Answer to Question #96574 in Mechanics | Relativity for AbdulRehman

Question

Following the framework of Newtonian physics, suppose that that earth is
moving through the luminiferous ether at a speed v. Suppose that you are
Michelson (of the famous experiment) and that you have an interferometer
with two arms, each of length L as shown below. If one arm is directed
along the direction of the earth’s motion and one arm is directed in a
perpendicular direction,
i How long will it take a bit of light to make a complete circuit (out and
back) along each arm??
ii How big is the difference between these two times?

Answer 1

If the Earth has velocity v and the ether has speed c, the time along the direction of the earth's motion (out and back) will be

$$t_1=\frac{L}{v+c}+\frac{L}{c-v}=\frac{2cL}{c^2-v^2}=\frac{2L}{c}\frac{1}{(1-v^2/c^2)}.$$

because the 4th power of v/c is insignificantly small.

Calculate the time along the perpendicular axis. Consider a triangle ABC. The distance required for the device to travel from A to B is $vt_2.$

The hypotenuse will be therefore:

$$AB=BC=\sqrt{L^2+(vt_2/2)^2},$$

and the total distance from A to C through B, i.e. the distance of the path ABC is

$$ct_2=2\sqrt{L^2+(vt_2/2)^2}.$$

Hence:

$$t_2=\frac{2L}{c}\frac{1}{\sqrt{1-v^2/c^2}},$$

The difference is

$$\Delta t=\frac{2L}{c\sqrt{1-v^2/c^2}}\bigg(1-\frac{1}{\sqrt{1-v^2/c^2}}\bigg).$$