Master Theory

[Alexander Masterov]

Actually, it really isn't sufficient. There are many ideas which can only be described mathematically. You are correct to say that this isn't one of them, but this is one of those ideas which is simpler to describe mathematically.

You just need to drop a hint.

If you are going to manipulate the Lorentz transformations, then: Lorentz transformations and absoluteness cross-scale are the result of [math]x^2 - (ct)^2 =x'^2 - (ct')^2=0[/math] whose validity requires proof.

You are trying to discuss physics. Matrices are not little-known to physicists - they are everywhere in physics.

Are you sure that the mathematical expression [math](A^{v^{-1}})_{0,0}[/math] should be clear to everyone?

 

I have the education an physics and applied mathematics. I have not seen such expressions before. Them meaning do not understand.

 

I'm sure: you do not have answer my question and (in order to confuse the debate) you juggle with obscure mathematical expressions.

The basis was the assumption that the speed of light was absolute - that it was the same in every frame. It is possible to derive that the "cross-scale is absolute" from that assumption.

It's not true.

 

I showed that from absoluteness of the speed of light follow infinite much of consequences and not follow absolute cross-scales. (See above, post 2)

Why would it be impossible for each of them to see a different sequence of frames? What precisely is wrong with that? And where does causality, which you invoked, come into it?

On the screen can be mounted photographic recorder, which responds to the sequence of frames. The result of registration can not be individualized for each observer. Edited September 19, 2011 by Alexander Masterov

[uncool]

You just need to drop a hint.

If you are going to manipulate the Lorentz transformations, then: Lorentz transformations and absoluteness cross-scale are the result of [math]x^2 - (ct)^2 =x'^2 - (ct')^2=0[/math] whose validity requires proof.

That equation is a result of the absoluteness of the speed of light. What it means to say that the speed of light is absolute is that any observer in any frame will observe the speed of light to be c. You are saying that the speed of light is different in different frames, or in other words that all forms of relativity are false. That relativity is true comes entirely from experiment, so the absoluteness of the speed of light is an assumption.

Are you sure that the mathematical expression [math](A^{v^{-1}})_{0,0}[/math] should be clear to everyone?

Anyone who has studied college level and even most high-school level math, yes. It denotes the {0, 0} element of the inverse matrix of A^v, where I have already defined what A^v is.

I have the education an physics and applied mathematics. I have not seen such expressions before. Them meaning do not understand.

Well, we can go through one by one. A^v is the matrix I defined earlier. Then (A^v)^-1 is the inverse of A^v. Finally, (A^v)^-1_0,0 denotes the 0,0 element of A - that is, the element in the upper left hand corner of A.

I'm sure you do not have answer my question and (in order to confuse the debate) to do juggling by means of obscure mathematical expressions.

Matrices are by no means obscure, and taking one element of a matrix is the entire point of having matrices in the first place. If you think they are obscure then you have not learned enough. And yes, I have an answer to your question; you are refusing to even think about the math behind that answer.

It's not true.

 

I showed that from absoluteness of the speed of light follow infinite much of consequences and not follow absolute cross-scales. (See above, post 2)

That is because you have misunderstood what it means for the speed of light to be absolute. The speed of light being absolute means that for every observer in every inertial frame of reference, the speed of light is the same. In other words, the speed of light being absolute is the statement that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0.

On the screen can be mounted photographic recorder, which responds to the sequence of frames. The result of registration can not be individualized for each observer.

Which screen? You are talking about 2 screens here, not 1. Why must the two observers agree on ordering?

=Uncool-

Edited September 19, 2011 by uncool

[Alexander Masterov]

That equation is a result of the absoluteness of the speed of light. What it means to say that the speed of light is absolute is that any observer in any frame will observe the speed of light to be c. You are saying that the speed of light is different in different frames, or in other words that all forms of relativity are false. That relativity is true comes entirely from experiment, so the absoluteness of the speed of light is an assumption.

It's not true.

I use absoluteness of lite speed for Master Theory and not use [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math].

 

Formuls of Master Theory satisfy absoluteness of lite speed, but not satisfy [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math].

 

[math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math] - Unfounded limitation (restriction).

Anyone who has studied college level and even most high-school level math, yes. It denotes the {0, 0} element of the inverse matrix of A^v, where I have already defined what A^v is.

No.

 

[math] A^{v^{-1}}=A^{\frac{1}{v}} [/math] - It can be understood?

