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Thread: Exact differential

  1. #1
    Member Ruun's Avatar
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    Exact differential

    Hi. Given equations of constraint for the motion of a particle in the form:

    $\displaystyle \sum_{i=1}^{n} g_{i}(x_{1},...,x_{n} ) dx_{i} = 0$

    will be holonomic only if exists a function $\displaystyle f(x_{1},..,x_{n})$ such that:

    $\displaystyle \frac{\partial fg_{i}}{\partial x_{j}} = \frac{\partial fg_{j}}{\partial x_{i}}$ for all $\displaystyle i \neq j$.

    In the particular case of:

    $\displaystyle dx - a sin(\theta) d\phi =0$
    $\displaystyle dy + a cos (\theta) d\phi = 0$

    I should find an integrating factor of the form $\displaystyle f(x,y,\theta,\phi)$, but I don't know wich are exactly the $\displaystyle g_{i}$ functions. There are four $\displaystyle g_{i}$.

    Thanks for your time.
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  2. #2
    MHF Contributor

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    Well, your problem is that those two equations are NOT of the form $\displaystyle \sum_{i=1}^{n} g_{i}(x_{1},...,x_{n} ) dx_{i} = 0$ and can be put in that form in a number of different ways. For example, just adding them gives $\displaystyle dx+ dy+ a(cos(\theta)- sin(\theta))d\phi= 0$. In that case $\displaystyle g_x= 1$, $\displaystyle g_y= 1$, $\displaystyle g_\theta= 0$, and $\displaystyle g_\phi= a(cos(\theta)- sin(\theta))$.

    If you were to subtract the two equations rather than add, you would get $\displaystyle dx- dy- a(sin(\theta)+ cos(\theta))d\phi$ and that would a different f.

    Personally, what I would do is integrate $\displaystyle dx= a sin(\theta)d\phi$ and $\displaystyle dy= -acos(\theta)d\phi$, treating $\displaystyle \theta$ as a constant. That will give x and y as functions of $\displaystyle \theta$ and $\displaystyle \phi$.
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