Frame of Reference

The system relative to which the position or motion of a body is specified is

called the frame of reference. The position of any object is specified relative to
the frame of reference, which is assumed to be connected with the rigid body,
called the origin of frame of reference and position of body is specified by the
position vector of the body from the origin. Generally we consider the observer
situated at the origin.
The most simple frame of reference is the cartesian coordinate system in which
the position of a moving particle at any instant is expressed by (x, y, z) or by the
position vector, r = i x + j y + k z from the origin (0, 0, 0).
The velocity of particle, v = dr/dt = idx/dt + jdy/dt + kdz/dt and
acceleration a = dv/dt = d2r/dt2 = i d2x/dt2+ j d2y/dt2+ k d2z/dt2
 Realtivity
Inertial frame of reference:- The frame of reference for which Newton's law
holds are called inertial frames of reference. According to Newton's law, a body
not acted upon by any external force has an unaccelerated motion. In other words
we can say that a body remains in its position of motion or at rest unless an
external force is applied i.e., F=0 or ma=0, it means, acceleration,
A = dv/dt = d2r/dt2 = i d2x/dt2 + j d2y/dt2 + k d2z/dt2 = 0.
An observer in two inertial frames S & S’ measures the same value of
acceleration in both the frames then frames will be inertial if the value of
acceleration is same in both the frames.
Thus all the frames of reference which are either stationary relative to each other
or are in uniform motion, are called inertial frames provided that one of these
frames is inertial. Inertial frames are necessarily the unaccelerated frames.
 Galilean transformation
A particle has different coordinates in different frames of reference at any instant
of time. The equations which relate the coordinates of two frames of reference are
called the transformation equations. Similarly the equations relating the
coordinates of a particle in two inertial frames are called the Galilean
transformations. If S & S’ are two frames in which S is an inertial frame, an
event occurs at any instant ‘t’ at a point ‘P’ then the coordinates of the point P will
be (x, y, z, t) and (x’, y’, z’, t’) respectively. If frame S’ moving with velocity ’v’
relative to frame S along x-axis and the coordinate axes of ‘S’ are parallel to
frame S with their origins O & O’ coincident at t = t’ = 0 then the coordinates of
both frames are related as, x’ = x - vt or x = x’ + vt . . . . . . . . (1)
y’ = y y =y’ . . . . . . . . . . . . (1)
z’ = z z = z’ . . . . . . . . . . . (1)
and t’ = tt = t’ . . . . . . . . . . . . . . .(1)
 Galilean transformation
If the position vectors of point ‘P’ at any instant in frames S & S’ are r & r’ then
we can write from vector triangle △OO’P
OP = OO’ + O’P or r = vt + r’ or r’ = r - vt . . . . . . . . . . . . (2)
The equations (1) and (2) are called the Galilean transformations.
From equation (2) we can write,
dr’/dt = dr/dt +dvt/dt or dr’/dt = dr/dt +v or V’ = V + v . . . . . . . (3)
Where V’ & V are the velocities of the particles observed in frames S’ & S
respectively.
This equation (3) is called the Galilean law of addition of velocities.
Similarly dV’/dt = dV/dt +dv/dt or a’ = a + 0 (as velocity v of frame S’ relative to
S is constant)
So acceleration of the particle in S & S’ is the same or both the frames are
inertial.
Galilean transformation
 Postulates of Galilean transformation
Galilean transformation are based on the following three postulates:
i) Time of occurrence of an event at any point does not depend on the frame of
reference (t = t’)
ii) The distance between two points does not depend on the frame of reference. If
the coordinates of the end points of a rod in the two frames are (x1, y1, z1), (x2, y2,
z2) and (x’1, y’1, z’1), (x’2, y’2, z’2) then the length of rod in frames will be,
L=(x2- x1)2+(y2- y1)2+(z2- z1)2 and L’=(x2'- x1')2+(y2'- y1')2+(z2'- z1')2
But from Galilean transformations, x2'- x1' = x2- x1, y2'- y1' = y2 - y1
and z2'- z1' = z2 - z1.
