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
\frac{\partial s}{\partial \phi}= sin(\theta)[/itex]<br /> <br /> That is, <br /> V_\phi= \frac{\partial V}{\partial \phi}= cos(\theta)V_\sigma+ sin(\theta)V_s[/itex]&lt;br /&gt; &lt;br /&gt; To find V_{\phi\phi}, for example, do that again:&lt;br /&gt; V_{\phi\phi}= \frac{\partial}{\partial \phi}\left(cos(\theta)V_\sigma+ sin(\theta)V_s\right)&lt;br /&gt; = 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

 

[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.