a bc in Visual Studio .NET

Painting QR Code JIS X 0510 in Visual Studio .NET a bc

We can calculate the nonzero Christoffel symbols for this metric using 1 ad g ( b gdc + c gdb d gbc ). For example, 2

1 = g ud ( x gd x + x gd x d gx x ) 2 1 = v g x x 2 1 d (1 v (v))2 = 2 dv d ( v (v)) = (1 v (v)) dv d = (1 v (v)) (v) v dv = (1 v (v)) ( (v) + v (v)) = (1 v (v)) (v)

To move to the last line, we used the fact that f (v) (v) = f (0) , so v (v) = 0. It is a simple exercise to nd all of the nonzero Christoffel symbols using equation (4.16), which turn out to be

= (1 v (v)) (v), (v) , = 1 v (v)

= (1 + v (v)) (v)

(v) = 1 + v (v)

(13.48)

The wave is propagating along v, and so choosing la to be along this direction we set la = (0, 1, 0, 0). We can nd the other basis vectors in the tetrad using gab = la n b + lb n a m a m b m b m a (13.49)

The vector m a is spacelike, and so considering guv we can write guv = lu n v + lv n u = lv n u = n u nu = 1 Since n u is null, we can take this to be the only nonzero component and write n a = (0, 1, 0, 0). Moving to the spacelike vector, we take m x to be real and so using (13.49) we have gxx = l x n x + l x n x m x m x m x m x = 2m 2 x Using gx x = (1 v (v))2 this gives mx = 1 v (v) = mx 2

A similar exercise, taking m y to be complex, leads us to choose my = i 1 + v (v) 2 and m y = i 1 + v (v) 2

Summarizing, for this metric we choose the following null tetrad: la = (0, 1, 0, 0) , m a = 0, 0, m a = 0, 0, n a = (1, 0, 0, 0)

1 v (v) 1 + v (v) , ,i 2 2 1 v (v) 1 + v (v) , i 2 2 (13.50)

Raising indices with the metric, we nd l a = (1, 0, 0, 0) , m a = 0, 0, m a = 0, 0,

n a = (0, 1, 0, 0) 1 1 (13.51)

1 , i 2 (1 v (v)) 2 (1 + v (v)) 2 (1 v (v)) ,i 1 2 (1 + v (v))

Now that we have the null tetrad, we can compute the spin coef cients. First noting that the only nonzero component of the rst null vector is the constant lv =1, looking at the Christoffel symbols (13.48) we can see that la;b = b la v ab lv = 0. This is, of course, what we expect for a pp-wave spacetime. This requirement dictates that several of the spin coef cients will vanish, in particular, = = = = 0. An exercise also shows that = = = = = = 0. Let s compute the two remaining spin coef cients. Starting with = n a;b m a m b , looking x y at (13.51) we observe that only terms involving m and m will be nonzero. Expanding the sum with this in mind, we nd = n a;b m a m b = n x;x m x m x + n x;y m x m y + n y;x m y m x + n y;y m y m y Recall that the covariant derivative is n a;b = b n a