James, Crisp, Thed, Chroot, Q, Lethe, C'est Moi, Overdoze, and anyone else interested,

As most people on this forum know, I am a firm non-believer in relativity. People keep telling be that relativity is counter-intuitive, but I find it illogical as well.

I understand that the only way that I can prove that relativity is wrong, is to prove that the Michelson-Morley experiment was flawed (since the the results of M-M experiment are what relativity was founded on).

The attachment I included with this post is an illustration of the Michelson-Morley experiment. In short, the Michelson-Morley apparatus generates two perpendicular round trips of light. If the apparatus is moving forward in the aether, the roundtrip of one of the beams of light should take longer than the other beam of light:

t1=2wc/(c^2-v^2)

t2=2w/sqrt(c^2-v^2)

(w is the distance between the beam splitter and the mirrors, v is the speed of the apparatus through the aether)

This would result in the two beams of light in being out of phase when they are recombined at the beam splitter. As a result, the out-of-phase beams of light should produce an interference pattern at the target.

However, since no interference pattern was detected, Einstein (and Lorentz) assumed that the two beams of light arrived at the target at the same time. In order to explain this, Einstein stated that a moving object experiences time dilation and length contraction that compensates for the time delay I stated above.

A few weeks ago I suggested that red-shifting and blue-shifting of the beams of light may compensate for the time delay, without having to introduce time dilation or length contraction. As I started performing calculations to prove my theory, I stumbled upon some troubling results. I am posting this thread in hopes that someone with more mathematical prowess than I, can find if I did something wrong in my calculations. So, here it goes:

First, let's forget about any doppler shifting of light in the Michelson-Morley experiment. For the sake of simplicity in the remainder of the post, let's assume that the frequency for all the beams are the same.

Michelson and Morley thought that their device would measure even small speeds in the aether, since their device was based on interference. They assumed that the Earth's orbit around the Sun is fast enough to give positive results in their experiment. However, I found out that this is not the case.

To calculate whether the two perpendicular waves of light are in phase at the target, we must find the total number of wavelengths of light for both beams. We do this by multiplying the time it takes the beam of light to make the round trip, by the frequency of the light:

wavelengths=t*f

For the first roundtrip beam of light:

wavelengths1=t1*f

wavelengths1=2wcf/(c^2-v^2)

Now let's take w (the distance between the beam splitter and the mirror) to equal 1m. v is the speed of the Earth's orbit around the sun. Let's round off v to 30,000 m/s. The wavelength of the beam is equal to 694 nm, which equals 4.3 * 10^14 Hz. Let's round this to 4 * 10^14 hz. Now if we include these values in the formula, we get:

wavelengths1=2666666.6933333336000000026

Now let's get the total number of wavelengths in the other roundtrip beam of light:

wavelengths2=t2*f

wavelengths2=2wf/sqrt(c^2-v^2)

If we use the same values as above, we get:

wavelengths2=2666666.6800000001

Now to find out how much the two beams of light would be out-of-phase as the result of the motion of the M-M apparatus through the aether, we subtract wavelengths1 by wavelengths 2:

wavelengths(difference)= wavelengths1-wavelengths2

wavelengths(difference)=0.0133333335000000026

That means that the difference between the wavelengths is just 0.013 wavelengths. Total interference occures at 180 degrees, or a 0.5 wavelength difference. As a result of the Earth's motion around the Sun, the two beams of light are out of phase by:

phase difference = 0.013333333 * 360 degrees

phase difference = 4.800000060000000936 degrees

That means as the device is rotated, the difference in the phases of the two beams of light will be between 0 and 4.8 degrees. Since interference patterns occur between 91 and 269 degrees (where total interference occures at 180 degrees), anyone can see that a phase difference of 4.8 degrees max will show no interference pattern.

Conclusion: The speed of the Earth's orbit around the Sun is too small to create any interference patterns in the Michelson-Morley experiment.

If my math is wrong, please indicate any mistakes. Thanks.

