## modphysnoob Group Title 2. In the 1887 experiment by Michelson and Morley, the length of each interferometer arm was 11m. The experimental limit on the measurable fringe shift was 0.005 fringes. If sodium light was used (λ = 589nm), what upper limit did the null experiment place on the speed of the Earth through the expected ether? one year ago one year ago

1. modphysnoob Group Title
2. modphysnoob Group Title

t1= l/ (c-v )+ l/ (c+v)

3. Jemurray3 Group Title

Right, so you need to find the phase difference as a function of velocity. If a river flows downstream at speed u, and I can paddle my boat and a speed c relative to the river, how long does it take me to go a distance L and back?

4. Jemurray3 Group Title

at a speed c*

5. modphysnoob Group Title

l/c+v

6. Jemurray3 Group Title

To summarize, the guy who has to go upstream and downstream takes a time $t = \frac{2L c}{c^2-u^2}$ and goes through a number of cycles $\frac{t}{T} = \frac{2L}{\lambda} \cdot \frac{c^2}{c^2 - u^2}$ The guy who is unaffected takes time $t' = \frac{2L}{c}$ And goes through a number of cycles $\frac{t'}{T} = \frac{2L}{\lambda}$ Everything else was a typo -- I got my c's and u's mixed up, sorry.

7. modphysnoob Group Title

so we are trying to find number of oscillation?

8. Jemurray3 Group Title

We're more interested in the difference between the two.

9. Jemurray3 Group Title

The phase difference is the difference between those two quantities times 2 pi (if the difference is zero, that's fine. If it's one, that means they are 1 whole wavelength or 2pi radians out of phase). So, the phase difference is $\Delta \phi = 2\pi \frac{2L}{\lambda} \frac{u^2}{c^2-u^2}$

10. Jemurray3 Group Title

But the main point of your question is this: The "fringe shift" is just the difference in number of cycles between the two. Therefore, $N = \frac{2L}{\lambda} \frac{u^2}{c^2-u^2}$ if we assume that the speed of light is immensely greater than the speed with which we move through the aether, $N \approx \frac{2L}{\lambda} \frac{u^2}{c^2}$ given N, which for you is .005, we can find u: $u = c \sqrt{ \frac{N\lambda}{2L} }$ Which is a few hundred meters per second, if I'm estimating that correctly.

11. modphysnoob Group Title

Thanks you for you help, we just started relativity this week and it's very tricky

12. Jemurray3 Group Title

No problem. Just as a caveat -- I lied just a little bit during that derivation (in particular, the part where I calculated the travel time for the guy traveling across, rather than up or downstream) but in the end it doesn't meaningfully affect the calculation and you'd get the same answer.

13. modphysnoob Group Title

but wouldn't guy going across be affected by river current as well since he have to go diagnally? since|dw:1365735827635:dw|