Square root in Laplace operator
Can some explain to me why we devide by the square root of the distances to compute the second derivative in the LaplaceOperator?
https://gitlab.mpcdf.mpg.de/ift/NIFTy/blob/nifty2go/nifty/operators/laplace_operator.py#L99
If I see it correctly, it appears first in commit https://gitlab.mpcdf.mpg.de/ift/NIFTy/commit/05eeeabb347754445190675af2ab761f89a6ed60 by @theos. I do not see why this makes sense already because of the denominator's dimensions.