- #1
P J Strydom
- 68
- 1
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, let's 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.
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, let's 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.