Ah, Dingle. Classical.
Master Theory (edition 2)
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Ah, Dingle. Classical.
Then you admit you are anti-science today? You can not deprive anyone of a livelihood when you are just spreading untrue and useless ideas based on your personal ignorance of math and physics. Every time you make an untrue math claim you ignore requests to supply a mathematical proof and every time you make a useless physics claim you ignore requests to provide fact-based arguments for the claim. You are anti-science and anti-truth.
Infinite light propagation speed is not necessary for Gauss's and Stokes's theorems to be correct. They also apply in other fields of physics like fluid dynamics and do not imply that the speed of flowing water is infinite.
It is obviously untrue that they require static fields since there are time derivatives in the integral representation.
Where is the fact-based argument to support this ridiculous claim? You have asserted it, but not given a reason to believe it.
Yes, they used Jefimenko's equations and the continuity equation of charge to prove that Maxwell's equations result. Nowhere have you attempted to actually calculate the left and right sides of any equation to demonstrate that Maxwell's equation is untrue.
You are confused. I was merely translating your question into better English. Lorentz transformations are the subject of your question. You have failed to provide a reason for your claim that: $$\begin{pmatrix} \cosh \varphi & \quad & \sinh \varphi \\ \sinh \varphi & \quad & \cosh \varphi \end{pmatrix}$$ is any more or less rational than the well-known Euclidean rotation transform: $$\begin{pmatrix} \cos \theta & \quad & - \sin \theta \\ \sin \theta & \quad & \cos \theta \end{pmatrix}$$
This has nothing to do with your question or my answer. Further, it is another paragraph filled with unsupported claims. Are you the pope? No? Then you are not allowed to make unsupported claims based on your personal authority and expect people to believe as you do.
Further, Tach seems to recognize this a plagiarized claim. I wish he would expand on that cryptic reference to "Dingle".
This belongs after my translation and answer to your following question.
Where is the fact-based argument to support this claim that time can not slow down? You have asserted it, but not given a reason to believe it.
Further, as I show below we do have many reasons to believe time can slow down.
How, when it is based in facts and logic and math, is it not scientific? Where is the fact-based argument to support this claim that evidence that time can slow down is not a scientific argument for the proposition that time can slow down? You have asserted it, but not given a reason to believe it.
Untrue. Physics supports the Lorentz transformation. Physics does not support your transformation. In fact, I do not see that if you use your transformation in 1 direction and then a transformation in the opposite direction that you can ever get the identity transformation, therefore your transformation cannot describe a group.
Galileo said $$v_3 = v_1 \oplus_{G} v_2 = v_1 + v_2$$ so $$v \oplus_{G} (-v) = 0$$ and so $$ \frac{c}{2} \oplus_{G} \frac{c}{2} = c$$
Einstein said $$v_3 = v_1 \oplus_{E} v_2 = \frac{v_1 + v_2}{1 + \frac{v_1 v_2}{c^2}}$$ so $$v \oplus_{E} (-v) = 0$$ and so $$ \frac{c}{2} \oplus_{E} \frac{c}{2} = \frac{4}{5} c$$
Masterov says $$v \oplus_{M} (-v) = \cdots$$ ?
Masterov says $$\frac{c}{2} \oplus_{M} \frac{c}{2} = \cdots$$ ?
In general, there is no velocity parameter which makes the Masterov transformation on the right side equal to the product of the two Masterov transformations on the left side.
Herbert Dingle spent a lot of his life, time and effort fighting the reciprocity of time dilation. Masterov shares the same misconceptions with Dingle (among many other misconceptions).
No, they don't. They are not statements about some physically varying system where the (for instance) material contained within a volume must communicate with the surface so that they are all updated on what bit of stuff is doing what, it is a statement about mathematical functions in non-trivial (or sometimes trivial) spaces.
