## MM experiment- swimmers’ model

MM experiment actually disproves the absurd notion of constant SOL if interpreted correctly.

If the photons (swimmers/ bullets) motion is unaffected by the mirrors/ interferometer, the two photons will be out of phase when they meet finally and hence should result in interference pattern.

Conversely, lack of interference can be easily explained if we abandon the idiotic notion of constant SOL and by assuming the photons as classical projectiles whose velocity gets affected by that of the source/ mirrors. This holds true whether one believes in ether or not. One doesn’t need ether drag to explain the lack of interference. One just has to remove the ‘relativity spectacles’ and use the classical physics- any projectile’s motion does get affected by that of the source.

Let’s ‘replay’ the experiment using any of the two models used by the relativists

Swimmers (photons) in still water (ether)

or

Bullets (photons) in vacuum (no ether)

If swimmer’s (photon’s) velocity is not influenced by the west-ward motion of the launch pad (mirror),

To summerise the above 3 illustrations in the reference frame of the platforms

Swimmer ‘x’ / Outward leg: As his velocity is 10m/sec and the distance between the moving platforms is 10 meters, swimmer ‘x’ reaches the opposite platform in 1sec. But by that time, the platform would have moved 1meter towards west, so he goes to point ‘D’ instead of going ‘straight’ to point B.

Swimmer ‘x’ / Return leg: Similarly, he swims back the same 10 meters and returns to the first platform in 1sec but again instead of returning ‘straight’, he reaches point ‘E’.

Swimmer ‘x’/ Total journey: He travels 20 meters in 2sec but reaches point ‘E’ instead of the original starting point ‘A’ because of the westward motion of the platform.

Swimmer ‘y’/ Outward leg: As his velocity is 10meters/sec, he reaches point ‘C’ in just 0.9sec. That’s because point ‘C’ on the platform is moving at 1m/sec, he needs to swim only 9meters to reach this point.

Swimmer ‘y’ / Return leg: To meet swimmer ‘x’, he needs to come to point ‘E’. He has to swim back 9meters to come to this point. So he takes again 0.9sec.

Swimmer ‘y’/ Total journey: He reaches point ‘E’ in just 1.8sec by travelling a distance of 18meters. He has to ‘wait’ here to join swimmer ‘x’ who will reach the same point in 2sec by travelling 20meters. Hence this must result in phase difference between the photons and hence must result in loss of coherence between the final beams.

This would not happen if we imagine that the motion of swimmers/ photons get affected by that of the platforms/ mirrors. This straight away debunks the obviously silly theory of relativity.

So the problem is not lack of maths here. One needs a human mind ‘uninfected by the relativity virus’ to interpret the data. That is what is lacking with the relativists whose minds are occupied by the relativity demon and ‘programmed’ by its weird maths. The demon having captured the scientific minds doesn’t allow them to think straight, because that will be suicidal for the demon itself.

• Bart Klein, Netherlands  On August 3, 2013 at 12:59 am

When swimmer y swims to the right (what you call “outward leg”), he observes point C approaching him with a velocity of 11 m/s. So one could say y’s velocity with respect to point C is 11 m/s. Therefore it will take y 10/11=0.90909 (repetitive) seconds to reach point C (not 0.9 seconds as you say). This follows from simple vector mathematics. It can equally be deduced that on the return trip y’s relative velocity with respect to A is 9 m/s. What I want to make clear with this is that one does not simply “debunk” relativity when your grasp of Galilean relativity is non existing. I leave it to you to find the other flaws in your interpretation of your model.

Furthermore, you cannot prove relativistic mechanics wrong by using classical (Galilean /Newtonian) mechanics, in the very same way that you can not “debunk” the existence of steam ships by just looking at wind propelled ships, ignoring any evidence of the contrary in the process. To disprove relativity, you need to at least incorporate relativity in your model. Only then can you demonstrate where it fails.

