# Galilean transformation equations derivation

Let there are two inertial frames of references S and S’. S is the stationary frame of reference and S’ is the moving frame of reference. At time t=t’=0 that is in the start, they are at the same position that is Observers O and O’ coincides. After that S’ frame starts moving with a uniform velocity v along x axis.

Let an event happen at position P in the frame S’. The coordinate of the P will be x’ according to the observer in S’ and it will be x according to O in S.

The frame S’ has moved a distance “vt” in time t (refer figure).

What should be the relation between x and x’. As we can see from the figure that

x = x’ + vt’

But here the t = t’ thus

x = x’ + vt (1)

Where t and t’ are the time measured from S and S’ frames respectively.

But what should be x’ = ?

Yes you are right, it is x’ = x – vt (2)

It can be achieved by just exchanging the sides of the equation (1).

But there is more to it. It is just not by exchanging the sides.

If we see equation 1, we will find that it is the position measured by O when S’ is moving with +v velocity. But if the same thing is measured by O’ then velocity of S should be –v. (For example, when we travel in a train, then according to the outside observers, we are travelling in x direction (suppose), but the outside objects, according to me travel in the opposite direction with the same but negative velocity).

What should be the relation of y with y’?

It will be

y = y’ (3)

or y’ = y (4)

because there is no movement of frame along y-axis.

Similarly z = z’ (5)

And z’ = z (6)

And here t = t’ (7)

And t’ = t (8)

Equations 1, 3, 5 and 7 are known as Galilean inverse transformation equations for space and time.

Equations 2, 4, 6 and 8 are known as Galilean transformation equations for space and time.

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