Consider something simpler, Cauchy's residue theorem. Given a holomorphic function then the integral along a simple closed curve is proportional to the sum of residues of the function at the poles surrounded by the curve. This is in fact a form of Stoke's theorem (or vice versa). There's no communication, no physical properties, no signals, no propogation, no light or sound, nothing other than a function with a very specific property, it is holomorphic.
Consider the function $$f(z) = \frac{1}{z}$$ and a contour integral on $$|z| = R$$ for some huge R. It is what it is, regardless of how large R is made. If you say "Make R 20 times bigger!" there's no change in the integral, no need to wait for signals to propagate out that far. It is a statement about how holomorphicity constraints certain properties of functions. Stokes' theorem is likewise, a statement about how certain properties, which are stated in the definition of the theorem, lead to certain other properties. It is a relationship between mathematical structures, nothing to do with physical properties.
If you don't grasp this when you failed to understand a critical conceptual point in mathematics and its application to physics. Just like you obviously don't understand how Lorentz transforms work in physics or their mathematical structure. To someone living in $$\mathbb{C}^{2}$$ rather than Minkowski or Euclidean space-time there is no distinction between standard rotations and Lorentzian boosts, they are just special cases of a more general set of transformations. Ask Penrose, he loves $$\mathbb{C}^{2}$$ twistors and generalised rotations.
I'm a pea brain and simply reading along, but I can't tell if this low-confidence guess as to what Masterov was asking in his second question has been confirmed to be what he meant to ask, given the language barrier.
Given that Masterov responded:
Are you referencing the earlier arguments about atmospheric muon decay time dilation and the atomic clocks getting out of sync when one is flown at high altitude and speed (this post) as the scientific evidence that time can slow down? I'm just asking if that is the evidence or if I missed other evidence that you are referring to?
Define $$G_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$ then the Galiean transform for a space-time interval $$\begin{pmatrix} c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = G_x(v)\begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ \Delta x - v \Delta t \\ \Delta y \\ \Delta z \end{pmatrix} $$ .
Note that $$G_x(0) = I$$ which is the identity transform.
So we wish to solve $$ G_x(v_2) G_x(v_1) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} = G_x(v_3) \begin{pmatrix} c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix} $$ for all space-time intervals.
Obviously $$ G_x(v_2) G_x(v_1) = G_x(v_3)$$
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_1}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_2}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -\frac{v_3}{c} & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix}$$
Expanding the left half, we have :
$$\left( I - \frac{v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) \left( I - \frac{v_1}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \right) = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} + \frac{v_1 v_2}{c^2} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}^2 = I - \frac{v_1 + v_2}{c} \begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} = G_x(v_1 + v_2) = G_x(v_3)$$.
So $$v_3 = v_1 + v_2$$ as Galileo and Newton would expect.
Define $$\Lambda_x(v) = \begin{pmatrix} \cosh \, \tanh^{-1} \frac{v}{c} & \quad & - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ - \sinh \, \tanh^{-1} \frac{v}{c} & \quad & \cosh \, \tanh^{-1} \frac{v}{c} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ \frac{-v}{c \sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & 0 \\ 0 & \quad & 0 & \quad & 1 & 0 \\ 0 & \quad & 0 & \quad & 0 & 1 \end{pmatrix}$$ Note that $$\Lambda_x(0) = I$$.
We wish to solve $$\Lambda_x(v_2)\Lambda_x(v_1) = \Lambda_x(v_3)$$
and since $$\begin{pmatrix}\cosh a & \quad & \sinh a \\ \sinh a & \quad & \cosh a \end{pmatrix} \begin{pmatrix}\cosh b & \quad & \sinh b \\ \sinh b & \quad & \cosh b \end{pmatrix} = \begin{pmatrix}\cosh a \cosh b + \sinh a \sinh b & \quad & \cosh a \sinh b + \sinh a \cosh b \\ \cosh a \sinh b + \sinh a \cosh b & \quad & \cosh a \cosh b + \sinh a \sinh b \end{pmatrix} = \begin{pmatrix}\cosh (a + b) & \quad & \sinh (a + b) \\ \sinh (a + b) & \quad & \cosh (a + b) \end{pmatrix} $$,
we have the solution $$v_3 = c \tanh \left( \tanh^{-1} \frac{v_1}{c} \; + \; \tanh^{-1} \frac{v_2}{c} \right) = c \frac{ \frac{v_1}{c} + \frac{v_2}{c} }{1 + \frac{v_1}{c} \frac{v_2}{c}} = \frac{ v_1 + v_2 }{1 + \frac{v_1 v_2}{c^2}}$$.
