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asp net display barcode Shadow cast by converging rays in Visual Studio .NET
Shadow cast by converging rays Scan QR Code JIS X 0510 In VS .NET Using Barcode Control SDK for .NET framework Control to generate, create, read, scan barcode image in Visual Studio .NET applications. QR Code Creator In .NET Using Barcode maker for VS .NET Control to generate, create QR Code ISO/IEC18004 image in VS .NET applications. Size of object
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Drawing EAN13 In None Using Barcode printer for Online Control to generate, create EAN13 Supplement 5 image in Online applications. Draw GS1  13 In ObjectiveC Using Barcode creator for iPad Control to generate, create EAN / UCC  13 image in iPad applications. Fig. 138. The shadow of the object with no twist shown with the solid line. The dashed
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Painting GTIN  128 In VB.NET Using Barcode creation for .NET framework Control to generate, create GS1128 image in .NET framework applications. Generating Barcode In None Using Barcode encoder for Software Control to generate, create bar code image in Software applications. twist, and shear of a null congruence: Re ( ) expansion Im ( ) twist   shear These quantities are interpreted in the following way. For the sake of understanding, we think of the null congruence as a set of light rays. Now imagine that an object is in the path of the light rays and it casts a shadow on a nearby screen. (see Fig. 138) The expansion can be understood as seeing the cast shadow either larger or smaller than the object. That is, if the shadow is larger, the light rays are diverging while a smaller shadow indicates that the light rays are converging. The twist is described in this thought experiment by a rotation of the shadow. (13.27) Fig. 139. Shear distorts the shadow of the object. We show the shadow with zero shear
as a circle with the solid line. Shear distorts the circle into the ellipse, shown with the dashed line. Gravitational Waves
To understand shear, we consider an object that casts a perfect circle as a shadow if no shear is present. If the shear is nonzero, the shadow will be cast as an ellipse (see Fig. 139). The spin coef cient used to de ne shear is a complex number. The magnitude   determines the amount of stretching or shrinking of the axes that de ne the ellipse while the phase of de nes the orientation of the axes. pp Gravity Waves
We now consider a more formal and generalized study of gravitational waves abstracted from sources and propagating at the speed of light. In particular, we study vacuum ppwaves where pp means plane fronted waves with parallel rays. Considered to travel in a at background spacetime described by the Minkowski metric, a ppwave is one that admits a covariantly constant null vector eld. We now consider this de nition. Let k a be a null vector such that the covariant derivative vanishes; i.e., ka,b = 0 (13.28) We say that a vector k a that satis es (13.28) is covariantly constant. Recalling that we can de ne a null tetrad for a given metric by the vectors (l a , n a , m a , m a ), in the case of a ppwave spacetime it is possible to take l a as the covariantly constant null vector. This null vector eld is taken to correspond to the rays of gravitational waves. In 9 we learned that the covariant derivative of l a shows up in the de nitions of several spin coef cients. Speci cally, we have = la,b m a m b and = la,b m a m b (13.29) This tells us that if l a is covariantly constant, then the gravitational wave of a ppwave spacetime has no expansion or twist, and the shear vanishes.
Physically, the fact that null congruence has zero expansion means that the wave surfaces are planes. Furthermore, noting that the spin coef cient is given by = la,b m a n b (13.30) Gravitational Waves
we see that in this case we also have = 0. This implies that the null rays are parallel. Plane waves are only a special class of the more general ppwaves. However, for simplicity we will focus on plane waves for the time being. First we will introduce two null coordinates U and V , which are de ned as follows: U =t z Inverting these relations, we have t= Therefore we have dt = nd U +V 2 V =t +z
(13.31) (U V ) 2 (dU + dV ) and dz = 1 (dU dV ). Squaring, we 2 1 dU 2 + 2 dU dV + dV 2 4 1 dU 2 2dU dV + dV 2 4

