Michelson-Morely experiment?

Hello Hala1666,

the MICHELSON MORLEY experiment (MMX) should demonstrate the movement of the Earth relative to a hypothetical, omnipresent supersubstance called ether with an interferometer. The rays migrate the armhin and back and overlap what gives stripes.

According to MAXWELL’s ideas, the light always spreads with c relative to the ether.

a) Explain how to get to these values…

The idea is that the light relative to the interferometer in movement direction with c − v =: c(1 − β) and against in this direction with c(1+β) and therefore overall

(1.1) tp = d/c(1 − β) + d/c(1 + β)

need. In order to add, we must extend the first summand with 1+β and the second with 1−β; with the third. Binomic formula

(1.2) tp = 2d/c(1 − β2).

When the light signal is moved transversely to the direction of movement, it must be clear that the light signal relative to the hypothetical ether is oblique and its speed in the direction of movement has a component of the size cβ. c√{1 −β2} remains for the component transverse to the direction of movement, so that the signal back and forth

(2) tn = 2d/c√{1 − β2}

needs a little less time (the denominator is a little bigger, there is always something smaller than 1.

The difference is

(3.1) tp − tn = 2d⁄c(1−β2) − 1⁄√{1−β2}),

where we need to expand with √{1 − β2}:

(3.2) tp − tn = 2d(1 − √{1−β2}) /c(1−β2)

Actually, this can no longer be simplified. For β < 1, however, only β2 < < < < 1, and so we can approximate

(N1) √{1 − β2} ≅ 1 −1⁄2β2

and the approximation

(N2) 1⁄(1 − β2)≅1 + β2

apply and receive to (3.2)

(3.3) tp − tn ≅d(β2 + β4)⁄c ≅dβ2⁄c.

…and what one should have seen this difference in the duration of the experiment!

When turning the interferometer, his arms change the rollers: The previously longer runtime becomes shorter, the shorter longer. As a result, the interference strips should move.

(b) A point on the earth’s surface is already moving at high speed due to the earth’s rotation. At the same time, the earth moves around the sun and the entire solar system circles around the center of the Milky Way. The sum of the different speed components of our solar system derived from measurements is about 370 kilometers per second. How large should the time difference be at a distance of the mirrors A and B from the center S from one meter when one assumes this speed for the earth?

You set the numbers in (3.3). Very coarsely calculated β is somewhat less than 5⁄4∙10-3, i.e. β2 ≅25⁄16∙10-6 and d/c is somewhat more than 10⁄3∙10-9 s, so that in this case

(4.1) tp − tn ≅ 250⁄48∙10–15 s;

so a little more than 5×10-15 s comes out.

PS: I calculated it and got 5,07075×10-15 s, right?

Apparently, it’s pretty close.

By the way, the difference in gear of light waves is surprisingly large, namely

(4.2)

clearly above the wavelength of visible light, which lies at as much as 3.9×10-7m to 7.8×10-7m.