Special Theory of Relativity
3
For Galilean transformations, it is easy to show that the velocities and
accelerations in the two frames are related by
u′ = u–v
(1.2)
where v is along the x-direction, and
a′ ′ ′
′ ′ = a
(1.3)
respectively. Then, if the interaction potential V is a function of only the distances
between particles, Newton’s equations in the two frames are:
m i a i = ∇ i V
m i a i ′ = ∇ i ′V
(1.4)
where the subscript i is the particle index. These equations are related by the
transformations (1.1) and are of the same form. However, it was observed that
the Galilean transformations are not consistent with the dynamical theory of
electromagnetic fields as formulated by Maxwell (1865).
1.3 VELOCITY OF LIGHT
It follows from Maxwell’s equations for electromagnetic fields that
electromagnetic waves travel in vacuum with a speed equal to the ratio of the
electromagnetic unit to the electrostatic unit of charge. This ratio is essentially
equal to the speed of light so that light itself is taken as a form of electromagnetic
radiation.
Now, how does the velocity of light transform from one inertial frame to
another? According to Galilean transformations, the velocities are different in
different frames and are related by Eq. (1.2). However, Maxwell’s equations
have no reference to the velocity of the inertial frame and hence imply that the
speed of light is independent of the velocity of the inertial frame. Observationally
also, the Michelson-Morley experiment (1887) analysed below suggests that
the speed of light is independent of the velocity of the inertial frame.
Suppose, the earth is moving with velocity v in the x-direction with respect
to the ‘standard’ frame in which the velocity of light is c in all directions. Then
according to Eq. (1.2), the velocity of light with respect to an observer on Earth
is c–v. The time taken for light to travel along the limb AB of the interferometer
(Fig. 1.1), from A to B and back is
1
1
1
l
l
t c v c v
=
+
−
+
(1.5)
While travelling from A to C and back, the velocity c–v is parallel to AC and
hence perpendicular to v. Therefore
c.v = v
2
(1.6)
3
For Galilean transformations, it is easy to show that the velocities and
accelerations in the two frames are related by
u′ = u–v
(1.2)
where v is along the x-direction, and
a′ ′ ′
′ ′ = a
(1.3)
respectively. Then, if the interaction potential V is a function of only the distances
between particles, Newton’s equations in the two frames are:
m i a i = ∇ i V
m i a i ′ = ∇ i ′V
(1.4)
where the subscript i is the particle index. These equations are related by the
transformations (1.1) and are of the same form. However, it was observed that
the Galilean transformations are not consistent with the dynamical theory of
electromagnetic fields as formulated by Maxwell (1865).
1.3 VELOCITY OF LIGHT
It follows from Maxwell’s equations for electromagnetic fields that
electromagnetic waves travel in vacuum with a speed equal to the ratio of the
electromagnetic unit to the electrostatic unit of charge. This ratio is essentially
equal to the speed of light so that light itself is taken as a form of electromagnetic
radiation.
Now, how does the velocity of light transform from one inertial frame to
another? According to Galilean transformations, the velocities are different in
different frames and are related by Eq. (1.2). However, Maxwell’s equations
have no reference to the velocity of the inertial frame and hence imply that the
speed of light is independent of the velocity of the inertial frame. Observationally
also, the Michelson-Morley experiment (1887) analysed below suggests that
the speed of light is independent of the velocity of the inertial frame.
Suppose, the earth is moving with velocity v in the x-direction with respect
to the ‘standard’ frame in which the velocity of light is c in all directions. Then
according to Eq. (1.2), the velocity of light with respect to an observer on Earth
is c–v. The time taken for light to travel along the limb AB of the interferometer
(Fig. 1.1), from A to B and back is
1
1
1
l
l
t c v c v
=
+
−
+
(1.5)
While travelling from A to C and back, the velocity c–v is parallel to AC and
hence perpendicular to v. Therefore
c.v = v
2
(1.6)