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Relating 2nd order partial derivatives in a coordinate transformation.

  1. Jun 30, 2010 #1
    1. The problem statement, all variables and given/known data

    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 [tex]\phi(t)[/tex] and [tex]\chi(t)[/tex] and the potential V which is a function of these two. Suppose I introduce new variables [tex]\sigma[/tex] and [tex]s[/tex] such that

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


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

    where the overdot represents the derivative wrt t.

    Denote partial derivatives as follows: [tex]A_x \equiv \frac{\partial A}{\partial x}[/tex].

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

    2. Relevant equations
    3. The attempt at a solution

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

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

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

    [tex]V_{\sigma\sigma} = V_{\sigma\phi}\cos\theta + V_{\sigma\chi}e^{-b(\phi)}\sin\theta[/tex] and taking the apropriate derivatives of (5)
    [tex]\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[/tex]

    What makes me suspicious is the last term which arises because of the [tex]\phi[/tex] dependence of b. For example it makes mixed derivatives non-symmetric i.e. [tex]V_{\sigma s} \neq V_{s \sigma}[/tex]. 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.
  2. jcsd
  3. Jun 30, 2010 #2


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    Science Advisor

    Use the chain rule.

    If [itex]dV= V_{\phi}d\phi+ V_\chi d\chi[/itex] and you are replacing [itex]d\phi[/itex] by [itex]d\phi= cos(\theta)d\sigma- sin(\theta)ds[/itex] and [itex]d\chi= sin(\theta)d\sigma+ cos(\theta)ds[/itex] (essentially, just a rotation), then
    [tex]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}[/tex]

    Since [itex]d\sigma= cos(\theta)d\phi- sin(\theta)d\chi[/itex]
    [tex]\frac{\partial \sigma}{\partial \phi}= cos(\theta)[/tex]
    and since [itex]ds= sin(\theta)d\phi+ cos(\theta)d\chi[/itex]
    [tex]\frac{\partial s}{\partial \phi}= sin(\theta)[/itex]

    That is,
    [tex]V_\phi= \frac{\partial V}{\partial \phi}= cos(\theta)V_\sigma+ sin(\theta)V_s[/itex]

    To find [itex]V_{\phi\phi}[/itex], for example, do that again:
    [tex]V_{\phi\phi}= \frac{\partial}{\partial \phi}\left(cos(\theta)V_\sigma+ sin(\theta)V_s\right)[/tex]
    [tex]= 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]
  4. Jun 30, 2010 #3
    That's pretty much what I did as I explained in the previous post. Except in the other direction since I need to find [itex]V_{\sigma\sigma},V_{\sigma s}[/itex] and [itex]V_{\sigma s}[/itex] and not [itex]V_{\phi\phi}, V_{\phi\chi}[/itex] and [itex]V_{\chi\chi}[/itex]. However, you left out of your post the [itex]e^{b(\phi)}[/itex] terms which is where my difficulty lies because they introduce additional terms (like the term [itex]- b_{\phi}V_{\chi}e^{-b(\phi)}\sin\theta\cos\theta[/itex] in [itex]V_{\sigma\sigma}[/itex]) which according to my calculations make mixed derivatives non-symmetric.
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