Four Vectors in .NET framework

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In special relativity we work with a uni ed entity called spacetime rather than viewing space as an arena with time owing in the background. As a result, a vector is going to have a time component in addition to the spatial components we are used to. This is called a four vector. There are a few four vectors that are important in relativity. The rst is the four velocity, which is denoted by u and has components u= dt dx dy dz , , , d d d d
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We can differentiate this expression again with respect to proper time to obtain the four acceleration a. The norm or magnitude squared of v v tells us if a vector is timelike, spacelike, or null. The de nition will depend on the sign convention used for the line element. If we take ds 2 = c2 dt 2 dx 2 dy 2 dz 2 , then if v v > 0 we say that v is timelike. If v v < 0, we say v is spacelike. When v v = 0, we say that v is null. The four velocity is always a timelike vector. Following this convention, we compute the dot product as v v = (v t )2 (v x )2 v y
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The dot product is an invariant, and so has the same value in all Lorentz frames. If a particle is moving in some frame F with velocity u, we nd that energy and momentum conservation can be achieved if we take the energy to be E = m 0 c2 and de ne the four momentum p using p = m 0u (1.27)
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where m 0 is the particle s rest mass and u is the velocity four vector. In a more familiar form, the momentum four vector is given by p = E/c, px , p y , pz .
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Special Relativity
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For two frames in the standard con guration, components of the momentum four vector transform as px = px E/c py = py pz = p z E = E cpx Using our sign convention for the dot product, we nd
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2 2 p p = E 2 /c2 px p 2 pz = E 2 /c2 p 2 y
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Remember, the dot product is a Lorentz invariant. So we can nd its value by calculating it in any frame we choose. In the rest frame of the particle, the momentum is zero (and so p 2 = 0) and the energy is given by Einstein s famous formula E = m 0 c2 . Therefore p p = m 2 c2 0 Putting these results together, we obtain E 2 p 2 c2 = m 2 c4 0 (1.29)
Relativistic Mass and Energy
The rest mass of a particle is the mass of the particle as measured in the instantaneous rest frame of that particle. We designate the rest mass by m 0 . If a particle is moving with respect to an observer O with velocity v then O measures the mass of the particle as m= m0 1 v 2 /c2 = m0 (1.30)
Now consider the binomial expansion, which is valid for |x| < 1, (1 + x)n 1 + nx
For n = 1/2,
Special Relativity
1 (1 x) 1/2 1 + x 2 Setting x = v 2 /c2 in (1.30), we obtain m= m0 1 v 2 /c2 m0 1 + 1 v2 2 c2 1 v2 = m0 + m0 2 2 c
Multiplying through by c2 we obtain an expression that relates the relativistic energy to the rest mass energy plus the Newtonian kinetic energy of the particle: 1 mc2 = m 0 c2 + m 0 v 2 2
Quiz
1. An inertial frame is best described by (a) one that moves with constant acceleration (b) a frame that is subject to constant forces (c) a frame that moves with constant velocity (d) a frame that is subject to galilean transformations
The proper time d 2 is related to the interval via (a) d 2 = ds 2 (b) d 2 = ds 2 (c) d 2 = c2 ds 2 2 2 (d) d = ds 2 c 3. The principle of relativity can be best stated as (a) The laws of physics differ only by a constant in all reference frames differing by a constant acceleration. (b) The laws of physics change from one inertial reference frames to another. (c) The laws of physics are the same in all inertial reference frames. 2. 4. Rapidity is de ned using which of the following relationships (a) tanh = v c (b) tan = v c (c) tanh = v c (d) v tanh = c
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