# Relative motion

###### Sections

- Galilean relativity
- River and boat problem

##### Galilean relativity

We observe phenomena, make measurements, formulate laws and verify laws. We happen to be on a spaceship (called earth) that is hurtling through space at a tremendous 29.8 km/s (imagine going from Adyar to Mahabalipuram under 1.5 seconds). Yet this motion is not felt by us. Our experiments reveal that Newton’s laws are perfectly valid from our point of view. On the earth we find the same laws are operative as the ones that successfully predict planetary motion or galactic motions.

The earth is also accelerating - after all, it is not moving in a straight line. But this acceleration is not felt by us. The 'earth' frame of reference (neglecting rotation for now) is an inertial frame. The reason for this is discussed in a later chapter.

For now, we treat the 'Ground-frame' as an inertial frame without further explanation.

One of the great findings of scientists is that the laws of motion are the same in frames of reference that move with constant velocity with respect to each other. This is called the principle of relativity.

The principle of relativity was discovered by Galileo.

Figure 1

What Galileo did was have one of his cronies standing on the top of the mast of a fast-moving ship drop a ball. By the time the ball landed, the ship had moved some distance. Did the ball drop behind the mast? Nope. It dropped right under the release point, just as it would have had the ship not been moving. That is, from the point of view of Galileo’s friend, the ball just dropped down, just as if the ship was stationary.

Galileo stood on the bank watching. From his point of view, the ball moved in a parabolic arc, landing further down, as shown by the dotted line in Figure 1. From his point of view (shortened to POV from now on), the ball had an initial velocity, it moved in the correct trajectory predicted by the same equations.

We have what are called two frames of reference, Galileo’s and his friend’s. In Galileo’s frame of reference there is an x-y-z-coordinate system that is stationary with respect to Galileo. He also has clocks that are stationary with respect to him. He makes measurements using this coordinate system and clocks.

There is another frame of reference, that of Galileo’s friend, which is moving with respect to Galileo’s. Imagine that there is another set of x-y-z-axes that are moving with the ship’s velocity. All the positions of objects are being given in this coordinate system by G’s friend. This is not very difficult to imagine since we on earth measure everything with respect to us, and we are moving with respect to the sun.

G’s friend would say( if he believes only what he sees ) that the tree on the bank was moving with the velocity –**v** and that his ship was stationary. G on the other hand, would say that the tree was stationary and the ship was moving with the velocity **v**.

In this section we are going to learn how to find the velocity of an object in one frame when we know it in another.

Consider two frames of reference, S and S'; S' moving with the velocity **u** with respect to S.

Figure 2

For simplicity we assume that they coincide at t = 0. At time t the displacement of S' is **u**t from S, see Figure 2. Again, for simplicity, the drawing shows **u** pointing along the x-direction, but this need not be so.

Figure 3

On the left in Figure 3 we see a ball at time t=0, located at the position shown. At this time frames S and S' are coincident. During time t, the frame S' moves by **u**t. If the ball is stationary with respect to S', it would be located at the position displaced by **u**t from its original location. This is shown with a dotted line. The ball, however, moves with the velocity **v**’ w.r.t. the frame S' . So in the S' frame it suffers a displacement of **v**’t. Seen from the *stationary* frame S, its displacement during t is **v**t. You can see that

(1)

Therefore

(2)

The velocity of a particle in a frame S is the vector sum of the particle’s velocity in another frame S' with the velocity of S' with respect to S.

Consider an example in one dimension:

If a train is moving with the velocity of 20 m/s and a passenger inside the train rolls a ball along the floor of the train with a velocity of 6 m/s (in the same direction as the train’s velocity), the velocity of the ball as seen from the outside is 26 m/s.

Consider an example in two dimensions:

A car has the velocity 20**i** + 30**j** m/s as seen from the *stationary* earth frame. A train has the velocity –30**i** + 40**j** m/s as seen from the same frame. The velocity of the car as seen by an observer on the train is: **v**’ = **v** –** u** = (20**i** + 30**j **)** -** (** **–30**i** + 40**j **)** = **50**i** – 10**j**.

Figure 4