Master Theory

[uncool]

???Yes. But reality has it

Where?

and reality has not double time dilation.

Please demonstrate this.

Similar ambiguity is impossible.

This is gibberish.

#140

Please quote the precise part of your post where you show that the equation is incorrect. I don't want an assertion; I want a specific proof.

=Uncool-

 

Again, Alex, your entire argument comes down to the assertion that changes in time-ordering are impossible. You have not given any reason to believe this. You have not shown that relativity violates causality in any way.

 

As an analogy to Galilean mechanics that you might understand:

 

Do you agree that you can order events - that is, points in space-time - according to x-position, given a choice of frame?

 

Further, do you agree that this ordering changes depending on your frame?

=Uncool-

[Alexander Masterov]

This is gibberish.

To unambiguously only.

Please quote the precise part of your post where you show that the equation is incorrect. I don't want an assertion; I want a specific proof.

Look:

Time must be diferent.

 

[math] x^2-(ct)^2=(x')^2-(ct')^2=0 [/math] - Time is not diferent. It's error.

 

 

[math] x^2-(ct)^2=(x'-vt')^2-(ct')^2=0 [/math] - Time is diferent:

 

[math]t_1'=x'/(c-v)[/math]

[math]t_2'=x'/(c+v)[/math]

 

It's correct.

Edited September 30, 2011 by Alexander Masterov

[uncool]

To unambiguously only.

This is still gibberish.

Look:

Time must be diferent.

 

[math] x^2-(ct)^2=(x')^2-(ct')^2=0 [/math] - Time is not diferent. It's error.

What do you mean by "Time is not different"?

=Uncool-

Edited September 30, 2011 by uncool

[Alexander Masterov]

This is still gibberish.

Time dilation can not be responsible for (to cause) two the physical phenomena simultaneously: acceleration time and deceleration time. Must be one thing.

What do you mean by "Time is not different"?

 

 

[math] x^2-(ct)^2=(x')^2-(ct')^2=0 [/math]

 

[math] x^2-(ct)^2=0 [/math]

[math]t_1=x/c[/math]

[math]t_2=-x/(-c)=x/c[/math]

Time is not different

 

[math] x'^2-(ct')^2=0 [/math]

[math]t_1'=x'/c[/math]

[math]t_2'=-x'/(-c)=x'/c[/math]

Time is not different

 

[math] (x'-vt')^2-(ct')^2=0 [/math]

[math]t_1'=x'/(c-v)[/math]

[math]t_2'=-x'/(-c-v)=x'/(c+v)[/math]

Time is different

Edited October 1, 2011 by Alexander Masterov

[uncool]

Time dilation can not be responsible for (to cause) two the physical phenomena simultaneously: acceleration time and deceleration time. Must be one thing.

Why not?

 

[math] x^2-(ct)^2=(x')^2-(ct')^2=0 [/math]

 

[math] x^2-(ct)^2=0 [/math]

[math]t_1=x/c[/math]

[math]t_2=x/c[/math]

Time is not different

Which it shouldn't be, because light is traveling at the same speed in both directions, and is covering the same distance.

[math] x'^2-(ct')^2=0 [/math]

[math]t_1'=x'/c[/math]

[math]t_2'=x'/c[/math]

Time is not different

Because you specified the same distance to be traveled. The time should not be different - because the speed of light should be the same in all directions in all frames. Remember, time and space mix - space becomes time and time becomes space.

=Uncool-

Edited September 30, 2011 by uncool

[Alexander Masterov]

Why not?

Ambiguity requires justification.

Ambiguity requires additional physical law, which will determine the choice of ambiguity. No such law.

Which it shouldn't be, because light is traveling at the same speed in both directions, and is covering the same distance.

Dog's rate is constant always also.

Because you specified the same distance to be traveled. The time should not be different - because the speed of light should be the same in all directions in all frames. Remember, time and space mix - space becomes time and time becomes space.

Dog's rate is constant always also.

===================================

 

Look at the following animation to see this:

 

 

You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. The first expression allege that these times are identical.

 

Let solve one a school's puzzle:

 

1. Two travelers do walking on the road with equal speed () in one direction at a distance () from each other.

 

2. Between them runs a dog (speed of it ).

 

QUESTION: How much time a dog runs forward, and how many - back?

 

ANSWER: and

 

But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression.

Edited October 1, 2011 by Alexander Masterov

[uncool]

Ambiguity requires justification.

