Null result of the michelson morley experiment

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Homework Help Overview

The discussion revolves around the Michelson-Morley experiment and its null result, specifically exploring how this can be explained through the concept of length contraction as described by Lorentz transformations. Participants are examining the implications of these transformations on the fringe shift observed in the experiment.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to relate the fringe effect to Lorentz transformations and questioning whether the calculations maintain consistency with the original experiment's results. There is also exploration of the time taken for light to travel different paths in relation to Earth's motion.

Discussion Status

Some participants have provided guidance on using binomial approximations and have confirmed aspects of the calculations. However, there are still questions regarding the correctness of the original poster's approach and the implications of length contraction on the time calculations.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the depth of exploration into the problem. There is an ongoing discussion about the assumptions related to length contraction and its effects on the experiment's outcomes.

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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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