 Postulates of Galilean transformation

So L = L’, hence the length of a rod (distance between two points) is invariant
under the Galilean transformations.
iii) The time interval of an event occurring at a point does not depend on the
frame of reference. If the time interval of an event in frames S & S’ are t 2-t1= Δt
and y2'-y1' = Δt' respectively then Δt' = Δt.
Since two inertial frames are related to Galilean transformations then we can say
that the basic laws of Physics are invariant in frames which are related by the
Galilean transformations.
 Absolute inertial frame
According to Newton, time is an invariant quantity that is time does not depend
upon the frame of reference. Indeed no frame of reference connected with a
material body, can be assumed to be a stationary inertial frame because there is no
stationary material body.
According to Maxwell’s electromagnetic theory the speed of light in vacuum is c,
but its related frame of reference was unknown. if we consider a frame of
reference relative to which the speed of light is c, then according to Galilean
transformation the speed of light will be different in different directions relative
to another inertial frame moving with a constant velocity ‘v’ relative to the first
frame.
 Absolute inertial frame
Thus it is essential to consider the frame of reference (inertial) in which the speed
of light is c in all directions. such a frame is called the absolute frame of
reference. The unaccelerated and non-rotatory absolute frame of reference is
called the absolute inertial frame of reference. for this a rigid, homogenous
transparent medium it was assumed and physical bodies can pass through it
without any obstruction. It was thought that a coordinator system attached with
the Ether medium can be assumed to be the absolute inertial frame and thus
assuming the Ether medium to be stationary and we can determine the absolute
velocity of any body.
 Michelson-Morley’s experiment
This experiment was used to determine the velocity of light relative to Ether. In
this experiment a very sensitive interferometer was used.
Experimental arrangements:- In Michelson Morley’s experiment interferometer
in which a monochromatic light from source ‘S’ falls on a partially silvered glass
plate ‘P’ which is kept inclined at 45o with the vertical. The light from P is
partially reflected towards a mirror M1 and is transmitted towards the mirror M2.
M1 and M2 are mutually perpendicular and equidistant from plate P. The light
reflected from M1 and M2 reunite at P, due to superposition of which interference
fringes are formed and seen by telescope T. Light ray coming from M1 cross
twice the plate P so another plate of the same medium and thickness is introduced
in the path of light coming from the M2 to compensate the optical path. The
whole apparatus is kept in Mercury so that it can be easily rotated to change its
direction.
Michelson-Morley’s experiment
Michelson-Morley’s experiment
 Michelson-Morley’s experiment
Theory:- Let the apparatus be kept such that direction PM2 is along the direction
of motion of earth relative to Ether. If the orbital velocity of earth relative to Ether
is v (=3✕104 m/sec) then apparatus along with earth will have velocity v along
PM2 relative to Ether. Therefore during the time, the light reaches M1 after
reflecting from P, the mirror M1 will move to position M1’ and light reflects M1’
instead of M1. If time taken by light to travel from P to M1 is tu and perpendicular
distance between P and M1 is l1,
Then M1M1’= v.tu, PM1= l1 and PM1’= c.tu, is the speed of light ‘c’ in
absolute frame of reference (i.e., Ether). So in △PM1M1’
(PM1’)2 = (PM1)2 + (M1M1’)2
Or (c.tu)2 = (l1)2 + (v.tu)2
Or (c2 - v2).tu2 = l12 or tu2 = l12/(c2 - v2) or tu = l1/[(c2-v2)1/2]
 Michelson-Morley’s experiment
Similarly time taken by light to travel after reflecting from M1 to P will be the
same as the new position of P will now be P’ due to motion of Earth.
Hence total time of travel of light from P to M1 and back to P from M1 in
perpendicular direction of motion of apparatus (or Earth),
t1 =2.tu = 2. l1/[(c2-v2)1/2] = 2.l1/[c(1- v2/c2)1/2] = [2.l1/c].[1- v2/c2]-1/2
By Binomial expansion ignoring the higher terms we can write,
t1 = [2. l1/c].[1+ v2/2c2] . . . . . . . . . . . . . . . . . (1)
Now the light ray travelling towards M2 from P the speed of light relative to the
earth (or apparatus) will be (c - v) because direction of travel of light and the
apparatus (or Earth) is the same. If the distance between M2 and P is l2 then time
of travel of light from P To M2,
tf = l2/(c - v).
