- #1

RyanP

- 18

- 0

## Homework Statement

The aether-wind theory of the Michelson-Morley experiment was discussed in class

for the special case where the arms of the interferometer (each of equal length L) are

parallel and perpendicular to the wind. In this problem we consider the general case

for an angular setting. The aether wind in moving with velocity v in the -x direction, while the arms of the interferometer are each offset clockwise by an angle

**θ**from the x and y axes. Prove that to a good approximation, the time difference between light pulses from each arm is, t = (v

^{2}L/c

^{3}) * |cos

**θ**|

Hint: Do NOT solve the problem in the Lab's frame. Instead look at it from the point

of view of the aether frame, where the instrument is moving toward the right with

velocity +v, and where the speed of light is always c.

## Homework Equations

distance traveled by a beam of light = ct

## The Attempt at a Solution

I calculated the distance light travels in the first arm (the one that is offset from the +y axis). In the first portion of the light's path, it covers an x displacement of Lsin

**θ**+ vt

_{1}and a vertical displacement of Lcos

**θ**. Using the Pythagorean theorem, I get the total distance traveled by the light beam in the first part to be d

_{1}= sqrt(L

^{2}+ 2Lvt

_{1}*sin

**θ**+ v

^{2}t

_{1}

^{2}). Then using the same approach I find that the light's return trip distance d

_{2}= sqrt(L

^{2}- 2Lvt

_{2}*sin

**θ**+ v

^{2}t

_{2}

^{2})

I use the same idea to get the distance traveled by the light beam in the second arm (offset from the +x axis by the same angle). The result is very similar, just with some cosines instead of sines, etc. I understand that the time for the light to travel there and back along an arm is (d

_{1}+d

_{2})/c , but I don't know how to get from here to the result above.