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visual basic print barcode label Mathematical Aside: The Euler Characteristic in Java
Mathematical Aside: The Euler Characteristic QR Scanner In Java Using Barcode Control SDK for Java Control to generate, create, read, scan barcode image in Java applications. Painting QR Code ISO/IEC18004 In Java Using Barcode generation for Java Control to generate, create QR Code image in Java applications. The Euler characteristic is a number which describes the shape of a topological space. Consider a polyhedron, and let V be the number of vertices, E be the number of edges, and F be the number of faces. Then the Euler characteristic is Recognize QR Code In Java Using Barcode scanner for Java Control to read, scan read, scan image in Java applications. Drawing Bar Code In Java Using Barcode maker for Java Control to generate, create barcode image in Java applications. =V E+F
Decode Barcode In Java Using Barcode decoder for Java Control to read, scan read, scan image in Java applications. QRCode Printer In C# Using Barcode creation for .NET framework Control to generate, create Quick Response Code image in Visual Studio .NET applications. (2.28) Encoding QR Code JIS X 0510 In VS .NET Using Barcode generator for ASP.NET Control to generate, create QRCode image in ASP.NET applications. QRCode Printer In VS .NET Using Barcode maker for .NET framework Control to generate, create QR Code JIS X 0510 image in .NET framework applications. In string theory, we often want to know whether or not two geometric shapes or topologies are similar to one another in a speci c way. In particular, we want to know if we can continuously deform one shape into another (imagine working with clay and deforming one shape into another without breaking the clay apart, or introducing or losing any holes). Formally, a homeomorphism is a deformation of a geometric object into a new shape by stretching or compressing and being it, without tearing or breaking it. For instance, the quintessential example is a donut and a coffee cup (conveniently paired for police of cers). You could use a continuous deformation to transform one into the other or vice versa. So we say that a coffee cup and a donut are homeomorphic. On the other hand, a sphere and a donut are not Painting QR In Visual Basic .NET Using Barcode drawer for .NET Control to generate, create QRCode image in .NET framework applications. GS1  12 Generation In Java Using Barcode printer for Java Control to generate, create Universal Product Code version A image in Java applications. CHAPTER 2 Equations of Motion
Encoding Bar Code In Java Using Barcode generator for Java Control to generate, create bar code image in Java applications. Linear Printer In Java Using Barcode generation for Java Control to generate, create Linear Barcode image in Java applications. homomorphic the donut has a hole but a sphere does not. The bottom line is there is no way to transform the donut into the sphere. If a geometric shape is homeomorphic to a sphere, then the Euler characteristic is (2.29) Many shapes have an Euler characteristic which vanishes. Some examples of this include a torus, a m bius strip, and a Klein bottle. Another example is a cylinder, which also has = 0 (see Fig. 2.2). Why is this interesting for us If the worldsheet of a string has a vanishing Euler characteristic, then it is possible to write the auxiliary eld h as a twodimensional at space metric. That is, we take [using the choice of coordinates for the worldsheet as ( , )] 1 0 h = 0 1 (2.30) Create 2 Of 7 Code In Java Using Barcode drawer for Java Control to generate, create Code27 image in Java applications. Bar Code Encoder In C#.NET Using Barcode printer for Visual Studio .NET Control to generate, create barcode image in .NET framework applications. =V E+F = 2
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Cylinder
Figure 2.2 Some surfaces with a vanishing Euler characteristic. When the Euler characteristic vanishes, we can de ne the auxiliary eld such that it has a representation of the at space Minkowski metric. String Theory Demysti ed
Now notice that with this choice, h = det h = 1. We also have h X X = X X + X X = X 2 + X 2 In this case, we are able to write the Polyakov action in the remarkably simple form SP = T 2 d ( X 2 X 2 ) 2 (2.31) EXAMPLE 2.3 Find the equations of motion using the Polyakov action as written in Eq. (2.27) when the auxiliary eld takes the form of the at space metric. SOLUTION In this case we have SP = = = T 2 d h h X X 2 T 2 d ( X X + X X ) 2 T 2 d ( X X + X X ) 2 So, we can write the lagrangian as L = X X + X X = X X + X X Therefore, L = X X + X X = X = X X X L = X X + X X = X = X X X
CHAPTER 2 Equations of Motion
The EulerLagrange equations are L L + =0 X X Hence, we nd that the equations of motion for the relativistic string are 2 X 2 2 X 2 =0 (2.32) (2.33) LightCone Coordinates
It will be convenient to call upon lightcone coordinates in string theory. First, let s look at how lightcone coordinates can be de ned in Minkowski spacetime in general and then consider having them in the context of the worldsheet and the equations of motion of the string. As we will see, this will simplify the way we write the action and the resulting equations of motion. For simplicity, let s take ordinary (3 + 1) dimensional spacetime. The contravariant coordinates are x = ( x 0 , x1 , x 2 , x 3 ) where x 0 = ct and x 1 = x , x 2 = y, x 3 = z say. We form lightcone coordinates by choosing one spatial direction, which in this case we take to be x1, and forming linear combinations of it with x0 as follows: x+ = x 0 + x1 2 x = x 0 x1 2 (2.34) These are two null or lightlike coordinates, but you can think of x+ as a timelike coordinate and x as a spacelike coordinate. Hence when we use indices and summations, we will treat + as a 0 index and as a 1 index. The other coordinates x2 and x3 are left alone. It is easy to derive the inverse relationship using Eq. (2.34). We have x0 = x+ + x 2 x1 = x+ x 2

