# 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