We don't know what Masterov's equivalent of $$\Lambda_x(v)$$ because he only defines: $$L' = L ( 1 - v^2/c^2) \\ H' = H \sqrt{ 1 - v^2/c^2} \\ T' = T$$ so we can define:
$$\mathcal{L}_x(v) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ -\frac{v A(v)}{c} & 1 - \frac{v^2}{c^2} & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ such that
$$\begin{pmatrix} 0 \\ \Delta x' \\ 0 \\ 0 \end{pmatrix} = \mathcal{L}_x(v)\begin{pmatrix} 0 \\ \Delta x \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} 0 \\ (1 - \frac{v^2}{c^2}) \Delta x \\ 0 \\ 0 \end{pmatrix} $$
This suggests that $$ 1 - \frac{v_3^2}{c^2} = (1 - \frac{v_2^2}{c^2}) (1 - \frac{v_1^2}{c^2})$$ or $$v_3^2 = v_1^2 + v_2^2 - \frac{v_1^2 + v_2^2}{c^2}$$
But to solve $$v_3 = 0$$ we have the relation $$v_2 = \pm \frac{v_1}{\sqrt{\frac{v_1^2}{c^2} -1}}$$ which if $$0 < \left| v_1 \right| < c$$ requires $$v_2$$ to be imaginary.
Likewise we can define:
$$\mathcal{H}_x(v) = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) & -\sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \sin B(v) & \sqrt{1 - \frac{v_2^2}{c^2}} \cos B(v) \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} & 0 \\ 0 & 0 & 0 & \sqrt{1 - \frac{v_2^2}{c^2}} \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & \cos B(v) & - \sin B(v) \\ 0 & 0 &\sin B(v) & \cos B(v) \end{pmatrix}$$ such that
$$\begin{pmatrix} c \Delta t' \\ 0 \\ \Delta y' \\ \Delta z' \end{pmatrix} = \mathcal{H}_x(v)\begin{pmatrix} c \Delta t \\ 0 \\ \Delta y \\ \Delta z \end{pmatrix} = \begin{pmatrix} c \Delta t \\ 0 \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta y \cos B(v) - \Delta z \sin B(v) \right) \\ \sqrt{1 - \frac{v_2^2}{c^2}} \left( \Delta z \cos B(v) + \Delta y \sin B(v) \right) \end{pmatrix} $$
Similar problems exist, and the physical requirement that v be a observed velocity indicates this cannot describe the physics of this universe.
I'm referencing the entire experimental record, which since 1859 has included more precision and faster experimental speeds than in the pre-1859 record and so highlights that Newtonian and Galilean low-precision, low-velocity experiments do not describe nature as well as Special Relativity, which works at all speeds.
http://math.ucr.edu/home/baez/physics/Relativity/SR/experiments.html
But I linked to a description of one particular 1859 observation.
Thank you, great link too.
You made a point early in the thread about SR being about the fact that the physics are the same in all reference frames. To me that and the fact that the math works are the keys. I follow the threads on SR to some extent and the "controversy" about evidence seems to focus on the spherically expanding light wave front being observed as spherical in the rest frame and spherical to observers in all inertial frames as opposed to being seen as an oblate spheroid in the non-rest frame. Also, to see the light wave front as spherical the point of origin of the light must move with the observer in his frame. I guess that all frames can be considered the rest frame and so all frames must see the spherical light wave front. It is just that someone said that when you do the transformations the wave front doesn't come out spherical in the transformed frame. Can you address that for me too?