• drgsrinivas  On August 8, 2013 at 1:27 pm

Platforms A, B, and C represent the mirrors in the MM experiment – let me remind you that they are fixed relative to each other. Hence if the velocity of ‘swimmer Y’ is 10m/sec with respect to platform A, that remains the same even with respect to platform C.
One more thing to note is that swimmers X and Y here represent light photons. Because your witch theory of relativity proposes that speed of light is constant to all observers, the speed of swimmer Y should remain 10m/sec with respect to platform C whether it moves or not. After all this, don’t argue that swimmers can’t represent light photons – your own masters of the witch theory have done that already.

“you can not “debunk” the existence of steam ships by just looking at wind propelled ships, ignoring any evidence of the contrary in the process”
I am not ignoring the evidence of the contrary; I am actually examining the evidence. That is the reason why I am talking about Michelson’s and other experiments. I am not arguing that relativity is false because Archimedes is true. Your ignorance is again obvious here.

“To disprove relativity, you need to at least incorporate relativity in your model”
Though it is a common practice amongst relativists; that actually amounts to stupidity. If the weird assumptions of a theory were to be believed and taken into account while interpreting experimental data, then every stupid theory will be proven as right.

When everything can be explained by classical reasoning, why is there a need for a weird theory like relativity? We can explain the result of MM experiment by just imagining light photons as any other classical particles (whether or not you believe in the existence of ether).

• robinpike  On August 6, 2013 at 8:39 pm

True, relativistic mechanics cannot be proved wrong by using Newtonian mechanics. But the converse is also true – that simply because relativistic mechanics works, does not prove that Newtonian mechanics itself is wrong.

• pyramidal18  On January 3, 2014 at 8:37 am

In the MMX, the transversal time is TT=2D/SQUARE ROOT(C*C-V*V) then the transversal speed of light is CT=SQUARE ROOT(C*C-V*V).
In special relativity, nothing happens transversally so CT will remain equal to SQUARE ROOT(C*C-V*V).
CT is different from C so the speed of light is not constant.

• pyramidal18  On January 3, 2014 at 8:54 am

In the MMX, the transversal time is TT=2D/SQUARE ROOT(C*C-V*V) then the transversal speed of light is CT=SQUARE ROOT(C*C-V*V).
In Special Relativity, nothing happens transversally so TT will remain equal to SQUARE ROOT(C*C-V*V) wich is different from C.
Thus, the speed of light is not constant.

• drgsrinivas  On January 7, 2014 at 5:53 pm

If we bring in the concept of vectors, the stupidity of relativity becomes immediately obvious.
If a light beam travels at speed ‘c’ in one direction, it can’t obviously travel with the same speed in a perpendicular direction. Though relativists are obsessed with vectors when they talk about momentum etc in their delusional 4D space-time, they forget that velocity is a vector in our 3D universe.
Is it not ridiculous that a theory which talks so much about space and time doesn’t say what happens in transverse direction/ plane. (As if ours is only a 1-dimentional universe)

• Jerry  On January 8, 2014 at 3:41 am

You seem to be confused. To say “speed” is constant is to say the magnitude of the velocity is constant. Obviously, light can be reflected (by a mirror, etc.), so it’s direction and hence velocity changes, but not its speed. I’m beginning to think this site is an exercise in satire.

• drgsrinivas  On January 8, 2014 at 4:40 pm

That’s exactly how the religious followers of the nude Emperor would describe the His majesty’s marvellous costume. Each one would give a different description of the costume because there was no costume at all. And each one will change their description of the costume as we start exposing their hallucination.
http://debunkingrelativity.com/2013/05/09/relativity-and-the-nude-emperor/

• pyramidal18p  On January 9, 2014 at 10:37 pm

In the twins paradox, for the twin in the rocket, the longitudinal part of his body will be aging less than the transversal part of his body. The transversal part of his body will be aging the same way as the twin on the earth.
For example: the longitudinal part of the twin in the rocket will have 20 years old while his transversal part will have 100 years old when the factor gamma=5.

• pyramidal18  On January 9, 2014 at 10:45 pm

In the twins paradox, the longitudinal part of the twin in the rocket will have for example 20 years old while his transversal part will have 100 years old when gamma factor is 5.
The twin in the rocket will be at the same time young (longitudinally) and old (transversally).