The document discusses fundamental concepts of special relativity, including the constancy of the speed of light, Lorentz transformations, and their consequences such as length contraction and time dilation. It illustrates these concepts through various scenarios, including the muon decay problem and the twin paradox. The document emphasizes that measurements of time and distance are relative to the observer's frame of reference.

1. Content to be covered
 Constancy of the Speed of Light
 Lorentz Transformation
 Consequence of Relativity
 Length Contraction
 Time Dilation
 Muon decay problem
 Twin Paradox
Lecture #17,18
Resources to be consulted
Concepts of Modern Physics by Arthur Beiser
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Physics-1

2. • Consider the fixed system S and the moving system S’.
• At t = 0, the origins and axes of both systems are coincident with system
S’ moving to the right along the x axis.
• A flashbulb goes off at both origins when t = 0.
• According to postulate 2, the speed of light will be c in both systems and
the wave fronts observed in both systems must be spherical.
S
The Constancy of the Speed of Light
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3. Spherical wavefronts in S:
Spherical wavefronts in S’:
S S’
2 2 2 2 2 2 2 2 2 2
( 2 v v )
x y z x x t t y z c t
   
       
There are a couple of extra terms
(-2xvt +v2t2) in the primed frame.
Note that this cannot occur in Galilean transformations.
x’ = x – vt
y‘ = y
z‘ = z
t’ = t
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4. The special set of linear transformations that:
preserve the constancy of the speed of light (c)
between inertial observers; with the assumption
that the time is not an absolute quantity.
Therefore t  t
known as the Lorentz transformation equations
The Lorentz Transformation
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5. Consider a system of two inertial frames of references S and S .
S: is at rest and S : is a moving frame with constant velocity v relative to S
(x,y,z,t) coordinates of event P for an observer O in




the stationary frame S
(x ,y ,z ,t ) coordinates of event P for an observer O in the moving frame S
Let x = (x-vt) ....(1) γ : proportionality constant.
According to first
     

 
postulate of STR
Therefore x= (x +vt ) ....(2) where t t
According to second postulate of STR
The wavefront along x, x axis must satisfy
x=ct and x =ct
Thus
  
 

 
substitute the value of x and x in eq (1)and (2) respectively.
ct = (c-v)t and ct= (c+v)t
Multiply these equations and solve it for

 
 
 2 2
1
1 v / c
 

The Lorentz Transformation: Derivation
Here γ is a factor of
proportionality which does not
depend upon either x or t but
may be a function of v. γ must
be same in both the reference
frames as there is no difference
between S and S’ other than
the sign of v.
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6. 2
x
x [ (x - vt) vt ] (x - vt) vt
x x x 1
t t (1 )
v v v
x = (x-vt) ....(1) x= (x +vt ) ....(2)
From eq(1) and eq(2)
or
t = or t =
similarly t
  
 
 
      

 
 
      
 
2
x 1
t (1 )
v
=



  

2
2 2
2
xv
t
vx c
t t
c v
1
c

 
    
 
 

2
2 2
2
x v
t
x v c
t t
c v
1
c

 

 

   
 
 

2 2 2 2 2 2 2 2 2 2
y z c t x y z c t
Imp:Prove that
x
under Lorentz Transformation
   
      
The Lorentz Transformation: Derivation
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7. Lorentz Transformation
•Measurements of time and position depend upon the frame of reference of observer so
that two events which occur simultaneously in one frame at different places need not be
simultaneous in another.
•Lorentz transformation reduces to ordinary Galilean transformation when v<<c.
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8. 2 2
v
1 v /
x t
x
c

 

2
2 2
v /
1 v /
t x c
t
c

 

y y
 
z z
 
If v << c, i.e.  ≈ 1, yielding the familiar Galilean transformation.
Space and time are now linked, and the frame velocity cannot exceed c.
2 2
v
1 v /
x t
x
c
 