That is because you have misunderstood what it means for the speed of light to be absolute. The speed of light being absolute means that for every observer in every inertial frame of reference, the speed of light is the same. In other words, the speed of light being absolute is the statement that if x^2 - (ct)^2 = 0, then x'^2 - (ct')^2 = 0.

Formuls of Master Theory satisfy absoluteness of lite speed, but not satisfy [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math].

Which screen? You are talking about 2 screens here, not 1. Why must the two observers agree on ordering?

the screen - singular.

 

Projectors - pair (couple).

Edited September 19, 2011 by Alexander Masterov

[uncool]

It's not true.

I use absoluteness of lite speed for Master Theory and not use [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math].

No, you did not, because you do not know what it means for the speed of light to be absolute.

 

Show me precisely your definition of the absoluteness of the speed of light. What, precisely, does the statement that c is absolute mean?

[math] A^{v^{-1}}=A^{\frac{1}{v}} [/math]

You are placing the parentheses wrong. Read as follows:

[math](A^v)^{-1}[/math]. I wrote it with the brackets so that I could separate the superscript sections, as v and -1 should be considered completely separate, as v is an index, not an exponent.

Formuls of Master Theory satisfy absoluteness of lite speed, but not satisfy [math] x^2 - (ct)^2 =x'^2 - (ct')^2=0 [/math].

Again, what precisely does "absoluteness of light speed" mean in your "theory"?

=Uncool-

Edited September 19, 2011 by uncool

[Alexander Masterov]

No, you did not, because you do not know what it means for the speed of light to be absolute.

 

Show me precisely your definition of the absoluteness of the speed of light. What, precisely, does the statement that c is absolute mean?

Again, what precisely does "absoluteness of light speed" mean in your "theory"?

 

 

 

In these two images shows a clock (one the clock).

The first picture corresponds to a stationary observer.

The second image corresponds to a moving observer.

It is assumed that the velocity of the red dot (the speed of light) in both pictures (both ways) is identical. (The images inappropriately shows it. I intend to fix it.)

 

The constancy of the speed of red dots (the independence of this velocity on the speed of the observer) make a demonstration of the absoluteness of the light speed.

 

You are placing the parentheses wrong. Read as follows:

[math](A^v)^{-1}[/math]. I wrote it with the brackets so that I could separate the superscript sections, as v and -1 should be considered completely separate, as v is an index, not an exponent.

 

Expression A^{v^{-1}} is not similar to (A^v)^{-1}. Is'nt it?

 

Expression A^{v^{-1}} is equivalency A^{1/v}.

_________________________________________________

 

You want to show me: how you do know how to do reverse Lorentz transformation? I understand you correctly?

 

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

 

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

 

TO:

 

[math]x'=\frac{x+vt}{\sqrt{1-v^2/c^2}}[/math]

 

[math]t'=\frac{x+vt/c^2}{\sqrt{1-v^2/c^2}}[/math]

______________________________________________________

 

[math]\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}[/math]

 

TO:

 

[math]\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}[/math]

 

What does that prove?

Edited September 20, 2011 by Alexander Masterov

[ajb]

We need to know what you mean by constant speed of light.

 

The difficulty here is that we know in special relativity what this means. In particular it is in reference to as being measured in any inertial reference frame. These frames can be defined via local coordinates to be the frames in which the metric takes on the canonical Minkowski form. These frames (equivalently coordinates) form a privileged class in the theory.

 

Now, in Master Theory we no longer have a Minkowski metric, if I understand what you are saying. The two related questions them become

 

1) What is an inertial frame in Master Theory?

2) What is meant by constant speed of light?

[Alexander Masterov]

We need to know what you mean by constant speed of light.

 

The difficulty here is that we know in special relativity what this means. In particular it is in reference to as being measured in any inertial reference frame. These frames can be defined via local coordinates to be the frames in which the metric takes on the canonical Minkowski form. These frames (equivalently coordinates) form a privileged class in the theory.

 

Now, in Master Theory we no longer have a Minkowski metric, if I understand what you are saying. The two related questions them become

1. I have shown that the Minkowski metric is non-unique consequence of the absoluteness of the light speed.

 

2. Minkowski metric implies the relativity of time, contrary to experience. (Time is absolute.)

 

3. I decided the one case (between the infinite number) in which time is absolute and called it - "Master Theory".

1) What is an inertial frame in Master Theory?

A reference system is inertial, if a stationary object in it does not feel an acceleration.

2) What is meant by constant speed of light?

