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Relativity Refutation


Masanov
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I looked into Longman, yes it's old usage. In Hornby obsolescence is not mentioned. Thank you, and thank you, Mart, for support, too.

 

Misapplication, ... he used to evaluate everything without thorough analysis. And the reason for that is his mood, I think. We know relativity only when the light impulse is moving 90 degrees to the spaceship's movement, and to imagine three light pulses between Einstein mirrors and apply in all these 3 cases Einstein's postulates, for him is mundane. He would have received 3 different times (see above, I explained many times) in one spaceship, which is a real paradox, sensational, not that mundane, and promises great discussions. Maybe you suggest me how to convince these guys. To suggest EXCEL table, with the collumns to fill in the cells step by step as the theory of relativity is being explained. I am lost in doubts of what to do next.

 

When I reveal simple mistakes to the public, which is entertaining, I simplify lifes, try to gain contacts ..., as anybody would do in my place in this situation, they treat my discussions as avatarism, martyrism, dullsville, and the latest hit is pseudo science. They are so keen on relativity that would rather fly into outer space for good. If I was a flight organizer, I would be quite happy to give them life time supply of food. In the film RAT RACE there was a nice episode "buy the squirrel". I propose to apply postulates to three pulses of light moving along their relative distances to the spaceship and receive three different times, and understand how dangerous is to mistrust Newton.

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No, this is not true.

 

Between parallel mirrors the light pulse moved not :confused: at 90 degrees inside the spaceship to the speed of the spaceship? Open pls the textbook and try to

 

[study pls the picture (paramount picture)]

http:\\http://www.rainbow-calendar.hotmail.ru/7

The theory of relativity was explained only to the black observer' date=' because the light pulse which he sees moves perpendicular :mad: to the direction of the spaceship's movement. His time is T90= T[imath']

\sqrt{1-(\frac{v90^o}{c})^2}

[/imath], or T sin alpha.

Grey observers with times Tbeta2=T90(alpha2)/sin beta2, and Tbeta3= T90(alpha3)/sin beta3 were not discussed in the theory at all.

So, Einstein's relative time in the spaceship depends upon positions of the observers in this spaceship, and what is really stunning upon the turn of their head, or eyes! You [are an observer] turn your eyes to the right - you have time B, to the left - time A, turn backwards - time C. This is relativity of time.

.

The same, if you turn your eyes to the right - you have relative speed B, to the left - relative speed A, turn backwards - relative speed C. This is relativity of speed by Isaac Newton. The same as relativity of time but without haveing different times in one spaceship! ;)

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You may be correct. However, an explanation would make it seem less like an edict.

 

You are free to analyze the Michelson-Morley experiment at any angle you choose. Masanov needs to provide the refutation as it's his claim. Something rigorous, like math along with some well-defined terms, rather than the hand-waving he's provided thus far. Preferably without inventing new terminology.

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Between parallel mirrors the light pulse moved not :confused: at 90 degrees inside the spaceship to the speed of the spaceship? Open pls the textbook and try to

 

 

The same' date=' if you turn your eyes to the right - you have relative speed B, to the left - relative speed A, turn backwards - relative speed C. This is relativity of speed by Isaac Newton. The same as relativity of time but without haveing different times in one spaceship! ;)[/quote']

 

 

You show an observer outside of the spaceship. It appears that you are mixing two different reference frames. You have to analyze each reference frame by itself. In the spaceship frame, the pulse is emitted in the center of the circle, not over on the left.

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You are free to analyze the Michelson-Morley experiment at any angle you choose. Masanov needs to provide the refutation ... hand-waving ...
I see :eek: it's all vain to explain.

 

I told you many times and you once agreed that this was not Michelson and Morley experiment. This is the parallel mirrors experiment [the opposite mirror where pulses are going to be stricken back is not drawn]. All three onservers are in the spaceship looking through three viewports.

 

I see that you have quite an imagination. PLS don't mix in here MM experiment.

 

As you argued that there could not be relative speed of light I propose to the analysis OF EVERYBODY my experiment, that can be easily done.

http:\\http://www.rainbow-calendar.hotmail.ru\Relativity_Experiment.htm

I would like to WARN that this is not Michelson and Morley experiment!!!!!

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As you argued that there could not be relative speed of light I propose to the analysis OF EVERYBODY my experiment, that can be easily done.

 

So do some analysis already. Explain a measurement that has been made or could be made, and what you predict should be the result.

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In order to examine this claim that light emmited at different angles somehow changes the results let's try the the folowing exercise.

 

Let's call the frame of the sphere frame S' and the frame in which the sphere moves at v, S

In S the light travels a distance of ct1 while our sphere travels a distance of vt1.

