Green's functions for translationally invariant systems

["Don't panic!"]

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?!)

 

[Orodruin]

Yes, it is as easy as that. Why are you uncomfortable with setting $y = -a$?

 

["Don't panic!"]

Yes, it is as easy as that. Why are you uncomfortable with setting y=−a?

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.

 

[Orodruin]

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.

 

["Don't panic!"]

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 aa to a particular value.

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?

 

[Orodruin]

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?

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.

 

["Don't panic!"]

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.

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?!

 

[fzero]

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?!

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.

 

["Don't panic!"]

If you don't like that, then you can look at the variation in GG under an infinitesimal transformation (x,y)→(x+a,y+a)(x,y)\rightarrow (x+a,y+a), which is
δG=G(x+a,y+a)−G(x,y)=G(x,y)+δx∂xG(x,y)+δy∂yG(x,y)−G(x,y)=a(∂xG(x,y)+∂yG(x,y)). \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±=x±yx_\pm = x \pm y and see that this is ∂x+G(x+,x−)\partial_{x_+} G(x_+,x_-), so that G=G(x−)G = G(x_-).

So here we are setting δG=0\delta G=0. while the approach you mention is equivalent to requiring that δG\delta G is independent of aa. In this case, they lead to the same constraint.

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 $$G(x,y)=G(x+a,y+a)=G(x+(-y), y+(-y))=G(x-y,0)\equiv G(x-y)$$

 

[fzero]

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 $$G(x,y)=G(x+a,y+a)=G(x+(-y), y+(-y))=G(x-y,0)\equiv G(x-y)$$

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 $G(x,y)$.

 

[Orodruin]

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 $G(x,y)$.

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 $a$.

 

[Ben Niehoff]

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 $G(x,y)$.

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.

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 $a$.

 

[fzero]

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 $a$.

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.

 

["Don't panic!"]

But the entire point is that it holds regardless of whether you select a=−y1a=-y_1 or a=−y2a=-y_2. You can do the translation for any aa.

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 $(x,y)$?