Ambiguity requires additional physical law, which will determine the choice of ambiguity. No such law.Dog's rate is constant always also.Dog's rate is constant always also.

===================================

 

Look at the following animation to see this:

 

 

You should see that the transit time of red spots are different in different directions, despite the fact that the rate of red spots is equal(same) in both directions. The first expression allege that these times are identical.

No, it doesn't. It alleges that the times are identical in another frame. You forget that time is frame-dependent in relativity.

Let solve one a school's puzzle:

 

1. Two travelers do walking on the road with equal speed () in one direction at a distance () from each other.

 

2. Between them runs a dog (speed of it ).

 

QUESTION: How much time a dog runs forward, and how many - back?

 

ANSWER: and

 

But Einstein, Lorentz, Minkowski and others (who entrust to them) say that time will be the same. So is was alleged by the first expression.

You are mixing measurements from different frames. Relativity says that whether the transit times are the same is frame-dependent - in this frame, where the objects are moving, the transit time will be different. However, in the frame where neither are moving, the transit time will be the same.

 

Relativity also gets that T1 = L/(c + v) and T2 = L/(c - v); but those are only applicable in this frame. In other frames, the time could be different.

=Uncool-

Edited October 1, 2011 by uncool

[Alexander Masterov]

No, it doesn't. It alleges that the times are identical in another frame. You forget that time is frame-dependent in relativity.

Time dilation can not align intervals of time.

[uncool]

Time dilation can not align intervals of time.

This is gibberish. I think I can guess what you're saying, but I would like to make sure, so please try to reexplain.

=Uncool-

[Alexander Masterov]

This is gibberish. I think I can guess what you're saying, but I would like to make sure, so please try to reexplain.

Slowing down time can not change the fact: [math]T_1\neq T_2[/math] Edited October 1, 2011 by Alexander Masterov

[uncool]

Slowing down time can not change the fact: [math]T_1\neq T_2[/math]

But relativity is more than just slowing time - again, space in one frame becomes time in another. Therefore, the fact that the spatial coordinates are different can change that.

 

In this example:

 

From the frame where the walls are standing still, the events we see are:

 

Light rays emitted: [math](0,0,0,0)[/math]

Light ray absorbed on the left: [math](\frac{L}{c}, -L, 0, 0)[/math]

Light ray absorbed on the right: [math](\frac{L}{c}, L, 0, 0)[/math]

 

From the frame where the walls are moving, according to the standard transformations:

 

Light rays emitted: [math](0,0,0,0)[/math]

Light ray absorbed on the left: [math](\frac{\gamma L (c - v)}{c^2}, -\frac{\gamma L (c - v)}{c}, 0, 0)[/math]

Light ray absorbed on the right: [math](\frac{\gamma L (c + v)}{c^2}, \frac{\gamma L (c + v)}{c}, 0, 0)[/math]

 

which can alternatively be written as:

 

Light rays emitted: [math](0,0,0,0)[/math]

Light ray absorbed on the left: [math](\frac{L}{(v + c)\gamma}, -\frac{cL}{(v + c)\gamma}, 0, 0)[/math]

Light ray absorbed on the right: [math](\frac{L}{(v - c)\gamma}, \frac{cL}{(v - c)\gamma}, 0, 0)[/math]

 

Which is exactly what you got, up to the factor of [math]\gamma[/math], which is there due to the spatial dilation (the barriers are closer together than in the non-moving frame).

 

You have to include the effects of spatial displacements.

=Uncool-

[Alexander Masterov]

But relativity is more than just slowing time - again, space in one frame becomes time in another. Therefore, the fact that the spatial coordinates are different can change that.

 

In this example:

 

From the frame where the walls are standing still, the events we see are:

 

Light rays emitted: [math](0,0,0,0)[/math]

Light ray absorbed on the left: [math](\frac{L}{c}, -L, 0, 0)[/math]

Light ray absorbed on the right: [math](\frac{L}{c}, L, 0, 0)[/math]

...

Master Theoey:

 

[math](\frac{L(1-v^2/c^2)}{c+v}, -L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

 

and

 

[math](\frac{L(1-v^2/c^2)}{c-v}, L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

 

OR:

 

[math](\frac{L(c+v)}{c^2}, -L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

[math](\frac{L(c-v)}{c^2}, L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

===================================================

 

Which is exactly what you got, up to the factor of , which is there due to the spatial dilation (the barriers are closer together than in the non-moving frame).

 

You have to include the effects of spatial displacements.

