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Homework Help: Simple act of taking derivatives, I suppose

  1. Aug 22, 2010 #1
    I am having bit of a problem proving Eq. (0.2):

    [tex](0.1) \text{ } G_{\omega}(t-t^\prime)=\theta(t-t^\prime) \frac{e^{-i\omega(t-t^\prime)}}{2\omega}+\theta(t^\prime-t) \frac{e^{i\omega(t-t^\prime)}}{2\omega}[/tex]
    [tex] (0.2) \text { }\left (-\partial^2_{t}-\omega^2 \right ) G_\omega (t-t^\prime)=i\delta(t-t^\prime)[/tex]

    The problem is dealing with the first and second order derivatives of the theta function in Eq. (0.1); they don't match the right hand side of (0.2).

    MODs: It's not a homework problem.
     
  2. jcsd
  3. Aug 22, 2010 #2

    mathman

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    I am totally unfamiliar with these equations. However when I see a delta function in one equation and a complex exponential in the other I think of Fourier transforms.
     
  4. Aug 22, 2010 #3
    The derivative of [itex]\Theta(t)[/itex] is [itex]\delta(t)[/itex]. You can then use this result, along with the fact that the second derivative of [itex]\Theta(t)[/itex] is peaked at [itex]t=0[/itex] (so that these terms cancel) to find [itex]\partial^2_t G[/itex] and substitute to find the answer.

    If you want any more details please show what you have worked out so far.
     
  5. Aug 22, 2010 #4
    Thanks for your response. I knew about the derivative of the delta function, but the problem is when you take first and second derivatives of the theta function, you get one term with delta function and [tex]cos\left (\omega(t-t^\prime)\right )[/tex] (the other derivative is carried over the exp function) and for the second derivative, the derivative of delta function multiplied by sine of the same variable (all with some additional constant coefficients). It's these cosine and sine functions that I couldn't get them equal to one and zero respectively (modulo those presumably manageable coefficients).
     
    Last edited: Aug 22, 2010
  6. Aug 23, 2010 #5
    The sin and cos terms are multiplied by delta functions, right?

    So you can take them to be at t=t' and so they give the required values.
     
  7. Aug 23, 2010 #6
    Alright, I see that the cosine term that is multiplied by a delta function could be taken as one, but the sine term is multiplied by the derivative of delta function and I can't see how you could set it equal to zero.
     
    Last edited: Aug 23, 2010
  8. Aug 23, 2010 #7
    The derivative of a delta function has the property
    [tex]\int_{-\infty}^\infty\delta'(t)f(t)dt=-f'(0)[/tex]

    In your equation you should have a term like
    [tex]\delta'(\tau)(e^{i\omega\tau}+e^{-i\omega\tau})[/tex]
    which at [itex]\tau=0[/itex] has a derivative which goes away so this term gives no contribution.

    Hope that clears things up.
     
  9. Aug 23, 2010 #8
    Thanks for the explanations CPT, but due to opposite variables in the theta function, when you take the first derivative, a relative negative sign between exponential functions appears which means it's more like [tex]\delta'(\tau)(e^{i\omega\tau}-e^{-i\omega\tau})[/tex] which as I mentioned is proportional to a sine function. Plus, even if it was a cosine function you should take this as an identity, which means if I multiply [tex]\delta'(\tau)\cos(\omega\tau)[/tex] with another function [tex]f(\tau)[/tex] in a more general context, then that integral identity would not help to make the whole expression vanish taking [tex]f'(\tau) \neq 0[/tex] -integration implied.
     
    Last edited: Aug 23, 2010
  10. Aug 24, 2010 #9
    I agree with all of that, and am now stumped, sorry :frown:

    If you work it out, let me know!
     
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