Now taking a = , b = r, c = , we nd d2 1 dr d =0 + ds 2 r ds ds in Visual Studio .NET

Creating QR Code ISO/IEC18004 in Visual Studio .NET Now taking a = , b = r, c = , we nd d2 1 dr d =0 + ds 2 r ds ds

Now taking a = , b = r, c = , we nd d2 1 dr d =0 + ds 2 r ds ds
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Finally, since
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= 0 for all possible indices, we obtain d2 z =0 ds 2
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The geodesics are obtained by solving for the functions r = r (s), = (s), and z = z (s). Since we are in at space we might guess these are straight lines. This is easy to see from the last equation. Integrating once, we get dz = ds where is a constant. Integrating again we nd z(s) = s + where is another constant. Recalling y = mx + b, we see this is nothing other than your plain old straight line. The geodesic equation provides a nice shortcut that can be used to obtain the Christoffel symbols. Following D Inverno (1992), we make the following de nition: K = 1 gab x a x b 2 (4.35)
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where we are using the dot notation to refer to derivatives with respect to the af ne parameter s. We then compute K d = a x ds K xa (4.36)
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for each coordinate and compare the result to the geodesic equation. The Christoffel symbols can then simply be read off. We demonstrate the technique with an example. EXAMPLE 4-10 The metric for Rindler space (see Fig. 4-3) can be written as ds 2 = 2 d 2 d 2 Use the geodesic equation and (4.35) to nd the Christoffel symbols.
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SOLUTION 4-10 For the Rindler metric, using (4.35) we nd K = 1 2 2 2 2 (4.37)
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Now we wish to nd (4.36), which for the coordinate gives K =0 and so we have d ds K =0 (4.38)
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Using this result together with (4.37), we obtain d ds K = 2 + 2 (4.39)
Setting this equal to zero (due to (4.38)) and comparing with the geodesic equation, we nd 2 + = 0 2 = =
A similar procedure applied to K using the coordinate shows that
(4.40)
The Riemann Tensor
The nal piece of the tensor calculus puzzle that we need to nd the curvature for a given metric is the Riemann tensor (sometimes called simply the curvature tensor). In terms of the metric connection (Christoffel symbols) it is given by R a bcd = c
a bd
d
(4.41)
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Fig. 4-3. Rindler coordinates describe the motion of observers moving in Minkowski
spacetime that have constant acceleration. In the solved example, we consider an observer undergoing constant acceleration in the +x direction. The metric shown is the metric for region 1.
We can lower the rst index with the metric, which gives
e Rabcd = gae Rbcd
This allows us to write (4.41) as Rabcd = c
d
(4.42)
Using (4.15), it follows that we can write the Riemann tensor in terms of the metric as Rabcd = 1 2 + 2 gbc 2 gac 2 gbd 2 gad + a d b d a c x b x c x x x x x x
e ade bc
(4.43)
The Riemann tensor has several important symmetries. We state them without proof (they are easy, but tedious to derive using the de nitions) Rabcd = Rcdab = Rabdc = Rbacd Rabcd + Racdb + Radbc = 0 (4.44)
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The Riemann tensor also satis es the Bianchi identities a Rdebc + c Rdeab + b Rdeca = 0 (4.45)
All together, in n dimensions, there are n 2 n 2 1 /12 independent nonzero components of the Riemann tensor. This fact together with the symmetries in (4.44) tells us that in two dimensions, the possible nonzero components of the Riemann tensor are R1212 = R2121 = R1221 = R2112 while in three dimensions the possible nonzero components of the Riemann tensor are R1212 = R1313 = R2323 = R1213 = R1232 = R2123 , R1323 = R3132
Computation of these quantities using a coordinate basis is extremely tedious, especially when we begin dealing with real spacetime metrics. In the next chapter we will introduce noncoordinate bases and a preferred way of calculating these components by hand. Calculations involving (4.41) are very laborious and should be left for the computer (using GR tensor for example). Nonetheless, it is good to go through the process at least once, so we shall demonstrate with a simple example. EXAMPLE 4-11 Compute the components of the Riemann tensor for a unit 2-sphere, where ds 2 = d 2 + sin2 d 2 SOLUTION 4-11 The nonzero Christoffel symbols are
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