Lorentz transformations are the result of the expression:

 

[math]x^2-(ct)^2=(x')^2-(ct')^2[/math]

 

Coordinate transformations must to derive from this expression:

 

[math]x^2-(ct)^2=(x'-vt')^2-(ct')^2[/math]

Edited October 2, 2011 by Alexander Masterov

[uncool]

Master Theoey:

 

[math](\frac{L(1-v^2/c^2)}{c+v}, -L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

 

and

 

[math](\frac{L(1-v^2/c^2)}{c-v}, L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

 

OR:

 

[math](\frac{L(c+v)}{c^2}, -L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

[math](\frac{L(c-v)}{c^2}, L(1-v^2/c^2), H\sqrt{1-v^2/c^2}, W\sqrt{1-v^2/c^2})[/math]

At the moment, we're not talking about your idea. We're talking about relativity. You still have yet to show that there is a problem with it; I have shown you why your objection to it is wrong.

 

Additionally, in context, your post makes no sense. There is no H; there is no W.

=Uncool-

[Alexander Masterov]

At the moment, we're not talking about your idea. We're talking about relativity. You still have yet to show that there is a problem with it; I have shown you why your objection to it is wrong.

 

Additionally, in context, your post makes no sense. There is no H; there is no W.

=Uncool-

 

 

How do you get [math]t_1'\neq t_2'[/math] from [math] x^2-(ct)^2=(x')^2-(ct')^2 [/math]?

Edited October 2, 2011 by Alexander Masterov

[uncool]

How do you get [math]t_1'\neq t_2'[/math] from [math] x^2-(ct)^2=(x')^2-(ct')^2 [/math]?

Because we start with x_1 = x_2. We do not get that x_1' = x_2'.

=Uncool-

[Alexander Masterov]

Because we start with [math]x_1 = x_2[/math]. We do not get that [math]x_1' = x_2'[/math].

[math]x_1 = -x_2[/math] and [math]x_1' = -x_2'[/math] always.

 

Do you intend to contest this?

[uncool]

[math]x_1 = -x_2[/math] and [math]x_1' = -x_2'[/math] always.

 

Do you intend to contest this?

EDITED:

 

If you define x_1' as being the distance between source and wall, then that is true. However, in this frame, light does not have to cover that entire distance - the wall is moving towards the light already.

 

If you're defining x_1' as the distance between the origin and the wall, then yes, I am contesting that.

=Uncool-

Edited October 3, 2011 by uncool

[Alexander Masterov]

The observer at rest:

[math] x_1 [/math] - it's path of light from right to left = distance between the mirrors.

[math] x_2 [/math] - it's path of return.

[math] -x_2 [/math] - the distance between the mirrors. ([math] x_2=-x_1 [/math])

[math] path_1 = -path_2 = x_1 [/math]

[math] t_1=path_1/c = x_1/c [/math]

[math] t_2=-path_2/(-c) = x_1/c [/math]

[math] t_1=t_2 [/math]

 

 

 

The observer move ([math] v>0 [/math]) for Master Theory:

[math] x_1' [/math] - distance between the mirrors for moving observer. ([math] x_1'=x_1(1-v^2/c^2) [/math])

[math] path_1' = x_1'/(1-v/c) [/math] - it's path of light from right to left for moving observer.

[math] path_2' = -x_1'/(1+v/c) [/math] - it's path of return.

[math] t_1'=path_1'/c = x_1'/(c-v) [/math]

[math] t_2'=-path_1'/(-c) = x_1'/(c+v) [/math]

[math] t_1\neq t_2 [/math]

 

Tese pathes is roots of this equation: [math](path'-vt)^2-(ct)^2=0[/math] and [math] path_1' \neq -path_2' [/math]

 

SRT pathes is roots of this equation: [math](path')^2-(ct')^2=0[/math] and [math] path_1' = -path_2' [/math]. (It's not correct.)

Edited October 3, 2011 by Alexander Masterov

[uncool]

Alex, I'd like you to define exactly what you think x_1', x_2', t_1', and t_2' mean.

 

So far, I've been defining the x_i' s as the distance between the origin (as seen in the second frame) and the walls, and the t_i' s as the times to get to those walls in the second frame. As far as I can tell, you are defining them differently.

=Uncool-

Edited October 3, 2011 by uncool

[Alexander Masterov]

Alex, I'd like you to define exactly what you think x_1', x_2', t_1', and t_2' mean.

I added to up.

