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I'm having some conceptual misunderstandings of the Michelson-Morley experiment. The time for the beam going perpendicular ($t_{across}$) of the aether wind I am getting:

$$(ct)^{2}=d^{2}+(vt)^{2}\\ (ct)^{2}-(vt)^{2}=d^{2}\\ c^{2}t^{2}-v^{2}t^{2}=d^{2}\\ t^{2}(c^{2}-v^{2})=d^{2}\\ t^{2}=\frac{d^{2}}{(c^{2}-v^{2})}\\ t_{across}=\frac{d}{\sqrt{(c^{2}-v^{2})}}\\ t_{acrossTotal}=\frac{2d}{\sqrt{(c^{2}-v^{2})}}\\ t_{acrossTotal}=\frac{2d\sqrt{(c^{2}-v^{2})}}{c^{2}-v^{2}}$$

For the time of the beam going with the aether wind (t_along) I get: $$t_{1}=\frac{d}{c+v}\\ t_{2}=\frac{d}{c-v}\\ t_{alongTotal}=\frac{d}{c+v}+\frac{d}{c-v}\\ t_{alongTotal}=\frac{d(c-v)}{(c+v)(c-v)}+\frac{d(c+v)}{(c+v)(c-v)}\\ t_{alongTotal}=\frac{dc-dv+dc+dv}{(c+v)(c-v)}\\ t_{alongTotal}=\frac{2dc}{c^{2}+v^{2}}\\$$

Now from what I'm reading about the experiment, what I need to be showing is that $$t_{along} \geq t_{across}$$ The equations I have now doesn't really show that so I'm confused. From what I'm reading they did the experiment multiple times; by rotating the interferometer, and by waiting for the earth to be in a different position around the sun to detect a .6 fringe difference in the interference patten.

Am I assuming correctly that in order to prove this mathematically, c should greater than or equal to $\sqrt{c^2-v^2}$? Is the experiment trying to show that $t_{along} \geq t_{across}$ when comparing if the aether wind speed is $v=0$ or $v!=0$?

Any assistance in clearing up my misunderstanding is appreciated.

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Say that light relative to the aether at rest moves at speed $c$, and say that the ether is moving in the $x$ direction at speed $v_E$.

Let's consider a simplified version of the Michelson Morley experiment where the interferometer consists of two arms of length $L$ pointed along the $x$ and $y$ directions. Since the aether's velocity is assumed to be purely along the $x$ direction, it is very easy to analyze the motion of light along the $y$ arm. For the $y$ arm, the light travel time $t_y$ is given by $$ t_y = \frac{2L}{c} $$ Meanwhile, the light travel time along the $x$ axis needs to be analyzed in two parts. First, when light travels with the aether, then by addition of velocities we have that light is traveling at $c+v_E$. When light reflects off the mirror and comes the reverse direction, it travels at speed $c-v_E$. Therefore, the light travel time $t_x$ along the $x$ direction is $$ t_x = \frac{L}{c+v_E} + \frac{L}{c-v_E} = \frac{2L c}{c^2 - v_E^2} $$ If the light used is monochromatic with angular frequency $\omega$, then the observable is the phase difference $$ \Delta \phi = \omega \Delta t = - \frac{2L\omega}{c} \frac{v_E^2}{c^2-v_E^2} $$ Therefore, Michelson and Morley were hoping to detect $\Delta \phi$, which would tell them about $v_E$. Of course, we now know that in special relativity, the aether doesn't exist so $v_E=0$ and $\Delta\phi=0$, consistent with what Michelson and Morley observed.

In the real experiment, Michelson and Morely could not assume the arms were perfectly aligned with the aether motion, but they could use the motion of the Earth around the sun to argue that after a few months, the alignment of the arms relative to the aether's velocity would change, which would lead to a difference in $\Delta \phi$. You can carry that through in the above analysis by introducing the $x$ and $y$ components of the aether, $v_{E, x}$ and $v_{E, y}$, then $$ t_x = \frac{2Lc}{c^2 - v_{E, x}^2} $$ and $$ t_y = \frac{2Lc}{c^2 - v_{E, y}^2} $$ so $$ \Delta \phi = \frac{2\omega L}{c} \frac{c^2\left(v_{E,x}^2 - v_{E,y}^2\right)}{\left(c^2 - v_{E,x}^2\right)\left(c^2 - v_{E,y}^2\right)} \approx \frac{2\omega L}{c} \left(\frac{v_{E,x}^2}{c^2} - \frac{v_{E,y}^2}{c^2}\right) $$ where the last approximate equality holds if $v_E \ll c$. You can see that if $v_{E,x}$ and $v_{E,y}$ change due to the Earth's rotation, then $\Delta \phi$ will change as well. Again, it's easy to see that since $v_{E,x}=v_{E,y}=0$ in special relativity, that $\Delta\phi=0$, which is the result Michelson and Morely obtained when they did the experiment.

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  • $\begingroup$ Thank you so much for this explanation, it helped. I do have a question about the phase shift you mentioned. Where did the expected .6 fringe shift come from in the math? $\endgroup$
    – Qubit
    Commented May 8 at 16:31
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    $\begingroup$ Visible light has a wavelength of about $\lambda=500\ {\rm nm}$. Their original apparatus had $L=11\ {\rm m}$, and they expected $v_E\approx 30\ {\rm km/s}$. Plugging that into the formula above (using $\omega=2\pi f=2\pi c/\lambda$)... $$ \frac{\Delta \phi}{2\pi} = \frac{2L}{\lambda}\frac{v_E^2}{c^2} \approx 0.4 $$ I guess your source got 0.6 by making a different assumption about the wavelength of light used; it wasn't a laser but white light, so there is some error bar there. $\endgroup$
    – Andrew
    Commented May 8 at 16:59

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