# Null result of the michelson morley experiment

1. Sep 6, 2007

### Benzoate

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

3. 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: Sep 6, 2007
2. Sep 6, 2007

### genneth

You can use a binomial approximation to calculate sqrt(1-v^2/c^2)=1-1/2*v^2/c^2+...

3. Sep 6, 2007

### Benzoate

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

4. Sep 6, 2007

### learningphysics

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?

5. Sep 7, 2007

### Benzoate

if length is contracted during parrallel path, would L = L(0)/gamma ?

6. Sep 7, 2007

### learningphysics

Yes, the length along the parallel path would be L/gamma. So substitue L/gamma instead of L in the T2 formula...