In this system the frames were taken as inertial one, which obey Newton ‘s laws. Also any reference frame moving with uniform velocity with respect to another inertial frame is also an

inertial frame. Our experience shows that any experiment performed in a uniformly moving

vehicle will be the same if we perform the same experiment in any stationary laboratory. It implies the universality of every inertial frame.

Galileo’s principle of relativity tells that laws of mechanics are the same in all inertial frames of reference. The transformation from one inertial frame of reference to another is called the Galilean Transformation.

Consider two frames of reference S and S’ as in Figure 8.1.

**FIGURE 8. 1** Stationary and moving names of reference

The Galilean transformation establishes a relationship between the coordinates (x, y, z, t)

of an event at Sand the coordinates (x’, y’, z, t’) of the same event at S’. We assume S’ moves in such a way that x and x’ are in one straight line and have some positive directions. For convenience, we are restricting the motion of the moving frame in x-direction only. Also 0 and O’ coincide at t = 0 and S’ moves with a uniform velocity ‘v’ with respect to S.

**I. Coordinate Transformation**

From Figure 8.1 above,

x = x’ +v t =} x’ = x – v t

y’ = y; z’ = z; t’ = t

So the Galilean transformations for Sand S’ frame are

x’ = x- v t

y’ = y

z ‘ = z

t’ = t

** **

**II. Velocity Transformation**

Consider the movement of a moving body. Let the body be at the position P_{1} at t_{1} and P_{2} at t_{2}. Now at P_{1} the coordinates in system S are (x_{1}, y_{1}, z_{1}, t_{1}) and P_{2} the coordinates in system S’ are (x_{2}, y_{2}, z_{2}, t_{2})

Hence, for S’ system, using coordinate transformation as above

x_{1} = x_{1} –v t _{1} ; Y_{1}’ = Y_{1}; z’_{1} = z_{1}

and , x’_{2} = x_{2}– vt_{2}; y_{2} = y_{2}; z_{2} = z_{2} .

so , x’_{2} – x i = (x_{2}– x_{1})- v(t_{2}-t_{1})

à x’_{2}-x’_{1}/t_{2} – t _{l} = x_{2}– x _{l} / t_{2}-t _{l} -u

Let , u ’_{x}= x_{2}‘-x’_{1} / t _{2}– t _{l} =; u _{x} = x_{2} –x _{l} / t _{2}-t _{l}

So the Galilean transformation for velocity is,

U’ _{x} = u _{x} -v

U’ _{x} = u _{y}

u’ _{z} = u _{z}

**III. Acceleration Transformation**

We now consider the body in 5 moves along x-axis with acceleration ‘a’ and a’ with respect to S’. So at some moment of time,

u = u’ + v

After time ∆t, the velocity in both systems increased and the new velocities are

u + ∆u = (u’ + ∆u’) + v

Using equation (1) we have,

[(u’ + v) + ∆u]= (u’ + ∆u’) + v

∆u = ∆u’

∆u /∆t =∆t / ∆t’

a= a’

This is Galilean transformation for acceleration

If F and f’ are the forces in S-frame and S’ frame respectively, then, F =ma= ma’ = F’.

It shows that the acceleration and forces do not change from one inertial frame to another inertial frame. Hence force and acceleration are invariant under Galilean transformation