Note: In doing my calculations, I needed to find a very accurate calculator that displays as many digits as possible. After searching the net for hours, I finally found a calculator called BCalc that can display up to 5000 digits!! This was perfect for subtracting very large numbers with very small numbers (like c^2-v^2). The calculator is small and it's free. It also has some constants like the speed of light that physicists will find useful. You can get it at http://www.home52365.fsnet.co.uk/bcalc.htm .

Tom

Problem 1: Your analysis of the trip time through the experiments' legs is wrong.

Problem 2: Your calculation of the number of wavelengths in each leg is wrong.

Problem 3: Your understanding of interference is wrong.

Have you really not bothered yet to go get a book on relativity which actually explains the M-M experiment? Are you really still insisting that you can prove us all wrong with poor math skills and senseless premises?

Here's the deal: relativity predicts a lot of weird effects, almost all of which have been measured to incredible precision (the few that remain in question will require very expensive satellites to fly, and those satellites have had funding problems). IF relativity is wrong, its successor will still have to explain all of those experimental results, but some additional ones that relativity got wrong. Therein lies the rub: so far, we haven't found any experimental results that relativity got wrong.

So the best you can hope to do is to make your model match all of the experimental evidence. Relativity does this. Therefore, any alternative theory you produce, which also predicts all the same results, will be nothing more than a new representation for relativity. You'll discover that your new theory and relativity are mathematically equivalent.

It's pointless to go on and on about how you don't like relativity --it predicts the results of every known experiment -- and that's all that matters. Frankly, you continue to use the same tired and incorrect arguments over and over and over again... and I don't know why.

- Warren

Hi Tom,

"However, since no interference pattern was detected, Einstein (and Lorentz) assumed that the two beams of light arrived at the target at the same time. In order to explain this, Einstein stated that a moving object experiences time dilation and length contraction that compensates for the time delay I stated above."

This is not (entirely) correct. The Michelson-Morely experiment indicates that there is no directional dependence for the speed of light, i.e. regardless of how we move (with respect to any other observer), the speed of light appears to be the same. This led Einstein to postulate that the speed of light is a constant for every observer, regardless of their own motion. But semantics aside, there are some other comments I would like to make ;):

"phase difference = 4.800000060000000936 degrees"

Here you should use:
<font face="symbol">Df</font> = 2<font face="symbol">p D</font>x / <font face="symbol">l</font>, where <font face="symbol">D</font>x is the difference in lengths (you can interpret the assumed change in lightspeed as if the light travels a longer distance at constant velocity c). Denote the length of both the arms of the interferometer as L.

The formula is obtained as follows:

(phase)
= (change of phase per distance) * (distance)
= ( 2<font face="symbol">p</font> / <font face="symbol">l</font> ) * (distance)

Since a wave cycles exactly 2<font face="symbol">p</font> in one wavelength <font face="symbol">l</font>. The difference is then easily obtained:

<font face="symbol">f</font>₁ = ( 2<font face="symbol">p</font> / <font face="symbol">l</font>₁ ) x₁
<font face="symbol">f</font>₂ = ( 2<font face="symbol">p</font> / <font face="symbol">l</font>₂ ) x₂

Using that x₁ = x₂ = L (both lightbeams travel the same distance), and taking the difference:

<font face="symbol">Df</font> = 2<font face="symbol">p</font>L<font face="symbol">D</font>(1/<font face="symbol">l</font>)

The difference if wavelengths can be calculated. You assumed the frequencies of both lightbeams to remain constant. Since a beam-splitter does not change the frequency, both lightbeams have the same frequency as the original beam. Denoting this frequency as f, we have that:

v = <font face="symbol">l</font> f

Or equivalently:

1/<font face="symbol">l</font> = f/v

So we find

<font face="symbol">D</font>(1/<font face="symbol">l</font>) = f <font face="symbol">D</font>(1/v)

*IF* one lightbeam would a speed of v = 30.000 m/s faster, then we have that the speed of that lightbeam is (c + 30.000) = 1.1c, or:

