# B Measuring Michelson–Morley light beams

#### P J Strydom

I was away for a few months and thought I might pose this thought I am working on, on this forum when I return.
There was a very help full person who assisted me in the finer explanation of the Lorenz transformation a few months ago, and I really learned a lot about SR and GR.
A few weeks ago, I thought to myself, Myself, can one measure the path the light beams on the Michelson Morley inferometer?
I played on Google and found some beautiful mathematical equations that could actually show the expected difference in length if a light beam when split and sent in opposing directions as what MM at first anticipated.

Then I thought, what if I follow the paths and measure the length and time and see if I can get a difference that might explain why MM did not find the fringe interference.

Now, it is a fact that I am not a scientist, but I would like to know what you guys think about my calculations.
The simple way math of coarse.
Michelson Morley's contraption allowed a 11 m path for the light to travel.
I decided to make it 12 m for ease.
This will give us a pathway for Beam A to travel 3m up to the beam splitter, 3m Left to the mirror, 3m right to the beam splitter again, and 3 m right to get to the end.
Beam B will travel 3 M Up, and 3 m up, and 3m down, and 3m right.
Therefore, if there is no (theoretical) movement on the contraption, the beams A and B will travel 12m each.
this will result in each beam traveling at 0.00000001 sec from start to finish.

However, when we assume that the contraption is moving at say 340Km/sec (Earth moving in space)in the parallel direction of Beam B, things started to get interesting.
Beam A was traveling Up = 340000m/s+3m/s
This gave a time of 0.00113334333 UP
I had to calculate using Pythagoras the distance of 3m sq+340000m sq= sqr root to get the distance the light would travel from the splitter mirror to the left mirror
this gave a distance of left=340000.000013235 m
Time traveled = 0.00113333333 sec.
The same distance will go for the Right and Right to arrive at the end.
The totals is as such for A
Up = 340003m/s time 0.00113334333s
Left = 340000.000013235 m time = 0.00113333333 sec
Right = 340000.000013235 m time = 0.00113333333 sec
Right = 340000.000013235 m time = 0.00113333333 sec

Therefore, the total time traveled for A was
1 360 003.00 m 0.00453334333 sec

B traveled
Up = 340003m/s time 0.00113334333s
Up = 340003m/s time 0.00113334333s
Down = -339 997.00 m time -0.00113332333 (this is a huge mistake because it would mean that the light beam arrives at the mirror, before it was reflected from the splitter)
Right = 340000.000013235 m time = 0.00113333333 sec

Therefore B traveled 680 009.00 m Time 0.00226669667 sec.
But looking at the problem we have on beam B when it reached the top of the path, and had to return back to the splitter to deflect it to the end, I decided to increase Michelson Morley's contraption to one c.
We made it 300 000 (theoretical c for this purpose)Km wide.

This made it work much better.
Beam A traveled
Up 1.00113333333 Sec 300 340 000.00 m
Horizontal 1.00000064222 Sec 300 000 192.67 m
Horizontal 1.00000064222 Sec 300 000 192.67 m
Horizontal 1.00000064222 Sec 300 000 192.67 m
Total 4.00113526000 Sec 1 200 340 578.00 m

Beam B traveled
Up 1.00113333333 Sec 300 340 000.00 m
Up 1.00113333333 Sec 300 340 000.00 m
Down 0.99886666667 Sec 299 660 000.00 m
Horizontal 1.00000064222 Sec 300 000 192.67 m
Total 4.00113397556 Sec 1 200 340 192.67 m

Now I am thinking.
If I worked the path and time out correct, using an inferometer with a width of 600 000 Km, and there is a difference of 0.0000012844 sec, will there be any fringe interference visible?
This is a addition of 0.0000000007645500% on yellow light at 540THz.

I then thought, lets go for 0.5c on the movement of the Earth and look at that difference.
The difference between A and B is now 0.2360679775 sec.
This will give a difference of 0.0001405166532737% on the wave of 540THz.

I explicitly wanted to get away from well known formulas, and do it as simple as possible.

Does this not say that MM would not have seen interference at all?
I hope my thinking was correct.

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#### Nugatory

Mentor
In problems of this sort, it's a good idea to do the algebra first to get the general formula (in this case, the travel times and distances for the two paths as functions of $D$ (the arm length), $v$ (the hypothetical speed through the ether) and $c$. You can check that for reasonableness (for example, does your formula show the travel time increasing with $D$?) and dimensional consistency, and only after you're sure you have the formula right should you plug in numbers to get a quantitative result.

Google for "Michelson Morley calculations" and you'll find several good examples of this problem being worked. Bear in mind that an interferometer will easily detect differences that are a fraction of a single wavelength, and you'll find that the MM experiment is quite sensitive enough to justify the claim of a negative result.

