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The Schwarzschild Solution in VS .NET
The Schwarzschild Solution QRCode Recognizer In .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in .NET framework applications. Paint QR Code JIS X 0510 In .NET Using Barcode creator for Visual Studio .NET Control to generate, create QR Code image in VS .NET applications. radius. In this case, we have Veff (r = 2m) = l2 l2 m l 2m 1 l 2m 1 + = + = 2 3 2 3 2m 2 (2m) 2 8m 8m 2 (2m) Decode QR In .NET Framework Using Barcode scanner for VS .NET Control to read, scan read, scan image in .NET framework applications. Draw Bar Code In Visual Studio .NET Using Barcode generation for VS .NET Control to generate, create bar code image in VS .NET applications. For large r , the potential reduces to the newtonian potential Veff (r ) m r
Bar Code Decoder In VS .NET Using Barcode reader for VS .NET Control to read, scan read, scan image in .NET applications. Encode QR Code In C# Using Barcode creation for .NET framework Control to generate, create Quick Response Code image in Visual Studio .NET applications. Looking once again at (10.51), notice that the term under the square root becomes negative when the angular momentum l 2 is less than 12m 2 . We then get the unphysical result where the radius is a complex number. This tells us that in this case no stable circular orbit can exist. Physically speaking, if l 2 is less than 12m 2 , this indicates that the orbiting body will crash into the surface of the star. If the body happens to be approaching a black hole under these conditions, it will simply be swallowed up by the black hole. These results are illustrated in Figs. 101 and 102. By comparing the curves in the two gures, you can see that as r gets large the newtonian and relativistic cases converge. At small r , the differences are quite dramatic. This tells us that the best place to look for relativistic effects in the solar system is near the Sun. QR Code 2d Barcode Encoder In .NET Using Barcode maker for ASP.NET Control to generate, create QR Code ISO/IEC18004 image in ASP.NET applications. Creating QR In VB.NET Using Barcode drawer for Visual Studio .NET Control to generate, create QR Code image in .NET framework applications. Fig. 101. Plots of the effective potential for general relativity. For simplicity, we used a
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There are four standard or classical tests of general relativity that have been applied within the solar system. These include the precession of the perihelion of Mercury, the bending of light passing near the Sun, the travel time of light in a Schwarzschild eld, and the gravitational red shift. These phenomena are covered in all the major textbooks. We will consider two tests involving light and begin by considering the derivation of an equation for the trajectory of a light ray (see Fig. 103). This derivation will follow that of the previous section with some small differences. We can assume once again that the motion will take place in a plane and set = /2. Moreover, from special relativity we know that the path of a light ray lies on a light cone, and so can be described by the case ds 2 = 0. These considerations mean that the equation following (10.44) becomes 2m r 2m r t2 1 r 2 r 2 2 = 0
(10.52) The Schwarzschild Solution
Proceeding as we did for massive particles, we use de nitions (10.47) and (10.48) to write (with the difference that differentiation is not with respect to the proper time , instead we take the derivative with respect to some parameter we denote by ) 2m e = 1 r t2 and l 2 = r 4 2
where we have used dots to denote differentiation with respect to . Then (10.52) becomes 1 2m r
e2 1 2m r
r2 l2 =0 r2
(10.53) To nd the trajectory, we are interested in obtaining an expression for r = r ( ). We will now use primes to denote differentiation with respect to . With this in mind, let s rewrite r as r= dr dr = d d d d = dr d d d =r Now we can use r 2 = l to write rl r =r = 2 r To write the equation in a more convenient form, we introduce a new variable u = 1/r . Note that u = and so, we can write r= l rl = r 2 u 2 = lu 2 r r 1 r = 1 r r2 Returning to (10.53), we multiply through by (1 2m/r ) and set u = 1/r to obtain e2 l 2 u 2 l 2 u 2 (1 2mu) = 0

