Intrinsic Antagonistic Inconsistency in the Special Relativity

March 2020
Jan Onderco

Abstract

A train embankment thought experiment is being used to explain the intrinsic antagonistic inconsistency of the Special Relativity (SR).
$L’=3.4641cs’$ long train car moves at constant speed $v=0.866c$ through a train station. The Lorentz factor $\gamma=2$ between the train car reference frame $S’$ and a station platform reference frame $S$. A train conductor at the front of the train car travels along the train station platform that is $L=1.732cs$ long in the platform reference frame and reaches the end of the station platform. The platform observer at the end of the platform is aligned with the front of the train car and the front conductor. This is an event $A$ where $x=0cs=x’=0cs’$ and when $t=0s=t’=0s’$. The second train conductor at the end of the train car has synchronized clocks with the front conductor.
The first question – what is the $x$ position and what is the time $t$ on the ground that the second conductor at the end of the train car sees outside at train car time $t’=0s’$? The Lorentz Transformation (LT) gives us the answer $x=-6.9282cs$ and the time is $t=-6s$. This is an event $B$.
The second question – what is the $x$ position and what is the time $t’$ when the second conductor at the end of the train car sees time $t=0s$ on the platform? The LT gives us the answer again $x=-1.732cs$ and the time on the train is $t’=3s’$. This is an event $C$ at the beginning of the station platform aligned with the end of the train car.
The third question – what kind of Special Relativity ‘fairy tale magic’ will shrink the back of the train car from the event $B$ to the event $C$ when the front of the train car is not moving, it is ‘stuck in a moment’ of the event $A$?
There is no ‘magic’, the events $A, B, C$ create an ‘impossibility triangle’ based on the current understanding of the SR; the SR as we know it is inherently broken!

Introduction

The abstract is a half a pager that is enough to take down the Special Relativity. Lorentz transformation calculations, one figure and one space-time diagram show the Special Relativity problem in a very clear light.

The Lorentz transformation calculations

The following calculations are straight forward. The two inertial reference frame systems are the ground $S$ and the moving train car $S’$. The train car moves at constant speed $v=0.866c$ so the Lorentz factor$^{[1]}$ is:
$\gamma=1/\sqrt{1-\frac{v^2}{c^2}}=2$

The Lorentz transformation$^{[1]}$ between the $S$ and $S’$ frames are:

$t=\gamma(t’+(v/c^2)x’)$
$x=\gamma(x’+vt’)$
$y=y’$
$z=z’$
$t’=\gamma(t-(v/c^2)x)$
$x’=\gamma(x-vt)$
$y’=y$
$z’=z$

The train car length is $L’=3.4641cs’$ in the $S’$ frame.
The first question – what is the $x$ position and what is the time $t$ on the ground that the second conductor at the end of the train car sees outside at train car time $t’=0s’$?

$t=\gamma(t’+(v/c^2)x’)=2(0+(0.866c/c^2)(-3.4641cs’)=2*0.866*(-3.4641)s’=-6s$
$x=\gamma(x’+vt’)=2*(-3.4641cs’+0.866c*0s)=-6.9282cs$

This is the event $B$.

The second question – what is the $x$ position and what is the time $t’$ when the second conductor at the end of the train car sees time $t=0s$ on the platform?

The moving train car is contracted due to the Lorentz contraction.
The solution of the equation $x’=\gamma(x-vt)$ gives us
$x=x’/\gamma=-3.4641cs’/2=-1.732cs$

for $t=0s$.

Now we can calculate time $t’$.
$t’=\gamma(t-(v/c^2)x)=2(0-(0.866c/c^2)(-1.732cs)=2*0.866*1.732s=3s’$

This is the event $C$.

The third question – what kind of Special Relativity ‘fairy tale magic’ will shrink the back of the train car from the event $B$ to the event $C$ when the front of the train car is not moving, it is ‘stuck in a moment’ of the event $A$?

Figure 1: The ‘impossibility triangle’ of the events $A, B, C$ based on the current understanding of the Special Relativity.

Figure 2: The space-time diagram of the events $A, B, C$ – the ‘impossibility triangle’ based on the current understanding of the Special Relativity.

Conclusion

The Special Relativity is undefendable.

Feedback

The content of the paper is crazy, arrogant, ridiculous and it demonstrates lack of understanding of the Special Relativity simultaneity by the author.‘ a Special Relativity expert.

Yes! The paper is outrageous and it is meant to be in order to shake the whole relativity world!

What does stuck in a moment mean?‘ question asked by my friend Michael.

Simultaneity and ‘stuck in a moment’

Michael, Einstein’s famous 1905 paper$^{[2]}$ starts with the definition of the Simultaneity. Imagine a grid of inertial observers that are stationary to each other. Each observer has his own atomic clocks with a counter displaying clocks’ ticks. You are at the front of the train car and I am at the end of the train car, seating, stationary $3.4641cs$ apart and we have an agreement. You sent a light beam towards me and you record the number of ticks on the clock display when you sent the beam. I reflect the beam straight back to you and you receive it. You record the number of ticks on the clock display when you received the beam back. I recorded the number of ticks at the time of the reflection. Our agreement is resetting the clocks. I change my clock counter in such a way that the time of the reflection is $0s’$ and you change your clock so your $0s’$ is in the middle of the emission, reception time interval. Our zeros identify simultaneity on our respective time axes. There is nothing special about $s’$ – “second prime”. It measures the time in the moving train car reference frame. The single $s$ measures time in the platform reference frame. This is done for better readability.