2
2 2
v /
1 v /
t x c
t
c
 



y y

z z

The complete Lorentz Transformation
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9. The Major Consequences To This Theory
are:-
Length Contraction
Time Dilation
Mass Expansion
Consequence of Relativity
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10. A fast-
moving
plane at
different
speeds.
v = 10% c
v = 80% c
v = 99.9% c
v = 99% c
Lorentz Contraction
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11. A rod is lying along the x’ axis of the moving
frame S’. If an observer determines the
coordinates of its ends to be x’1 and x’2, then
the length of the rod w .r . t. O’ is: L0 = x2’ – x1’
Therefore, L0 is the actual length of the rod
(w.r.t .to O’) in a frame in which the rod is at
rest. According to Lorentz transformation,
length measured in S w.r.t. O will be the
apparent length (L = x2– x1).
)
( 1
1
'
vt
x
x 
  )
( 2
2
'
vt
x
x 
 
2
2
1
)
(
)
(
)
( 1
2
1
2
0
c
v
L
x
x
vt
x
vt
x
L







 


Length Contraction
As ends of the rod is
measured at the same
time, so; t1 = t2 in S.
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12. 2 2 o
o 1 v / c
 

L
Therefore L=L
The length (L0) of an object in motion with respect to an
observer (​in S Frame) ​appears to be shorter, this phenomenon is
known as Lorentz FitzGerald contraction or Length Contraction.
Length Contraction
o
Case I : if v=0, L= ?
Case II: if v=c, L=?
Exercise : Find L when L =1m and v=0.9c.
Faster means shorter
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13. distance traveled 2
speed
  
p
d
t
c
2
2 2
2 p
p
2
t
t t
1
c t v t
2
c
2


 
v
d 
   
 
   
  
 



 p
γ >1,thu Δt Δt
s
Time Dilation
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14. Time Dilation
Assume a clock is placed at x’ in the moving frame S’. When an observer in S’
measures a time interval
The observer in S, will measure this interval as
2 1
2 2
2 1 2 2
2 2
0
2 1
2 2
2 2
vx vx
t t
c c
t t t
v v
1 1
c c
t
t t
v v
1 1
c c
 
 
 
   
 
 

 
  Thus improper time will be longer
than proper time; called Time dilation.
o 2 1
t t t
 
  Two measurements
of time t2’ & t1’ were
taken at same position.
proper time
apparent time
or improper time
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15. 0
t
t 

Thus a stationary clock measures a longer time interval between events occurring in
a moving frame of reference – time dilation.
Since γ ≥ 1, t will be larger than t0.
Thus if the time interval between two beats (tick) of the clock in
the proper frame S’ is 1 second, then this time interval will be
observed more than 1 second in the frame S i.e a moving clock
appears to go slow.
Time Dilation
A moving clock ticks more
slowly than a clock at rest.
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16. 6.6102 m
Muon is created
Muon
decays
The life time of μ mesons is 2.2 μs and their
speed 0.998c, so that they can cover only a
distance of 0.998c x 2.2 μs or 0.66km in their
life time, and yet they are found in abundance
at sea level, i.e., at a depth of 10 km from the
upper atmosphere where they are produced.
How may this be explained on the basis of (i)
length contraction (ii) time dilation?
Example of Time Dilation
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17. Muon is created
Muon decays
2 2
(i) Length contraction:
t 2.2 , L =2.2 0.998 0.66
Distance on the earth,
L L 1 / L 10.42 10
p
o o
s s c km
v c km km
 
  
    
2 2
(ii) Time Dilation
t
t= 34.8
1 /
distance travelled on the earth
34.8 0.998 10.42 10
p
t s
v c
s c km km


 


   
Example of Time Dilation
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18. 1. An observer in the laboratory sees two particles colliding at
x = 20.5 m, y=0, z=0 and t= 7.2 s. What are the co ordinates
of this event in a frame moving at 30.5 m/s with respect to
the laboratory frame?
2. A particle with a proper life time of 1 µs moves through the
laboratory at 0.9c. What is its lifetime as measured by
observers in the laboratory? Ans.: 2.294 μs.
Problem-2
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19. Twin Paradox
A longer life but it will not seem longer!
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