Michelson Morley Experiment for Arbitrary Angle

[hawaiianair]

Homework Statement

The ether wind theory of the Michelson-Morley experiment is discussed in the text for the special case where the arms of the interferometer are parallel and perpendicular to the wind. Consider the general case for an angular setting θ. Prove that, for equal arms of length L, the time difference for the two paths is given to a good approximation by Δt=[v^2l.cos2θ]/c^3 where v is the velocity of the ether wind.

Homework Equations

relative velocity problem where t=distance/velocity

The Attempt at a Solution

I tried to work out the velocity of the light going through each arm and got
t1=t2=L/(c^2+(c-v)^2)^1/2+L/(c^2+(c+v)^2)^1/2
which is obviously wrong. I think I'm meant to split the velocity of the wind into vcosθ and vsinθ but I'm not sure how to apply this to the problem?

 

[anathema]

To solve this problem, we need to consider the path difference between the two beams of light in the interferometer. Let's label the two paths as path 1 and path 2.

For path 1, the light travels a distance of L along the arm parallel to the ether wind, and a distance of Lcosθ along the arm perpendicular to the wind. This gives us a total distance of L+Lcosθ for path 1.

For path 2, the light travels a distance of L along the arm perpendicular to the wind, and a distance of Lcosθ along the arm parallel to the wind. This gives us a total distance of L+Lcosθ for path 2.

Now, we can use the formula for time taken for a given distance and velocity: t=d/v. For path 1, the time taken is:
t1 = (L+Lcosθ)/c

For path 2, the time taken is:
t2 = (L+Lcosθ)/(c+v)

The time difference between the two paths is:
Δt = t2 - t1 = [(L+Lcosθ)/(c+v)] - [(L+Lcosθ)/c]

Simplifying this expression, we get:
Δt = [Lcosθv/(c+v)] - [Lcosθv/c]

Using the formula for relative velocity, we can rewrite this as:
Δt = Lcosθv[(c-v)/(c+v)(c)]

Simplifying further, we get:
Δt = Lcosθv[(c^2-v^2)/(c^3+vc^2)]

Since we are assuming that the velocity of the ether wind is much smaller than the speed of light (v<<c), we can neglect the term v^2/c^2. This gives us:
Δt = Lcosθv/c^3

Which is the desired result:
Δt = [v^2Lcosθ]/c^3

Note: This derivation assumes that the arms of the interferometer are of equal length. If the arms are not of equal length, the time difference will be slightly different, but the general form of the equation will remain the same.