"On the Development of the Cuckoo Malformations created by Albert Einstein."

Note: The black text is Einstein's, taken verbatim from http://www.fourmilab.ch/etexts/einstein/specrel/www/

If we place x'=x-vt, it is clear that a point at rest in the system k must have a system of values x', y, z, independent of time.

 

(All that means is the length of the train (or system k, in this illustration the distance between the wheels) doesn't change from day to day.)

 We first define t as a function of x', y, z, and t. To do this we have to express in equations that t is nothing else than the summary of the data of clocks at rest in system k, which have been synchronized according to the rule given in § 1.

(The rule given in § 1 was 'we establish by definition that the "time" required by light to travel from A to B equals the "time" it requires to travel from B to A', like this:

As we all know, 4 = 12, of course, because Einstein says so. Don't question it, he was a genius, even though in 1895 he failed an examination that would have allowed him to study for a diploma as an electrical engineer at the Eidgenössische Technische Hochschule in Zurich (couldn't even pass the SATs).

As we shall soon see, Einstein does not mean "time" in quotation marks is the same as time. Clocks on the train are synchronised, clocks on the track are synchronized, but train clocks and track clocks do not agree. Hence the use of t and t.) 

From the origin of system k (back of the train) let a ray be emitted at the time t0 along the X-axis to x', and at the time t1 be reflected thence (front of the train, end of fence) to the origin of the co-ordinates (back of the train but not the start of the fence, the train has moved), arriving there at the time t2;

we then must  have ½(t0 + t2)  = t1 , (which it is clear and in agreement with experience means ½(16+4)  = 16 in system K) or, by inserting the arguments of the function t and applying the principle of the constancy of the velocity of light in the stationary system:­

 

Or, in our system of chosen numbers with the superfluous y, z and t removed "and where for brevity it is assumed that at the origin of k, $\tau =0$, when t=0."

And where for brevity, according to Einstein,  half of 20 is 16, the other half is 4 because 4+16 =20.

Hence, if x' be chosen infinitesimally small,

(It is to be noted the train's length is now infinitesimal, virtually a single point)

\begin{displaymath}\frac{1}{2}\left(\frac{1}{c-v}+\frac{1}{c+v}\right)\frac{\par... ...au}{\partial x'}+\frac{1}{c-v}\frac{\partial\tau}{\partial t}, \end{displaymath}

or

\begin{displaymath}\frac{\partial\tau}{\partial x'}+\frac{v}{c^2-v^2}\frac{\partial\tau}{\partial t}=0. \end{displaymath}

 

It is to be noted that instead of the origin of the co-ordinates (deliberately vague, the origin of k no longer coincides with the origin of K) we might have chosen any other point for the point of origin of the ray, and the equation just obtained is therefore valid for all values of x', y, z.

It is to be noted that Einstein was an incompetent idiot who didn't know what a half is or when the origin of k coincides with the origin of K, but as he later claims, the velocity of light in our theory plays the part, physically, of an infinitely great velocity, which it is clear and in agreement with experience would return the reflected light to both the origin of k and the origin of K simultaneously.

An analogous consideration--applied to the axes of Y and Z -- it being borne in mind that light is always propagated along these axes, when viewed from the stationary system, with the velocity $\sqrt{c^2-v^2}$gives us

\begin{displaymath}\frac{\partial\tau}{\partial y}=0, \frac{\partial\tau}{\partial z}=0. \end{displaymath}

Since $\tau$ is a linear function,

t(0,0) = 0 (emission)

t(80,16) = 8 (reflection)

t(60,20) = 16 (reception)

It is to be noted that t is not a linear function of two variables, there is a discontinuity at reflection. dt/dt one way does not equal dt/dt   the other.

it follows from these equations that

\begin{displaymath}\tau=a\left(t-\frac{v}{c^2-v^2}x'\right) \end{displaymath}

where a is a function $\phi(v)$ at present unknown, and where for brevity it is assumed that at the origin of k, $\tau =0$, when t=0.

It is to be noted that it follows in the argument below that  f(v) = 1, hence we easily determine the quantity dt = dt and dt/dt = 1.

With the help of this result we easily determine the quantities $\xi$, $\eta$, $\zeta$ by expressing in equations that light (as required by the principle of the constancy of the velocity of light, in combination with the principle of relativity) (which are irreconcilable)  is also propagated with velocity c when measured in the moving system (which it is not). For a ray of light emitted at the time $\tau =0$ in the direction (note direction) of the increasing $\xi$

\begin{displaymath}\xi=c\tau\ {\rm or}\ \xi=ac\left(t-\frac{v}{c^2-v^2}x'\right). \end{displaymath}

But the ray moves relatively to the initial point of k, when measured in the stationary system, with the velocity c-v (which is also supposedly measured as c in all frames of reference), so that

\begin{displaymath}\frac{x'}{c-v}=t. \end{displaymath}

And the ray moves relatively to the reflecting point of k, when measured in the stationary system, with the velocity c+v, so that

Since the time t for light to go each way is the same we then must have v = 0. BUT x' is infinitesimally small (= dx') and it will take an infinitesimal time for light to travel dx', hence dx'/(c-v) = dt or dx'/dt = c-v and dx'/dt  = c+v which is in agreement with the statement but not in agreement with the definition that the "time" required by light to travel from A to B equals the "time" it requires to travel from B to A.

