The Schwarzschild Solution in .NET

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The Schwarzschild Solution
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Squaring both sides and solving for dr 2 , we nd that dr 2 = 1 C 1+ r dC 2C dr
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d 2
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1 r Now let s rede ne the coef cient function B = C 1 + 2C dC B so that dr 2 2 we have Bdr = B d . With these conditions in place, we can rewrite (10.4) as
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ds 2 = A dt 2 B d 2 2 d 2 + sin2 d 2 where A is a function of . Up to this point, other than the requirement that the coef cient functions go to unity as the radial coordinate gets large, we have not imposed any requirements on the form that they must take. Therefore we ve basically been using nothing but labels. So we can relabel everything again, dropping primes on the coef cient functions and changing r , allowing us to write the line element as ds 2 = Adt 2 Bdr 2 r 2 d 2 + sin2 d 2 Now we impose one last requirement. In order to correspond to the metric (10.2) at large r , we also need to preserve the signature. This can be done by writing the coef cient functions as exponentials, which are guaranteed to be positive functions. That is, we set A = e2 (r ) and B = e2 (r ) . This gives us the metric that is used to obtain the Schwarzschild solution: ds 2 = e2 (r ) dt 2 e2 (r ) dr 2 r 2 d 2 + sin2 d 2 (10.5)
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The Curvature One Forms
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We will proceed to nd the solution, using the orthonormal tetrad method. By now the reader is aware that the rst step along this path is to calculate the curvature one forms and the Ricci rotation coef cients. To do so, we de ne the following basis one forms: t = e (r ) dt, r = e (r ) dr, = r d ,
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= r sin d
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(10.6)
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Therefore, we have dt = e (r ) t ,
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The Schwarzschild Solution
dr = e (r ) r ,
1 d = , r
d =
1 r sin
(10.7)
The exterior derivative of each of the basis one forms is given by d t = d e (r ) dt = d r = d e (r ) dr =
d (r ) d (r ) r t e dr dt = e dr dr d (r ) e dr dr = 0 dr e (r ) r r
(10.8) (10.9) (10.10)
d = d (r d ) = dr d =
d = d (r sin d ) = sin dr d + r cos d d = cot e (r ) r + r r (10.11)
Once again recall Cartan s rst structure equation, which for our coordinates will assume the form
d a = a b
b =
a t
t
a r
r
a
a
Taking each basis one form in turn gives d t = d r = d =
t t r t t t
t t t t
t r r r r r
r r r r
t r
t r
(10.12) (10.13) (10.14) (10.15)
d =
Comparing (10.12) with (10.8), note that the only nonzero term in (10.8) of the form r t . Therefore, we conclude that
t t
t
t
t r
d (r ) t e dr
(10.16)
The Schwarzschild Solution
Using ab = abc c , we set t r t = d e (r ) . Since (10.9) vanishes and gives no dr direct information, we move on to compare (10.14) and (10.10). Again, with only a single term which this time involves r , we take the nonzero term in (10.14) to be r r and conclude that t
r
e (r ) r
(r )
(10.17)
Again using ab = abc c , we conclude that r = e r . Finally, We compare (10.15) with (10.11) using the same procedure, which gives t
(r )
= 0,
r
e (r ) , r
cot r
(10.18)
cot and where r = e r = r . Now we return to the case of (10.9), which gave us no information since it vanished. To nd the curvature one forms in this case, we will use symmetry considerations. Looking at the metric (10.5), we see that in this case we can de ne 1 0 0 0 0 1 0 0 a b = 0 0 1 0
to raise and lower indices. Therefore, we have the following relationships:
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