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Derivative with respect to a function

  1. Dec 30, 2008 #1
    I'm learning about Lagrangian and Hamiltonian mechanics and I'm running into issues with the math.

    It seems like there are instances where you have to take a derivative or a partial derivative of a variable with respect to another variable. My issue is that the variable to which you pay respect is often a function itself.

    For example, the Hamiltonian is a function of position and momentum. Position and momentum are both functions themselves, dependent on time. It seems very similar to the idea of a Riemann-Stieltjes integral. Only what I'm looking for would be differentiation, not integration.

    So, in the general case, a Hamiltonian looks like this:

    [tex]H(x, v, t) = V(x(t)) + K(v(t))[/tex]

    where V is a function from position to (potential) energy and K is a function from velocity to (kinetic) energy.

    So the partial derivative of H with respect to x is such an example.

    It seems like there should be some method that looks just like the standard derivative. Something like:

    [tex]\lim_{h \to 0} \frac{H(x + h, v, t)}{h}[/tex]

    Where h is an function which approaches the zero function in the limit.

    However, what I end up with is an expression

    [tex]\lim_{h \to 0} \frac{V(x(t) + h(t)) - V(x(t))}{h}[/tex]

    And I'm not sure how to take that limit. I can vaguely see the form of it. You let t be some constant, x(t) becomes a constant, and h(t) becomes a very small constant, infinitesimal in the limit. The answer should be V' or something like that. But I don't know how to word it to make the argument solid and rigorous (... or at least I don't know how to convince myself beyond a doubt this is correct).

    EDIT: I found this inconsistency shortly afterwards. In the denominator, I'm dividing by a function. One might jump to the assumption that it should be h(t), not just h, but that doesn't follow from the definition of the derivative.

    The hamiltonian as I've stated it has an unusual type: (R->R) -> (R->R) -> R -> R, and maybe the strange results I'm getting are a result of not treating it as a function-valued operator of two arguments instead: (R->R) -> (R->R) -> (R -> R).

    Doing this, instead of saying H(x, v, t) = V(x(t)) + K(v(t)), I'd restate it as H(x, v) = V*x + K*v, where * is function composition, and addition and division are acting as they would in a function space. However, my algebra here is a bit weak.

    \frac{\partial H}{\partial x}(x, v) &= \lim_{h \to 0} \frac{H(x+h, v) - H(x,v)}{h} \\
    &= \lim_{h \to 0} \frac{V*(x+h) + K*v - (V*x + K*v)}{h} \\
    &= \lim_{h \to 0} \frac{V*(x+h) - V*x}{h}

    At this point, I'm a little lost. It seems that if V is differentiable and h is infinitesimal, V(x+h) = V(x) + V(h), but I don't know how to justify that completely using limits in the context of this equation. But assuming It's justified, that leaves me with:

    [tex]\frac{\partial H}{\partial x}(x, v) = \lim_{h \to 0} \frac{V*h}{h}[/tex]

    And I don't know what property allows me to reduce this to V'.
    Last edited: Dec 30, 2008
  2. jcsd
  3. Dec 30, 2008 #2
    Your denominator should have h(t) and your limit should be as h(t) -> 0 right? (just as a notational issue)
  4. Dec 30, 2008 #3
    Perhaps my update might make that more clear. What I intended was a limit as the function approached the zero of a real function space.

    To be honest, all my qualms are notational issues when it comes to the mathematics in physics =-) Integrals are easy. Figuring out what kind of integral the author means is usually the hard part.
  5. Dec 30, 2008 #4


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  6. Dec 30, 2008 #5


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    There is nothing peculiar about differentiating one function with respect to another function- after all, "x" itself is a function.

    Use the chain rule: If f and g are both functions of x, then
    [tex]\frac{df}{dg}= \frac{df}{dx}\frac{dx}{dg}= \frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex]
  7. Dec 30, 2008 #6
    This isn't quite a functional (the codomain is a function, not a scalar). I don't believe it has to do with directional derivatives.

    I'm trying to formulate this in a type-consistent way, and I want to be very careful about that. I'm not just looking for the 'right answer' here. Otherwise, I'd be all over the chain rule =-)
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