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Consider the following:

  • The Euclidean plane has $3$ translation symmetries and $3$ rotation symmetries.

  • Any physical quantity $K(x,y)$ on the Euclidean plane, where $x$ and $y$ are two arbitrary spacetime points, depends only on $(x-y)^{2}$ because $(x-y)$ is translation invariant and $(x-y)^{2}$ is rotation invariant.

But I am not sure how the dependence changes if we have a Euclidean disk (that is, a plane with a boundary). My intuition is that

  • $K(x,y)$ now depends not only on the spacetime points $x$ and $y$, but also on the 'border.'
  • The dependence is such that $K(x,y)$ for the Euclidean disk tends to $(x-y)^2$ as the 'border' tends to infinity.

But I am not able to carry my intuition any further and write down an explicit form for the dependence of $K(x,y)$ for the Euclidean disk.

It would be really helpful if you share some thoughts here.

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  • $\begingroup$ Doesn't the Euclidean plane have 2 translation and 1 rotation symmetries ($\mathbb{R}^2$ and $SO(2)$)? $\endgroup$ – coconut Dec 12 '17 at 20:10
  • $\begingroup$ by plane, i mean spacetime $\endgroup$ – nightmarish Dec 12 '17 at 20:42
  • $\begingroup$ 4-dimensional Euclidean space, then? That'd be 4 translations and 6 rotations ($\mathbb{R}^4$ and $SO(4)$), right? $\endgroup$ – coconut Dec 12 '17 at 20:45

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