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Dr.R.Hepzi Pramila Devamani,
Assistant Professor of Physics,
V.V.Vanniaperumal College for Women,
Virudhunagar
The Michelson-Morley
Experiment
Experimental arrangement
 A beam of light from a monochromatic source S falls on a half
silvered glass plate P, placed at an angle of 45" to the beam.
 The incident beam is split up into two parts by P (Fig. 53).
 The reflected portion travels in a direction at right angles to the
incident beam, falls normally at B on the plane mirror M, and is
reflected back to P. It gets refracted through P and enters the
telescope T.
 The transmitted portion travels along the direction of the initial
beam, falis normally on mirror M2 at A and is reflected back to P.
 After reflection from the back surface of P, it enters the telescope T.
 The two reflected beams interfere. The interference fringes are
viewed with the help of the telescope T.
The Michelson-Morley
Experiment
The Michelson-Morley
Experiment
 The beam reflected upwards to M1 traverses the thickness
of plate P thrice. But the beam refracted on to mirror M2
traverses P only
once.
 The effective distance of the mirrors M and M, from the
plate P is made to be the same by the use of a
compensating plate not shown
in figure.
The Michelson-Morley
Experiment
 The whole apparatus was floating on mercury.
 One arm (PA) was pointed in the direction of earth's
motion round the sun.
 The other (PB) was pointed perpendicular to this motion.
 The paths of the two beams and the positions of their
reflections from M1 and M2 are shown by the dotted lines.
The Michelson-Morley
Experiment
Calculation
 Assume that the velocity of the apparatus (or earth)
relative to fixed ether is v in the direction PA.
 The relative velocity of a light ray travelling along PA is
(c-v) while its value would be (c + v) for the returning ray.
 Let PA = PB = d.
Time taken by light to travel from P to A
 = d/(c - v)
The Michelson-Morley
Experiment
 Time taken by light to travel from A to P
 = d/(c + v)
 .: Total time taken by light to travel from P to A and back

The Michelson-Morley
Experiment
 Now, consider the ray moving upwards from P to B.
 It will strike the mirror M, not at B but at B' due to the
motion of the earth.
If t1 is the time taken by the ray starting from P to reach
M1, then
PB' = ct1 and BB' = vt2 ,
The Michelson-Morley
Experiment
The Michelson-Morley
Experiment
 This is the path difference between the two parts of the
incident beam.
 If the apparatus is turned through 90°, the path difference
between the two beams becomes
 2 dv2/c2.
 Michelson and Morley expected a fringe shift of about 0.4
in their apparatus when it was rotated through 90°.
 They believed that they could detect a shift as small as
0.01.of a fringe.
The Michelson-Morley
Experiment
 But, in the experiment, no displacement of the fringes was
observed.
 They repeated the experiment at different points on the
earth's surface and at different seasons of the year without
detecting any measurable shift in fringes.
 This negative result suggests that it is impossible to
measure the speed of the earth relate to the ether.
 Therefore, the effects of ether are undetectable. Thus, all
attempts to take ether as a fixed frame of reference failed.
Explanation of the Negative
Result
 The proper explanation for the negative result of the
Michelson-Morley, experiment was given by Einstein.
 He concluded that the velocity of light in space is a
universal constant.
 This statement is called the principle of constancy of the
speed of light.
 The speed of light is c rather than (c + v) in any frame of
reference.
Postulates of Special Theory of Relativity
The laws of Physics are the same in
all inertial frames of reference.
The velocity of light in free space is
constant.
It is independent of the relative
motion of the source and the
observer.
Explanation
 It is impossible to perform an experiment which measures
motion relative to a stationary ether.
 So no unique stationary frame of reference can be
discovered.
 The first postulate expresses this fact.
 Since there is no favoured 'rest' frame of reference, all
systems moving with constant velocity must be on equal
footing.
 We cannot discuss absolute motion. We can discuss only
relative motion
Explanation
 We know that the velocity of light is not constant under
Galilean transformations.
 But according to the second postulate, the velocity of light
is the same in all inertial frames.
 Thus the second postulate is very important. Only this
postulate is responsible to differentiate
the classical theory and Einstein's theory of relativity.
Lorentz-Transformation
Equations
 Let S and S' be two reference frames (Fig. 5.4).
 Frame S is at rest.
 Frame S' is moving along x-axis with a constant velocity 'V' with
respect to frame S.
 Initially the origins of two frames coincide.
 (x, y, z, 1) represent the space and time co-ordinates of an event
for an observer in frame S.
 (x', y', z', t') are the space and time co-ordinates of the same event
with respect to an observer in frame S'.
 Let the time be counted from the instant when the origins O and
O'
coincide.
Lorentz-Transformation
Equations
Lorentz-Transformation
Equations
Lorentz-Transformation
Equations
Length Contraction
Derivation
 Consider two coordinate systems S and S' with their X
axes coinciding at time t = 0.
 S' is moving with a uniform relative speed v
with respect to S in the positive X-direction.
 Imagine a rod (AB), at rest relative to S' (Fig. 5.5).

