigma\sigma},V_{\sigma s}$ and $V_{\sigma s}$ and not $V_{\phi\phi}, V_{\phi\chi}$ and $V_{\chi\chi}$. However, you left out of your post the $e^{b(\phi)}$ terms which is where my difficulty lies because they introduce additional terms (like the term $- b_{\phi}V_{\chi}e^{-b(\phi)}\sin\theta\cos\theta$ in $V_{\sigma\sigma}$) which according to my calculations make mixed derivatives non-symmetric.