# Why the michelson-morley experiment is not reliable

#### johanen

Why the Michelson-Morley experiment is not reliable, and why Einsteins (I dont want to reveal my identity over the internet, but i will include a number which is connected to my identity in a highly complicated way, and i have also taken other precautions, so that i can prove that i was the one that wrote this document and that i was the one that figured it all out.
Here is my number: 5097330 Sweden )

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#### Tide

Homework Helper
this short extra time it takes for the beam is not detectable with an atomic clock
And that is precisely why Michaelson used an interferometer! :)

You didn't make clear the relationship between the sources, detectors and the observer but I don't see that you have made any comparison with Michaelson-Morley. They were interested in comparing the speed of light both parallel and perpendicular to the "ether" and they were very careful to account for time differences related to both relative positions and orientation.

#### LeonhardEuler

Gold Member
johanen said:
Therefore the distance that the beam has to go from the mirror to the receiver will be a little bit shorter than the distance between the mirror and the receiver. And because the distance between the light source and the mirror is almost exactly the same as the distance between the mirror and the receiver, the increase of the distance will be the same as the decrease of the distance. So it even out.
This is not true, the changes in distance don't even out. (More exactly, they don't even out in the frame of the beam of light using classical transformations. Of course there is no difference in the distance in the instrument's frame since the length doesn't change) On the first trip the target is moving towards the beam. If v is the speed of the target and $t_0$ is the time taken for this trip, then the change in distance is $-vt_0$. Simmilarly, when the target is moving away, the change is $vt_1$, where $t_1$ is the length of time for this trip. So why do these not cancel? Well, the target is moving away in the first trip, so it takes a longer time to complete this trip than the first trip. Therefore $t_1>t_0$. The diference in distance is $v(t_1-t_0)$, and therefore it is positive.

How positive? Using distance equals rate times the time in the first trip gives: (keeping in mind that the distance is shrunk)
$$(d - vt_0) = ct_0$$
$$d = t_0(c+v)$$
$$t_0 = \frac{d}{c+v}$$
For the second trip:
$$(d + vt_1) = ct_1$$
$$d = t_1(c-v)$$
$$t_1 = \frac{d}{c-v}$$
The change in distance is therefore:
$$\frac{d}{c-v} - \frac{d}{c+v} = \frac{2dv}{(c+v)(c-v)}$$

A final note. The Michelson Morley experiment did not depend on accurate measurements of time. It exploited the fact that the wavelength of visible light is quite small, so a small change in distance can be the difference between constructive and destructive interferance. The beam used to interfere with this one was one sent perpendicular to it. For more details look at http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html

#### Doc Al

Mentor
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"Why the michelson-morley experiment is not reliable"

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