qr code reader c# .net Comparison with (5.30) gives R = 1 in VS .NET

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Comparison with (5.30) gives R = 1
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Another application of the symmetries of the Riemann tensor shows us that R = R = +1
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Cartan s Structure Equations
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R = 0
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R = R = R ,
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The second line holds true because Rabcd = Rbacd and if a = a a = 0. This also holds for R . At this point, we can calculate the Ricci tensor and the Ricci scalar. Since the latter quantity is a scalar, it is invariant and it is not necessary to transform to the global coordinates. From the previous argument, we nd
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R = R a a = R + R = 0
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The other terms are
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R = R a a = R + R = 0 + 1 = 1 R = R a a = R + R = 1 + 0 = 1
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(5.31)
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From this, it is a simple matter to calculate the Ricci scalar
R = a b Ra b = R + R = R + R = 1 + 1 = 2
The Ricci scalar can be used to give a basic characterization of the intrinsic curvature of a given geometry. The value of the Ricci scalar tells us what the geometry looks like locally. If the Ricci scalar is positive, as it is in this case, the surface looks like a sphere. If it is negative, then the surface looks like a saddle. We can think about this in terms of drawing a triangle. If R > 0, then the angles add upto more than 180 degrees, while if R < 0 they add up to less than 180 degrees. These observations lead to the designations of positive curvature and negative curvature . Now, if R = 0, then the geometry is at and the angles of a triangle add upto the expected 180 degrees. EXAMPLE 5-3 The Robertson-Walker metric ds 2 = dt 2 + a 2 (t) dr 2 + a 2 (t)r 2 d 2 + a 2 (t)r 2 sin2 d 2 2 1 kr
describes a homogeneous, isotropic, and expanding universe. The constant k = 1, 0, 1 depending on whether the universe is open, at, or closed. Find the components of the Riemann tensor using tetrad methods.
Cartan s Structure Equations
SOLUTION 5-3 Looking at the metric, we see that we can de ne the following orthonormal basis of one forms: t = dt, r = a(t) 1 kr 2 dr, = ra(t)d ,
= ra(t) sin d
(5.32)
From this point on to save a bit of writing we will write a(t) = a. Using (5.9) to calculate the curvature, one forms gives d t = d (dt) = 0 a adt dr a dr = = r t d r = d a 1 kr 2 1 kr 2 d = d (rad ) = r adt d + adr d
a 1 kr 2 ra a t dr d = + a 1 kr 2 ra 1 kr 2 r ra 1 kr 2 a t = r a ra a = t + a d = d (ra sin d )
= r a sin dt d + a sin dr d + ra cos d d cot a 1 kr 2 t = r a ra ra Using the following relations found by writing out the right-hand side of (5.9) d r = d =
r t t t
t t t
r r
r r r
d =
Cartan s Structure Equations
we read off the following curvature one-forms:
a r t = , a
a t = , a
a r = t = , a 1 kr 2 , r = ra
1 kr 2 ra
(5.33)
cot = ra
We can raise and lower indices using a b . Notice the form of the metric, which indicates that in this case we must set a b = diag( 1, 1, 1, 1). Let s see how this works with the curvature one forms:
r t r
= r r =
rt r
rt
= =
tr
= t t = r r
t r
r
t r
r
r
Now we calculate the curvature two forms using (5.27). We explicitly calculate r:
r
=d =d =d
r r r
+ + +
c t t
t r t r
r
r r
Now, for the rst term in this expression we have
1 kr 2 1 kr 2 (rad ) = = = ra ra
1 kr 2 d
Therefore d
r
=d =
1 kr 2 d = k r a2
kr 1 kr 2
dr d
For the other two terms, we obtain
t
t r
a a r a2 = 2 r a a a
Cartan s Structure Equations
cot = ra
1 kr 2 =0 ra
Therefore, the curvature two form is
= =
k r a2 + 2 r 2 a a
k a2 r + 2 r 2 a a (5.34)
a 2 + k r = 2 a
Using (5.28), we can obtain the components of the Riemann tensor. First, using the symmetries of the Riemann tensor together with a b = diag( 1, 1, 1, 1), note that R r r = R r r = R r r = R r r = R r r = R r r
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