Special Relativity in .NET framework

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Special Relativity
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where L is a 4 4 matrix. Given that the two frames are in standard con guration, the y and z axes are coincident, which means that y =y and z =z
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To get the form of the transformation, we rely on the invariance of the speed of light as described in postulate 2. Imagine that at time t = 0 a ash of light is emitted from the origin. The light moves outward from the origin as a spherical wavefront described by c2 t 2 = x 2 + y 2 + z 2 Subtracting the spatial components from both sides, this becomes c2 t 2 x 2 y 2 z 2 = 0 Invariance of the speed of light means that for an observer in a frame F moving at speed v with respect to F, the ash of light is described as c2 t 2 x 2 y 2 z 2 = 0 These are equal, and so c2 t 2 x 2 y 2 z 2 = c2 t 2 x 2 y 2 z 2 Since y = y and z = z, we can write c2 t 2 x 2 = c2 t 2 x 2 (1.16) (1.15)
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Now we use the fact that the transformation is linear while leaving y and z unchanged. The linearity of the transformation means it must have the form x = Ax + Bct ct = C x + Dct We can implement this with the following matrix [see (1.14)]: D B L= 0 0 C A 0 0 0 0 0 0 1 0 0 1 (1.17)
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Special Relativity
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Using (1.17), we rewrite the right side of (1.16) as follows: x 2 = (Ax + Bct)2 = A2 x 2 + 2ABct x + B 2 c2 t 2 c2 t 2 = (C x + Dct)2 = C 2 x 2 + 2CDct x + D 2 c2 t 2 c2 t 2 x 2 = C 2 x 2 + 2CDct x + D 2 c2 t 2 A2 x 2 2ABct x B 2 c2 t 2 = c2 D 2 B 2 t 2 A2 C 2 x 2 + 2 (CD AB) ct x This must be equal to the left side of (1.16). Comparison leads us to conclude that CD AB = 0 CD = AB D2 B 2 = 1 A2 C 2 = 1 To obtain a solution, we recall that cosh2 sinh2 = 1. Therefore we make the following identi cation: A = D = cosh (1.18)
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In some sense we would like to think of this transformation as a rotation. A rotation leads to a transformation of the form x = x cos y sin y = x sin + y cos In order that (1.17) have a similar form, we take B = C = sinh With A, B, C, and D determined, the transformation matrix is cos h sin h L= 0 0 sin h cos h 0 0 0 0 0 0 1 0 0 1 (1.19)
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(1.20)
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Now we solve for the parameter , which is called the rapidity. To nd a solution, we note that when the origins of the two frames are coincident;
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that is, when x = 0, we have x = vt. Using this condition together with (1.17), (1.18), and (1.19), we obtain x = 0 = x cosh ct sinh = vt cosh ct sinh = t (v cosh c sinh ) and so we have v cosh c sinh = 0, which means that v cosh = c sinh v sinh = tanh = cosh c
(1.21)
This result can be used to put the Lorentz transformations into the form shown in elementary textbooks. We have x = cosh x sinh ct ct = sinh x + cosh ct Looking at the transformation equation for t rst, we have lct = sinh x + cosh ct = cosh = cosh ( tanh x + ct) v = cosh ct x c v = c cosh t 2 x c v x c2 sinh x + ct cosh
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