<font face="symbol">D</font>(1/v) = 1/c - 1/1.1c = 11/(11c) - 10/(11c) = 1/(11c)

Inserting all this into the formula for the phase difference, we get:

<font face="symbol">Df</font> = 2<font face="symbol">p</font>Lf<font face="symbol">D</font>(1/v) = 2<font face="symbol">p</font>Lf/(11c)

Using a wavelengh of <font face="symbol">l</font> = (c/f) = 670 nm (red laser ---- note that you use the speed of light here) and a interferometer of L = 1 m, yields:

<font face="symbol">Df</font> = 2<font face="symbol">p</font>Lf<font face="symbol">D</font>(1/v) = 2<font face="symbol">p</font>L/(11 <font face="symbol">l</font>) = 852535 rad

Ofcourse this number should be taken modulus 2<font face="symbol">p</font>. We have that:
852535 = 135685*2<font face="symbol">p</font> + 1,32143 rad

Since 1,32 rad is nearly <font face="symbol">p</font>/2 (or 90°), this phase difference surely would have been noted.

Disclaimer: there probably are some errors in the derivation above, be sure to point em out (too lazy to reread it actually ;)).

"That means as the device is rotated, the difference in the phases of the two beams of light will be between 0 and 4.8 degrees. Since interference patterns occur between 91 and 269 degrees (where total interference occures at 180 degrees), anyone can see that a phase difference of 4.8 degrees max will show no interference pattern.
"

This is total non-sense: from the moment you combine two (laser) lightbeams (techy talk: coherent light) you have interference. Even if there is a phase difference of 10^-30 there is an interference pattern, but ok, you won't see it ;). Also, if the phase difference is 180°, you have what is called "destructive interference" (the combination of the light will be ... zero, i.e. no light).

Anyway, perhaps it helps...

Bye!

Crisp

Tom,

I would like to add that this is probably (most certainly) not the derivation Michelson and Moreley used to predict the phase difference (I should look it up, but too lazy ;)). This is just what you would get if you assume that the frequency doesn't change (no blueshifts/redshifts)...

Bye!

Crisp

Coupla things to add to the otherwise excellent comments made.

What matters is only phase difference. You make sure the apparatus takes into account differences in optical paths (number of wavelengths). Bad experiment otherwise. Too many rounding errors, a difference of 0.3*10^14 Hz is huge. Knowing when to use orders of magnitude is an important skill. When dealing with possible phase variations of 1 degree with 10.4 Ghz sources, this matters.

Interference can and does occur at very small phase variations, less than one degree. It normally does not matter but it can do.

Crisp,

Since 1,32 rad is nearly p/2 (or 90°), this phase difference surely would have been noted.

I'll have to take a better look at the formulas you used. It seems very unlikely that my results are so much different than yours. Do you see any major errors in my calculations??

Tom

Thed,

Too many rounding errors, a difference of 0.3*10^14 Hz is huge. Knowing when to use orders of magnitude is an important skill. When dealing with possible phase variations of 1 degree with 10.4 Ghz sources, this matters

I used the BCalc calculator to obtain my results. It is accurate to 5000 digits. You should try it. It's awsome. Here is where you can get it:

http://www.home52365.fsnet.co.uk/bcalc.htm

Interference can and does occur at very small phase variations, less than one degree. It normally does not matter but it can do.

The problem is that back in the late 1800's, the small phase variations you are talking about were unmeasurable. It was unwise for Michelson and Morley to conclude that there was no phase difference when, if there was a small phase difference, they wouldn't have noticed it.

Tom

Originally posted by Prosoothus
I'll have to take a better look at the formulas you used. It seems very unlikely that my results are so much different than yours. Do you see any major errors in my calculations??
Essentially everything in your treatment was wrong. You don't understand wave phenomena very well.
I used the BCalc calculator to obtain my results. It is accurate to 5000 digits. You should try it. It's awsome. Here is where you can get it:
Why did you repeat this endorsement? Do you somehow think it makes your argument stronger?
The problem is that back in the late 1800's, the small phase variations you are talking about were unmeasurable. It was unwise for Michelson and Morley to conclude that there was no phase difference when, if there was a small phase difference, they wouldn't have noticed it.
This is W R O N G Tom.