Now considering image 1:

image1.jpg

 

Assuming the light is emitted at at an angle of alpha from the center of the sphere as measured in S' , then the point at which the light hits the sphere is as measured in S is:

[math] ( Y \sin \alpha , vt_1+ \gamma Y \cos \alpha ) [/math]

where Y is the radius of the sphere as measured in S'.

and that the center of the sphere at the instant of impact is at (0,0)

[math]\gamma = \sqrt{1- \frac{v^2}{c^2}}[/math]

 

The gamma factor is needed because the sphere is length contracted in S

 

Thus:

 

[math]c^2t_1^2 = Y^2 \sin^2 \alpha + ( vt_1 + \gamma Y \cos \alpha)^2 [/math]

 

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \gamma^2 Y^2 \cos^2 \alpha

[/math]

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( 1- \frac{v^2}{c^2} \right)Y^2 \cos^2 \alpha

[/math]

 

[math]c^2t_1^2 = Y^2 \sin^2 \alpha +v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha + \left( \cos^2 - \frac{ \cos^2 v^2}{c^2} \right)Y^2\alpha

[/math]

 

re-arranging:

 

[math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( \sin^2 \alpha + \cos^2 - \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

 

[math]c^2t_1^2 = v^2t_1^2 + 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

Gathering the t1^2 terms together:

[math]c^2t_1^2-v^2t_1^2 = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

 

[math]t_1^2 (c^2-v^2) = 2 \gamma Y v t_1 \cos \alpha +Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right)

[/math]

 

Moving all the terms to the left side:

 

[math]t_1^2 (c^2-v^2) - 2 \gamma Y v t_1 \cos \alpha -Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) =0

[/math]

 

We get an equation that can be solved for t1 using the quadriatic equation thusly:

 

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4 \gamma^2 Y^2 v^2 \cos^2 \alpha + 4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm \sqrt{ 4\left( 1- \frac{ v^2}{c^2} \right)Y^2 v^2 \cos^2 \alpha -+4 (c^2-v^2) Y^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + (c^2-v^2) \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \frac{(c^2-v^2)}{c^2} c^2 \left( 1- \frac{ \cos^2\alpha v^2}{c^2} \right) } }{2(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ 2 \gamma Y v \cos \alpha \pm 2 Y \sqrt{ \left( 1- \frac{ v^2}{c^2} \right) v^2 \cos^2 \alpha + \left( 1- \frac{ v^2}{c^2} \right) \left( c^2- \cos^2\alpha v^2 \right) } }{2(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + \left( c^2- v^2 \cos^2\alpha \right) } }{(c^2-v^2) }[/math]

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ v^2 \cos^2 \alpha + c^2- v^2 \cos^2\alpha } }{(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm Y \sqrt {1- \frac{ v^2}{c^2}} \sqrt{ c^2 }} {(c^2-v^2) }[/math]

 

[math]t_1 = \frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } [/math]

 

Now consider image 2 which shows the path the light takes on its return to the center of the sphere:

 

image2.jpg

 

 

Here we can see the light travels a distance of ct2 in frame S, while the sphere moves a distance of vt2.

 

As a result we can see that:

 

[math]c^2 t_2^2 = Y^2 \sin^2 \alpha + ( \gamma Y \cos \alpha - vt_2 )^2 [/math]

 

and following the same steps as above we find that solving for t2 we get:

 

[math]t_2 = \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 }[/math]

 

The total time T it takes for the light to leave the emitter and return is

 

[math]T = t_1 + t_2[/math]

 

Thus

[math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {(c^2-v^2) } + \frac{- \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

 

[math]T=\frac{ \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}} - \gamma v Y \cos \alpha \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

 

[math]T=\frac{ \pm cY \sqrt {1- \frac{ v^2}{c^2}} \pm cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

 

atr this point we have a choice to make. There is only one possible answer for T, but there are four ways to solve this equation due to the "plus or minus" signs. Luckily we can automatically elliminate 3.

If both are minus we end up with a negative time for an answer, so that doesn't work. And if one is minus and the other plus we end up with a answer of 0 for T and that makes no sense. this leaves us with both being plus giving us:

 

[math]T=\frac{ cY \sqrt {1- \frac{ v^2}{c^2}} + cY \sqrt {1- \frac{ v^2}{c^2}}} {c^2-v^2 } [/math]

 

[math]T=\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } [/math]

 

Now let's relate T to T' the total round trip time in S'

 

In S' the light travels a distance of Y to the sphere and then travels a distance of Y back again.