A family of spherical light fronts is emitted at times $$t_i$$, i=1,n in frame S. Their equation is (in one dimension, for simplicity):
$$x^2-(ct_i)^2=0$$
Now, consider frame S', moving away from S at speed v along the x axis. Applying the Lorentz transforms between S and S':
$$x'=\gamma(x-vt)$$
$$t'=\gamma(t-vx/c^2)$$
you can easily prove that :
$$x'^2-(ct'_i)^2=0$$
That is, the spherical fronts are also concentric in S' and they are centered around O', the origin of the system of axes in S'.
Repeat the calculations in a frame S" moving towards S:
$$x"=\gamma(x+vt)$$
$$t"=\gamma(t+vx/c^2)$$
If you do them right, you should get:
$$x"^2-(ct"_i)^2=0$$
As an added bonus, if you consider that the time separation between two spherical fronts is $$t_{i+1}-t_i$$ in S, you can easily deduce the relativistic Doppler effect in S' and in S".
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Points in space need three numbers (coordinates) to describe them. A popular choice are Cartesian coordinates x, y and z since each of them by themselves is a real number line just like any other, and each of the number lines are at right angles to the other two.
The definition of distance in space between any two points, distance(A,B), is given by the square root of the sums of the squares of each difference of coordinates, $$\textrm{distance}(A,B) = \sqrt{ (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 }$$. This follows by using the famous theorem named for Pythagoras.
The definition of a sphere is that all points, P, have the same distance, R, from the center, $$O = (x_O,y_O,z_O)$$ so $$R = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }$$.
The definition of something spherically expanding is that at any instant in time, t, the points at that instant in time, P, form a sphere about the center, O, and the radius of that sphere is a function that increase as time goes on, $$\frac{\partial R(t)}{\partial t} > 0$$. The simplest motion is a constant increase of R at the same rate and so we can describe the radius growing with the speed of light as: $$R(t) = c(t - t_0)$$ where the radius is zero at time = $$t_0$$.
Putting the two sides together, we have $$c(t-t_0) = R(t) = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }$$. Since the distance is a non-negative number and we don't care about times prior to $$t_0$$, we can square both sides.
$$c^2(t-t_0)^2 = (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2$$
So for all times we care about, we have $$(x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 - c^2(t-t_0)^2$$.
Or concisely, $$(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 = 0$$
Relativity says that every coordinate system where the laws of physics are the same are equally valid. In Special Relativity, we are talking about inertial coordinate systems. They don't rotate, they don't accelerate, in them Newton's law of inertia is valid. So two different inertial coordinate systems can have a velocity difference between them.
In what follows, lets assume that x,y,and z of one coordinate system was instantly aligned with the x', y' and z' of another at some instant and that the only motion is along the x axis. Then the simplest version of the Lorentz transformation applies.
$$\begin{pmatrix}c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ 0 & \quad & 0 & \quad & 1 & \quad & 0 \\ 0 & \quad & 0 & \quad & 0 & \quad & 1 \end{pmatrix} \begin{pmatrix}c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$
This allow us to calculate:
$$\begin{eqnarray} c \Delta t' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( c \Delta t - \frac{v}{c} \Delta x \right) \\ \Delta x' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( \Delta x - \frac{v}{c} c \Delta t \right) \\ \Delta y' \quad & = & \Delta y \\ \Delta z' \quad & = & \Delta z \end{eqnarray} $$
So $$ (c \Delta t')^2 = \frac{(c \Delta t)^2 + \frac{v^2}{c^2}(\Delta x)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ (\Delta x' )^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ ( \Delta y')^2 = ( \Delta y)^2 \\ ( \Delta z')^2 = ( \Delta z)^2 $$
And therefore $$ (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t) - (c \Delta t)^2 - \frac{v^2}{c^2}(\Delta x)^2 + \frac{v}{c} (\Delta x)(c \Delta t) }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 \\ = \frac{(\Delta x)^2 - \frac{v^2}{c^2}(\Delta x)^2 }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 - \frac{ (c \Delta t)^2 - \frac{v^2}{c^2}(c \Delta t)^2 }{1 - \frac{v^2}{c^2}} \\ = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2$$
This is the most important property of the Lorentz transform:
$$ (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2$$
Because we are dealing with a sphere expanding at the speed of light, the right side equals zero, and therefore the left side equals zero. Therefore both coordinate systems agree that the light is expanding spherically at the speed of light.
