# Is particle i included in the summation?

Hi

I found someone (https://pmocz.github.io/manuscripts/pmocz_sph.pdf) writing the momentum equation as:

In this case it does not m atter since r*i - r*i = 0, and the gradient kernel will produce 0, but in non gradient terms this would matter. Should the summation not include particle i?

Kind regards

## Comments

Hi,

For most SPH kernels which are symmetric (even) functions (i.e. Wendland kernels) the gradient is zero (odd function) for i-i interaction. For the kernel itself (even) the i-i interaction is has a finite non-zero value and should be included in the summation.

do a quick check/exercise, use the Wendland kernel with rij=0 and calculate the kernel and its gradient.

Thanks,

George

Hi !

Thank you very much, it makes sense now!

Kind regards