 

this gives us:

 

[math]T' =\frac{ 2Y}{c}[/math]

and

 

[math]\frac{T'}{T}= \frac{\frac{ 2Y}{c}}{\frac{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } {c^2-v^2 } }[/math]

 

[math]\frac{T'}{T}= \frac{ 2Y}{c}} \frac {c^2-v^2 }{ 2cY \sqrt {1- \frac{ v^2}{c^2}} } [/math]

 

[math]\frac{T'}{T}= \frac {c^2-v^2 }{ c^2 \sqrt {1- \frac{ v^2}{c^2}} } [/math]

 

divide top and bottom by c²:

 

[math]\frac{T'}{T}= \frac {1- \frac{v^2}{c^2} }{ \sqrt {1- \frac{ v^2}{c^2}} } [/math]

 

simplify and re-arrange:

 

[math]T = \frac{T'}{\sqrt{1- \frac{v^2}{c^2}}}[/math]

 

And we see that no matter what angle alpha the light is aimed, the stardard time dilation holds, thus falsifying the claim.

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And we see that no matter what angle alpha the light is aimed, the stardard time dilation holds, thus falsifying the claim.

 

And imagine everybody's shock...

 

 

Thanks, Janus, for slogging through the math that Masanov declined to do.

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I will think over during a week, but this logic reminds me relative horizontal distance proofs, when after a lot of eqivalent formulas transformations, gamma factor was inserted without proofs.

I promise to reread carefully this and say my word about it, though I ask you to be specific and reconsider the same but to all three pulses at once, because my logical experiment with three pulses between parallel mirrors is at stake here. And we should consider absolute distances also. The scheme of my experiment was this scheme:

---------------------------------------------------------------------------

Pulses shot from absolute points____Pulse1_______Pulse2_________Pulse3

____________________________(considered in

_____________________________the theory)_ (not considered)(not considered)

---------------------------------------------------------------------------

Absolute Distances Covered

for the time the spaceship

have gone the distance

C * cos ALPHA * T________________CT___________CT____________CT

---------------------------------------------------------------------------

Relative distances to one

observer moving in a spaceship_CT*sinALPHA__CT*sinALPHA1__CTsinALPHA2

___________________________________________sin BETA1_____sin BETA2

--------------------------------------------------------------------------

!!!!!!! According to Einstein we should apply to absolute and relative distances mentioned postulates, that is we should divide the distances by C. As a result we receive these time

equations:___________________T*sinALPHA___T*sinALPHA1___T*sinALPHA2

________________________________________ ------------___------------

__________________________________________sin BETA1______sin BETA2

---------------------------------------------------------------------------

 

Your explanation does not differ much from experiment with horizontal parallel mirrors, where relative distances were proven. Gamma equation was inserted there like a bolt from the blue. I cannot agree with you here because the main point in the theory of relativity is relative distances passed by light with the same speed. Distance cT2 in your explainations is not relative, it is not CTsinALPHA/sinBETA. It's the same as Tom said. I do not fight with it. In schools and other books explanation goes at the beginning by presenting gamma factor as sin ALPHA=V 1-v2/c2 in famous logical experiment with parallel mirrors where absolute distance is presented as L and relative distance between parallel mirrors as Lo or LsinALPHA, or Lgamma-factor. And this gamma-factor was derived by dividing relative distance Lo and L by C, receiving time equation. In your explanation, there's no need to apply quite a number of equal formulas for S and then for S', because you have already insterted gamma-factor at the beginning. Your Ct1 and Ct2 are the same... and you return to the equation you insterted from the very beginning. Distance Ct2 is not relative, it's the same Ct1, it's not Ct1*sin AlPHA.

(In a week I will tell for sure..)

As the result, according to your thinking :P your do not deny relative distances CTsinALPHA1/sinBETA1 and CTsinALPHA2/sinBETA2. You never knew of them. To see these relative distances, pls visit again:

http:\\http://www.rainbow-calendar.hotmail.ru/7

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IMAGE3.PNG

In your image you did not mention the reflector and misapplied direction of vector in vt2: the arrow should point to the left, not to the right. Yellow distance is misleading distance, nobody sees along this path anything, and this distance shown at angle BETA, not ALPHA: or else in your calculations BETA angle should be mentioned. Anyway when the light pulse is at the end of the yellow distance, reflector moves together with the sphere S to a position of the sphere S'. So spheres S and S' is the same spaceship!

 

Distances Ct1 and Ct2 and Y are seen together at one common angle ALPHA. To say, that they are different, means fun or mischief. So, if gamma-contraction is assigned to distances ct1 and ct2, then assign it to the distance Y: contraction is a squeeze, and a squeeze is a squeeze from the point of view of anybody... or God.

 

Ha-ha: the same spaceship! Einstein was not good at school, he could allow himself any misapplication of vectors and mundane|vain calculations. [Your picture showed "a spider got entangled in his own web". It is not a shock, Swansont, it's this spider dizziness of exhaustion].