[Alexander Masterov]

I think that a neutrino detector can be constructed from a balloon with compressed hydrogen. (The mass of the hydrogen atom is 28 times lighter than the nitrogen molecule and 32 times lighter than oxygen, which means that hydrogen is more than five faster than any of them.)

 

This balloon should be lowered to great depths in the ocean. (Near Mariana Trench.) At 11 km depth, under pressure of 1,100 atmospheres, the density of hydrogen would be 1,000 times more. (100kg/m3)

 

In this case, hydrogen can be obtained (by electrolysis of distillate) directly into the cylinder, while immersing the container into the depths. (This - as an option.)

 

Can be in a different way: using a cascade of pumps at different depths.

 

And it is possible: instead of pumps can be used with a balloon+plunger. Hanging on the ropes these balloons (in the form of two garlands), dropping and picking up by turns - you compel them inject hydrogen into the depths.

 

And you can still way: Balon Dewar (with liquid hydrogen) set cylinder. The hydrogen to do evaporation - balon go down. (This option seems most appropriate.)

==========================================

 

Let me explain:

 

Neutrinos with matter should not have to interact with, and (if their rate significantly greater than the speed of light) - so be it.

 

Methods which try to detect neutrinos today, you can catch those neutrinos, whose rate is only slightly greater than the speed of light (more on the rate of Brownian motion).

 

Of the neutrino flux filtered out by those whose speed exceeds the speed of light is greater than the rate of Brownian motion of granite. Therefore, the number of neutrinos passing through the Earth increases sharply if their speed exceeds the speed of light is greater than the rate of Brownian motion in the interior of the earth.

 

 

Water molecules are lighter molecules granite. Therefore, the velocity of the Brownian motion of water molecules in two - two and a half times (the square root of mass ratio) of greater. Therefore, water molecules move twice (at least) faster than the molecules of granite. This means that there is a chance to catch the water in those neutrinos which are not extinguished in the interior of the earth, but the speed is (still) not much greater than the speed of light.

 

If used as a medium for hydrogen, the sensitivity will increase many times, since the rate of Brownian motion of the atoms of hydrogen is ten times higher than that of granite. Hence: the hydrogen is able to catch those neutrinos, which are almost not able to catch the granite (and water). Should be a lot of neutrinos (for which the thickness of the Earth is transparent).

 

And in Baykal lake to catch a cosmic neutrinos is even more difficult, because Brownian speed of motion of atmosphere much more than in water. Atmosphere do swallow up those neutrinos which could be fixed in the lake.

 

 

 

Edited October 3, 2011 by Alexander Masterov

[Alexander Masterov]

Electron neutrino is a positron, whose rate is higher than the speed of light. And an electron antineutrino is an electron whose velocity is greater than the speed of light.

We can calculate the velocity of neutrinos, since the mass of the electron and positron are known.

 

My arguments:

 

1. Electron is born in a pair with a positron.

2. Mass on the velocity-independent.

3. Neutrinos move faster than the speed of light (otherwise it would not neutrinos).

 

Matter, whose speed is greater than the speed of light does not interact with EMF, if the source field is static. More than that, the matter (whose speed is greater than the speed of light) does not interact with a static matter. Positron seems neutral, since his speed than the speed of light. Positron does not interact with static EMF and a static matter, since has speed greater than the speed of light.

Edited October 3, 2011 by Alexander Masterov

[Alexander Masterov]

Neutrino has an unlimited ability to penetrate into matter.

 

Relativistic neutrons seek to do so if their rate tends to the speed of light.

 

Cherenkov's effect does not interfere for a neutrons.

 

The charged particles would behave the same way as the neutron, if Cherenkov effect were not interfere.

 

However, all particles (electrons and positrons - too) get unlimited ability to penetrate into matter, if their speed more than the speed of light.

 

Cherenkov's effect does not operate at speeds exceeding the speed of light in a vacuum.

 

Any (hyper-light-speed) matter has unlimited ability to penetrate into any statical matter.

 

All converted into neutrinos, if moving faster than light.

 

 

[Alexander Masterov]

Results of experiments with the beta-spectrometer, which (allegedly) demonstrates the relativistic increase in mass, are interpreted incorrectly:

 

Deviation of the trajectories of relativistic electrons from the trajectories (defined by the classical theory of electromagnetism and Newtonian mechanics) is due to weakening of the magnetic field effect on the relativistic particle, but does not increase the mass of a relativistic particle. In other words, should adjust classic expression of Lorentz force, since mass are an absoluteness (speed-independent).

