Null result of the michelson morley experiment

AI Thread Summary
The discussion centers on explaining the null result of the Michelson-Morley experiment through the concept of length contraction as described by Lorentz transformations. The user attempts to demonstrate that the fringe shift remains consistent despite the effects of motion, using calculations involving the speed of Earth and the Lorentz factor. They derive the time taken for light to travel in both perpendicular and parallel paths to Earth's motion, emphasizing the impact of length contraction on the parallel path. The conclusion drawn is that the length along the parallel path should be adjusted by the Lorentz factor, confirming the expected outcomes of the experiment. Overall, the discussion highlights the relationship between relativistic effects and the experiment's null result.
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Homework Statement



Show that the null effect of the michelson morley experiment can be accounted for if the interferometer arm parallel to motion is shortened by a factor of
(1-v^2/c^2)^.5

Homework Equations



V(earth)=29.8 km/s , fringe in MM experiment is .4 ; gamma =sqrt(1-v^2/c^2)

The Attempt at a Solution



In the MM original experiment , .4 fringe/(29.8 km/s)^2 = .00045043

I was supposed to show that the fringe effect from the original experiment is still the same even if I used lorentz transformations right?>

.4 fringe/(v(earth) * gamma)=.4 fringe/((29.5)*^2*sqrt(1-(29.5)^2/c^2))= .00045043

My calculator approximates sqrt(1-v^2/c^2 ) to 1
 
Last edited:
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You can use a binomial approximation to calculate sqrt(1-v^2/c^2)=1-1/2*v^2/c^2+...
 
genneth said:
You can use a binomial approximation to calculate sqrt(1-v^2/c^2)=1-1/2*v^2/c^2+...

you kinda didn't answer my first question

Original question:
I was supposed to show that the fringe effect from the original experiment is still the same even if I used lorentz transformations right?>

Other than that, Is everything else I solved correct
 
The total time taken light was expected to take, in the path perpendicular to Earth's motion was:

T_1 = \frac{2L}{\sqrt{1-v^2/c^2}}

The total time expected parallel to Earth's motion was:

T_2 = \frac{2Lc}{c^2-v^2}

So if length contraction happens during the parallel path, what happens T2... and hence what does T2 - T1 become?
 
learningphysics said:
The total time taken light was expected to take, in the path perpendicular to Earth's motion was:

T_1 = \frac{2L}{\sqrt{1-v^2/c^2}}

The total time expected parallel to Earth's motion was:

T_2 = \frac{2Lc}{c^2-v^2}

So if length contraction happens during the parallel path, what happens T2... and hence what does T2 - T1 become?

if length is contracted during parrallel path, would L = L(0)/gamma ?
 
Yes, the length along the parallel path would be L/gamma. So substitue L/gamma instead of L in the T2 formula...
 
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