As usual. (How would you answered your question usually?) Edited September 20, 2011 by Alexander Masterov

[ajb]

What is really confusing me is that the speed of light postulate in special relativity is deeply tied to the notion of an inertial frame. You need to carefully characterise an inertial frame before you can state that the speed of light is constant. Because of this, once you specify that speed of light is constant in all inertial frames, understood in the Einsteinian/Minkowskian sense then you are lead to the Poincare group being the automorphism group of space-time. Thus we have Minkowski space-time.

 

Or you could end up with the de Sitter space-time, which also allows a constant speed of light. (I am far less familiar with this).

 

To my knowledge, these are the only two choices consistent with Einstein's ideas.

[uncool]

 

 

In these two images shows a clock (one the clock).

The first picture corresponds to a stationary observer.

The second image corresponds to a moving observer.

It is assumed that the velocity of the red dot (the speed of light) in both pictures (both ways) is identical. (The images inappropriately shows it. I intend to fix it.)

 

The constancy of the speed of red dots (the independence of this velocity on the speed of the observer) make a demonstration of the absoluteness of the light speed.

That is not a definition. I am asking for a definition of what it means for the speed of light to be absolute.

 

Expression A^{v^{-1}} is not similar to (A^v)^{-1}. Is'nt it?

 

Expression A^{v^{-1}} is equivalency A^{1/v}.

Again, only if you read it wrong.

_________________________________________________

 

You want to show me: how you do know how to do reverse Lorentz transformation? I understand you correctly?

That is a part of what I want to show you, yes.

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

 

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

 

TO:

 

[math]x'=\frac{x+vt}{\sqrt{1-v^2/c^2}}[/math]

 

[math]t'=\frac{x+vt/c^2}{\sqrt{1-v^2/c^2}}[/math]

______________________________________________________

 

[math]\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}[/math]

 

TO:

 

[math]\begin{pmatrix} ct' \\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}\begin{pmatrix} ct \\ x \\ y \\ z \end{pmatrix}[/math]

 

What does that prove?

So your objection is equivalent to the statement that if the (0, 0) element of A^v (that is, the upper left element (which for this transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (A^v)^-1 must be less than 1. Do you agree?

 

Additionally, what is your transformation matrix?

=Uncool-

[Alexander Masterov]

What is really confusing me is that the speed of light postulate in special relativity is deeply tied to the notion of an inertial frame. You need to carefully characterise an inertial frame before you can state that the speed of light is constant. Because of this, once you specify that speed of light is constant in all inertial frames, understood in the Einsteinian/Minkowskian sense then you are lead to the Poincare group being the automorphism group of space-time. Thus we have Minkowski space-time.

 

Or you could end up with the de Sitter space-time, which also allows a constant speed of light. (I am far less familiar with this).

???

 

To my knowledge, these are the only two choices consistent with Einstein's ideas.

Einstein's problems has an infinite number of solutions.

 

This infinite number of solutions came after I took away the absoluteness from a cross-scale, which Einstein presented to it (without reason).

 

[math]L' = L (1 - v^2/c^2)^{0.5+\alpha}[/math]

 

[math]H' = H(1 - v^2/c^2)^\alpha[/math]

 

[math]T' = T/(1 - v^2/c^2)^{ 0.5-\alpha}[/math]

 

[math]\alpha[/math] - free parameter. (Light speed absoluteness for any [math]0\le\alpha\le 0.5[/math])

 

[math]\alpha=0[/math] - Einsteinian theory (cross-scale absoluteness).

 

[math]\alpha=0.5[/math] - Master Theory (time absoluteness).

 

Additionally, what is your transformation matrix?

 

Master Theory is valid in all reference frames.

 

Non inertial reference frame

 

Real coordinates:

 

Real speed:

 

Visual coordinates:

Visual time:

 

[math] t' = t - \frac{|\vec r(t')|}{c} [/math]

 

Visual speed:

 

[math]\vec v(t)=\frac{2\vec V(t')}{(\sqrt{1+4V^2/c^2}+1)(1+\vec n\vec V/c)}[/math]

 

[math] \vec n = \frac{\vec r}{|\vec r|} [/math]

Edited September 20, 2011 by Alexander Masterov

[uncool]

 

Master Theory is valid in all reference frames.

 

Non inertial reference frame

 

Real coordinates:

 

Real speed:

 

Visual coordinates:

Visual time:

 

[math] t' = t - \frac{|\vec r(t')|}{c} [/math]

 

Visual speed:

 

[math]\vec v(t)=\frac{2\vec V(t')}{(\sqrt{1+4V^2/c^2}+1)(1+\vec n\vec V/c)}[/math]

 

[math] \vec n = \frac{\vec r}{|\vec r|} [/math]

Not what I asked. I asked you: What is your transformation matrix?