The definition of distance in space between any two points, distance(A,B), is given by the square root of the sums of the squares of each difference of coordinates, $$\textrm{distance}(A,B) = \sqrt{ (x_A - x_B)^2 + (y_A - y_B)^2 + (z_A - z_B)^2 }$$. This follows by using the famous theorem named for Pythagoras.
The definition of a sphere is that all points, P, have the same distance, R, from the center, $$O = (x_O,y_O,z_O)$$ so $$R = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }$$.
The definition of something spherically expanding is that at any instant in time, t, the points at that instant in time, P, form a sphere about the center, O, and the radius of that sphere is a function that increase as time goes on, $$\frac{\partial R(t)}{\partial t} > 0$$. The simplest motion is a constant increase of R at the same rate and so we can describe the radius growing with the speed of light as: $$R(t) = c(t - t_0)$$ where the radius is zero at time = $$t_0$$.
Putting the two sides together, we have $$c(t-t_0) = R(t) = \textrm{distance}(P,O)= \sqrt{ (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 }$$. Since the distance is a non-negative number and we don't care about times prior to $$t_0$$, we can square both sides.
$$c^2(t-t_0)^2 = (x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2$$
So for all times we care about, we have $$(x_P - x_O)^2 + (y_P - y_O)^2 + (z_P - z_O)^2 - c^2(t-t_0)^2$$.
Or concisely, $$(\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2 = 0$$
Relativity says that every coordinate system where the laws of physics are the same are equally valid. In Special Relativity, we are talking about inertial coordinate systems. They don't rotate, they don't accelerate, in them Newton's law of inertia is valid. So two different inertial coordinate systems can have a velocity difference between them.
In what follows, lets assume that x,y,and z of one coordinate system was instantly aligned with the x', y' and z' of another at some instant and that the only motion is along the x axis. Then the simplest version of the Lorentz transformation applies.
$$\begin{pmatrix}c \Delta t' \\ \Delta x' \\ \Delta y' \\ \Delta z' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ \frac{-v/c}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} & \quad & 0 & \quad & 0 \\ 0 & \quad & 0 & \quad & 1 & \quad & 0 \\ 0 & \quad & 0 & \quad & 0 & \quad & 1 \end{pmatrix} \begin{pmatrix}c \Delta t \\ \Delta x \\ \Delta y \\ \Delta z \end{pmatrix}$$
This allow us to calculate:
$$\begin{eqnarray} c \Delta t' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( c \Delta t - \frac{v}{c} \Delta x \right) \\ \Delta x' \quad & = & \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \left( \Delta x - \frac{v}{c} c \Delta t \right) \\ \Delta y' \quad & = & \Delta y \\ \Delta z' \quad & = & \Delta z \end{eqnarray} $$
So $$ (c \Delta t')^2 = \frac{(c \Delta t)^2 + \frac{v^2}{c^2}(\Delta x)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ (\Delta x' )^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t)}{1 - \frac{v^2}{c^2}} \\ ( \Delta y')^2 = ( \Delta y)^2 \\ ( \Delta z')^2 = ( \Delta z)^2 $$
And therefore $$ (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = \frac{(\Delta x)^2 + \frac{v^2}{c^2}(c \Delta t)^2 - \frac{v}{c} (\Delta x)(c \Delta t) - (c \Delta t)^2 - \frac{v^2}{c^2}(\Delta x)^2 + \frac{v}{c} (\Delta x)(c \Delta t) }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 \\ = \frac{(\Delta x)^2 - \frac{v^2}{c^2}(\Delta x)^2 }{1 - \frac{v^2}{c^2}} + (\Delta y)^2 + (\Delta z)^2 - \frac{ (c \Delta t)^2 - \frac{v^2}{c^2}(c \Delta t)^2 }{1 - \frac{v^2}{c^2}} \\ = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2$$
This is the most important property of the Lorentz transform:
$$ (\Delta x')^2 + (\Delta y')^2 + (\Delta z')^2 - (c \Delta t')^2 = (\Delta x)^2 + (\Delta y)^2 + (\Delta z)^2 - (c \Delta t)^2$$
Because we are dealing with a sphere expanding at the speed of light, the right side equals zero, and therefore the left side equals zero. Therefore both coordinate systems agree that the light is expanding spherically at the speed of light.
