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Variation of a derivative?

  1. Jul 21, 2013 #1
    Suppose we are taking the variation of a multiple integral and the integrand contains some terms with [itex]\frac{\partial g}{\partial x}[/itex]. When is it ok to put

    [itex]\delta\frac{\partial g}{\partial x}=\frac{\partial}{\partial x}(\delta g)[/itex]

    ?
     
    Last edited: Jul 21, 2013
  2. jcsd
  3. Jul 21, 2013 #2
    This is a trivial way of showing that what you have written is valid, but I'm guessing you already know it. I'm assuming [itex]\delta[/itex] is dirac, so for [itex]\delta ≠ 0[/itex]:

    Since [itex]g_x \rightarrow \frac {\partial g}{\partial x}[/itex], [itex]\delta \cdot g_x = \delta[/itex]([itex]\frac {\partial g}{\partial x}[/itex]) and [itex]\frac {\partial (\delta g)}{\partial x} = \delta g_x[/itex] [itex]\leftrightarrow \delta \frac {\partial g}{\partial x} [/itex], is implied.

    Simply put, if you rewrite the terms on either side of your equation you get:

    [itex]\delta g_x = \delta g_x[/itex]


    Showing it formally would be a little more involved I suppose. The elemental basis of [itex]g_x[/itex] would have to be somewhat disambiguated.
     
    Last edited: Jul 21, 2013
  4. Jul 21, 2013 #3
    Thank you for the reply. I'm not sure what you mean here. But my delta is not a dirac function. It means here a first order variation.
     
  5. Jul 22, 2013 #4
    Any time. You asked "when is it ok to do this", and I assumed that it was always ok to do that since both terms can be rewritten and equate to each other by definition. I'm not sure if if I'm oversimplifying the question, but unless there is a bound-problem with [itex]g_x[/itex] this seems to be more a matter of semantics than theory/concept.
     
  6. Sep 12, 2013 #5
    I am still interested in this question. Spelling it out a little better:

    Suppose we have a functional

    [tex]J[q]=\int^{t_2}_{t_1}{L\left(q(t),\dot{q}(t)\right)dt}[/tex]

    where q can be any function in some appropriate domain of functions. And we take the first variation [itex]\delta J[/itex] by varying the function [itex]q\rightarrow q+\delta q[/itex]. Under what conditions can we put [itex]\delta \dot{q}=\frac{d}{dt}(\delta q)[/itex] ?
     
    Last edited: Sep 13, 2013
  7. Sep 13, 2013 #6

    pasmith

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    Always.

    If [itex]q \to q + \delta q[/itex] then [itex]\dot q \to \frac{d}{dt} (q + \delta q) = \dot q + \frac{d}{dt} (\delta q)[/itex]. The change in [itex]\dot q[/itex] is then by definition [itex]\delta \dot q = \frac{d}{dt}(q + \delta q) - \dot q = \frac{d}{dt}(\delta q)[/itex].
     
  8. Sep 13, 2013 #7
    But if we also are making a variation of the parameter [itex]t \rightarrow t+\delta t[/itex] this can complicate the relation between [itex]\delta q(t)[/itex] and [itex]\delta \dot{q}(t)[/itex], right? I just wanted to be sure that as long as the endpoints are not being varied, then [itex]\delta \dot{q}(t)=\frac{d}{dt}\delta q[/itex] and that I wasn't overlooking some other possible complication.
     
  9. Sep 13, 2013 #8

    pasmith

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    No.

    If you vary a function [itex]q \to q + \delta q[/itex] then the variation in [itex]\dot q[/itex] is fixed as soon as you choose [itex]\delta q[/itex]. Differentiation is a function of a function; if [itex]D(q) = q'[/itex] then [itex]D(q + \delta q) = (q + \delta q)' = q' + (\delta q)'[/itex] because the derivative is linear.

    Remember that [itex]\delta q[/itex] is the difference between the new [itex]q[/itex] and the old; thus if you have [itex]q \to q \circ f + r[/itex] then [itex]\delta q = q \circ f + r - q[/itex]. Then [itex]q' \to (q \circ f)' + r'[/itex] and [itex]\delta (q') = (q \circ f)' + r' - q' = (\delta q)'[/itex].
     
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