momenergy 4-vector For each of the following cases, write down the four componen
ID: 1476493 • Letter: M
Question
momenergy 4-vector For each of the following cases, write down the four components of the momentum - energy (momenergy) 4-vector in the given frame in the form {E, px, py, Pz}. Assume that each particle has mass m. You may use square roots in your answer. A particle moves in the positive x-direction in the laboratory with total energy equal to five times its rest energy. Same particle as observed in a frame in which it is at rest. Another particle moves in the z-direction with momentum equal to three times its mass. Yet another particle moves in the negative y-direction with kinetic energy equal to four times its mass. Still another particle moves with total energy equal to ten times its mass and x-, y-, and z-components of momentum in the ratio 1 to 2 to 3.Explanation / Answer
(a) Answer: E = 5m0c^2 , pc = 5^2- 1^2(m0c^2 ) = 24m0c^2 .
The particle moves in the positive x direction, so px = p = 24m0c, py = pz = 0.
The Energy-momentum four vector ( E/c, p x , p y , p z ) = (5 m0c, 24m0c,0, 0).
(b) Answer: E = m0c^2 , p = 0, since it is at rest in the frame.
( E/c, px , py , pz ) = ( m0c,0,0,0)
(c) Answer: pc = 3m0c^2 , E = p^2c^2 + ( m0c^2 )^2 = 3^2 + 1^2m0c^2
= 10m0c^2.
Therefore, ( E/c, p x , p y , p z ) = ( 10m0c,0,0,3m0c)
(d) Answer: Kinetic energy is equal to four times its mass i.e. K = 4m0c^2 ,
so E = m0c^2 + K = 5m0c^2 and pc = m0c^2 5^2- 1^2 = 24m0c^2 .
Since the particle moves in the negative y-direction,
( E/c, p x , p y , p z ) = (5 m0c,0,-24m0c, 0)
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