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Such an arguments are worthy of a medieval Inquisition, but not a scientist.
I asked the question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis?
Einstein had entitled (bestowed the absoluteness to time), but no did it.
Why?
I do asked a question only. Why do you accuse me of blasphemy and heresy?
Check Stokes's theorem by a simple example: Two loops of different radii and an electric conductor of alternating current pierces through which.
Stokes's theorem return equal results for both loops. Is it true?
I do not get it.
No.
My question: Einstein bestowed the absoluteness to the cross-scale (but not for time) - on what basis?
Einstein had entitled (bestowed the absoluteness to time), but no did it.
Why?
I do not get it.
It to do causality principle to incorrect or violates the equality of inertial reference frames.
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Seriously? You're asking where that comes from? It's the integrated (for constant motion) space-time interval for something moving through Minkowski space-time at the speed of light.
Have you even read a book on special relativity?
OK, let's try differently, what do you think that the following expression means:
$$x^2+y^2+z^2-(ct)^2=0$$
Hint: a spherical wavefront propagating at light speed.
I see. So you could deduce the same in S'' and S'''? Or is that taking it too far?
And therefore the math works perfectly as I knew it did
. And 0' moves relative 0 along one axis, or is that my misconception?Apologies to Masterov for jumping in but I wanted to clarify that rpenner was correct as to what you were asking. No responses needed to this post. Thanks for the help.
works in any inertial frame
It is boundary between events that may be a cause and consequence of each other of Einsteinian theory, that is also a consequence of no-proven assumption that nothing can travel faster than light.
$$x^2-(ct)^2=0$$ - special case (y,z=const)
No, it isn't. It is a statement about trajectories. If $$v^{\mu}v_{\mu} = -v^{0}v^{0} + \mathbf{v}\cdot \mathbf{v} = 0$$ then it means $$v = v^{\mu}\partial_{\mu}$$ is a null vector. A vector which corresponds to something moving faster than light locally has (for this metric signature) $$v^{\mu}v_{\mu} > 0$$.
A consequence of the Lorentzian signature of space-time is that no object with mass can go faster than the local speed of light, not the reverse.
For any constant null velocity vector $$v = v^{\mu}\partial_{\mu}$$ there is always a frame where $$v^{\mu}v_{\mu} = 0$$ takes the form $$-v^{0}v^{0} + v^{x}v^{x} = 0$$. Simply use an SO(3) rotation (which is contained within the SO(3,1) symmetry group of flat Lorentzian space-time metrics) to align the x axis with the velocity vector and you're done.
Seriously, learn some special relativity.
For example: from target of elementary particle accelerator generate unstable particles emitted from the known lifetime (tau leptons - 5 10^-13 seconds) sometimes. During life, they manage to overcome the distance that can not be overcome if the move at the speed of light.
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