Galilean Relativity

Question 1

What is a frame of reference?

Answer 1

An array of rigid, massless, interpenetrable rods defining a grid over space

Question 2

How is time viewed in Galilean relativity?

Answer 2

Time is seen as universal. An absolute time, that is the same is all reference frames, with an event occurring simultaneously in all frames.

Question 3

Imagine your two inertial frames, one moving w.r.t S and a particle P. What are the related vectors?

Answer 3

r = R + r’
r’ = r - R

Where R is the distance between the two origins, if R is constant, so are r & r’.

A frame moving at constant velocity relative to an inertial frame is itself inertial.

Question 4

List off the Galilean transformation.

Answer 4

r’ = r - R
R = vt
v = constant
r’ = r - vt [ x’ = x - vt , y = y’ , z = z’ ]
u’ = u - v
t’ = t
a = a’ = universal

Question 5

Considering the travelling of the speed of sound. Think about the derivations of time intervals between two clappers, horizontally and diagonally.

Answer 5

Horizontally:
Sound would take longer to reach observer on far end of moving train, and the opposite on the way back after the second clap. Considering platform length L, total time,
t = 2Lv(s)/(v(s)^2 - v(p)^2)

Diagonally (transverse to length of platform) :
Now we have a vertical and horizontal component. Sound travels vertically upwards at sqrt(v(s)^2 - v(p)^2).
Total time,
t = 2L/sqrt(v(s)^2 - v(p)^2)

• sanity check with v(p) = 0

Question 6

Implication of hand clapping example (horizontal and transverse)

Answer 6

The horizontal time along the length of platform is always greater than simply the transverse motion.

Difference in time between the two is extremely small when the platform speed is far less than the speed of sound.

Question 7

The Michelson - Morley experiment explanation?

Answer 7

Proving the existence of a luminiferous ether.
Speed of the Earth w.r.t the ether.

Challenge is how large c is. Using monochromatic source and beam splitter, depending on the orientation of the Earth, when light recombined after being reflected perpendicularly, interference fringes should be shifted, but weren’t!

Equal path lengths. Interferometer shows fringes. Using logic of hand clapping, light should travel at different speeds at different orientations.

Maths on the diagonal time vs horizontal time yields the Lorentz factor. Contraction factor needed.
T1 = yT2

Consequence? Speed of light is always ‘c’.