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Integral equality

  1. Nov 29, 2013 #1
    Can anyone help med understand the manipulations done to the integral operator equation attached? θ(t) is the heaviside function so:
    θ(t1-t2) = {1 for t1>t2 0 t1<t2}
    The t1 and t2 are dummy variables and I don't see how the rewriting takes place.
     

    Attached Files:

  2. jcsd
  3. Nov 29, 2013 #2
    2nd line: just split the thing into two equal parts and swap the dummy index names on the 2nd half.

    3rd line: the 2nd integral (whose upper bound is a dummy variable) can be rewritten over the full range, if you add the step function to remove the bits that are not included in the original integration domain.
     
  4. Nov 29, 2013 #3
    2nd line I understood of course :)

    What I don't understand on the third line is why the bit that is included in the integration from t0 to t and not in t0 to t1 is related to when t1<t2?
     
  5. Nov 30, 2013 #4
    so, let a<b<c be reals and f be a function. [tex] \int_a^b f(x) dx = \int_a^c f(x) \theta(b-x) [/tex] is what you're trying to see. It might be easiest to see if you split the integral on RHS into integrals from a to b and from b to c.
     
  6. Nov 30, 2013 #5
    Okay I get it now - don't know why I got confused by something so simple. One thing that still confuses me though is that the rewriting you see there is actually the time ordering operator in quantum field theory. And it seems that it is related to the Feynman diagrams describing events happening at different physical times t1, t2. But I don't see how this parallel is drawn since the times are really just dummy variables and from my view you can't give them any physical value. I mean should t1>t2 or the opposite?
     
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