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Stupid chain rule

  1. Nov 21, 2008 #1

    malawi_glenn

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    This is stuff I do in order to understand analytical mechanics better, I encounter the followin thing:

    [tex]\frac{\partial L}{\partial \dot{\phi}} = \text{?}[/tex]

    Where [tex]\dot{\phi} = \frac{\partial \phi}{\partial q} \frac{dq}{dt} = \frac{\partial \phi}{\partial q} \dot{q}[/tex]

    I should know this! It is embarrasing :-(

    p.s take notice of the "dots", what's wrong with the latex generator?
     
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  3. Nov 21, 2008 #2

    gabbagabbahey

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    I'm not sure why you would want to express [tex]\frac{\partial L}{\partial \dot{\phi}}[/tex] in terms of [tex]\dot{q}[/tex] and [tex]\frac{\partial \phi}{\partial q}[/tex] , but if you really want to; I suppose you could use:

    [tex]\frac{\partial L}{\partial \dot{\phi}}=\frac{1}{\frac{\partial \dot{\phi}}{\partial L}}[/tex]

    ...and just take the partial derivative of your expression for [tex]\dot{\phi}[/tex] with respect to your Lagrangian and invert it (easier said than done).

    ...out of curiosity, why do you want to do this?
     
  4. Nov 21, 2008 #3

    malawi_glenn

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    Ok will try later.

    Iam trying to derive an expression for conserved quantities under a generelized coordinate transformation.

    [tex]q \rightarrow \phi(q,\epsilon )[/tex]

    epsilon is a differential quantity.

    Minimizing the action w.r.t to epsilon:

    [tex]\frac{d}{d\epsilon}\int L(\phi(q,\epsilon), \dot{\phi}(q,\epsilon), t) dt= 0[/tex]

    As you will see, one will get things like:
    [tex] \frac{\partial L}{\partial \phi} \qquad \frac{\partial L}{\partial \dot{\phi}} [/tex]

    Which I don't know how to handle..

    according to some of my sources, this is the result:
    [tex]\int ( \frac{\partial L}{\partial q} + \frac{\partial L}{\partial \dot{q}} \frac{\partial \phi}{\partial \epsilon} )dt= 0[/tex]

    But I only come to:
    [tex]\int ( \frac{\partial L}{\partial \phi} + \frac{\partial L}{\partial \dot{\phi}} \frac{\partial \phi}{\partial \epsilon} )dt= 0[/tex]

    So I conclude that
    [tex] \frac{\partial L}{\partial\phi} = \frac{\partial L}{\partial q} \qquad \frac{\partial L}{\partial\dot{\phi}} = \frac{\partial L}{\partial \dot{q}} [/tex]

    But HOW?

    Should I perhaps Taylor-expand the Lagrangian and only keeping 1st order terms?
     
  5. Nov 21, 2008 #4

    gabbagabbahey

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    I don't get what you are getting; if you post the first few lines of your derivation, I should be able to point out where you are going wrong.
     
  6. Nov 21, 2008 #5

    malawi_glenn

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    sure mate, I think I did some stupid latex errors in my latest post, here it is in more detail:

    [tex]\frac{d}{d\epsilon}\int L(\phi(q,\epsilon), \dot{\phi}(q,\epsilon), t) dt= 0[/tex]

    [tex]\frac{dL}{d\epsilon} = \frac{\partial L}{\partial \phi}\frac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{\phi}}\frac{\partial \dot{\phi}}{\partial \epsilon} [/tex]

    Now:
    [tex]\frac{\partial \dot{\phi}}{\partial \epsilon} = \frac{\partial ^2 \phi}{\partial q \partial \epsilon}\dot{q}[/tex]
    since:
    [tex]\dot{\phi} = \frac{d\phi}{dt}= \frac{d\phi}{dq}\frac{dq}{dt} = \frac{d\phi}{dq}\dot{q}[/tex]

    so:
    [tex]\frac{dL}{d\epsilon} = \frac{\partial L}{\partial \phi}\frac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{\phi}}\frac{\partial ^2\phi}{\partial q \partial \epsilon} [/tex]

    But the thing with:
    [tex]\frac{\partial L}{\partial \phi} \qquad \frac{\partial L}{\partial \dot{\phi}}[/tex]

    How to tackle those?

