roineust said:
I think that would be:
root((5kph^2)-(2kph^2))=root(21)kph=~4.6kph
1km/4.6kph=~13minutes, one way straight line.
13minutes*2=~26minutes, round trip from bank to bank straight lines.
Is that correct?
Yes that is correct!
And if the river were 21 km wide, the round trip would take about 2x4.6= 9.2 hours.
Now compare that with your earlier calculation for the round trip over the same distance downstream and upstream; you will see that that differs. For that trip you calculated 10 hours.
That is the basis of Michelson's experiment.
He assumed that the Earth is flowing through a light ether, and that it would be possible to detect that motion by comparing the return trip of light in two different directions. In his time it was impossible to directly measure the round trip time over such a short distance, but a change in the roundtrip time could be made visible with his interferometer.
However, he did not find a significant difference. So, he had made a mistake, but where?
Lorentz and Fitzgerald proposed that perhaps matter contracts because it is held together by electromagnetic fields. Heaviside had calculated for charges in motion that those fields contract in the direction of motion, by what is now know as the Lorentz factor.
For the calculation above, the Lorentz factor is (I write SQRT where you write root):
1 / SQRT(1 - 4/25) = 1/0.92.
So (here the boat example falls flat on its face but just consider the calculation!), if the distance of the embankment would shrink like that due to the motion of the river, then the return time as you first calculated would not be 10 hours but 9.2 hours. That is exactly the same as what you get for a trip to the other side of the river and back!
As a matter of fact, it would be the same in all directions.
If you now look again at my post #131, you may be able to understand it this time.
There I only discuss the one-dimensional problem. You may recognize c+v as the light going in the opposite direction as the apparatus, and c-v as the apparatus running ahead of the light, so that the light must catch up - and that takes longer.
Here follow the same formula's with the numbers of the boat example plugged in.
Michelson thought:
t1 = L/(c+v) t1 = 21/(5+2)
t2 = L/(c-v) t2 = 21/(5-2)
t1+t2 = T = [L(c-v)+L(c+v)] / [(c+v)(c-v)]
T = 2L * c /(c² - v²)
T = 2L/c * 1/(1 -v²/c²) T = 2x(21/5) x 1/(1-4/25) = 10
Cheers,
Harald