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Now, the covariant derivative of the metric tensor vanishes, c gab = 0. The covariant derivative of the metric tensor is given by c gab = c gab d ac gdb d bc gad . Since this vanishes, we can write c gab =
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Let s use this to rewrite the Lie derivative of the metric tensor. We have L X gab = X c c gab + gcb a X c + gac b X c = Xc = gdb
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ac X
+ gad
Some manipulation can get this in the form we need to write down covariant derivatives of X . Note that b X c = b X c +
In our expression we just derived for the Lie derivative the last term we have is gac b X c . Can we nd the other term needed to write down a covariant derivative Yes, we can. Remember, a repeated index is a dummy index, and we are free to call it whatever we want. Looking at the last line we got for the Lie derivative, consider the second term gad
The indices c and d are repeated, so they are dummy indices. Let s switch them c d and rewrite this term as gac
First, let s write down the result for the Lie derivative and rearrange the terms so that they are in the order we want for a covariant derivative. LX gab = gdb
d ac X c
+ gad
bc X c c
+ gcb a X c + gac b X c
d bc X c
= gac b X c + gdb = gac b X c + gad
ac X
+ gad
+ gcb a X c
d ac X c
bc X
+ gcb a X c + gdb
At this point, we put in the change of indices we used on what is now the second term in this expression. LX gab = gac b X c + gad = gac b X c + gac = gac b X c +
c d c bc X bd X d c d
+ gcb a X c + gdb + gcb a X c + gdb
d d ac X c
ac X
ac X c
bd X
+ gcb a X c + gdb
ac X
= gac b X c + gcb a X c + gdb
Now we switch indices on the last term and get L X gab = gac b X c + gcb a X c + gdb = gac b X c + gcb a X c + gcb = gac b X c + g cb a X c + = gac b X c + gcb a X c
c d c
Killing Vectors
ac X ad X d
ad X
Since the covariant derivative of the metric vanishes, we move the metric tensor inside the derivative and lower indices. For the rst term we nd gac b X c = b (gac X c ) = b X a and for the second term we obtain gcb a X c = a (gcb X c ) = a X b Therefore, we have LX gab = gac b X c + gcb a X c = b X a + a X b Since we are given that L X gab = 0, this implies (8.1). Often, we need to nd the Killing vectors for a speci c metric. We consider an explicit example by nding the Killing vectors for the 2-sphere. EXAMPLE 8-2 Use Killing s equation to nd the Killing vectors for the 2-sphere: ds 2 = a 2 d 2 + a 2 sin2 d 2 SOLUTION 8-2 Killing s equation involves covariant derivatives. Therefore we need to recall the af ne connection for this metric. In an earlier chapter we found
= cot
= sin cos
Now, we recall that the covariant derivative is given by b V a = b Va
c ab Vc
Killing Vectors
Starting with a = b = in Killing s equation, we nd X + X = 0 X = 0 Using the equation for the covariant derivative, and recalling the Einstein summation convention, we obtain X = X Since
c
Vc = X
V
= 0, this reduces to the simple equation X = 0
More explicitly, we have X =0 Integrating, we nd that the X component of our Killing vector is some function of the variable: X = f ( ) Next, we consider a = b = . Using Killing s equation, we obtain X = 0 Working out the left-hand side by writing out the covariant derivative, we nd X = X Now, = 0 and becomes
c
(8.3)
= X
X
X
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