 

 

 

Repetition of the said long ago at the beginning of my discussions:

image4.PNG

 

 

If the light pulse is seen between parallel mirrors on the relative distance at some angle BETA we see its different side, as if we see different side of the moon. According to M.Gardner in "The Theory Of Relativity For The Millions" we catch up with the light pulse and see its another side at angle BETA equal 90 degrees. So if somebody quotes Gardner here in short, we can step by step prove additional BETA times, presented in the table. In the picture is shown that light pulses move along BETA distances by their different sides askewed.

IMAGE5.PNG

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IMAGE3.PNG

In your image you did not mention the reflector and misapplied direction of vector in vt2: the arrow should point to the left' date=' not to the right. Yellow distance is misleading distance, nobody sees along this path anything, and this distance shown at angle BETA, not ALPHA: or else in your calculations BETA angle should be mentioned. Anyway when the light pulse is at the end of the yellow distance, reflector moves together with the sphere S to a position of the sphere S'. So spheres S and S' is the same spaceship!

 

Distances Ct1 and Ct2 and Y are seen together at one common angle ALPHA. To say, that they are different, means fun or mischief. So, if gamma-contraction is assigned to distances ct1 and ct2, then assign it to the distance Y: contraction is a squeeze, and a squeeze is a squeeze from the point of view of anybody... or God.

 

Ha-ha: the same spaceship! Einstein was not good at school, he could allow himself any misapplication of vectors and mundane|vain calculations. [Your picture showed "a spider got entangled in his own web". It is not a shock, Swansont, it's this spider dizziness of exhaustion'].

 

 

Of course it is the same space ship! The oval to the left is the spaceship at the instant when the light hits the mirror. The oval to the right is the Same spaceship at the instant when the light returns to the center. This is as seen by an observer outside of the ship, watching it go by at v. (to keep the image uncluttered, I omitted the spaceship at the instant the light is emitted from the center from the image)

 

To someone in the ship, the light simply travels out at angle alpha (wrt the relative motion) for a distance Y, hits the mirror and returns along the same path to the center. Since the light returns to the center of the ship in S' (the frame of the ship), it must also return to the center in frame S (the frame in which the ship is moving at v) Thus in S' the light follows the path shown in my images.

 

Alpha is the angle the light is launched as measured by someone traveing with the ship. Y is the radius as measured by someone traveling with the ship The length of the ship is only contracted along the direction of motion when measured by someone to which the ship has a relative motion. The ship has zero relative motion wrt to someone traveling with the ship, thus it has no length contraction for that person.

 

To say otherwise shows a complete and utter lack of understanding of what Special Relativity is about. To refute Relativity you must first understand it.

 

The simple fact is that Relativity is a completely self-consistant theory. It contains no mathematical or logical errors. It makes numerous predictions which have held up under every real experimental test put to it.

 

No thought experiment you can ever dream up will ever refute it. If you think you've come up with one, all it means is that you have made an assumption that is contradictory to one of the postulates of Relativity. All you've done in this case is show that Relativity is not compatible with your assumption. This in of itself provides nothing in the way of proving which is correct, your assumption or Relatiivity.

 

The only way to refute Relativity is to produce results from a Actual Physical Experiment that gives results that differ from that which Relativity predicts.

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Contraction is a squeeze to both observers.

Show me, how the outside observer is going to measure contraction, by looking to reference books? If contraction occurs it should be real, not surreal. How can a squeeze be not a squeeze? Actual experiment is when we have actual squeeze, real to all observers inside and outside the spaceship, or else no experiments can be made.

Onserver outside the spaceship cannot see ALPHA angle. Distance Y is seen by inside observer at ALPHA angle, ct1 and ct2 are also seen at ALPHA by inside observer. As these distances the same, C the same, and time should be the same. If you introduce another observer, first show relative angle.

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Contraction is a squeeze to both observers.

Show me' date=' how the outside observer is going to measure contraction, by looking to reference books? If contraction occurs it should be real, not surreal. How can a squeeze be not a squeeze? Actual experiment is when we have actual squeeze, real to all observers inside and outside the spaceship.[/quote']

 

Time and space are relative and not absolute and thus depend on who is making the measurement.

 

Here we have an example of what I was talking about. You have made the assumption that length is absolute (that observers in and out of the space ship will measure the same length.) Relativity says this is not the case. Thus your assumption and Relativity contradict each other. But you have not provided any actual evidence that your assumption is correct (no matter how much you believe it to be so) and Relativity wrong.

 

On the other hand, Relativity has made numerous predictions that have been confirmed by actual physical experiments.

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