    PLEASE note that the "dot" looks very very tiny here.. don't know why the latex generator is like this..

    I have found a wiki article on this which is quite similar to the procedure which I would like to learn. But they neither motivate why

    [tex] \frac{\partial L}{\partial\phi} \rightarrow \frac{\partial L}{\partial q} \qquad \frac{\partial L}{\partial\dot{\phi}} \rightarrow\frac{\partial L}{\partial \dot{q}} [/tex]

    nor that they evaulate at epsilon = 0...

    http://en.wikipedia.org/wiki/Noether's_theorem

    see section "Derivations -> One independent variable"

    please enlighten me guys and girls
     
  7. Nov 22, 2008 #6

    gabbagabbahey

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    Hmmm... First of all, if phi is a function of q and epsilon (i.e. [tex]\phi=\phi(q,\epsilon)[/tex] ), I think in general phi-dot should be a function of q, epsilon and their time derivatives (i.e. [tex]\dot{\phi}=\dot{\phi}(q,\dot{q},\epsilon)[/tex]

    Secondly, you need to be more mindful about which derivatives are partials and which are normal derivs....for example, since you are taking the normal derivative of your Lagrangian you should have:

    [tex]\frac{dL}{d\epsilon} = \frac{\partial L}{\partial \phi}\frac{d \phi}{d \epsilon} + \frac{\partial L}{\partial \dot{\phi}}\frac{d \dot{\phi}}{d \epsilon}[/tex]

    The way you had it written, it would simplify under the usual rules for partial derivatives to [tex]\frac{dL}{d\epsilon} = \frac{\partial L}{\partial \epsilon}+ \frac{\partial L}{\partial \epsilon}=2\frac{\partial L}{\partial \epsilon}[/tex] which is clearly nonsense.

    Edit- a better method is the following....
     
    Last edited: Nov 23, 2008
  8. Nov 23, 2008 #7

    gabbagabbahey

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    First notice that;

    [tex]\frac{d \dot{\phi}}{d\epsilon}=\frac{d}{d\epsilon}\left( \frac{d \phi}{dt}\right)=\frac{d}{dt}\left( \frac{d \phi}{d\epsilon}\right)[/tex]

    [tex]\Rightarrow \frac{dL}{d\epsilon} = \frac{\partial L}{\partial \phi}\frac{d \phi}{d \epsilon} + \frac{\partial L}{\partial \dot{\phi}}\left[ \frac{d}{dt}\left( \frac{d \phi}{d\epsilon}\right) \right]
    [/tex]

    Now, what do you get when you use the product rule on the following expression:

    [tex]\frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right][/tex] ?

    What does that make [tex]\frac{\partial L}{\partial \dot{\phi}}\left[ \frac{d}{dt}\left( \frac{d \phi}{d\epsilon}\right) \right][/tex] ? :wink:
     
  9. Nov 23, 2008 #8

    malawi_glenn

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    Now I do this in fully, using the fact that q and epsilon are independent variables.

    [tex]\dfrac{dL}{d \epsilon} = \dfrac{\partial L}{ \partial \phi}\dfrac{d \phi}{d \epsilon} + \dfrac{\partial L}{ \partial \dot{\phi}}\dfrac{d \dot{\phi}}{d \epsilon}[/tex]

    [tex]\dfrac{d\phi}{d\epsilon} = \dfrac{\partial \phi}{\partial q}\dfrac{\partial q}{\partial \epsilon} + \dfrac{\partial \phi}{\partial \epsilon}\dfrac{\partial \epsilon}{\partial \epsilon} = \dfrac{\partial \phi}{\partial \epsilon}[/tex]

    [tex]\dfrac{d \dot{\phi}}{d\epsilon} = \dfrac{\partial \dot{\phi}}{\partial q}\dfrac{\partial q}{\partial \epsilon} + \dfrac{\partial \dot{\phi}}{\partial\dot{ q}}\dfrac{\partial \dot{q}}{\partial \epsilon} + \dfrac{\partial \dot{\phi}}{\partial \epsilon}\dfrac{\partial \epsilon}{\partial \epsilon} = \dfrac{\partial \dot{\phi}}{\partial \epsilon}[/tex]

