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Aug 14, 2020 at 17:44 vote accept anbhadane
Aug 14, 2020 at 17:44 vote accept anbhadane
Aug 14, 2020 at 17:44
Aug 14, 2020 at 11:43 history edited Qmechanic
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Aug 14, 2020 at 11:39 answer added xyzrggong timeline score: 1
Aug 14, 2020 at 9:00 comment added anbhadane @NathanaelNoir Thank you for your help, now i got something.
Aug 14, 2020 at 9:00 comment added anbhadane @Philip oh, got it, Thank you for your help.
Aug 14, 2020 at 8:54 comment added Philip @anbhadane This is explained in the paragraph just above the equation that's bothering you in Jackson. I would suggest that you first read that paragraph carefully. If you still have trouble understanding it, I would suggest including it in your question. That would also explain why this question is essentially a duplicate of the one posted by Nathanael. :)
Aug 14, 2020 at 8:47 comment added anbhadane @NathanaelNoir Thank you, but i got something now, but still have doubt about the term $O(a^2)$, from where it comes?
Aug 14, 2020 at 8:44 history edited anbhadane CC BY-SA 4.0
deleted 12 characters in body
Aug 14, 2020 at 8:43 comment added anbhadane @Philip i got it, sorry. I messed up it, will modify the question.
Aug 14, 2020 at 8:42 comment added Philip $x-x' \neq r^\frac{1}{2}$ ;) What I expect you mean is $r \nabla \rho$...
Aug 14, 2020 at 8:41 comment added anbhadane @Philip but in taylor's expansion the second term is $(x-x')\nabla{\rho}$, so if this is not, then what should be?
Aug 14, 2020 at 8:39 comment added Philip The second term in your proposed Taylor expansion is dimensionally inconsistent. It can't be $r^{\frac{1}{2}}$.
Aug 14, 2020 at 8:38 comment added Nathanael Noir I think that will help: physics.stackexchange.com/q/441818
S Aug 14, 2020 at 8:33 history suggested Nathanael Noir CC BY-SA 4.0
in the title was an typo
Aug 14, 2020 at 8:32 review Suggested edits
S Aug 14, 2020 at 8:33
Aug 14, 2020 at 8:26 history asked anbhadane CC BY-SA 4.0