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The Einstein Field Equations
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This well-known equation has the solution R = A cos ( t) + B sin ( t) where A = A (r ) and B = B (r ). Turning to G , using = 1 and T = 0, we nd 2b + t 2 b t
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+ 2 = 0
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(6.42)
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To nd a solution to this equation, we let f = eb(t,r ) . Therefore f b b = e t t 2 f = 2 t t b b e t 2b = 2 eb + t b t
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2b + t 2
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b t
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And so we can write 2b + t 2 and (6.42) becomes 2 f + 2 f = 0 t 2 Once again, we have a harmonic oscillator-type equation with solution eb = C cos ( t) + D sin ( t) b t
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1 2 f f t 2
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where, as in the previous case, the constants of integration are functions of r .
The Einstein Field Equations
There are two more Einstein s equations in this example that could be used for further analysis. We simply state them here: 2 R R b =0 r t t r e 2b(t,r ) R b R 2b(t,r ) 2 R R b 2 + e t t r r r + 2 =
Energy Conditions
Later, we will have use for the energy conditions. We state three of them here:
The weak energy condition states that for any timelike vector u a , Tab u a u b 0. The null energy condition states that for any null vector l a , Tab l a l b 0. The strong energy condition states that for any timelike vector u a , Tab u a u b 1 c d T cu ud . 2
Quiz
1. Using the Bianchi identities, a Rdebc + c Rdeab + b Rdeca = 0, it can be shown that the contracted Bianchi identities for the Einstein tensor are (a) b R ab = 0 (b) b G ab = T ac (c) b G ab = 0 (d) b G ab = Consider Example 6-2. Using Cartan s equations 1 2 (a) R t t = R 2 tR 2
(b) R t t = (c) R t t = 3.
1 2 R R t 2 1 2 R 2 tR 2
The best statement of the strong equivalence principle is (a) the laws of physics are the same in an accelerated reference frame and in a uniform, static gravitational eld
The Einstein Field Equations
(b) tidal forces cannot be detected (c) inertial and accelerated reference frames cannot be differentiated 4. The Einstein equations are (a) 2 = 4 G (b)
D 2 a D 2
= R a bcd u b u c d
(c) b T ab = 0 (d) G ab = Tab Consider the following metric: ds 2 = dt 2 + L 2 (t, r ) dr 2 + B 2 (t, r ) d 2 + M 2 (t, r ) dz 2 5. The Ricci rotation coef cient 1 (a) L L t
1 (b) L L t
trr
(c) (d) 6.
1 L L 2 t B t
r
is given by
(a) (b)
1 B B t 1 M M t
(c) L1B B r
B (d) L1B t
Taking Tt t = and setting the cosmological constant equal to zero, show that the Einstein equation for G t t becomes 7. (a) (b) (c) B
L B L BL 3
M L M L ML 3
BM B M L2
BL BL
ML ML
BM BM
=
B L B L BL 3
M L M L ML 3
=
BM B ML 2
L B L BL 3
M L M L ML 3
BL BL BL
ML ML
BM BM
The Einstein Field Equations
The Ricci scalar is given by 2 2 2 2 2 2 R = L L + B tB + M tM + 2 2 t 2 (a) (b) R = (c) R =
M + B2 t M
+ L 32M L r
2 2 L L t 2 2 2 L L 2 t 2
2 LB 2 B L 2B B L 22M 2 t r 2 M M L 2 2 M B r r B r
L B M + L2 L t t t M t 2 M + L 2B L B 3 r 2 r r
2 2 B B t 2 2 2 B B t 2
2 2 M M t 2 2 2 M M t 2
2 L B L B t t 2 L B L B t t
2 L M L M t t 2 L M L M t t
CHAPTER
The EnergyMomentum Tensor
In general relativity, the stress-energy or energy-momentum tensor T ab acts as the source of the gravitational eld. It is related to the Einstein tensor and hence to the curvature of spacetime via the Einstein equation (with no cosmological constant) G ab = 8 GT ab The components of the stress-energy tensor can be arranged into a matrix with the property that T ab = T ba ; i.e., the stress-energy tensor is symmetric. In the following and throughout the book, we will use the terms stress-energy tensor and energy-momentum interchangeably. Let s see how to describe each component of the energy-momentum tensor. To understand the meaning of the component T ab , consider the surface de ned by constant x b . Then T ab is the ux or ow of the a component of momentum
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