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Differentiation under integral sign - one parameter family

  1. Mar 3, 2013 #1

    WannabeNewton

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    Hi guys! Let [itex]\left \{ B_{t} \right \}_{t\in \mathbb{R}}[/itex] be a one - parameter family of compact subsets of [itex]\mathbb{R}^{3}[/itex] with smooth (manifold) boundary (e.g. one - parameter family of closed balls). In my context, each [itex]B_{t}[/itex] belongs to a different constant time slice of Minkowski space - time. How does one compute [itex]\frac{\partial }{\partial t}\int_{B_{t}}f(t,\mathbf{x})dV[/itex]? Again in my context, [itex]f(t,\mathbf{x})[/itex] happens to be [itex]T_{00}(t,\mathbf{x})[/itex], the time - time component of the stress energy tensor and as such this component is assumed to be a smooth scalar field.
     
    Last edited: Mar 3, 2013
  2. jcsd
  3. Mar 4, 2013 #2
    Isn't your derivative a Calc I style total derivative as opposed to a partial? Replace the first t by a u and the second t by a v: (the integral over Bu of the function f(v,x).)

    Now to calculate the derivative, you set u=t, v=t and use the multivariable chain rule:
    d/dt F(u,v) = d/du F (u=t) + d/dv F (v=t)
    Sorry about the notation, my keyboard is broken and half the keys don't work :(

    This is a simple but handy technical device to separate the two instances of t in the formula.

    To account for the change in the domain, you could try to express the change in the domain Bt as B0 + s(t,x)n, where n is the unit outward normal vector on the boundary of B0, and s(t,x) depends on time t and the point x on the boundary of B0. Then when you differentiate with respect to the change in boundary, you will basically be integrating the function s(0,x)*f(0,x) over the boundary surface.
     
  4. Mar 5, 2013 #3

    WannabeNewton

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    Thanks Vargo. That is exactly what wiki does as well apparently. So if instead my integration regions were constant time slices of R^4 itself (so the integration regions would just be R^3 for each time slice) then could I just ignore the boundary integral and simply pull the time derivative inside? Again, thanks for the response I really appreciate it and I might have to ask more once this one is answered if that's ok =D (although they might be a bit physicsy? idk =p)
     
  5. Mar 5, 2013 #4
    Yes, if your region of integration is constant, then you can just move the differentiation inside. Well, there is a condition that must be met. If, when you move the derivative inside the integral, the resulting integral is absolutely convergent, then the operation is valid. I think that's the rule. I'd have to look it up to be sure.
     
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