 

Additionally, do you agree that I have correctly summarized your objection?

=Uncool-

[ajb]

If we want an isotropic and homogeneous Universe then we are really restricting ourself to rotations and translations as automorphisms of our space-time. Insisting that we have a constant speed of light produces the the Poincare group and the de Sitter group.

 

You can use these groups, and their stabaliser the Lorentz group to "create" Minkowski space-time and de Sitter space-time. (One employs standard methods of homogeneous space etc. )

 

The Minkowski space-time gives us standard special relativity, while the de Sitter gives a deformed version which turns out to be a kind of doubly special relativity.

 

As far as I know these two are the only possible groups, under the assumption that we want an isotropic and homogeneous Universe. It may be possible to cook up something more exotic, loosing some of our usual assumptions.

 

This is why I am confused about your notion of "constant speed of light". It is deeply tied down to our notion of an inertial frame, which itself is really tied down to our notion of the group of automorphisms acting on our space, and then of course the space itself.

[uncool]

If we want an isotropic and homogeneous Universe then we are really restricting ourself to rotations and translations as automorphisms of our space-time. Insisting that we have a constant speed of light produces the the Poincare group and the de Sitter group.

 

I'm not sure that I see this. It seems to me that isotropy guarantees rotations, and homogeneous guarantees translations as automorphisms (in fact, that is pretty much the definition of the two) - but that other automorphisms are not out of the question. For example, say you had R^3 as your universe - that universe would still be isotropic and homogeneous, but it would also have scale invariance for distance.

=Uncool-

[ajb]

For example, say you had R^3 as your universe - that universe would still be isotropic and homogeneous, but it would also have scale invariance for distance.

 

Equipped with the standard metric? The group that preserves this is the Euclidean group. That is the group that preserves our notion of distance, it is known as an isometry group. We have just rotations and translations here.

 

You can generalise this to conformal symmetries, here we weaken the notion and preserve the metric only up to scale. Angles remain unchanged, but our lengths can be stretched.

[uncool]

Equipped with the standard metric?

I would say that equipping it with the standard metric is then what restricts the automorphisms. Isotropy and homogeneity are only guarantees that those automorphisms exist.

 

It just sounded like you were saying that any isotropic, homogeneous space would have no other symmetries.

=Uncool-

[Alexander Masterov]

If we want an isotropic and homogeneous Universe then we are really restricting ourself to rotations and translations as automorphisms of our space-time. Insisting that we have a constant speed of light produces the the Poincare group and the de Sitter group.

 

You can use these groups, and their stabaliser the Lorentz group to "create" Minkowski space-time and de Sitter space-time. (One employs standard methods of homogeneous space etc. )

 

The Minkowski space-time gives us standard special relativity, while the de Sitter gives a deformed version which turns out to be a kind of doubly special relativity.

 

As far as I know these two are the only possible groups, under the assumption that we want an isotropic and homogeneous Universe. It may be possible to cook up something more exotic, loosing some of our usual assumptions.

 

This is why I am confused about your notion of "constant speed of light". It is deeply tied down to our notion of an inertial frame, which itself is really tied down to our notion of the group of automorphisms acting on our space, and then of course the space itself.

I'm not sure that I see this. It seems to me that isotropy guarantees rotations, and homogeneous guarantees translations as automorphisms (in fact, that is pretty much the definition of the two) - but that other automorphisms are not out of the question. For example, say you had R^3 as your universe - that universe would still be isotropic and homogeneous, but it would also have scale invariance for distance.

Master Theory defines two types of coordinates: real coordinates and visual coordinates.

 

Real coordinates obey Newton's physics and Galilean transformations. These coordinates can be calculated by double integration of acceleration, but acceleration can be measured (weight on a spring).

 

Visual coordinates exist on an equal footing with coordinates, which has we determine by ear. Ie - the visual coordinates have no great physical sense. Therefore, there is no point to philosophize about the rules of transformation of visual coordinates.

 

Master Theory returns we to Newton and to Galileo.

Edited September 21, 2011 by Alexander Masterov

[ajb]

If we have the Galilean group as our automorphisms then we have the notion of absolute time and the speed of light is not constant.

[Alexander Masterov]

If we have the Galilean group as our automorphisms then we have the notion of absolute time and the speed of light is not constant.

It's not true.