Length Contraction
Length Contraction
 Let x and x be the coordinates of the ends of the rod at any
instant of time in S. Then
 L0 = x2’ – x1’ (1)
 Since the rod is at rest in frame S’.
 Similarly, let x1 and x2, be the coordinates of the ends of
the rod at the same instant of time in S.
 Then, l = x2 – x1 (2)
 l is the length of the rod, measured relative to S.
 According to Lorentz transformations,
Length Contraction
Length Contraction
Length Contraction
 The shortening or contraction in the length of an object
along its direction of motion is known as the Lorentz-
Fitzgerald contraction.
 There is no contraction in a direction perpendicular to the
direction of motion.
Explanation
 Let lo, be the length of a rocket on a launching pad.
 The length of the rocket, when it moves through space with a
velocity v, shall appear to an observer on the ground as given by

This equation shows that the length of an object in motion as
measured by an observer appears to him to be shorter than when it is
at rest with respect to him.
 This phenomenon is known as Lorentz-Fitzgerald contraction.
If v = c, then I = 0.
 A rocket moving with the speed of light appears to be reduced
to a point to a stationary observer.
Examples
 A square and a circle in one appear to the observer in the
other to be a rectangle and an ellipse respectively (Fig. 5.6
).
 A body which appears to be spherical to an observer at
rest relative to it, will appear to be an oblate spheroid to a
moving observer.
The Michelson-Morley Experiment

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The Michelson-Morley Experiment

