9 The Taub-NUT Solution

Chapter 9
The Taub-NUT Solution

Page 273, Eq. (9.19) the coefficient 16 should be 32:

$∫ ∫ ( ) d4x (F ± ⋆F )2 = 2 d4x F 2 ± F ⋆F = 8SEY M ± 32π2c2 ≥ 0 ,$
Page 273, Eq. (9.20) the coefficient 2 should be 4:

$SEY M ≥ 4π2|c2|.$
Page 280, Eq. (9.56), parenthesis and square missing in the second equation:

$ds2 = f (r)(dt + 2N cos θdφ)2 - f-1(r)dr2 - (r2 + N 2)dΩ2(2), 2 2 Ftr = 4q(r----N-)-, (⋆F )tr = ---8qN-r---, (r2 + N 2)2 (r2 + N 2)2 (r - r+ )(r - r- ) f(r) = ----r2 +-N-2----, 2 2 2 2 r± = M ± r0, r0 = M + N - 4q .$
Page 280, Eq. (9.57), the function should be raised to the power -1 in the second and third equations and should replace H in the fourth:

$2 ds2 = |H |- 2(dt + A ) - |H |2d⃗x23 , A = 2ℜe (eiαH -1) , A˜ = 2ℑm (eiαH -1) , t t H = 1 + M--+-iN- , |⃗x3| A = A dxi , ϵ ∂ A = ± ℑm (H-∂ H ) . i ijk i j k$
Page 280, Eq. (9.58), the function should be raised to the power -1 in the second and third equations

$2 -2 2 2 2 ds = |H| (dt + A ) - |H | d⃗x3 , At = 2Re H -1 , A˜t = - 2Re (iH- 1) , i (-- ) A = Aidx , ϵijk∂iAj = ±Im H ∂kH , ∂i∂iH = 0.$