t = cosh t We also nd in Visual Studio .NET

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t = cosh t We also nd
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x = cosh x sinh ct = cosh (x tanh ct) = cosh (x vt)
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Special Relativity
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Now let s do a little trick using the hyperbolic cosine function, using cosh = cosh cosh = = 1 1 = = = 1 (1/ cosh ) 1/ cosh2 1 1 tanh
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cosh cosh2 sinh2 1 cosh2 sinh2 1 cosh2 sinh2 = 1 1 v 2 /c2
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This is none other than the de nition used in elementary textbooks: = 1 1 v 2 /c2 = cosh (1.22)
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And so we can write the transformations in the familiar form: t = t v x/c2 , x = (x vt) , y = y, z =z (1.23)
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It is often common to see the notation = v/c.
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There are three physical consequences that emerge immediately from the Lorentz transformations. These are time dilation, length contraction, and a new rule for composition of velocities.
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TIME DILATION
Imagine that two frames are in the standard con guration so that frame F moves at uniform velocity v with respect to frame F. An interval of time t
as measured by an observer in F is seen by F to be t= 1 1 2 t = t
Special Relativity
that is, the clock of an observer whose frame is F runs slow relative to the clock of an observer whose frame is F by a factor of 1 2 .
LENGTH CONTRACTION
We again consider two frames in the standard con guration. At xed time t, measured distances along the direction of motion are related by x = 1 1 2 x
that is, distances in F along the direction of motion appear to be shortened in the direction of motion by a factor of 1 2 .
COMPOSITION OF VELOCITIES
Now imagine three frames of reference in the standard con guration. Frame F moves with velocity v 1 with respect to frame F, and frame F moves with velocity v 2 with respect to frame F . Newtonian physics tells us that frame F moves with velocity v 3 = v 1 + v 2 with respect to frame F, a simple velocity addition law. However, if the velocities are signi cant fraction of the speed of light, this relation does not hold. To obtain the correct relation, we simply compose two Lorentz transformations. EXAMPLE 1-1 Derive the relativistic velocity composition law. SOLUTION 1-1 Using = v/c, the matrix representation of a Lorentz transformation between F and F is 1 2 12 0 0 1 1 1 1 1 1 2 0 0 1 2 L1 = (1.24) 1 1 1 0 0 1 0 0 0 0 1
Special Relativity
The transformation between F and F is 1 2 2 2 L 2 = 1 2 0 0
2
2 1 2 1 2 1 2
0 0 1 0
(1.25)
0 0 1
We can obtain the Lorentz transformation between F and F by computing L 2 L 1 using (1.25) and (1.24). We nd 1 1 2 1 0 0 1
2 1 1
1
2 1 1 2 1 1
0 2 2 1 2 1 0 0 0 0 1
2 1 2
2
2 1 2 2 1 2
0 0
0 1 0 0 1
2 2 = (1 1 )(1 2 ) 0 0
1+ 1 2 2 2 (1 1 )(1 2 ) ( 1 + 2 )
( 1 + 2 ) 2 2 (1 1 )(1 2 ) 1+ 1 2 2 2 (1 1 )(1 2 )
0 0 1 0
0 0 1
This matrix is itself a Lorentz transformation, and so must have the form 1
2 1 3 3
2 L 3 = 1 3 0 0
3
2 1 3 1 2 1 3
0 0 1 0
0 0 1
We can nd 3 by equating terms. We need to consider only one term, so pick the terms in the upper left corner of each matrix and set 1 + 1 2
2 1 1 2 1 2
2 1 3
Let s square both sides:
Special Relativity
(1 + 1 2 )2 1 = 2 2 2 1 3 1 1 1 2 Inverting this, we get
2 1 1 2 1 2 2 2 = 1 3
(1 + 1 2 ) Now we isolate the desired term 3 :
2 3 = 1
2 1 1
2 1 2
(1 + 1 2 )2
(1+ 1 2 )2 , (1+ 1 2 )2
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