Emmy Noether

If there's no ether, Emmy, my friend, why light speed $c$ as a particle speed limit?

Entanglement!  Isn't that a real drag!

Unfortunately...I don't have the number of particles in the whole universe.  I do have however, the largest particle in the whole universe (post date 28 Sep 2014 "Primero Uno, Number One").  This is probably at the center of the Universe.  From which we can find the total number of particles in the Universe.  And so light speed,

$c^{ 2 }=A(\cfrac { 2^{ N } }{ 3 } -1)$

where velocity $c$, squared is proportional to the number of particles entangled and $A$ a constant.

The total number of particles after $N$ division is,

$N_p=2^{ N }$

Since the number of entangle pair is $3$, the number of particles entangled is,

$N_t=\cfrac { 2^{ N } }{ 3 }$

Given any singular particle, the number of particles entangled with it is,

$N_t-1=\cfrac { 2^{ N } }{ 3 }-1$

The expression is not useful, with the unknown constant $A$.

However, from an energy perspective, energy imparted on a particle, $mv^2$ is shared by $\cfrac { 2^{ N } }{ 3 }$ other particles.   The energy of one particle is,

$E_p=\cfrac { 3 }{ 2^{ N } }mv^{ 2 }$

The loss in energy, $E_{loss}$ for one particle is,

$E_{ loss }=mv^2-E_p=(1-\cfrac { 3 }{ 2^{ N } } )mv^{ 2 }$

Since this particles exist in the time dimension $t_x$ or $t$, and are travelling along time $t$, the hypothetical force resulting in $E_{loss}$ is given by,

$F_{drag}=- \cfrac { dE_{ loss } }{ dt } =-(1-\cfrac { 3 }{ 2^{ N } } )m2v\cfrac { dv }{ dt }$

$F_{drag}=-2v(1-\cfrac { 3 }{ 2^{ N } } )m\cfrac { dv }{ dt }$

If, $F=m\cfrac { dv }{ dt }$ is the applied force on the particle,

$F_{drag} =-2v(1-\cfrac { 3 }{ 2^{ N } } )F$

The negative sign reflects that the force resulting in energy loss act against $F$.

At terminal velocity, $v=c$, the applied force equals the force due to drag or the force resulting in energy loss,

$\sum F=F+F_{drag}=F-\cfrac { dE_{ loss } }{ dt } =F-2(1-\cfrac { 3 }{ 2^{ N } } )Fv=0$

This implies,

$v=c=\cfrac { 1 }{ 2 } (1-\cfrac { 3 }{ 2^{ N } } )^{ -1 }$

$c=\cfrac { 2^{ N-1 } }{ 2^{ N }-3 }$

which is known given $N$, exactly.

And I just derived light speed!  More correctly given the number of entangled particles $N_t$,

$N_t=\cfrac { 2^{ N } }{ 3 }$

not assuming that all particles that can entangle are entangled,

$c=\cfrac { 1 }{ 2 } (1-\cfrac { 1 }{ { N_t } } )^{ -1 }$

$c=\cfrac { 1 }{ 2 }(\cfrac{N_t}{N_t-1})$

Unfortunately,

$299792458=\cfrac{1}{2}(\cfrac{N_t}{N_t-1})$

$N_t=1.0000000016678205028$

Which is odd and suggests that entanglement is not sharing energy equally, but instead, an average fraction $\bar\alpha$ is shared, such that, given light speed,

$\bar\alpha N_t=1.0000000016678205028$

where $\bar\alpha$ is the mean of a probability distribution of the extend of entanglement, $\alpha$ given a population of entangled particles.  This way all particles are entangled and is subjected to the same light speed limit, but among themselves they do not share energy equally.

It is also possible that $\alpha$ is a constant that dictates the amount of sharing among all entangled particles.  For that matter, $\alpha$ or $\bar\alpha$ can be different for different types of entangled particles pairs.  This raises the possibility of three different light speeds for three types of entangled particles pairs.

Entanglement and $\alpha$ provides a possible reason for light speed limit without a pervasive medium, ether.  See you at the center of the Universe.