- Warren

chroot,

Problem 1: Your analysis of the trip time through the experiments' legs is wrong.

I obtained t1 and t2 from this webpage:

http://galileoandeinstein.physics.virginia.edu/lectures/michelson.html

Problem 2: Your calculation of the number of wavelengths in each leg is wrong.

If the total number of wavelengths is not equal to the frequency of the light times the amount of time it traveled, what is it???

Example1: 4*10^14 Hz * 1s= 4*10^14 wavelengths
Example2: 4*10^14 Hz * 2s= 8*10^14 wavelengths

Isn't that simple enough??

Problem 3: Your understanding of interference is wrong.

Here you may have a point. But you have to admit, if I'm right and there was only a 4.8 degree phase shift between the two beams, Michelson and Morley wouldn't have noticed it.

Tom

Tom,

I don't have time to analyse all your mistakes right now. But let me say that even if the phase shift was 4.8 degrees, that is still easily big enough to be observed by M&M. These days we can observe wavelength shifts of tiny fractions of a degree.

Originally posted by Prosoothus
Thed,

I used the BCalc calculator to obtain my results. It is accurate to 5000 digits. You should try it. It's awsome. Here is where you can get it:
http://www.home52365.fsnet.co.uk/bcalc.htm

Yabbut, and if you plug in the wrong numbers the answer is wrong to an accuracy of 5000 digits. But it's still wrong.

The problem is that back in the late 1800's, the small phase variations you are talking about were unmeasurable. It was unwise for Michelson and Morley to conclude that there was no phase difference when, if there was a small phase difference, they wouldn't have noticed it.

Tom

Not at all. They first used a Fabry-Perot interferometer easily capable of measuring these small phase differences. If memory serves they repeated the experiment a few years later, , with a vastly more accurate Fabry-Perot interferometer. Fabry-Perot's are still used today for high accuracy work. Engineering skill was just as good in the 1800's as in the 1900's.

Tom, a point to ponder. The person who can find a fundamental flaw in Einsteins' work will be upheld as the next Einstein. Don't you think that every student of Physics has not spent considerable time trying to find a flaw themselves. After all, the fame it brings is considerable. I think you'll find we all came away happy that the reasoning and evidence is very much for Relativity. That's why we accept it.

Hmmm...

The M-M experiment and Einstein's SRT proved that the speed of light must be a function of the medium and does not depend on the speed of the observer or source.

To get around a preferred frame of reference, Einstein introduced the concept that you can define time in a reference frame by setting the round-trip from A-B-A and using te speed of light as constant in both directions with the additional provision that the distance from A to B is the same as the distance from B to A. You then get the famous SRT transformations between different reference frames.

This is a mathematical trick which is perfectably legal as you actually link the time to distance using a speed. This allows you to transform between a source and observer without knowledge of speed relative to the medium. Interesting is that you can do exactly the same with the speed of sound in air (or water, etc). If you define the time in terms of the speed of sound, you end up with the exact same transformations, except that c is now the speed of sound in the medium, and nothing can travel faster than the speed of sound. This would off course exactly describe the world to a blind person, as anything travelling at more than the speed of sound will invert the cause effect: a bullet would hit you before you would hear it's cause (the gun firing).

What I find interesting is this: if the speed of light is a function of the medium (as it is), does this not have to imply there has to be a medium (ie ether)? This conforms with the latest use of a preferred reference frame (the frame where black body radiation at 2.7K is isotropic).

eekf,

You're confusing math and physics. There are many examples in physics in which different systems end up being described by the same equation. Some differential equations arise practically everywhere. This doesn't mean the physical mechanisms are the same, or that all of the solutions to the equation are physical for all of the systems.

- Warren