Physics Problems: Analysis of Gamma, Twin Paradox & Particle Decay

This assignment provides solutions to several physics problems related to special relativity. It includes calculating the relationship between velocity and the gamma factor, analyzing the twin paradox using a space-time diagram to explain the timing differences between observers in relative motion, and deriving equations for energy and momentum conservation in particle decay scenarios, specifically focusing on a particle decaying into two photons. The solutions demonstrate the application of relativistic principles and conservation laws to solve these problems, highlighting the differences between relativistic and non-relativistic kinetic energy calculations.

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Problem: 4 – 8.
Gamma γ=100
The relationship between v
c =β velocity of frame and gamma is shown below.
γ= 1
√1− v2
c2
Now,
1
√1− v2
c2
=100
1− v2
c2 = 1
104
v2
c2 =1− ( 1
104 )
v2
c2 = 9999
10000
v=0.9995 c
Write one particles’ speed of other is shown below.
v'= 0.9995 c +0.9995 c
1+
( 0.9995∗0.9995
c2 )c2
v'= 2∗0.9995 c
1+0.9995∗0.9995
1− ( v'
c )
2
=1− ( 1.999
1+0.9995∗0.9995 )
γ '= 1
√ 1− ( v'
c )
2
Here,
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γ' is Lorentz factor for one particle with respect to other particle. Hence,
β=1−ε=1−
( 1.999
1+ ( 0.9952 ) )
Problem 4-11.
The relevant diagram is shown below.
In the twin paradox, there is timing gap owing to the L and M being situated light years
owing to which relativity comes into picture. As a result, the messages sent by L who is on
earth would be received much earlier in the frame of L but this is not the same for M
considering that about 30 years would be taken for the message to be receive and then the
same needs to be come back.
From the above graph, it is apparent that over a 60 year span on the earth, the message
between L and M would be exchanged 14 times but it would be spread evenly across for L
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who is in space. However, with regards to M, who is on earth most of the messages would be
received when he would be about to get 60 years old as is represented in the above diagram.
Problem 4-10.
It is noteworthy that when a particle (which is not at rest) decays then it releases E amount of
energy where (E = mc^2), m = mass of the particle, c = velocity of light
Hence, the energy released by decay into photons would be ¿ M
√ 1−( v
c )
2
This would be same as combined energy of the two released photons. Further, the energy of
each of photon would be half of the combined energy which means
¿
M
√ 1− ( v
c )
2
2
Energy of photon is independent of direction in which the photon is moving.
Particle M is at rest and decays into two photons and hence, the equation can be derived.
M =m 1+ m2
By applying linear conservation of momentum, the initial and final momenta would be equal
p(i) =p(f)
The parent particle is at rest and hence, the initial momentum would be zero and hence,
p(i) =p(f)=0 (Momentum conservation)
The final momentum p(f) = p1+p2 = 0
Hence,
p1+p2 = 0
p 1=− p 2
Magnitudes would be the same. However, the directions of each of the photons would be
opposite to each other.
Energy conservation
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E2= p2 c2 +m2 c4
For the two photons (1) and (2)
E ( 1 )2= p(1)2 c2 +m(1)2 c4
E ( 2 ) 2= p(2)2 c2 +m(1)2 c4
Solving these two equations
E ( 1 )2−E ( 2 )2=0
p ( 1 ) 2 c2 +m ( 1 )2 c4 − { p ( 2 ) 2 c2 +m ( 1 )2 c4 } =0
Magnitudes would be the same which means m(1) and m(2) are same and would be cancelled
out.
p ( 1 ) 2 c2− p ( 2 ) 2 c2=0
p ( 1 )2 c2= p ( 2 )2 c2
p ( 1 ) 2= p ( 2 ) 2
p ( 1 ) = p(2)
Hence, it can be concluded that E ( 1 ) 2=E ( 2 ) 2, which means both the photons (particles) are
having same energy E (1) =E (2). In terms of mass of the particle (at rest)
E ( 1 )= ( M 2 c2+m ( 1 )2 c2 )
2 M
Problem 1:
(a) Kinetic energy of the particle
Rest mass energy E0 =m c2
Mass of particle = 2kg
Set of speeds (m/s) is shown below.
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The kinetic energy of the particle for the various speed is computed in excel and is shown
below.
The value of c = 3*10^8 m/sec (speed of light)
(b) A new column has been added to the existing table which represents the non-relativistic
kinetic energy.
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(c) A new column to represent the ration of relativistic kinetic energy to the non-relativistic
energy.
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