 Michelson-Morley’s experiment
Similarly time of travel of light from M2 To P will be, tb = l2/(c + v), in this case
direction of travel of light is opposite to the direction of motion of the apparatus.
Thus total time of travel of light from P to M2 and back to P from M2 along the
direction of of motion of the apparatus (or Earth),
t2 = tf + tb = [l2/(c - v)] + [l2/(c + v)] = 2l2c/c2- v2 = [2l2c]/[c2(1- v2/c2)]
Or t2 = [2.l2/c].[1- v2/c2]-1
By Binomial expansion ignoring the higher terms we can write,
Or t2 = = [2.l2/c].[1+ v2/c2] . . . . . . . . . . . . . . (2)
 Michelson-Morley’s experiment
Now from equation (1) & (2), difference in time of travel of light in two mutually
perpendicular directions, Δt = t2 - t1 = [2.l2/c].[1+ v2/c2] - [2. l1/c].[1+ v2/2c2]
Or Δt = [2.(l2-l1)/c] + [(v2/c3).(2l2- l1)] . . . . . . . . . . . (3)
Hence path difference (distance) for light rays travelling in two mutually
perpendicular directions is, Δ = c.Δt = 2.(l2- l1) + (v2/c2).(2l2- l1) . . . . . . . (4)
Now if the whole apparatus is rotated 90o such that the lengths L1 and L2 are
mutually interchanged. The time of travel of light from P to M1 and back to P
from M1 is, t1' = [2.l1/c].[1+ v2/c2]
And the time of travel of light from P to M2 and back to P from M2 is,
t2' = [2.l2/c].[1+ v2/2c2]
 Michelson-Morley’s experiment
Hence difference in time of travel of light in two mutually perpendicular
directions, Δt' = t2'- t1' = [2.(l2- l1)/c] + [(v2/c3).(l2- 2l1)]
And the path difference, Δ' = c.Δt' = 2.(l2- l1) + (v2/c2).(l2- 2l1) . . . . . . . . . . (5)
Now from equation (4) & (5), the change in path difference due to rotation of
apparatus by 90o is
x = Δ - Δ' = {2.(l2-l1) + (v2/c2).(2l2- l1)}-{2.(l2- l1) + (v2/c2).(l2- 2l1)}
Or x = (v2/c2).(l1+l2) . . . . . . . . . . . .. . . . . . (6)
In Michelson-Morley’s experiment putting l1=l2=11m=l, c=3✕108m/sec,
v=3✕104m/sec then
x = (v2/c2)(l1+l2) = 22.[(3✕104)/(3✕108)]2 = 22✕10-8m
 Michelson-Morley’s experiment
For yellow light (wavelength 5500Å), the number of fringes shifted (passing
through a reference mark) will be,
n = path difference change (x)/ wavelength of light (λ) used
Or n = (22✕10-8m/5.5✕10-7m)=0.4
Now there must be a shift of fringes by 0.4 times the fringe-width when the
apparatus is turned by 90o. But experimentally no fringe shift was found even
when the experiment was performed quite carefully and was repeated several
times. This shows that the velocity of the Earth relative to Ether was zero or the
earth was stationary relative to Ether or fringe shift, x = 0.
 Michelson-Morley’s experiment
From this negative result of Michelson-Morley’s experiment shows that:
1. The velocity of the Earth is zero relative to Ether (absolute frame of
reference). In other words, the speed of light must be c in all directions
relative to all inertial frames.
2. The speed of light does not depend on the motion of the source or on the
motion of the observer. The effect of frame of reference is not observable.
Thus Michelson-Morley’s experiment is opposite to the result of Galilean
transformations.
3. The concept of Ether to be stationary is found to be wrong.
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