Now let s plug this into the exponential and we get in Visual Studio .NET

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Now let s plug this into the exponential and we get
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2 e p( ) = e2 ln(sin ) = eln(sin ) = sin2
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From this we deduce that e p( ) = sin12 . Now we can use the integrating factor formula to write down a solution for the function g. The formula together with what we ve just found gives us g ( ) = sin2 ( k) sin t
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dt + C sin2
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In this case, t is just a dummy variable of integration. When we integrate we will write the functions in terms of the variable. It s the easiest to look up the
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integral in a table or use a program like Mathematica, and if you can t remember you ll nd out that 1 dt = cot sin2 t and so, we get the following: g ( ) = sin2 dt + C sin2 = sin2 k cot + C sin2 sin2 t = sin2 (k cot + C) ( k)
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The nal piece is to get a solution for the term. Earlier, we had f ( ) d + f ( ) = 2 cot g ( ) g ( ) and we decided to set this equal to some constant that we called k. So looking at the piece, we obtain f ( )d + f ( ) = k Getting a solution to this one is easy. Let s differentiate it. That will get us rid of the integral and turn the constant to zero, giving us the familiar equation d2 f + f ( ) = 0 d 2 This is a familiar equation to most of us, and we know that the solution is given in terms of trignometric functions. More speci cally, we have f ( ) = A cos + B sin . We quickly see that f = A sin + B cos . Even better, we can explicitly calculate the integral that has been hanging around since we started this example. We get the following: f d = (A cos + B sin ) d = A sin B cos
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This is a really nice result. Remember, we had de ned the constant k in the following way: f ( ) d + f ( ) = k
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What this tells us is that the constant k must be zero. Using what we have just found, we have f d + f ( ) = A sin B cos A sin + B cos = 0
This is a real headache, but we re almost done. Remember, for our function we found g ( ) = sin2 (k cot + C) Since k = 0, we obtain the simple result g ( ) = C sin2 Finally, at this point we have all the pieces we need to write down our Killing vector. Looking at (8.3), we recall that we had X = f ( ). With the results we ve obtained, we have X = A cos + B sin In (8.4) we determined that X = sin cos f ( ) d + g ( )
And using the results we ve obtained, we get the nal form for this component of the vector: X = sin cos (A sin B cos ) + C sin2 The contravariant components of the Killing vector can be found by raising indices with the metric. It turns out that X = X and sin2 X = X
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It is going to turn out that for the 2-sphere, we are going to be able to write this Killing vector in terms of the angular momentum operators. Let s start by writing the entire vector out instead of individual components, and then do some algebraic manipulation: X = X + X = (A cos + B sin ) + [C cot (A sin B cos )] = A cos A cot sin + B sin + A cot cos + C = AL x + BL y + CL z where the angular momentum operators are given by L x = cos + cot sin L y = sin + cot cos Lz =
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