MIT Department of Mechanical Engineering 2.25 Advanced Fluid Mechanics Kundu & Cohen 2.6, 2.7, and 2.20 These problems are from chapter 2 in “Fluid Mechanics” by P. K. Kundu, I. M. Cohen, and D.R. Dowling • (2.6) Show that the condition for the vectors a, b, and c to be coplanar is: εijk ai bj ck • (2.7) Prove the following relationships: δij δij = 3 εpqi εpqj = 2δij • (2.20) Use Stokes theorem to prove that \ × (\φ) = 0 for any single-valued twice-diﬀerentiable scalar (φ) regardless of the coordinate system. Solution: 2.6 Condition for three vectors a, b, and c to be coplanar is a · (b × c) = 0 (1) b × c = (εijk bj ck ei ) (2) ai ei · (εijk bj ck ei ) = 0 (3) εijk ai bj ck (4) δij = 0 if i = j and δij = 1 if i = j (5) δij δij = δii δii = 1 + 1 + 1 = 3 (6) Also we know that From (1) and (2): which leads to: 2.7 From the deﬁnition for Kronecker delta: Thus For proving the other statement (εpqi εpqj = 2δij ) we should refer to the deﬁnition of the alternating tensors: if there is any repeating index, i.e. i = j, or j = k, or i = k then εijk = 0. This means that in εpqi εpqj the only non-zero terms are the ones in which p,q,i, and j have four diﬀerent index values. Since we only have three values for any possible index (1,2, and 3) the mentioned condition for having non-zero terms is only true when i = j (one can easily pick two diﬀerent values of i and j (e.g. i = 1 and i = 3)and see that all the terms turn to be zero in εpqi εpqj . Thus we will have the following: εpqi εpqj = 0 if i = j (7) if i = j : εpqi εpqj = 0 + εpqi εpqi = 0 + ε2pqi (8) Knowing that for any value of i there are only two remaining non-zero terms in the right hand side of (8) which are either −1 or 1, we will have: 2 2 if i = j : εpqi εpqj = 0 + (−1) + (1) = 2 (9) εpqi εpqj = 2δij (10) From (7) and (9): 2.20 From Stokes’ theorem: (\ × u) · dA = A u.ds (11) C Replacing u by \φ will lead to: (\ × (\φ)) · dA = A \φ.ds (12) C Now considering the right hand side of (12), for the line integral of a gradient vector we have the following: b \φ · ds = φ (b) − φ (a) (13) a Using (13) for the closed integral over the curve C, we will have: \φ.ds = φ (a) − φ (a) C In which a can be any arbitrary point on the curve C. Using (14) and also having (12) for any arbitrary area, A, one can conclude that \ × (\φ) = 0 (14)