Comment on “Some novel delta-function identities” by Charles P. Frahm [Am. J. Phys. 51, 826–829 (1983)]

摘要

In Ref. 1 the equationnu0002iu0002ju00051nr u0006= − u00054u0001n3 u0006u0002iju0002u0001ru0002 +n3xixj − r2u0002ijnr5 u00011u0002nis proposed for the second partial derivative of 1/ r. Thisnresult has subsequently been used in more recent papers.2–5nThe purpose of this comment is to show that the derivationnof Eq. u00011u0002 in Ref. 1 is flawed and to present a direct derivationnof this second partial derivative. While Ref. 1 uses twonindirect methods to deduce Eq. u00011u0002, we employ dyadicnnotation6 to take the partial derivatives directly.nIn dyadic notation, the left-hand side of Eq. u00011u0002 can benoperated on asnu0001u0001u00051nr u0006= − u0001u0005 rnr3u0006 u00012au0002n=−nu0001rnr3 − r u0001 u00051nr3u0006 u00012bu0002n=−nIˆˆnr3 − r u0001 u00051nr3u0006, u00012cu0002nwhere Iˆˆ is the unit dyadic.nTo evaluate the term u0001u00011/ r3u0002, we start with the wellnknown identitynu0001 · u0005 rnr3u0006= 4u0001u0002u0001ru0002. u00013u0002nThe left-hand side of Eq. u00013u0002 contains no vector other than r.nTherefore the scalar functions on each side of the equationncan have no angular dependence. The identity of Eq. u00013u0002 isnusually proven by integrating the left-hand side over a volumenand applying the divergence theorem. The volume integralnover the left-hand side becomes an integral over thenbounding surface. Because the integrand has no angular dependence,nthe integral over the solid angle equals 4u0001 if thenvolume contains the origin. It equals zero if the volume doesnnot contain the origin. Thus, the right-hand side of Eq. u00013u0002nequals 4u0001u0002u0001ru0002 by the definition of the delta function, and thenidentity is proven.nWe can express the left-hand side of Eq. u00013u0002 asnu0001 · u0005 rnr3u0006=nu0001 · rnr3 + r · u0001u00051nr3u0006=n3nr3 + r · u0001u00051nr3u0006, u00014u0002nwhich isolates the term u0001u00011/ r3u0002. Because the function u00011/ r3u0002ndepends only on r, its gradient must be in the rˆ direction.nThus we can writenu0001u00051nr3u0006= rˆgu0001ru0002, u00015u0002nwhere the scalar function gu0001ru0002 is give byngu0001ru0002 = rˆ · u0001u00051nr3u0006. u00016u0002nWe combine Eq. u00016u0002 with Eqs. u00013u0002 and u00014u0002 to findngu0001ru0002 =n4u0001u0002u0001ru0002nrn−n3nr4 u00017u0002nand thennu0001u00051nr3u0006=n4u0001rˆu0002u0001ru0002nrn−n3rˆnr4 . u00018u0002nSubstituting Eq. u00018u0002 into Eq. u00012cu0002, we obtainnu0001u0001u00051nr u0006=n3rˆrˆnr3 −nIˆˆnr3 − 4u0001rˆrˆu0002u0001ru0002. u00019u0002nIn the Cartesian tensor notation in Ref. 1, Eq. u00019u0002 would benwritten asnu0002iu0002ju00051nr u0006=n3xixj − r2u0002ijnr5 −n4u0001xixju0002u0001ru0002nr2 , u000110u0002nwhich differs from Eq. u00011u0002 in the delta function term.