#### P J Strydom

I explicitly tried to work around mathematical formulas, and thought that if I could calculate the path of a beam split as the MM experiment, by actually measuring each increment from the beam to splitter, and splitter to mirror, and along the 2 different paths, back to the splitter and ending in the interference splits, I might see the picture on the reason why the inferometer did not show deviations.

Please, I might be a fool in science, but I just love these scientific interpretations on the speed of light and time dilation etc.
This is why I want to see the calculations in a more practical format.

Perhaps you might be able to tell me if I was wrong with my measurements?

#### P J Strydom

(for example, does your formula show the travel time increasing with DDD?)
What I did was to take the distance to the left and right, squared it, and added it to the distance of movement ahead, squared, and with square root came to the distance the beam supposedly had to travel to get to the reflecting mirror, and back to the splitter, and then to the end.
a normal Pythagoras calculation.
Am I correct in my assumption?

#### Nugatory

Mentor
a normal Pythagoras calculation.
Am I correct in my assumption?
That's the right general approach.

If you work through the algebra instead of plugging in the numbers from the start you will find that the difference in the length of the two paths is approximately $Dv^2/c^2$. With $D$ equal to 11 meters and $v$ set to the orbital speed of the earth that works out to about 100 nanometers, easily detectable by an interferometer working with visible light.

https://hepweb.ucsd.edu/ph110b/110b_notes/node40.html is a pretty good explanation.

Thank you.

#### P J Strydom

I see the assumption in this experimental measurement is to only measure from the splitter mirror to the reflecting mirror and back.
There is no measurements for
1. the light source to the splitter,
2. and a measurement from the splitter mirror to the telescope.
This simplifies the measurements a great deal.
Am I correct?
(Sorry, I will be a pesky itch until I feel I mastered the Lorenz transformation, and the MM experiment is the foundation to it all.)

#### Nugatory

Mentor
I see the assumption in this experimental measurement is to only measure from the splitter mirror to the reflecting mirror and back.
Yes, because those are the only distances that affect the difference between the lengths of the two paths.
until I feel I mastered the Lorenz transformation, and the MM experiment is the foundation to it all.
The Lorentz transforms folllow from the invariance of the speed of light. The MM experiment is just one of many that demonstrate that invariance. For more, you can try section 3 of http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html (which is also linked in a sticky post at the top of this forum).

#### P J Strydom

I spent some of my private time and went through the calculations on the expected Aether difference MM thought they would be able to measure, between a light beam travelling parallel to supposed Aether, and again a light beam travelling perpendicular to the Aether.

Now, there are a few thoughts I would like to clarify before I can continue.

1. the beam travelling parallel will take longer to return, than the beam travelling perpendicular to the Aether.
----this means the beam travelling perpendicular to the aether, was expected to arrive back on position first. followed by the one in parallel to the aether.
2. the difference is cancelled out due to the length that contracts as the Michelson Morley inferometer travels against the aether, thereby squeezing the length of the parallel distance into itself equal to the difference between the beam traveling lengths. This difference is the calculation factor of Lorenz.
3. Due to this Length contraction, a clock that travels parallel with this light beam against the supposed Aether wind, will also run slower. otherwise we will measure the distance to be shorter that the time traveled. Time should also be adjusted accordingly.

It took me a while to summarize it this way.
If I am wrong, please tell me where.

#### Ibix

1. the beam travelling parallel will take longer to return, than the beam travelling perpendicular to the Aether.
----this means the beam travelling perpendicular to the aether, was expected to arrive back on position first. followed by the one in parallel to the aether.
Yes
2. the difference is cancelled out due to the length that contracts as the Michelson Morley inferometer travels against the aether, thereby squeezing the length of the parallel distance into itself equal to the difference between the beam traveling lengths. This difference is the calculation factor of Lorenz.
This is written with the assumption that the ether exists and awareness of the null result of Michelson-Morley. So this is basically Lorentz and Fitzgerald's explanation for the result. It's not what Michelson and Morley expected, nor is it Einstein's complete understanding. It's relevant to a long-abandoned phenomenological patch to ether theory that ether somehow contracts material bodies passing through it.
3. Due to this Length contraction, a clock that travels parallel with this light beam against the supposed Aether wind, will also run slower. otherwise we will measure the distance to be shorter that the time traveled. Time should also be adjusted accordingly.
I don't think this is correct. In an ether model, I don't think one can naively build a light clock and expect it to keep time except in the ether rest frame. And I don't think it's immediately obvious that all clocks will be affected the same way by motion through the ether. So I'd be inclined to think that trying to bring time dilation into this is opening an enormous can of worms to no purpose. You'd have to end up with something mathematically identical to relativity, which is perfectly happy without an ether.