    [tex]\dfrac{d \phi }{dq} = \dfrac{\partial \phi}{\partial q}\dfrac{\partial q}{\partial q} + \dfrac{\partial \phi}{\partial \epsilon}\dfrac{\partial \epsilon}{\partial q} = \dfrac{\partial \phi}{\partial q} [/tex]

    [tex]\dfrac{\partial \dot{\phi}}{\partial \epsilon} = \dfrac{\partial}{\partial \epsilon}\left( \dfrac{d\phi}{dt} \right) = \dfrac{\partial}{\partial \epsilon}\left( \dfrac{\partial \phi}{\partial q}\dfrac{dq}{dt} \right) = \dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} + \dfrac{\partial \phi}{\partial q}\left( \dfrac{\partial}{\partial \epsilon}\left( \dot{q} \right) \right) = \dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} [/tex]

    Therefor:

    [tex] \frac{dL}{d\epsilon} = \frac{\partial L}{\partial \phi} \dfrac{\partial \phi}{\partial \epsilon}+ \frac{\partial L}{\partial \dot{\phi}}\dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} [/tex]

    How to get

    [tex] \frac{dL}{d\epsilon} = \frac{\partial L}{\partial q} \dfrac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{q}} \dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} [/tex]

    ??
     
    Last edited: Nov 23, 2008
  10. Nov 23, 2008 #9

    gabbagabbahey

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    Again, you should have [tex]\frac{d q}{d \epsilon}[/tex] and [tex]\dfrac{d \epsilon}{d \epsilon}[/tex] instead of [tex]\frac{\partial q}{\partial \epsilon}[/tex] and [tex]\frac{\partial \epsilon}{\partial \epsilon}[/tex]

    [tex]\Rightarrow \frac{d\phi}{d\epsilon} = \frac{\partial \phi}{\partial q}\frac{d q}{d \epsilon} + \frac{\partial \phi}{\partial \epsilon}\frac{d \epsilon}{d \epsilon} =\frac{\partial \phi}{\partial q}\frac{d q}{d \epsilon} + \frac{\partial \phi}{\partial \epsilon}[/tex]

    [tex]\Rightarrow \frac{\partial L}{ \partial \phi}\frac{d \phi}{d \epsilon}=\frac{\partial L}{ \partial \phi} \left[ \frac{\partial \phi}{\partial q}\frac{d q}{d \epsilon} + \frac{\partial \phi}{\partial \epsilon} \right]=\frac{\partial L}{\partial q}\frac{d q}{d \epsilon} + \frac{\partial L}{\partial \epsilon}[/tex]

    And similarly for [tex]\dfrac{\partial L}{ \partial \dot{\phi}}\dfrac{d \dot{\phi}}{d \epsilon}[/tex]


    However!...it turns out that this method gets rather ugly; so I posted a better method in my previous reply ^^^ :wink:
     
  11. Nov 23, 2008 #10

    malawi_glenn

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    yes but I like this method better, will try again, thanx!
     
  12. Nov 23, 2008 #11

    malawi_glenn

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    why is not [tex]\frac{d q}{d \epsilon} = 0[/tex] ?
     
  13. Nov 23, 2008 #12

    gabbagabbahey

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    It is....which gives you the useless identity [tex]\frac{d L}{d\epsilon}=\frac{d L}{d\epsilon}[/tex]...and so this method doesn't work :frown:

    But....look at my post #7....there is a much better method :smile:
     
  14. Nov 23, 2008 #13

    malawi_glenn

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    Ok, in your post #7:

    [tex]\frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right][/tex]


    but WHY should I take the time derivative, it just comes out of nothing?

    evaluating [tex]\frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right][/tex] gives me:

    [tex]\frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right] = (\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}} )) \frac{d\phi}{d\epsilon} + \frac{\partial L}{\partial \dot{\phi}} (\frac{d}{dt}(\frac{d\phi}{d\epsilon})) [/tex]

    Why do I want this?

