Editing Michelson–Morley experiment (section)

=== Observer resting in the aether ===
[[File:Michelson-morley calculations.svg|thumb|300px|Expected differential [[phase shift]] between light traveling the longitudinal versus the transverse arms of the Michelson–Morley apparatus]]
The beam travel time in the longitudinal direction can be derived as follows:<ref group=A name=Feynman /> Light is sent from the source and propagates with the speed of light <math display="inline">c</math> in the aether. It passes through the half-silvered mirror at the origin at <math display="inline">T=0</math>.  The reflecting mirror is at that moment at distance <math display="inline">L</math> (the length of the interferometer arm) and is moving with velocity <math display="inline">v</math>. The beam hits the mirror at time <math display="inline">T_1</math> and thus travels the distance <math display="inline">cT_1</math>. At this time, the mirror has traveled the distance <math display="inline">vT_1</math>. Thus <math display="inline">cT_1 =L+vT_1</math> and consequently the travel time <math display="inline">T_1=L/(c-v)</math>. The same consideration applies to the backward journey, with the sign of <math display="inline">v</math> reversed, resulting in <math display="inline">cT_2 =L-vT_2</math> and <math display="inline">T_2 =L/(c+v)</math>. The total travel time <math display="inline">T_\ell=T_1+T_2</math> is:

:<math>T_\ell=\frac{L}{c-v}+\frac{L}{c+v} =\frac{2L}{c}\frac{1}{1-\frac{v^2}{c^2}} \approx\frac{2L}{c} \left(1+\frac{v^2}{c^2}\right)</math>

Michelson obtained this expression correctly in 1881, however, in transverse direction he obtained the incorrect expression

:<math>T_t=\frac{2L}{c},</math>

because he overlooked the increase in path length in the rest frame of the aether. This was corrected by [[Alfred Potier]] (1882) and [[Hendrik Lorentz]] (1886). The derivation in the transverse direction can be given as follows (analogous to the derivation of [[time dilation]] using a [[Time dilation#Simple inference of velocity time dilation|light clock]]): The beam is propagating at the speed of light <math display="inline">c</math> and hits the mirror at time <math display="inline">T_3</math>, traveling the distance <math display="inline">cT_3</math>. At the same time, the mirror has traveled the distance <math display="inline">vT_3</math> in the ''x'' direction. So in order to hit the mirror, the travel path of the beam is <math display="inline">L</math> in the ''y'' direction (assuming equal-length arms) and <math display="inline">vT_3</math> in the ''x'' direction. This inclined travel path follows from the transformation from the interferometer rest frame to the aether rest frame. Therefore, the [[Pythagorean theorem]] gives the actual beam travel distance of <math display="inline"> \sqrt{L^2+\left(vT_3\right)^2}</math>. Thus <math display="inline"> cT_3 =\sqrt{L^2+\left(vT_3\right)^2}</math> and consequently the travel time <math display="inline"> T_3 =L/\sqrt{c^2-v^2}</math>, which is the same for the backward journey. The total travel time <math display="inline">T_t=2T_3</math> is:

:<math>T_t=\frac{2L}{\sqrt{c^2-v^2}}=\frac{2L}{c}\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\approx\frac{2L}{c} \left(1+\frac{v^2}{2c^2}\right)</math>

The time difference between <math>T_\ell</math> and <math>T_t</math> is given by<ref group=A>{{cite book |author=Albert Shadowitz  |title=Special relativity |url=https://archive.org/details/specialrelativit0000shad  |url-access=registration  |isbn=978-0-486-65743-1 |publisher=Courier Dover Publications |edition=Reprint of 1968 |year=1988|pages=[https://archive.org/details/specialrelativit0000shad/page/159 159–160]}}</ref>

:<math>T_\ell-T_t=\frac{2L}{c}\left(\frac{1}{1-\frac{v^2}{c^2}}-\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)</math>

To find the path difference, simply multiply by <math>c</math>;

<math>\Delta{\lambda}_1=2L\left(\frac{1}{1-\frac{v^2}{c^2}}-\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\right)</math>

