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Euler Lagrange Derivation (Taylor Series)

  1. Jan 15, 2016 #1
    Mod note: Moved from Homework section
    1. The problem statement, all variables and given/known data

    Understand most of the derivation of the E-L just fine, but am confused about the fact that we can somehow Taylor expand ##L## in this way:

    $$ L\bigg[ y+\alpha\eta(x),y'+\alpha \eta^{'}(x),x\bigg] = L \bigg[ y, y',x\bigg] + \frac{\partial L}{\partial y} \alpha \eta(x) + \frac{\partial L}{\partial y'} \alpha \eta^{'} + O^{2}(\alpha \eta) $$

    I'm really uncomfortable with the idea that you can treat whole functions as independent of each other (when they are clearly not). I've looked up a lot of derivations online & from textbooks and none of them seem to bother to explain this. Am I missing something obvious? Surely the fact that ## y=f(x,t,\alpha )## should be taken into account somewhere?

    Thanks in advance for the help!
     
    Last edited by a moderator: Jan 15, 2016
  2. jcsd
  3. Jan 15, 2016 #2

    mathman

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    The expansion looks ok. What dependence are you concerned about?
     
  4. Jan 16, 2016 #3

    Ssnow

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    Yes, it is exactly the Taylor expansion of ##L\left[y+\alpha\eta(x),y'+\alpha\eta'(x),x\right]## at the point ##(y,y',x)##, obviously the term with the partial derivative of ##x## there isn't because ##\frac{\partial L}{\partial x}\cdot(x-x)=0##...
     
  5. Jan 20, 2016 #4
    Yes, I know, but why are we allowed to treat ##y(x)## and ##\dot{y}(x)## as independent variables? (as we are required to do for the Taylor series?) They clearly aren't independent.
     
  6. Jan 20, 2016 #5

    PeroK

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    What's happening is that the symbols ##y, \dot{y}, x## are being used for two purposes. I'm only on my phone so it's difficult to type a lot of latex. Consider ##L## as any function of three variables. To be clear we will define ##L## using dummy variables ##X,Y,Z##

    E.g we could define ##L(X,Y,Z) = X^2 + XYZ##

    Now, we have three more functions by taking the partial derivatives of ##L##:

    ##L_X, L_Y, L_Z##

    Using this notation for simplicity.

    Finally, we can do the following Taylor expansion for ##L##

    ##L(X_0 + a, Y_0 + b, Z_0 + c) = L(X_0, Y_0, Z_0) + aL_X + bL_Y + cL_Z + \dots##

    And that hold for any numerical values we choose for ##X_0, Y_0, Z_0, a, b, c##. In other words, that equation holds for whatever expressions we plug in to replace the dummy variables we've used.

    In this case, the expressions happen to be expressions in ##x,y, \dot{y}, \eta##.
     
  7. Jan 20, 2016 #6

    PeroK

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    PS I forgot to mention:
    ## \frac{\partial L}{\partial y}##

    Think of this not as the partial derivative of ##L## with respect to ##y##, but the function formed by taking the partial derivative of ##L## with respect to its first argument.
     
  8. Jan 22, 2016 #7

    Ssnow

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    Geometrically ##y(x)## is a function of ##x## and you can think this as subset of the euclidean space, as the graph of a function on the plane for example. The derivative ##\dot{y}(x)## represent the tangent direction of the graph of the function at the point ##x##. So ##\dot{y}(x)## generate the tangent space of ##y(x)## at ##x##. In general they lives in two different spaces...
     
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