 

The finiteness of the light speed (and of sound) and visual relativistic effects (due to the specifications of EMF) have no relationship to the real coordinates.

 

Real coordinates - on their own. Relativism, any properties of light and sound - on their own.

Edited September 21, 2011 by Alexander Masterov

[ajb]

In a some specified inertial coordinates with respect to the Galilean group, you can write down the wave equation and consider how it transforms. The wave equation is not invariant. In fact you see that if you set the speed of the wave to c, you can derive the Lorentz transformations.

 

So now we need to understand how your coordinates fit in. I think they cannot be inertial coordinates. In particular the class of coordinates you use are not related by a Galilean transformation. This means we would have to take care in understanding theory physics here.

 

Galilean mechanics and Einsteinian mechanics can be examined by using any coordinate system you like, they need not be the inertial ones.

[Alexander Masterov]

In a some specified inertial coordinates with respect to the Galilean group, you can write down the wave equation and consider how it transforms. The wave equation is not invariant. In fact you see that if you set the speed of the wave to c, you can derive the Lorentz transformations.

 

So now we need to understand how your coordinates fit in. I think they cannot be inertial coordinates. In particular the class of coordinates you use are not related by a Galilean transformation. This means we would have to take care in understanding theory physics here.

 

Galilean mechanics and Einsteinian mechanics can be examined by using any coordinate system you like, they need not be the inertial ones.

Linearity is a special case of nonlinearity. (Linearity does not exist in reality.)

Inertia is a special case of acceleration. (External actions and acceleration are taking place always.)

Master Theory solves the problem in the general (non-inertial) case.

Visual (of relativism) and sound effects must be treated separately from physical processes, if the field (or environment) in the problem are not considered.

 

Go back to Newton and to Galileo. Forget Einstein.

[uncool]

Alex, you have not answered what I asked. I asked you: What is your transformation matrix?

 

Additionally, do you agree that I have correctly summarized your objection?

=Uncool-

[Alexander Masterov]

Alex, you have not answered what I asked. I asked you: What is your transformation matrix?

 

The transformation of real coordinates (Galileo):

 

[math]\begin{pmatrix}t\\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix}1&0&0&0\\-v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t'\\x' \\ y' \\ z' \end{pmatrix}[/math]

 

[math]\begin{pmatrix}t'\\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix}1&0&0&0\\v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t\\x \\ y \\ z\end{pmatrix}[/math]

 

Additionally, do you agree that I have correctly summarized your objection?

???

[uncool]

The transformation of real coordinates (Galileo):

 

[math]\begin{pmatrix}t\\ x \\ y \\ z \end{pmatrix}=\begin{pmatrix}1&0&0&0\\-v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t'\\x' \\ y' \\ z' \end{pmatrix}[/math]

 

[math]\begin{pmatrix}t'\\ x' \\ y' \\ z' \end{pmatrix}=\begin{pmatrix}1&0&0&0\\v & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}\begin{pmatrix}t\\x \\ y \\ z\end{pmatrix}[/math]

So these are your transformation matrices? You, too, have set the "cross-scales" to be absolute here.

 

Furthermore, what precisely is your definition of the speed of light being absolute?

???

So your objection is equivalent to the statement that if the (0, 0) element of A^v (that is, the upper left element (which for the special relativistic transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (A^v)^-1 must be less than 1. Do you agree?

=Uncool-

Edited September 21, 2011 by uncool

[Alexander Masterov]

So these are your transformation matrices? You, too, have set the "cross-scales" to be absolute here.

 

Furthermore, what precisely is your definition of the speed of light being absolute?

The transformation of real coordinates no related to light speed and sound speed.

 

Real coordinates are changed in consequence of the presence of acceleration, which can be measured.

 

Go back to Newton and to Galileo. Forget Einstein.

 

 

 

 

[uncool]

The transformation of real coordinates no related to light speed and sound speed.

 

Real coordinates are changed in consequence of the presence of acceleration, which can be measured.

 

Go back to Newton and to Galileo. Forget Einstein.

Again, in that case, what precisely does it mean to say that the speed of light is absolute?

 

You have not defined it. I have. And there is reason to agree with my definition. Furthermore, there are experiments that bear out my definition.

 

And again:

 

So your objection is equivalent to the statement that if the (0, 0) element of A^v (that is, the upper left element (which for this transformation matrix is [math] \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}[/math]) is greater than 1, then the (0, 0) element of (A^v)^-1 must be less than 1. Do you agree?

=Uncool-

Edited September 21, 2011 by uncool