#### P J Strydom

if I may elaborate on my poor understanding of the movement of light through a split mirror as MM did.
1. Why did they claim that the light moving parallel (with the direction of the Earth against supposed Aether) will travel slower to the reflecting mirror and back to the splitter mirror, than the one perpendicular with this movement?
...The way I see it is that the light will travel slower towards the reflecting mirror as it has to work against the Aether, but the return stroke will be quicker and will actually cancel this delay out.
2. I have been thinking about this for a long time and did quite a few measurements, and find that as the retarded length took longer, the advanced length is shorter by the exact time, but averages out equally to if the time it took to travel longer to the mirror, is equal to the time the return stroke took to travel shorter.

3. In my mind, If I may commute my thoughts, the light beam travelling at an angle perpendicular to the "aether movement" is the one that will take longer to return to the splitter mirror.

Shaacks, why do I have it the wrong way around?

#### jbriggs444

Homework Helper
he way I see it is that the light will travel slower towards the reflecting mirror as it has to work against the Aether, but the return stroke will be quicker and will actually cancel this delay out.
If you want to drive from Chicago to Des Moines (call it 300 miles) at speed of 30 miles per hour for half the trip and 60 miles an hour for the other half, what is your average speed for the entire trip?

[Or to phrase it differently: "trust the math, not your intuition"]

#### P J Strydom

If you want to drive from Chicago to Des Moines (call it 300 miles) at speed of 30 miles per hour for half the trip and 60 miles an hour for the other half, what is your average speed for the entire trip?

[Or to phrase it differently: "trust the math, not your intuition"]
OK, I think I lost your explanation, but lets see if I understand your explanation if I work through it.
1. I will travel 150 miles at 30 miles an hr Westerly direction.
2. then east to Chicago another 150 miles, but this time at 60 miles per hour.

West traveling time = time 5 hrs
East traveling time = 2.5 Hrs.
Total traveling time = 7.5 hours
Average speed 40 miles per hour.

Perhaps I got lost in the understanding of your experiment.

Should you not have said,
1. I will travel in the direction of Des Moines (300 miles away West) at an average speed of 40 miles an hour, but only for 150 miles.
2. and return to direction Chicago (East) which is now 150 miles away at an average speed of 40 Miles an hour?
Now I traveled 300 miles at an average of 40 miles an hour.
I traveled 7.5 hours.

But relative to a satellite observer over the North pole, who observes my position as when I left Chicago...
I will travel West from Chicago's position, but should include the rotation of the Earth.
1. Therefore, I am traveling at 1000 + 40 Miles an hour in an westerly direction.
2. Then back to Chicago at 1000 - 40 Miles an hour in an eastern direction.

Conclusion.
The satellite saw me travelling West as if I was traveling East at 1000-40 Miles per hour.
And when I traveled back to Chicago, It saw me traveling East at 1000+40 miles an hour.
I still traveled 300 miles, even if I was observed from the satellite.

Why will it be different than If I was traveling due south for 300 Miles to Highway 69?
Both will be the same distance, and the speed will also be the same.

Therefore, If I drove west and east, and you south to north, we will arrive at the same time upon return.
If there was this huge wind at 5 miles an hour pushing me back as I drove West and my energy was such that I could only deliver 40 miles an hour, I will slow down to 35 miles an hour, but it will assist me to achieve 45 miles an hour on my return.
You on the other hand will experience no drag or assistance on your speed for the entire journey.
In this instance we will again arrive together.

But this speeds we are talking about is very slow, compared to c, but to me it explains the result.

From the perspective of the satellite, There was a clear difference in the speed I traveled west and east, but the satellite will see us both arrive together.

The question is, if the cars were light beams, what would we see?

#### jbriggs444

Homework Helper
Conclusion.
The satellite saw me travelling West as if I was traveling East at 1000-40 Miles per hour.
And when I traveled back to Chicago, It saw me traveling East at 1000+40 miles an hour.
I still traveled 300 miles, even if I was observed from the satellite.
OK, let us use your version. How long does it take for the round trip at 1000-40 one way and 1000+40 the other way.
How long does it take for the round trip at 1000 one way and 1000 the other way?

#### P J Strydom

In short, it took me longer to travel west, but less to travel east.
If the wind blew east, the reverse will be the result, but the time traveled the same.

Then I thought about the following.
It there was an aether wind interfering with the parallel light beam, the theory would be that the light traveled faster with it in the same direction, and slower upon return.
It will not show any difference when measured with the perpendicular light beam.

Someone that travels on the Michelson Morley inferometer on its "North pole" will also observe that the 2 beams arrived at the same time, even if there was an Aether wind.

I still dont understand why the calculations would say the parallel beam will arrive after the perpendicular beam.