Essentially, the simultaneity means one time slice through space. The $x$ axis in the Figure 2 represents that for the platform reference frame and $x’$ represents one slice of time for the train car reference frame. The shortest time interval based on the current physics calculations is the Planck time interval $5.391^{-44}s$$^{[4]}$. The simultaneity time slice interval means there is no $dt$. The clocks are stuck, frozen. There is no Planck time tick. The space is stuck in a moment.

It follows from the LT that any tiny $dt$ has a corresponding $dt’$. This works both ways. If there is not $dt$ then there is no $dt’$!!!
The event $A$ in Figure 2 where $x=0cs=x’=0cs’$ and when $t=0s=t’=0s’$ can be true only and only when both frames are frozen, stuck in a moment. We cannot have one frame frozen and the other moving.
The relativity of simultaneity is a logical fallacy!!!
Why??? Because $B, C$ events are bi-located in space and time and the event $A$ cannot be true for both of them. When the second train conductor at the end of the train car has time $t’=0s’$ then the conductor sees $t=-6s$ in the platform frame. The platform frame grid of inertial observers sees time $t’=-3s’$ on the front conductor clock when $t=-6s$. This invalidates the initial condition $t=0s=t’=0s’$. This is a catch 22, the logical contradiction.

Figure 3: The space-time diagram of the events $A, B, C$. The full black horizontal lines represent train positions as seen from the platform frame at times $-6s, -4s, -2s, 0s$.

Feedback 2

Stuck in a moment? Isn’t it a song by U2 or something? Pure non-sense.
The author just defined an absolute space. The absolute space died in 1905.
… a couple of comments by relativity experts.

Yes, I did define the absolute space. The absolute space is well, it is the Special Relativity as we know it that is dying though. Einstein himself poisoned the SR from the start and he even did not know about it.

One other thing which would be nice to clarify is the length of the train and a platform in their respective frames.‘ a request from my friend Eugene.

Eugene, the length contraction is reciprocal the same way as the time dilation is reciprocal, the two sides of the same coin. This is the reason why the conductor at the back of the train car sees $x=-6.9282cs$ when clocks are synchronized on the train where $x=0cs=x’=0cs’$ and when $t=0s=t’=0s’$. This is the length contraction from $x=-6.9282cs$ on the platform to $x’=3.4641cs’$ on the train car frame. The LT length contraction is reciprocal as well. The reciprocity is a contradiction. The SR fought the contradiction with the relativity of the simultaneity. This is proven to be illogical above. There is nothing that can defend the SR.
The Special Relativity is undefendable.

What is that ‘kill pill’ of the Special Relativity?

The Special Relativity kill pill

Einstein says: ‘… that light is always propagated in empty space with a definite velocity $c$ which is independent of the state of motion of the emitting body.$^{[5]}$
Einstein went on to derive the Lorentz Transformation that encapsulates the light propagation frame dragging. This is the SR killer.

Figure 4: The top view of the moving train car.

The red arrow up is crossing y-direction at speed c in the train frame
… and because the light beam is frame dragged then the y-component speed in the platform frame is c/2.

Copy and paste, replacing red and blue, changing frames.

The blue arrow up is crossing y-direction at speed c in the platform frame
… and because the light beam is frame dragged then the y-component speed in the train frame is c/2.

There is no length contraction issue. The red arrow moves along $x’=0cs’$, it leads straight to time dilation definition.
The same goes for the blue arrow up, it moves along $x=0cs$ and it leads straight to time dilation definition in the reciprocal way.

Figure 5: The space-time diagram showing the reciprocity contradiction.

$AB$ – red arrow up on the train -> $1cs’$ distance travelled in $1s’$.
$AE$ – blue arrow to the right -> $2cs$ distance travelled in $2s$ – $AB$ time dilation.
$AC$ – blue arrow up on the platform -> $1cs$ in $1s$.
$AD$ – red arrow to the left -> $2cs’$ in $2s’$ – $AC$ time dilation.
$AE$ – blue arrow up and down on the platform -> $2cs$ in $2s$ – the full round trip on the platform.
$AF$ – $4cs’$ in $4s’$ – time dilation of the full platform round trip $AE$.

The reciprocal time dilation and the length contraction are two sides of the intrinsic antagonistic inconsistency coin of the Special Relativity.

The last nail to the Special Relativity coffin is the following statement.
A light beam roundtrip in any direction of an inertial reference frame will appear time dilated in any other moving inertial reference frame. The Figures 6, 7 capture one example of such a roundtrip.

Figure 6: The top view of the moving train car.

Figure 7: The space-time diagram showing the roundtrip.

The $AB$ events are $2s$ apart in the platform frame but only $1s’$ in the train car frame.
The $BG$ events are $2s$ apart in the platform frame but $7s’$ in the train car frame.
The $4s$ roundtrip in the platform frame takes $8s’$ in the train car frame.

Conclusion 2

The same as the above… undefendable!

References

[1] Lorentz Transformation from http://hyperphysics.phy-astr.gsu.edu/hbase/Relativ/ltrans.html#c2

[2] ON THE ELECTRODYNAMICS OF MOVING BODIES by Albert Einstein, pages 2, 3 from http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf

[3] Atomic Clock from https://en.wikipedia.org/wiki/Atomic_clock#Optical_clocks

[4] Planck Units from https://en.wikipedia.org/wiki/Planck_units

[5] ON THE ELECTRODYNAMICS OF MOVING BODIES by Albert Einstein, page 1 from http://hermes.ffn.ub.es/luisnavarro/nuevo_maletin/Einstein_1905_relativity.pdf