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= sin cos , and so using (8.3), this equation
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X = sin cos X = sin cos f ( )
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Integrating, we obtain
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Killing Vectors
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X = sin cos
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f ( ) d + g ( )
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(8.4)
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Now, returning to Killing s equation, by setting a = and b = , we have the last equation for this geometry, namely X + X = 0 Let s write down each term separately. The rst term is X = X
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= X
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X
X
Looking at the Christoffel symbols, we see that this leads to X = X cot X Now we consider the second term in X + X = 0. We obtain X = X
c Xc
= X
X
X
= X cot X
and so, the equation X + X = 0 becomes X + X 2 cot X = 0 X + X = 2 cot X (8.5)
We can re ne this equation further using our previous results. Using (8.3) together with (8.4), we nd X = sin cos = (sin2 cos2 ) We also have X = f ( ) f ( ) d + g ( ) f d + g( )
Killing Vectors
Now, we can put all this together and obtain a solution. Adding these terms together, we get X + X = (sin2 cos2 ) f ( ) d + g ( ) + f ( )
Next we want to set this equal to the right-hand side of (8.5). But let s work on that a little bit. We get 2 cot X = 2 cot sin cos Now we know that cot (sin cos ) = Therefore, we can write 2 cot X = 2 cos2 f ( ) d + 2 cot g( ) cos (sin cos ) = cos2 sin f d + g ( )
Finally, we equate both sides of (8.5) and we have (sin2 cos2 ) f ( ) d + g( ) + f ( ) f d + 2 cot g ( )
= 2 cos2
Our goal is to get all terms on one side and all terms on the other. We can do this by adding 2 cos2 f ( ) d to both sides and then move the g ( ) on the left-hand side over to the right. When we do this, we get this equation f ( ) d + f ( ) = 2 cot g ( ) g ( ) If you think back to your studies of partial differential equations, the kind where you used separation of variables, you will recall that when you have an equation in one variable equal to an equation in another variable, they must both be constant. So we will do that here. Let s call that constant k. Looking at the
equation, we have
Killing Vectors
g ( ) 2 cot g ( ) = k We multiplied through by 1 so that the derivative term would be positive. We can solve this kind of equation using the integrating factor method. Let s quickly review what that is. Consider your basic differential equation of the form dy + p (t) y = r (t) dt First, we integrate the multiplying term p(t): p (t) = p(s) ds
Then we can solve the ordinary differential equation by writing y (t) = e p(t) e p(s)r (s) ds + Ce p(t)
Here C is our constant of integration. Looking at our equation g ( ) 2 cot g ( ) = k, we make the following identi cations. We set p ( ) = 2 cot and r ( ) = k. First, we integrate p: p ( ) = 2 cot d = 2 cos d = 2 ln (sin ) sin
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