Prove Lorentz-Fitzgerald Contraction w/ Michelson-Morley Experiment at Any Angle

[Happiness]

I find this question rather difficult!

Q1. I take Fig 1-11 to be the laboratory frame and $v$ to be the velocity of the ether wind. Since Lorentz length contraction occurs only horizontally, the right mirror should be further to the right horizontally in the ether frame. This makes the angle the light beam makes with the horizontal smaller than $\phi$ when in the ether frame (fact 1). Similarly, the top mirror should be further to the left horizontally in the ether frame. This makes the angle the light beam makes with the vertical bigger than $\phi$ when in the ether frame (fact 2). But facts 1 and 2 means the angle between the two "perpendicular" light beams is larger than 90$^\circ$ when in the ether frame. But does or doesn't the angle need to be perpendicular too in the ether frame?

Q2. By resolving the ether wind into 2 components: $v\cos\phi$ and $v\sin\phi$, I got the time difference before rotation as $\Delta t=\frac{l_1\sqrt{c^2-v^2\sin^2\phi}-l_2\sqrt{c^2-v^2\cos^2\phi}}{c^2-v^2}$, and that after rotation as $\Delta t'=\frac{l_1'\sqrt{c^2-v^2\cos^2\phi}-l_2'\sqrt{c^2-v^2\sin^2\phi}}{c^2-v^2}$, where $l_1$ is the length between the center and the right mirrors in the laboratory frame, $l_2$ is the length between the center and the top mirrors in the laboratory frame, $l_1'$ is the length between the center and the "then top" mirrors in the laboratory frame (after rotation), $l_2'$ is the length between the center and the "then left" mirrors in the laboratory frame (after rotation; the source is now at the bottom). We know the rest length $l_1^\circ$ (at rest in the ether frame) is the same after rotation (simlarly true for $l_2^\circ$). But since the angle between the two "perpendicular" light beams is not perpendicular in the ether frame, I find it very difficult to solve the problem! I still can't prove $\Delta t=\Delta t'$, as the relation between $l_1$ and $l_1'$ is rather complicated!

Lorentz-contraction hypothesis:

Note that $\phi=0$ in (1.9), and $l_1$ is the length from the center mirror to the reflecting mirror in the direction along the ether wind before rotation, and it will be perpendicular to the ether wind after rotation. $l_2$ is always perpendicular to $l_1$.

 

[PeterDonis]

Moderator's note: moved to homework forum.

 

[TSny]

Q1. I take Fig 1-11 to be the laboratory frame and $v$ to be the velocity of the ether wind. Since Lorentz length contraction occurs only horizontally, the right mirror should be further to the right horizontally in the ether frame. This makes the angle the light beam makes with the horizontal smaller than $\phi$ when in the ether frame (fact 1). Similarly, the top mirror should be further to the left horizontally in the ether frame. This makes the angle the light beam makes with the vertical bigger than $\phi$ when in the ether frame (fact 2). But facts 1 and 2 means the angle between the two "perpendicular" light beams is larger than 90$^\circ$ when in the ether frame. But does or doesn't the angle need to be perpendicular too in the ether frame?

I'm not sure I quite follow your description here. But I agree that when the apparatus is "tilted" and moving through the ether, then the arms will not be perpendicular according to measurements made in the ether frame.

The figure above is for the ether frame. We imagine the apparatus initially at rest in the ether and then set into motion to the right. The dotted lines labeled $l_1^0$ and $l_2^0$ represent the arms when the apparatus is at rest. The solid lines marked $l_1$ and $l_2$ represent the arms when the apparatus is moving through the ether to the right with speed v. $\phi$ is the angle of tilt before the apparatus is set into motion. When at rest in the ether, the angle between the arms is a right angle. But, in motion, the angle between the arms according to the ether frame is less than a right angle due to the effect of length contraction parallel to the motion.

In the frame of the apparatus (when it is moving through the ether), the lengths of the arms as measured by meter sticks would still be $l_1^0$ and $l_2^0$ since the meter sticks also suffer contraction effects. So, in the apparatus frame, it would still appear as though the arms are perpendicular.

Q2. By resolving the ether wind into 2 components: $v\cos\phi$ and $v\sin\phi$, I got the time difference before rotation as $\Delta t=\frac{l_1\sqrt{c^2-v^2\sin^2\phi}-l_2\sqrt{c^2-v^2\cos^2\phi}}{c^2-v^2}$.

I don't think this is correct. There are different ways to to do the calculation. You could try to calculate the travel times of the light from the point of view of the ether frame or from the point of view of the apparatus. You should get the same result using either reference frame since time is absolute in the Fitzgerald-contraction theory. I think it's easier to do the calculation in the ether frame. First, try to find the time interval for a light pulse to leave the origin 0 and arrive at mirror M₁ using the ether as the frame of reference.

 

[drjack984]

hope this help,
this is pratically the same of the one in french special relativity 3.1.
is not a super perfecly formal solution but is samething