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(4.16)
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EXAMPLE 4-4 Find the Christoffel symbols for the 2-sphere of radius a ds 2 = a 2 d 2 + a 2 sin2 d 2 SOLUTION 4-4 The metric is given by gab = a2 0 0 a sin2
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1 a2
1 a 2 sin2
and g =
1 . a 2 sin2
The only nonzero derivative is (4.17)
g = a 2 sin2 = 2a 2 sin cos
We nd the Christoffel symbols using (4.16). Considering the rst nonzero term of g ab , we set a = d = in (4.16) and obtain
bc
1 g 2
gc gbc gb + c b x x
1 = g 2
gbc
We set gc = g cb = 0 because g = g = 0, g = a 2 , where a is a constant, x xb and so all derivatives of these terms vanish. The only possibility is
1 = g 2
g
1 a2
2a 2 sin cos = sin cos
The only other nonzero possibility for this metric is the term involving g , and so we set a = d = in (4.16). This gives
bc
1 g 2
gc gbc gb + c b x x
Tensor Calculus
Only a derivative with respect to is going to be nonzero. Therefore we drop the rst term and consider
bc
1 g 2
gbc gc b x
First we take b = , c = , which gives
1 = g 2 = 1 2
g g
1 g g 2 cos = cot sin
1 a 2 sin2
2a 2 sin cos =
A similar procedure with b = , c = gives fel symbols are zero.
= cot . All other Christof-
EXAMPLE 4-5 A metric that is used in the study of colliding gravitational waves is the KahnPenrose metric (see Fig. 4-1). The coordinates used are (u, v, x, y). With u 0 and v < 0, this metric assumes the form ds 2 = 2 du dv (1 u)2 dx 2 (1 + u)2 dy 2 Find the Christoffel symbols of the rst kind (using (4.15) for this metric.
v Region 4 Time
Region 2
Region 3
Region 1
Fig. 4-1. The division of spacetime for the Kahn-Penrose solution used in the study of
gravitational waves. We consider the metric in region 2.
SOLUTION 4-5 In this problem, we use the shorthand notation a = =
xa
Tensor Calculus
and so write (4.15) as
1 ( a gbc + b gca c gab ) 2
We can arrange the components of the metric in a matrix as follows, de ning x 0 , x 1 , x 2 , x 3 (u, v, x, y) 0 1 0 1 0 0 [gab ] = 0 0 (1 u)2 0 0 0 and so we have guv = gvu = 1, gx x = (1 u)2 , g yy = (1 + u)2 0 0 0 2 (1 + u)
One approach is to simply write down all of the possible combinations of coordinates in (4.15), but we can use the form of the metric to quickly whittle down the possibilities. First we note that a guv = a gvu = 0 The only nonzero derivatives are u gx x = u [ (1 u)2 ] = 2(1 u) u g yy = u [ (1 + u)2 ] = 2(1 + u) Consider terms involving x rst. Looking at abc = 1 ( a gbc + b gca 2 c gab ), the possible combinations of a,b,c that will give a term involving u gx x are a = b = x, c = u a = u, b = c = x a = c = x, b = u
Tensor Calculus
Taking the rst of these possibilities, we have
1 1 1 ( x gxu + x gux u gx x ) = u gx x = [2(1 u)] = 1 + u 2 2 2
In the second case, we obtain
ux x
1 1 1 ( u gx x + x gxu x gux ) = u gx x = [2(1 u)] = 1 u 2 2 2
You can write down the third permutation and see that it gives the same result as this one. The next step is to consider the terms involving u g yy . A similar exercise shows that
uyy yyu
= 1 u =1+u
It is easy to see that if the components of the metric are constant, all of the Christoffel symbols vanish. Note that the Christoffel symbols of the rst kind are symmetric in their rst two indices:
EXAMPLE 4-6 Show that if the metric is diagonal, then
a ab
xb
1 ln gaa 2
SOLUTION 4-6 If the metric is diagonal then the inverse components of the metric are easily found to be g aa = g1 . Using (4.15), we obtain aa
a ab
= g ac
= g aa
a bc
1 gaa
1 gaa
1 gaa 2 xb
xb
1 ln gaa 2
We say a connection metric vanishes:
is metric-compatible if the covariant derivative of the c gab = 0
(4.18)
EXAMPLE 4-7 Show that c gab = 0 using (4.16).
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