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dr = in .NET framework
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The next step is to rewrite the expression we ve derived in terms of the basis one forms. When doing these types of calculations, it is helpful to write the coordinate differentials in terms of the basis one forms. Looking at (5.16), we quickly see that dt = t , dr = e (t,r ) r , d = 1 , R(t, r ) d = 1 R(t, r ) sin (5.17) Therefore, we obtain d r = = = e t e t
(t,r ) dt dr = t e
(t,r ) e t
(t,r ) t dr
(t,r ) t r t
Now we use (5.13) to write this as d r = r t t (5.18) Now let s use (5.9) to write out what d r should be. We nd d r =
r b
b = t r Comparing this expression with (5.18), we guess that
r t
r t
(5.19) Cartan s Structure Equations
Moving to the next basis one form, we get d = d (R(t, r )d ) = Rdt d + R dr d = = R R dt Rd + e R R e dr Rd
R R t + e r R R
Let s reverse the order of the terms using (5.13). The result is R R d = t e r R R Using (5.9), we have d = b t
b t
r r Comparing this with the result we found just above, we conclude that
R R
R e R
(t,r ) (5.20) Finally, we tackle . This term gives
d = d (R sin d ) = R sin dt d + R sin dr d + cos Rd d R R = dt R sin d + e (t,r ) e (t,r ) dr R sin d R R cos R d R sin d + R sin Cartan s Structure Equations
Writing the differentials in terms of the basis one forms, and then reversing the order of each term, we get d = R cot R t + e (t,r ) r + R R R R R cot = t e (t,r ) r R R R Using (5.9), we have d =
t t r Comparing this with the result we obtained above, we conclude that
R , R
R e R
(t,r ) , cot R
(5.21) Earlier we noted that we could obtain no information by calculating d t . For the other terms, we have basically gone as far as we can using the guess method. Now we can use the symmetry properties shown in (5.12) to nd the other curvature one forms. Speci cally, we have r t R t t = = R R t t = = R t r
Now paying attention only to spatial indices, we conclude that
r = = = = = R e R R e R
(t,r ) (t,r ) cot R
Cartan s Structure Equations
Now that we have calculated the curvature one forms, we can obtain the Ricci rotation coef cients using (5.10), which we restate here: a b
bc
We start by considering t b . It turns out that term. Moving on, we have
t r
= 0, and so we can skip this
t r
t t rt
r t rr (t,r ) r . t
t r
t r
We noted above that conclude that
= Comparing the two expressions, we
t rr
= = (t, r ) t =
t r
(5.22) = 0. Moving to the next co In addition, we conclude that ordinate, we have
t t rt
t r
c t r R Above we found that t = t = R , and so the only nonzero term is that involving , and we conclude that R R
(5.23) Similarly, we nd
t R R
b. (5.24) First we have
r t
Now we move on to terms of the form
r t
tt
tr
t Cartan s Structure Equations
r t
Now, in (5.19), we found that us to conclude that
= (t, r ) r . Comparison of the two leads
tr
t Next, skipping the
r term since it vanishes by (5.12) (set i = j), we go to
c t r r Comparison with our earlier result, where we found that R e (t,r ) , leads us to conclude that R r R e R
(t,r ) Finally, we have
r c r
t r Earlier we found that be the case that
= R e R
(t,r ) . Comparing this to the above, it must
R e R
(t,r ) A similar procedure applied to terms of the form Ricci rotation coef cients
t b and
b gives the remaining
t R , R
r r R e R
(t,r ) cot R
We have nished calculating the Ricci rotation coef cients, which can be used to obtain the Christoffel symbols or as we will see later, to compute the curvature. At this point, we can demonstrate how to transform these quantities to the coordinate frame to give the Christoffel symbols. This is not strictly necessary but it may be desired. Let s consider a simple example; the procedure is the same when applied to all terms. We repeat the formula we need here, namely (5.14): a bc
1 a e f
Cartan s Structure Equations
First, we need to construct the transformation matrix. This is easy enough to do; we just read off the coef cients of the metric. To keep you from having to ip back several pages we restate it here: ds 2 = dt 2 e 2 (t,r ) dr 2 R 2 (t, r ) d 2 R 2 (t, r ) sin2 d 2
The diagonal elements of the matrix can just be read off the metric. Therefore, the transformation matrix is 1 0 0 e (t,r ) = 0 0 0 0 0 0 R(t, r ) 0 0 0 0 R(t, r ) sin 0 0 0

