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  • Its not that I question the conclusions reached concerning the Michelson–Morley experiment, however I would like to know how the following issue was addressed please?

If I could pass bob through a beam splitter, and have each copy of him pace out each leg of the interferometer at say $2~\rm{km/hr}$. However if along one of the legs, there was an escalator aiding his initial progress at $1~\rm{km/hr}$.

Okay, so they each leave the beam splitter at the same time, however aided by the escalator one of them moves at $3~\rm{km/hr}$, and reaches the mirror at the end of his leg earlier than the other. But on the return journey he is inhibited by progression of the escalator, and moves at $1~\rm{km/hr}$. So the other bob who travelled at $2~\rm{km/hr}$ the whole time, makes up ground on the other bob on the return journey, and they arrive home at the same time as each other.

  • With so many clever people working on this experiment over the years, I know there must be a contingency for this issue. If somebody can inform me please?
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  • $\begingroup$ Trouble with homework? Tell us what you have tried on your own and we can try to help you. $\endgroup$
    – CuriousOne
    Commented Mar 14, 2016 at 5:49
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    $\begingroup$ What trouble are you having with calculating this scenario? $v=ds/dt$, from which you get $dt=ds/v$. Substitute the velocities and you can calculate the total time that your Bob would take along the arms of your interferometer. $\endgroup$
    – CuriousOne
    Commented Mar 14, 2016 at 6:26
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    $\begingroup$ Please take what @CuriousOne says seriously and calculate the travel time of the scenarios you suggest. The two walkers don't take the same time in the course of the trip. Really. Do the math in detail. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Mar 14, 2016 at 6:58
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    $\begingroup$ @BillAlsept: The velocities are in the denominator, not in the numerator. 1/3+1/1=4/3, which is not the same as 1/2+1/2=1. $\endgroup$
    – CuriousOne
    Commented Mar 14, 2016 at 7:24
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    $\begingroup$ In your example set the escalator speed to 2km/hr. Bob now takes an infinite time to return on the leg with the escalator because his net return speed is zero. $\endgroup$
    – John Rennie
    Commented Mar 14, 2016 at 7:34

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So the other bob who travelled at 2 km/hr the whole time, makes up ground on the other bob on the return journey, and they arrive home at the same time as each other.

You have missed an important feature of the experiment.
The escalator (ether wind) is in action for all Bob's movements; both up and down the escalator as well as Bob walking "across" the escalator.

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