#### P J Strydom

OK, let us use your version. How long does it take for the round trip at 1000-40 one way and 1000+40 the other way.
How long does it take for the round trip at 1000 one way and 1000 the other way?
Are you messing with me?
Please explain why the light beam parallel with the Aether wind would have traveled longer than the one travelling at 90 deg?
I have it the other way around.

#### jbriggs444

Homework Helper
No, I am not messing with you. I am asking you to do a simple calculation. How much time elapses for a round trip which is slow by 40 mph one way and fast by 40 mph the other way? Does it match with the elapsed time for a round trip which is the same speed both ways?

Possibly we are talking at cross purposes, but I think you are trying to determine round trip time.

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#### Ibix

The way I see it is that the light will travel slower towards the reflecting mirror as it has to work against the Aether, but the return stroke will be quicker and will actually cancel this delay out.
As @jbriggs444 is trying to get you to calculate, the difference doesn't cancel exactly.
2. I have been thinking about this for a long time and did quite a few measurements, and find that as the retarded length took longer, the advanced length is shorter by the exact time, but averages out equally to if the time it took to travel longer to the mirror, is equal to the time the return stroke took to travel shorter.
Recheck your maths (or post it). You've gone wrong somewhere.
3. In my mind, If I may commute my thoughts, the light beam travelling at an angle perpendicular to the "aether movement" is the one that will take longer to return to the splitter mirror.
Both take longer. The parallel beam takes more longer - a factor of $\gamma^2$ for the parallel beam versus $\gamma$ for the perpendicular one, if my maths is correct.

#### jbriggs444

Homework Helper
Average speed 40 miles per hour.
Note that 40 is not the average of 60 (45 + 15) and 30 (45 - 15).

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• Ibix

#### P J Strydom

Damn, I get confused.
Perhaps it is due to me not discussing one point at a time.
Please permit me to do the following.
First, the MM experiment standing still, then influenced by Aether wind.
Which beam will arrive first. 1. Now, if this inferometer is at rest.
And we will assume there is no influence on the light beam such as na aether wind.
= the light beams will arrive at the same time back to the beam splitter.

However, If there is an aether wind and if it blows parallel from the right hand to the left,
2. The light beam will travel against the aether wind to the mirror at c-velocity of aether.
3. but upon its return it will travel at c +Velocity of the aether.
4. but the beam traveling up to the mirror perpendicular to the aether, will be affected by a triangular distance it has to travel, and the same when it travels down to the splitter.
= this is where I might be making my error.
To me it seems as if the beam paralel will travel at the same time as it did when if there was no aether wind.
But the beam crossing the aether wind will take longer.

This is where I am missing out on the logic.
I get the Up-Down beam arriving later, and not before the Left-Right beam.

Jeeeeez I hope you can grasp my stupidity in this thought.
because, everytime I check animations on You tube about this description, they say the Left-Right beam will arrive later.

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#### Ibix

The travel time for the parallel beam assuming an ether wind of speed $v$ is $$t=\frac{d}{c+v}+\frac{d}{c-v}$$Algebra should give you $$t=\frac{2dc}{c^2-v^2}=2d\gamma^2/c$$Therefore the travel time is longer for all non-zero $v$, since $\gamma$ takes its minimum value when $v=0$.

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#### P J Strydom

Nice, eventually I got it.
If I was traveling East to West for 150 miles at 50 miles and hour, and back, it will normally take me 6 Hours.
However, if a wind blew at 5 Miles an hour I will be travelling With it for 2.7272...Hours, but Against it for 3.3333...Hours.
Total is 6.0606...
I was silly.
$$t=\frac{d}{c+v}$$ Plus
$$t=\frac{d}{c-v}$$
TNX Ibix

I did not comprehend that the distance will stay the same, due to me thinking in a framework of continious movement of both co ordinates.
I will explain later.
Guys you helped me a lot.

#### P J Strydom

And if I work out a travelling distance where I travel perpendicular to the wind,
I will travel a length of 150 Miles square, and 5 miles length square, at the square root of the sum.
Times 2 for the return.
This will result in a total time traveled 6.00332 Hrs.
Shorter than the parallel traveling time of 6.0606...
Great stuff.
Hope I am correct now.

#### P J Strydom

Now I am again thinking.
If we change the scenario to no aether wind but we measure the movement of the Earth in relation to the MM inferometer, will the result be the same? lets call beam A the one parallel with the movement of the earth around it's axis. (so from the splitter to the mirror, the earth will move towards the splitter against the light beam.
And Beam B the beam perpendicular to the Earths movement around its axis.

Now we have the light beam travelling to the Mirror, but the mirror moves closer to the light beam shortening the distance.
Once the light beam reflects back, the splitter will move away from the light beam, lengthening the distance to the splitter for the Light beam.
on the beam, B, it will now travel to its mirror and back in a triangular Pythagoras calculation.
Will they arrive at the same time, if not, which one will arrive first?

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