  • 1. Dr.R.Hepzi Pramila Devamani, Assistant Professor of Physics, V.V.Vanniaperumal College for Women, Virudhunagar
  • 2. The Michelson-Morley Experiment Experimental arrangement  A beam of light from a monochromatic source S falls on a half silvered glass plate P, placed at an angle of 45" to the beam.  The incident beam is split up into two parts by P (Fig. 53).  The reflected portion travels in a direction at right angles to the incident beam, falls normally at B on the plane mirror M, and is reflected back to P. It gets refracted through P and enters the telescope T.  The transmitted portion travels along the direction of the initial beam, falis normally on mirror M2 at A and is reflected back to P.  After reflection from the back surface of P, it enters the telescope T.  The two reflected beams interfere. The interference fringes are viewed with the help of the telescope T.
  • 4. The Michelson-Morley Experiment  The beam reflected upwards to M1 traverses the thickness of plate P thrice. But the beam refracted on to mirror M2 traverses P only once.  The effective distance of the mirrors M and M, from the plate P is made to be the same by the use of a compensating plate not shown in figure.
  • 5. The Michelson-Morley Experiment  The whole apparatus was floating on mercury.  One arm (PA) was pointed in the direction of earth's motion round the sun.  The other (PB) was pointed perpendicular to this motion.  The paths of the two beams and the positions of their reflections from M1 and M2 are shown by the dotted lines.
  • 6. The Michelson-Morley Experiment Calculation  Assume that the velocity of the apparatus (or earth) relative to fixed ether is v in the direction PA.  The relative velocity of a light ray travelling along PA is (c-v) while its value would be (c + v) for the returning ray.  Let PA = PB = d. Time taken by light to travel from P to A  = d/(c - v)
  • 7. The Michelson-Morley Experiment  Time taken by light to travel from A to P  = d/(c + v)  .: Total time taken by light to travel from P to A and back 
  • 8. The Michelson-Morley Experiment  Now, consider the ray moving upwards from P to B.  It will strike the mirror M, not at B but at B' due to the motion of the earth. If t1 is the time taken by the ray starting from P to reach M1, then PB' = ct1 and BB' = vt2 ,
  • 10. The Michelson-Morley Experiment  This is the path difference between the two parts of the incident beam.  If the apparatus is turned through 90°, the path difference between the two beams becomes  2 dv2/c2.  Michelson and Morley expected a fringe shift of about 0.4 in their apparatus when it was rotated through 90°.  They believed that they could detect a shift as small as 0.01.of a fringe.
  • 11. The Michelson-Morley Experiment  But, in the experiment, no displacement of the fringes was observed.  They repeated the experiment at different points on the earth's surface and at different seasons of the year without detecting any measurable shift in fringes.  This negative result suggests that it is impossible to measure the speed of the earth relate to the ether.  Therefore, the effects of ether are undetectable. Thus, all attempts to take ether as a fixed frame of reference failed.
  • 12. Explanation of the Negative Result  The proper explanation for the negative result of the Michelson-Morley, experiment was given by Einstein.  He concluded that the velocity of light in space is a universal constant.  This statement is called the principle of constancy of the speed of light.  The speed of light is c rather than (c + v) in any frame of reference.
  • 13. Postulates of Special Theory of Relativity The laws of Physics are the same in all inertial frames of reference. The velocity of light in free space is constant. It is independent of the relative motion of the source and the observer.
  • 14. Explanation  It is impossible to perform an experiment which measures motion relative to a stationary ether.  So no unique stationary frame of reference can be discovered.  The first postulate expresses this fact.  Since there is no favoured 'rest' frame of reference, all systems moving with constant velocity must be on equal footing.  We cannot discuss absolute motion. We can discuss only relative motion
  • 15. Explanation  We know that the velocity of light is not constant under Galilean transformations.  But according to the second postulate, the velocity of light is the same in all inertial frames.  Thus the second postulate is very important. Only this postulate is responsible to differentiate the classical theory and Einstein's theory of relativity.
  • 16. Lorentz-Transformation Equations  Let S and S' be two reference frames (Fig. 5.4).  Frame S is at rest.  Frame S' is moving along x-axis with a constant velocity 'V' with respect to frame S.  Initially the origins of two frames coincide.  (x, y, z, 1) represent the space and time co-ordinates of an event for an observer in frame S.  (x', y', z', t') are the space and time co-ordinates of the same event with respect to an observer in frame S'.  Let the time be counted from the instant when the origins O and O' coincide.
  • 20. Length Contraction Derivation  Consider two coordinate systems S and S' with their X axes coinciding at time t = 0.  S' is moving with a uniform relative speed v with respect to S in the positive X-direction.  Imagine a rod (AB), at rest relative to S' (Fig. 5.5). 
  • 22. Length Contraction  Let x and x be the coordinates of the ends of the rod at any instant of time in S. Then  L0 = x2’ – x1’ (1)  Since the rod is at rest in frame S’.  Similarly, let x1 and x2, be the coordinates of the ends of the rod at the same instant of time in S.  Then, l = x2 – x1 (2)  l is the length of the rod, measured relative to S.  According to Lorentz transformations,
  • 25. Length Contraction  The shortening or contraction in the length of an object along its direction of motion is known as the Lorentz- Fitzgerald contraction.  There is no contraction in a direction perpendicular to the direction of motion.
  • 26. Explanation  Let lo, be the length of a rocket on a launching pad.  The length of the rocket, when it moves through space with a velocity v, shall appear to an observer on the ground as given by  This equation shows that the length of an object in motion as measured by an observer appears to him to be shorter than when it is at rest with respect to him.  This phenomenon is known as Lorentz-Fitzgerald contraction. If v = c, then I = 0.  A rocket moving with the speed of light appears to be reduced to a point to a stationary observer.
  • 27. Examples  A square and a circle in one appear to the observer in the other to be a rectangle and an ellipse respectively (Fig. 5.6 ).  A body which appears to be spherical to an observer at rest relative to it, will appear to be an oblate spheroid to a moving observer.