Why Does Rotating the Michelson-Morley Apparatus Alter the Time Delay?

[BareFootKing]

Homework Statement

It isn't really a homework problem rather

http://spiff.rit.edu/classes/phys200/lectures/mm_results/mm_results.html

If you search " After they rotated the device by 90 degrees, the arm which was sideways to the ether would now be going with-and-against the wind, while the other arm would switch from there-and-against to sideways. This would cause the light beams to switch roles, so that the previous winner would now come in second place by the same difference. If Michelson and Morley compared the results from the two orientations, they ought to find a total difference which was double the difference in a single run: " on the link you will find where it talks about rotating the apparatus 90 degrees.

Homework Equations

The Attempt at a Solution

I am having trouble understanding why the time delay when the apparatus is rotated 90 degrees is twice that of the time delay of the original setup. I've looked through various books like kleppner but found no helpful explanation. Is the time delay measuring the time as he is rotating the apparatus and light is traveling or is it after the rotates the apparatus. He sends out a beam of light and calculates the time delay.


I would think that the time would be the same if all there doing is switching roles.

 

[rude man]

They were comparing the difference between the two orientations. If I have a difference of x in one direction and a difference of -x in the other, comparing the two differences I get 2x.

It did not say that in one orientation they got twice the difference of the other.

 

[BareFootKing]

I understand that fact when we are talking about displacement, but when we are talking about time delay why wouldn't the difference be 0.
If the time delays are the same in both orientations than wouldn't the difference between the time delays of the two orientations be 0?

 

[rude man]

I understand that fact when we are talking about displacement, but when we are talking about time delay why wouldn't the difference be 0.
If the time delays are the same in both orientations than wouldn't the difference between the time delays of the two orientations be 0?

No.

ΔT - (-ΔT) = 2ΔT.

They were able to tell which direction the delays went for each of the two orientations.

 

[BareFootKing]

but why is it x- (-x). Where does the x represent? I thought we were considering time not displacement.

 

[BareFootKing]

Thank you for the explanation, but I have one more question. I prob don't have a deep understanding of the experiment which is probably why I am having trouble understanding the part where they rotate the apparatus.

Why to the time delays have different signs? What does a negative time delay mean and a positive"

 

[rude man]

A boat starts rowing downriver from point A. His speed is v0 + v1 where v0 is his speed in quiet water and v1 is the speed of the river. At the end of L miles his time is T1 = L/(v0 + v1).

Now he turns around and flows upriver the same distance L. Now his time is T2 = L/(v0 - v1).

And the total difference is T2 - T1 ~ 2T1 or 2T2.

But M and M saw that T2 - T1 = 0 which had to mean the flow rate was zero.

 

[rude man]

Thank you for the explanation, but I have one more question. I prob don't have a deep understanding of the experiment which is probably why I am having trouble understanding the part where they rotate the apparatus.

Why to the time delays have different signs? What does a negative time delay mean and a positive"

If leg 1 was faster than leg 2 they called that +. If leg 2 was faster than leg 1 they called that -.

So + - (-) = 2+.

 

[BareFootKing]

Thank you for helping me so far. I just have one more question now that you told me that. Why exactly would they calculate the time delay in that way. Where t= t1-t2

where t1 is the time to travel up and down leg 1
and t2 is the time to travel through leg 2

And then rotate the apparatus so that leg 1 is in the position of leg 2 and leg 2 is in the position of leg 1

And than they calculate the time delay t1 - t2 which is of the same magnitude as the first time delay but opposite in sign since leg 1 and leg 2 have effectively swapping positions.

My question is why would the calculate it in such a way? what is the consequence of doing it this way?

Thank you very much for helping me thus far btw.

 

[rude man]

OK, if you want more details there are a number of good Websites that can explain those details far better than I. You'll have to familiarize yourself with interference fringes etc. if you haven't already.

For me the gist of the experiment was to show that light travels the same speed in every direction, ruling out some kind of medium fixed to the Earth (the "ether" most believed in before M&M.) By pointing their interferometer with and against the speed of the Earth as it travels around the Sun they were not able to detect any difference in speed of the light going "with" vs "against" the "ether" so they concluded the ether did not exist.

 

[BareFootKing]

Thank you for your help. I will look at some more texts/online material . Thanks for your help!