Galilean Relativity D Hoult 2011 The velocities stated

“Galilean” Relativity © D Hoult 2011

The velocities stated above are, of course, velocities

The velocities stated above are, of course, velocities relative to third body, the ground

The velocities stated above are, of course, velocities relative to third body, the ground In one second, A moves 3 m to the right and G moves 2 m to the left

The velocities stated above are, of course, velocities relative to third body, the ground In one second, A moves 3 m to the right and G moves 2 m to the left To find the velocity of G relative to A, imagine yourself to be A

The velocities stated above are, of course, velocities relative to third body, the ground In one second, A moves 3 m to the right and G moves 2 m to the left To find the velocity of G relative to A, imagine yourself to be A A will see G moving at 5 ms-1 in the

The velocities stated above are, of course, velocities relative to third body, the ground In one second, A moves 3 m to the right and G moves 2 m to the left To find the velocity of G relative to A, imagine yourself to be A A will see G moving at 5 ms-1 in the negative sense

The velocities stated above are, of course, velocities relative to third body, the ground In one second, A moves 3 m to the right and G moves 2 m to the left To find the velocity of G relative to A, imagine yourself to be A Similarly, G will see A moving at 5 ms-1 in the positive sense

To find the velocity of G relative to A we simply subtract the velocity of G (relative to the ground) from the velocity of A (relative to the ground)

To find the velocity of G relative to A we simply subtract the velocity of G (relative to the ground) from the velocity of A (relative to the ground) VG relative to A = VG – VA

To find the velocity of G relative to A we simply subtract the velocity of G (relative to the ground) from the velocity of A (relative to the ground) VG relative to A = VG – VA VG relative to A = – 2 – 3 = – 5

We know the relative speed of A and G (magnitude 5 ms-1)

We know the relative speed of A and G (magnitude 5 ms-1) and that A has measured the speed of p to be of magnitude 10 ms-1 relative to himself

We know the relative speed of A and G (magnitude 5 ms-1) and that A has measured the speed of p to be of magnitude 10 ms-1 relative to himself We might want to transform the measurement made by A to find the speed of p relative to G

Imagine that p was stationary relative to A

Imagine that p was stationary relative to A Clearly, in this case, the speed of p relative to G is the same as the speed of p relative to A

In this case to find the speed of p relative to G, we must

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G Vp relative to G =

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G Vp relative to G = 5

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G Vp relative to G = 5 +

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G Vp relative to G = 5 + (– 10) =

In this case to find the speed of p relative to G, we must add the speed of p relative to A to the speed of A relative to G Vp relative to G = 5 + (– 10) = – 5 ms-1

Frames of Reference

Frames of Reference A frame of reference is simply

Frames of Reference A frame of reference is simply a set of axes and

Frames of Reference A frame of reference is simply a set of axes and a clock

Consider two observers, A

Consider two observers and B

at t = zero, A and B are very close together

point p is observed by A and B

In A’s frame of reference p has coordinates x, y, z

In B’s frame of reference p has coordinates x’, y’, z’

at t = 0

at t = 0 x’ 0 = x 0 y’ 0 = y 0 z’ 0 = z 0

A and B have a relative velocity of magnitude u directed parallel to their x axes

Some time later we have

The distance A B is equal to

Now A and B attribute different values to the x coordinate of point p

It is clear that x’ =

It is clear that x’ = x – ut

Between time zero and time t the change in position of p relative to A is

Between time zero and time t the change in position of p relative to A is

Between time zero and time t the change in position of p relative to A is x – x 0

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0 we saw that x’ 0 = x 0

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0 we saw that x’ 0 = x 0 and x’ = x – ut

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0 we saw that x’ 0 = x 0 and x’ = x – ut (x’ – x’ 0) = (x – x 0) – ut

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0 we saw that x’ 0 = x 0 and x’ = x – ut (x’ – x’ 0) = (x – x 0) – ut dividing by t gives

Between time zero and time t the change in position of p relative to A is x – x 0 Between time zero and time t the change in position of p relative to B is x’ – x’ 0 we saw that x’ 0 = x 0 and x’ = x – ut (x’ – x’ 0) = (x – x 0) – ut dividing by t gives v’ = v – u

v’ = v – u v is the velocity of p relative to A

v’ = v – u v is the velocity of p relative to A u is the velocity of B relative to A

v’ = v – u v is the velocity of p relative to A u is the velocity of B relative to A v’ is the velocity of p relative to B

v’ = v – u v is the velocity of p relative to A u is the velocity of B relative to A v’ is the velocity of p relative to B this equation confirms the relation we used instinctively when combining relative velocities

v’ = v – u v is the velocity of p relative to A u is the velocity of B relative to A v’ is the velocity of p relative to B this equation confirms the relation we used instinctively when combining relative velocities and…

v’ = v – u v is the velocity of p relative to A u is the velocity of B relative to A v’ is the velocity of p relative to B this equation confirms the relation we used instinctively when combining relative velocities and… is wrong !