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Question about notation: lowercase delta in what appears to be a derivative

  1. May 16, 2010 #1
    In the http://en.wikipedia.org/wiki/Hamilton's_principle#Mathematical_formulation", I ran across a notation I'm not familiar with. The part I'm unsure about is:

    [itex]\frac{\delta \mathcal{S}}{\delta \mathbf{q}(t)}=0[/itex]

    In the context of the article, what is the meaning of that equation?


    (My math background goes up to multivariable calculus.)
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. May 16, 2010 #2
  4. May 16, 2010 #3


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    The article you linked to explains it, well sort of. Just keep reading that article, reread until you understand. The answers are there.

    The sentence after the expression you quote talks about it.
  5. May 16, 2010 #4
    The rest of the article leads me to believe that it's just like a normal derivative, except that it's being taken with respect to a function instead of a variable.

    I should have mentioned this in my original post, but part of my confusion is also coming from a book I have (Classical Dynamics by Thornton and Marion), which says:

    [itex]\frac{\partial \mathcal{S}}{\partial \alpha} d \alpha \equiv \delta \mathcal{S}[/itex]


    [itex]\mathcal{S}(\alpha) = \int_{t_1}^{t_2} \! L(\mathbf{q}(\alpha, t), \dot{\mathbf{q}}(\alpha, t); t) \, dt[/itex]


    [itex]\mathbf{q}(\alpha, t) = \mathbf{q}(0, t) + \alpha \eta(t)[/itex]


    [itex]\eta(t)[/itex] is a function such that [itex]\eta(t_1) = \eta(t_2) = 0[/itex]​

    I'm having trouble seeing how these two definitions of the [itex]\delta[/itex] operator are equivalent. Thornton and Marion's [itex]\eta(t)[/itex] seems to be the same as Wikipedia's [itex]\boldsymbol\varepsilon(t)[/itex].

    Does the following look correct?

    Let f be a function [itex]f(x(t))[/itex].

    Approaching from the Wikipedia side:

    Let [itex]\varepsilon(t)[/itex] be a small perturbation from [itex]x(t)[/itex].

    [itex]\delta f = f(x(t) + \varepsilon(t)) = \varepsilon(t) \frac{\partial f}{\partial x(t)}[/itex]

    Approaching from the Thornton and Marion side:

    [itex]x(t) = x(0, t); x(\alpha, t) = x(0, t) + \alpha \eta(t)[/itex]

    [itex]\delta f = \frac{\partial f}{\partial \alpha} d \alpha = \frac{\partial f}{\partial x(t)} \frac{\partial x(t)}{\partial \alpha} d \alpha = \eta(t) \frac{\partial f}{\partial x(t)} d \alpha[/itex]​

    I'm not sure what to do with that [itex]d \alpha[/itex] in the Thornton and Marion approach. Should that be there?
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