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Fig. 6-6. Two curves and in a manifold. Suppose each curve is parameterized by an
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af ne parameter . The deviation vector V connects the curves at two points p and q such that ( 1 ) = p and ( 1 ) = q for some value of the af ne parameter = 1 .
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The Einstein Field Equations
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Now, consider a connecting vector a ,which points from the geodesic of an inertial particle to the geodesic of another, in nitesimally close inertial particle. We say that a is Lie propagated if its Lie derivative with respect to u a vanishes; i.e., L u a = ub b a b b u a = 0 We will also use the following result. EXAMPLE 6-1 Show that a b V c b a V c = R cdab V d SOLUTION 6-1 Recall from 4 that the covariant derivative of a vector is given by b V a = Also note that c T a b = c T a b + Proceeding, we have a b V c = a ( b V c +
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c eb V e a d cd T b
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(6.1)
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We can easily calculate the second derivative by treating b V c + c eb V e as a single tensor. Let s call it S cb = b V c + c eb V e . Then using (6.1), we have a S cb = a S c b + Back substitution of S cb = b V c + a ( b V c +
c eb V e c c d ad Sb e
c ba S d
eb V c
in this expression gives
) = a ( b V c +
d ba
eb V
ed V e
b V d +
eb V
( d V c +
A similar exercise shows that b a V c = b ( b V c +
d ab ( d V c c ea V e
a V d +
ea V
ed V
The Einstein Field Equations
Let s calculate a b V c b a V c term by term. Since partial derivatives commute, and therefore a b V c b a V c = 0, subtraction of the rst terms gives a ( b V c + c eb V e ) b ( b V c + c ea V e ) = a ( = V e ( a c eb b c ea ) + c eb a V e c ea b V e
c eb V e
) b (
ea V
(6.2)
For the remaining terms, we use the fact that we are using a torsion-free connection, and therefore a bc = a cb , so we nd
c ad
b V d +
c bd
eb V
d V c +
ed V
e c ed V e e e
= = =
a V d +
ea V e
d V c +
ad c
b V d +
eb V d
d V c +
ed V c
a V d +
ea V e d
d V c +
ed V e d
b V d +
eb V
a V d +
d eb V e
ea V c bd
ad b V
bd a V
ea V
We can go further by relabeling dummy indices and again using the fact that the Christoffel symbols are symmetric in the lower indices to rewrite this term as
c ad b V d
bd a V c c
d c c
eb V
d c c
ea V
e ea V ea V e e
ae b V ea b V
be a V eb a V
ad ad
eb V eb V
bd bd
To get the nal result, we add this expression to (6.2). However, notice that e c e ea b V eb a V will cancel similar terms in (6.2), and so we are left with
c eb
a b V c b a V c = V e a = a = ( a
c c eb
b
c c ea da
c ad ae c
ad eb db
eb V c c
c ea
ea V
b b
bd be
da ) V
In the last line, we swapped the dummy indices d e. From (4.41), we see that R a bcd = c
a bd
d
The Einstein Field Equations
and therefore we conclude that
c a b V c b a V c = Rdab V d
(6.3)
Furthermore, since the Lie derivative vanishes, we can write u b b a = b b u a (6.4)
Now c is a measure of the distance between two inertial particles. From elementary physics, you recall that velocity is the time derivative of position, i.e., v = dx/dt and acceleration is a = d2 x/dt 2 . For inertial geodesics with tangent vector u a and parameter , by analogy, we de ne the relative acceleration between two geodesics by D2 a = u b b (u c c a ) D 2 = u b b ( c c u a ) = u b ( b c c u a + c b c u a ) = u b b c c u a + u b c b c u a We can use (6.3) to write the last term as b c u a = c b u a + R a dbc u d to obtain D2 a = u b b c c u a + u b c c b u a + R a dbc u d 2 D = u b b c c u a + c u b c b u a + c u b u d R a dbc = b b u c c u a + c u b c b u a + c u b u d R a dbc Relabeling dummy indices, we set c u b c b u a = b u c b c u a and we have D2 a = b b u c c u a + b u c b c u a + c u b u d R a dbc D 2 Now the Leibniz rule b (u c c u a ) = b u c c u a + u c b c u a and so this becomes D2 a = b b u c c u a + b u c b c u a + c u b u d R a dbc D 2 = b ( b u c c u a + u c b c u a ) + c u b u d R a dbc = b ( b (u c c u a )) + c u b u d R a dbc
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