If we insert this value of t in the equation for $\xi$, we obtain

\begin{displaymath}\xi=a\frac{c^2}{c^2-v^2}x'. \end{displaymath}

In an analogous manner we find, by considering rays moving along the two other axes, that

\begin{displaymath}\eta=c\tau=ac\left(t-\frac{v}{c^2-v^2}x'\right) \end{displaymath}

when

\begin{displaymath}\frac{y}{\sqrt{c^2-v^2}}=t,\ x'=0. \end{displaymath}

Thus

\begin{displaymath}\eta=a\frac{c}{\sqrt{c^2-v^2}}y\ {\rm and}\ \zeta=a\frac{c}{\sqrt{c^2-v^2}}z. \end{displaymath}

Substituting for x' its value, we obtain

\begin{eqnarray*} \tau & = & \phi(v)\beta(t-vx/c^2), \ \xi & = & \phi(v)\beta(x-vt), \ \eta & = & \phi(v)y, \ \zeta & = & \phi(v)z, \ \end{eqnarray*}

where

\begin{displaymath}\beta = \frac{1}{\sqrt{1-v^2/c^2}}, \end{displaymath}

and $\phi$ is an as yet unknown function of v. If no assumption whatever be made as to the initial position of the moving system and as to the zero point of $\tau$, an additive constant is to be placed on the right side of each of these equations.

We now have to prove that any ray of light, measured in the moving system, is propagated with the velocity c, if, as we have assumed, this is the case in the stationary system; for we have not as yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity (and no such proof has been forthcoming or presented here).

At the time $t=\tau=0$, when the origin of the co-ordinates is common to the two systems, let a spherical wave be emitted therefrom, and be propagated with the velocity c in system K. If (x, y, z) be a point just attained by this wave, then

x2+y2+z2=c2t2.

Transforming this equation with the aid of our equations of transformation we obtain after a simple calculation

\begin{displaymath}\xi^2+\eta^2+\zeta^2=c^2\tau^2. \end{displaymath}

The wave under consideration is therefore no less a spherical wave with velocity of propagation c when viewed in the moving system. This shows that our two fundamental principles are compatible. (Except it does not. The wave under consideration must return to the vague "origin of coordinates" to complete the proof. Unfortunately those origins no longer coincide). In the equations of transformation which have been developed there enters an unknown function $\phi$ of v, which we will now determine.

For this purpose we introduce a third system of co-ordinates ${\rm K}'$, which relatively to the system k is in a state of parallel translatory motion parallel to the axis of $\Xi$, such that the origin of co-ordinates of system ${\rm K}'$, moves with velocity -v on the axis of $\Xi$. At the time t=0 let all three origins coincide, and when t=x=y=z=0 let the time t' of the system ${\rm K}'$ be zero. We call the co-ordinates, measured in the system ${\rm K}'$, x', y', z', and by a twofold application of our equations of transformation we obtain

\begin{displaymath}\begin{array}{lllll} t' & = & \phi(-v)\beta(-v)(\tau+v\xi/c^2... ... z' & = & \phi(-v)\zeta & = & \phi(v)\phi(-v)z.\ \end{array}\end{displaymath}

Since the relations between x', y', z' and x, y, z do not contain the time t, the systems K and ${\rm K}'$ are at rest with respect to one another, and it is clear that the transformation from K to ${\rm K}'$ must be the identical transformation. Thus

\begin{displaymath}\phi(v)\phi(-v)=1. \end{displaymath}

We now inquire into the signification of $\phi(v)$. We give our attention to that part of the axis of Y of system k which lies between $\xi=0, \eta=0, \zeta=0$ and $\xi=0, \eta=l, \zeta=0$. This part of the axis of Y is a rod moving perpendicularly to its axis with velocity v relatively to system K. Its ends possess in K the co-ordinates

\begin{displaymath}x_1=vt,\ y_1=\frac{l}{\phi(v)},\ z_1=0 \end{displaymath}


and
 

\begin{displaymath}x_2=vt,\ y_2=0,\ z_2=0. \end{displaymath}

The length of the rod (or train) measured in K (the fence) is therefore $l/\phi(v)$; and this gives us the meaning of the function $\phi(v)$. From reasons of symmetry it is now evident that the length of a given rod (or train) moving perpendicularly to its axis, measured in the stationary system, must depend only on the velocity and not on the direction and the sense of the motion (velocity has direction). The length of the moving rod (or train) measured in the stationary system does not change, therefore, if v and -v are interchanged. Hence follows that $l/\phi(v)=l/\phi(-v)$, or

\begin{displaymath}\phi(v)=\phi(-v). \end{displaymath}

It follows from this relation and the one previously found that $\phi(v)=1$, so that the transformation equations which have been found become

\begin{eqnarray*}\tau & = & \beta(t-vx/c^2), \ \xi & = & \beta(x - vt), \ \eta & = & y, \ \zeta & = & z, \ \end{eqnarray*}

where

\begin{displaymath}\beta=1/\sqrt{1-v^2/c^2}. \end{displaymath}

 

The time for light to travel from A to B is equal to the time is takes to travel from B to A and the velocity of light is the same in all inertial frames of reference, 5 units per microsecond, which is clear and in agreement with experience therefore why we see trains move by peristalsis.

Gee, I hope I didn't make any errors, I so much want to understand relativity.

'Really, this is what is meant by the Fourth Dimension, though some people who talk about the Fourth Dimension do not know they mean it. It is only another way of looking at Time. There is no difference between Time and any of the three dimensions of Space except that our consciousness  moves along with it.' -- Herbert George Wells - "The Time Machine" - 1895.

"The secret to creativity is knowing how to hide your sources." --Einstein

Next, some real science:

  http://www.androcles01.pwp.blueyonder.co.uk/Algol/Algol.htm

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