The path difference is denoted by <math>\Delta \lambda</math> because the beams are out of phase by a some number of wavelengths (<math>\lambda</math>). To visualise this, consider taking the two beam paths along the longitudinal and transverse plane, and lying them straight (an animation of this is shown at minute 11:00, [[The Mechanical Universe|The Mechanical Universe, episode 41]]<ref name=":0">{{Cite web|title=The Mechanical Universe, Episode 41|website = [[YouTube]]|url=https://www.youtube.com/watch?v=Ip_jdcA8fcw&t=632s| archive-url=https://ghostarchive.org/varchive/youtube/20211118/Ip_jdcA8fcw| archive-date=2021-11-18 | url-status=live}}{{cbignore}}</ref>). One path will be longer than the other, this distance is <math>\Delta \lambda</math>. Alternatively, consider the rearrangement of the speed of light formula <math>c{\Delta}T = \Delta\lambda</math> .

If the relation <math>{v^2}/{c^2} << 1</math> is true (if the velocity of the aether is small relative to the speed of light), then the expression can be simplified using a first order binomial expansion;

<math>(1-x)^n \approx {1-nx}</math>

So, rewriting the above in terms of powers;

<math>\Delta{\lambda}_1 = 2L\left(\left({1-\frac{v^2}{c^2}}\right)^{-1}-\left(1-\frac{v^2}{c^2}\right)^{-1/2}\right)</math>

Applying binomial simplification;<ref name="Cengage Learning">{{cite book|last1=Serway|first1=Raymond|url=https://books.google.com/books?id=-g_y6CMkZ0IC|title=Physics for Scientists and Engineers, Volume 2|last2=Jewett|first2=John|publisher=Cengage Learning|year=2007|isbn=978-0-495-11244-0|edition=7th illustrated|page=1117}} [https://books.google.com/books?id=-g_y6CMkZ0IC&pg=PA1117 Extract of page 1117]</ref>

<math>\Delta{\lambda}_1 = 2L\left( (1+\frac{v^2}{c^2}) - (1+\frac{v^2}{2c^2})\right)={2L}\frac{v^2}{2c^2}</math>

Therefore;

<math>\Delta{\lambda}_1={L}\frac{v^2}{c^2}</math>

The derivation above shows that the presence of an aether wind would produce a difference in optical path lengths between the two arms of the interferometer. This path difference depends on the orientation of the interferometer relative to the aether wind. Specifically, the derivation assumes that the longitudinal arm is aligned parallel to the presumed direction of the aether wind. If instead the longitudinal arm is oriented perpendicular to the aether wind, the resulting path difference would have the opposite sign.

The magnitude of the path difference can vary continuously and may represent any fraction of the wavelength, depending on both the angle between the apparatus and the aether wind and the wind's speed.

To detect the existence of the aether, Michelson and Morley aimed to observe a "fringe shift" in the interference pattern. The underlying principle is straightforward: when the interferometer is rotated by 90°, the roles of the two arms are exchanged, altering the path difference due to the aether wind. The fringe shift is determined by calculating the difference in path differences between the two orientations, and then dividing that value by the wavelength.<ref name="Cengage Learning"/>

: <math>n=\frac{\Delta\lambda_1-\Delta\lambda_2}{\lambda}\approx\frac{2Lv^2}{\lambda c^2}.</math>

Note the difference between <math>\Delta \lambda</math>, which is some number of wavelengths, and <math>\lambda</math> which is a single wavelength. As can be seen by this relation, fringe shift n is a unitless quantity.

Since ''L''&nbsp;≈&nbsp;11 meters and λ&nbsp;≈&nbsp;500 [[nanometer]]s, the expected [[fringe shift]] was ''n''&nbsp;≈&nbsp;0.44. The negative result led Michelson to the conclusion that there is no measurable aether drift.<ref name="michel2" /> However, he never accepted this on a personal level, and the negative result haunted him for the rest of his life.<ref name=":0" />