Edited October 19, 2011 by Alexander Masterov

[Alexander Masterov]

Master Theory (MT) is a theory of relativistic.

MT satisfies the same conditions, which satisfies Einstein's Special Theory of Relativity (SRT).

Experimental evidence, confirmative SRT, confirm MT.

The difference in interpretation and in a results of experiments that are not public.

For example: in scientific literature there is no experimental confirmation of a formulas:

[math]E=mc^2/\sqrt{1-v^2/c^2}[/math]

 

[math]p=mv/\sqrt{1-v^2/c^2}[/math]

In Master Theory these formulas is stale.

 

In MT time is absolute therefore heve not "Twin Paradox".

In MT light speed, acceleration and mass is absolute (do not depend on the speed and are the same in all Inertial Frame).

Inertial Frame (IF) are equal in rights.

 

You can ask me: Einstein's task has non-unique solution?

 

My answer: YES!

 

How can this be?

 

My answer: Please note that a transverse coordinates ([math]y[/math] and [math]z[/math]) of Lorentz Transformations are absolute (whereas Einstein asserted that in his theory of everything is relative).

 

Einstein (had no no reason for it) gave the absoluteness for a transverse coordinates, but not for - time.

 

I exempt the transverse coordinates ([math]y[/math] and [math]z[/math]) from Einstein's absoluteness.

I to do free from Einstein's absoluteness the transverse coordinates (I set relative it, I set dependence on velocity for it).

 

So I got a free parameter.

 

For each value of this parameter, you can build a individual theory of relativity, which will exist as scientific theory on equal terms with SRT.

Moreover, because the transverse scale also depends on the speed (diminishing) - it is possible to solve the paradox of Ehrenfest (the paradox of a rotating disk).

 

Thus, Einstein's task has an infinite number of possible solutions, and SRT - only one solution of this infinite set.

 

Among this infinite number of solutions I've found one, in which time is absolute.

I call this theory: "Master Theory".

Only this theory is correct because it have not SRT's paradoxes.

 

So: Master Theory have absolutely time, and this difference has profound implications. (For example: in Master Theory are absent "Twin Paradox" and "Ehrenfest's Paradox".)

 

Master Theory

Let us consider the light-clock with a pair of vertical mirrors (one on the left, the other - right) and photon between them:

[math]L[/math] - the distance between the mirrors.

 

Time's cycle:

[math]T=L/c+L/c=2L/c[/math]

 

Suppose that we (the observer) has a motion with velocity [math]v[/math].

Îscillogram of this motion:

Speed of light in all cases is well-known-constant.

Hence the transit time of a photon from mirror to mirror in different directions will be different.

This is because: moving in one direction - the photon (in the view of the observer) meet-moving to the mirror (flight time is less).

In the other direction - in pursuit of the mirror (flight time is major):

[math]T=L/(c+v)+L/(c-v)\neq 2L/c[/math]

 

Acceptably are three variants:

1. Happened a time-dilation [math]T'\neq T[/math]

2. Happened a curtailment the visual of the longitudinal scale [math]L'\neq L[/math] (Master Theory);

3. Happened all (both of the above) (SRT).

 

We consider the second variant (corresponding Master Theory):

 

[math]T=L'/(c+v)+L'/(c-v)=2L/c[/math]

 

The longitudinal scale of the rate is (for the second variant) calculated as follows:

 

[math]L'/L=1-v^2/c^2[/math]

 

This SRT's formula looked so: [math]L'/L=\sqrt{1-v^2/c^2}[/math]

 

 

 

We proceed to calculate the cross-scale.

 

For this we consider new light-clock, which have a pair of horizontal mirrors (one - from the bottom, the other - from the top):

 

Time's cycle: [math]T=2H/c[/math]

 

Suppose that we (the observer) has a motion with velocity [math]v[/math].

The trajectory of photon will change into sawtooth and elongated:

Photon has a fixed velocity and can not travel long distances over the same time.

 

 

 

We proceed to calculate the cross-scale.

 

For this we consider new light-clock, which have a pair of horizontal mirrors (one - from the bottom, the other - from the top):

 

Time's cycle: [math]T=2H/c[/math]

 

Suppose that we (the observer) has a motion with velocity [math]v[/math].

The trajectory of photon will change into sawtooth and elongated:

Photon has a fixed velocity and can not travel long distances over the same time.

Edited April 28, 2012 by hypervalent_iodine
Fixed LaTex codes