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1 Einstein Notation PH 108 : Electricity & Magnetism Einstein Notation Rwitaban Goswami 1 Einstein Notation 2.6. The gradient operator r behaves like a vector in \ some sense". For example, divergence of a curl (r:r × A~ = 0) for any A~, may suggest that it is just like A:~ B~ × C~ being zero if any two vectors are equal. Prove that r × r × F~ = r(r:F~ ) − r2F~ . To what extent does this look like the well known expansion of A~ × B~ × C~ ? Solution: We will be using Einstein summation convention. According to this convention: 1. The whole term is summed over any index which appears twice in that term 2. Any index appearing only once in the term in not summed over and must be present in both LHS and RHS Don't be confused by this. The convention only is to simply omit the P symbol. You can mentally insert the P symbol to make sense of what the summation will look like. Now let us go over some common symbols that are used extensively in Einstein's notation: • δij is the Kronecker-Delta symbol. It is defined as ( 1 if i = j δ = ij 0 if i 6= j • ijk is the Levi-Civita symbol. It is defined as 81 if ijk is a cyclic permutation of xyz, eg xyz, yzx, zxy <> ijk = −1 if ijk is an anticyclic permuation of xyz, eg xzy, zyx, yxz :>0 if in ijk any index is repeating, eg xxy, xyy etc Now, you can verify: ~ ~ A · B = AiBi ~ ~ A × B = ijkAjBk e^i wheree ^i is the unit vector in i direction In the cross product formula, when the P symbol is reinserted, this simply means X X X ijkAjBke^i i=x;y;z j=x;y;z k=x;y;z A few things to note: P • δii = 3. How? Because we need to remember to put back the symbol, after which P it becomes i=x;y;z δii = δxx + δyy + δzz = 3 P • δijAj = Ai. Remember to over j, and you will see that the only non-zero term is where j = i 1 of 3 PH 108 : Electricity & Magnetism Einstein Notation Rwitaban Goswami • We can resolve the product of two Levi-Civita symbols as δil δip δiq ijklpq = δjl δjp δjq δkl δkp δkq • When one of the indexes of the product in the previous property repeats: ijkilm = δjlδkm − δjmδkl You can find this out using the previous property, keeping in mind the P over i Now let us commence the proof of the identity given in the question. @ Now we will represent each component of r, @i by @i, ~ ~ r × r × F = ijk@j r × F e^i = ijkklm@j@lFm e^i = kijklm@j@lFm e^i k = (δilδjm − δimδjl)@j@lFm e^i (* ijkilm = δjlδkm − δjmδkl) = (@m@iFm − @l@lFi)e ^i (* δijAj = Ai) = @i(@mFm)e ^i − @l@l(Fi e^i)(* @m@i = @i@m) ~ 2 ~ = r(r · F ) − r F @2 @2 Notice that this proof works because @m@i = @i@m which is just fancy notation for @x@y = @y@x We know that A~ × B~ × C~ = B~ (A~ · C~ ) − (A~ · B~ )C~ . If we write A~ = r, B~ = r, C~ = F~ , we get r × r × F~ = r(r · F~ ) − (r · r)F~ = r(r · F~ ) − r2F~ . So this serves as a fake proof of the identity 2 of 3 PH 108 : Electricity & Magnetism Einstein Notation Rwitaban Goswami p 2.7. As a more involved example, show that the operator L = −i~r × r where (i = −1) satisfies L × Lf = iLf where f is an arbitrary test function. (Notice that the cross product of an operator with itself does not necessarily vanish, can you see why?) Solution: L × Lf = ijkLjLkf e^i = −iijkjlmrl@mLkf e^i = −ijkijlmrl@mLkf e^i = −i(δklδim − δkmδil)rl@mLkf e^i = −i(rl@iLkf − ri@kLkf)e ^i = −i(Ai − Bi)e ^i Ai = rk@iLkf = −ikpqrk@irp@qf Bi = ri@kLkf = −ikpqri@krp@qf Now we can't directly switch the order of @i & @q, we have to use the product rule: (@a(rb@c))f = (rb@a@c)f + ((@arb)@c)f : 0 =) A = −i r r@ @ f − i r (@ r )@ f = −i r (δ )@ f = −i r @ f i kpq k p i q kpq k i p q kpq k ip q kiq k q (* kpqrkrp@i@qf = −pkqrkrp@i@qf = −kpqrprk@i@qf = −kpqrkrp@i@qf =) kpqrkrp@i@qf = 0) The first term cancels because in the summation 1 term will have k,q = x,y and another term k,q = y,z. Both differ only in their sign, which is due to the kpq. So they cancel out : 0 =) B = −i r r @ @ f − i r (@ r )@ f = −i r (δ )@ f = −i r @ f = 0 i kpq i p k q kpq i k p q kpq k kp q ppq k q (* kpqrirp@k@qf = −qpkrirp@k@qf = −kpqrirk@q@kf = −kpqrirp@k@qf =) kpqrirp@k@qf = 0) Similar reasoning as above is used here =) L × Lf = −kiqrk@qf e^i = ikqrk@qf e^i = iLf Notice that we have used @arb = δab. This is because ra is simply x if a = x, y if a = y and so on. And if we differentiate that w.r.t b = x or y or z, then it is 1 if a = b, and 0 @ @ otherwise, because @x x = 1, @x y = 0 The cross product of the operator with itself doesn't vanish here because the operators act on different objects. One L acts on Lf, and the other on f. 3 of 3.
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