    I want (partial L / partial q) & (partial L / partial q-dot)
     
    Last edited: Nov 23, 2008
  15. Nov 23, 2008 #14

    gabbagabbahey

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    Well, look at the second term on the righthandside....:wink:
     
  16. Nov 23, 2008 #15

    malawi_glenn

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    [tex]\frac{\partial L}{\partial \dot{\phi}} (\frac{d}{dt}(\frac{d\phi}{d\epsilon})) = \frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right] - (\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}} )) \frac{d\phi}{d\epsilon} [/tex]


    how will this give me:

    [tex] \frac{dL}{d\epsilon} = \frac{\partial L}{\partial q} \dfrac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{q}} \dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} [/tex]

    which allows me to use Euler Lagrange equation?
     
  17. Nov 23, 2008 #16

    gabbagabbahey

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    Good...


    It doesn't....you got this expression through a slightly different means, and although it's correct, I don't think it's very useful...

    Intstead, substitute [tex]\frac{\partial L}{\partial \dot{\phi}} (\frac{d}{dt}(\frac{d\phi}{d\epsilon})) = \frac{d}{dt} \left[ \frac{\partial L}{\partial \dot{\phi}} \frac{d\phi}{d\epsilon} \right] - (\frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}} )) \frac{d\phi}{d\epsilon} [/tex] into the expression for [tex]\frac{d L}{d \epsilon}[/tex] that I gave in post #7....rearrange it a little and you should see that part of it is zero...
     
  18. Nov 23, 2008 #17

    malawi_glenn

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    But that is the point of my question, I want to know how to end up with
    [tex] \frac{dL}{d\epsilon} = \frac{\partial L}{\partial q} \dfrac{\partial \phi}{\partial \epsilon} + \frac{\partial L}{\partial \dot{q}} \dfrac{\partial ^2 \phi }{\partial \epsilon \partial q}\dot{q} [/tex]


    which allows me to use Euler lagrange formula and use canonical equations!

    Iam trying to follow the derivation on that wikipedia page, with T = 0... I only concern spatial transformation.
     
  19. Nov 23, 2008 #18

    gabbagabbahey

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    Hmmm...okay, you can get that after all....but first do what I suggested in my previous post...it will take an extra step to get to the final result ; but in the end it will work out nicely......the wikipedia article uses some shady integration by-parts instead from what I can gather from it (it would be easier to see exactly what they were doing if they formatted their LaTeX equations properly)....anyways, both my method and the wiki method work...if you like I'll go through the wiki method with you after.
     
  20. Nov 23, 2008 #19

    malawi_glenn

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    I will try your "method" first:

    I have:

    [tex]\int ( \frac{\partial L}{\partial \phi}\frac{d \phi}{d\epsilon} + \frac{d}{dt}(\frac{\partial L}{\partial \dot{\phi}}\frac{d\phi}{d\epsilon} ) - ( \frac{d}{dt}( \frac{\partial L}{\partial \dot{\phi}} ))\frac{d \phi}{d\epsilon} )dt [/tex]

    The third part [tex] ( \frac{d}{dt}( \frac{\partial L}{\partial \dot{\phi}} ))\frac{d \phi}{d\epsilon} = \frac{\partial L}{\partial \phi}\frac{d \phi}{d\epsilon} [/tex]

    right?

    Which gives me:
    [tex] \frac{\partial L}{\partial \dot{\phi}}\frac{d\phi}{d\epsilon} [/tex] is conserved.

    But I want to have [tex] \frac{\partial L}{\partial \dot{\phi}} [/tex] in terms of q-dot. Think that [tex] \phi (q,\epsilon) = q + \epsilon q [/tex], i.e infinitesimal spatial translation, for instance.
     
    Last edited: Nov 23, 2008
  21. Nov 23, 2008 #20

    malawi_glenn

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    their unformated LaTeX is just for the time-part, it is not needed.

    My biggest issue is to understand how to go from

    partial L / partial phi TO partial L / partial q

    and

    partial L / partial phi-dot TO partial L / partial q-dot

    if you could help me with this, I would love you for eternity :-)
     
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