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Green's functions for translationally invariant systems

  1. Oct 25, 2015 #1
    As I understand it a Green's function ##G(x,y)## for a translationally invariant differential equation satisfies $$G(x+a,y+a)=G(x,y)\qquad\Rightarrow\qquad G(x,y)=G(x-y)$$ (where ##a## is an arbitrary constant shift.)

    My question is, given such a translationally invariant system, how does one show that ##G(x,y)=G(x-y)##?
    Is it as simple as setting ##y=-a## such that $$G(x+a,y+a)=G(x-y,0)\equiv G(x-y)$$ or is it more to it than that? (I'm slightly uncomfortable with setting ##y=-a##, it doesn't seem correct?!)
     
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  3. Oct 25, 2015 #2

    Orodruin

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    Yes, it is as easy as that. Why are you uncomfortable with setting ##y = -a##?
     
  4. Oct 25, 2015 #3
    I think it's because we're originally treating it as a function of x and y and then we're simply setting one of the variables to a particular value.
     
  5. Oct 25, 2015 #4

    Orodruin

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    But it is still a function of x and y, it just depends on them in a very particular combination. It is also not that you are setting y to a particular value, you are setting the translation amount ##a## to a particular value.
     
  6. Oct 25, 2015 #5
    Isn't y more of a parameter now though as a is chosen to be a constant, right? Is it that the translation invariance places a constraint on the functional dependence?
     
  7. Oct 25, 2015 #6

    Orodruin

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    I do not understand your objection. You can select whatever value of a you want. In particular you can let it be -y. Whatever value of a you have chosen, the translation symmetry will remain true. In particular, it will remain true for a = -y regardless of the value of y.
     
  8. Oct 25, 2015 #7
    I guess I'm looking for issues that aren't there, sorry.

    So is the point that the translation symmetry is valid for arbitrary ##a## and so we can simply choose it such that ##a=-y##?

    Also, I've read notes in which, in order to show that ##G(x,y)=G(x-y)##, they take a derivative with respect to ##a##, i.e. $$\frac{d}{da}G(x+a,y+a)=\frac{\partial G}{\partial x}\frac{\partial (x+a)}{\partial a}+\frac{\partial G}{\partial y}\frac{\partial (y+a)}{\partial a}=\frac{\partial G}{\partial x}+\frac{\partial G}{\partial y}=0$$ Now, at least initially, this doesn't make any sense to me since ##a## is a constant and so taking a derivative with respect to it doesn't make sense. Is the point though that we treat ##a## as a parameter and then "ask" how ##G## changes as we vary it, i.e. we take the derivative of G with respect to ##a##, then since G is translationally invariant this derivative equals zero?!
     
    Last edited: Oct 25, 2015
  9. Oct 25, 2015 #8

    fzero

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    If you don't like that, then you can look at the variation in ##G## under an infinitesimal transformation ##(x,y)\rightarrow (x+a,y+a)##, which is
    $$ \delta G = G(x+a,y+a) - G(x,y) = G(x,y) + \delta x \partial_x G(x,y) + \delta y \partial_y G(x,y) - G(x,y) = a \left( \partial_x G(x,y) + \partial_y G(x,y) \right).$$
    To complete the proof nicely, you can change variables to ##x_\pm = x \pm y## and see that this is ##\partial_{x_+} G(x_+,x_-)##, so that ##G = G(x_-)##.

    So here we are setting ##\delta G=0##. while the approach you mention is equivalent to requiring that ##\delta G## is independent of ##a##. In this case, they lead to the same constraint.
     
  10. Oct 26, 2015 #9
    Thanks for your help.
    Going back to an earlier point, is it simply that ##a## is an arbitrary parameter and so we can always choose it such that ##a=-y## whatever the values of ##x## and ##y## are, hence $$G(x,y)=G(x+a,y+a)=G(x+(-y), y+(-y))=G(x-y,0)\equiv G(x-y)$$
     
  11. Oct 26, 2015 #10

    fzero

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    The only objection that I have to this argument is that we can do this at a single point by choosing ##a=-y_1##, so that ##G(x_1,y_1)=G(x_1-y_1,0)##, but it does not immediately follow that for some other point ##G(x_2,y_2)=G(x_2-y_2,0)##. Therefore I think it is better to use an argument that involves the derivatives of ##G(x,y)##.
     
  12. Oct 26, 2015 #11

    Orodruin

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    But the entire point is that it holds regardless of whether you select ##a=-y_1## or ##a=-y_2##. You can do the translation for any ##a##.
     
  13. Oct 26, 2015 #12

    Ben Niehoff

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    This was my first thought, but Orodruin is actually right. Remember that y is not a function; it's just a value. And if G(x,y) is translationally-invariant, then you're free to translate it by y.

     
  14. Oct 26, 2015 #13

    fzero

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    Yes, I realize that we can do a second transformation to prove it at another point. I didn't mean to imply that it was wrong, just that I prefer the other argument as a matter of taste.
     
  15. Oct 26, 2015 #14
    So am I correct in thinking that as ##G(x,y)## is translationally invariant for each set of values ##(x,y)##, then we can simply choose ##a## such that ##a=-y## at each set of values ##(x,y)##?
     
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