An electron is shot towards a target that is negatively charged. While the electron is traveling, th

Why the log? Is it there to make the growth of the function slower?
As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\mathrm{\infty })$ to $(-\mathrm{\infty },\mathrm{\infty })$ (For a particle emitted along the beam axis after collision $\theta =0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

if $u$ and $u^{\prime }$ are a velocity referred to two inertial frames with relative velocity $v$ confined to the $x$ axis, then the quantities $l$, $m$, $n$ defined by
$(l,m,n)=\frac{1}{|u|}(u_x,u_y,u_z)$
and
$(l^{\prime },m^{\prime },n^{\prime })=\frac{1}{|u^{\prime }|}(u_x^{\prime },u_y^{\prime },u_z^{\prime })$
are related by
$(l^{\prime },m^{\prime },n^{\prime })=\frac{1}{D}(l-\frac{v}{u},m{\gamma }^{-1},n{\gamma }^{-1})$
and that this can be considered a relativistic aberration formula. The author gives the following definition for $D$, copied verbatim.
$D=\frac{u^{\prime }}{u}(1-\frac{u_{x}v}{c^2})={[1-2l\frac{v}{u}+\frac{v^2}{u^2}-(1-l^2)\frac{v^2}{c^2}]}^{\frac{1}{2}}$
Why is that better than the second expression?
Also, in case it's not clear, $\gamma =1/\sqrt{1-\frac{v^2}{c^2}}$ and $|u|=|(u_x,u_y,u_z)|=\sqrt{u_x^2+u_y^2+u_z^2}$

instead of assuming that the velocity $c$ is a maximal velocity, proving that while assuming $E=m{c}^2$.

Under the Lorentz transformations, quantities are classed as four-vectors, Lorentz scalars etc depending upon how their measurement in one coordinate system transforms as a measurement in another coordinate system.
The proper length and proper time measured in one coordinate system will be a calculated, but not measured, invariant for all other coordinate systems.
So what kind of invariants are proper time and proper length?

Environmentally induced decoherence makes wave function collapse unnecessary. But the environment, usually taken to be some heat bath, introduces a preferred frame. (That in which the total (spatial) momentum vanishes.) So, doesn't then the decoherence time depend on the motion of the prepared state relative to the environment? And, doesn't the ultimate environment, all particles in the universe, introduce a preferred frame into quantum mechanics in the sense that the decoherence time is relative to this frame? And would this be measureable, at least in principle

Einstein said that the synchronization of two clocks is dependant on the velocity of the observer. But I feel a conceptual contradiction can be made:
There are two observers A and B. Observer 'A' faces direction X, and will be labeled "stationary." Another observer B faces direction X and is travelling rapidly in that direction as well on a collision course with 'A'. Both observers are holding two clocks; One in each hand, with hands held perpendicular to direction X. Both observers hit a "synchronization" button on the clocks before colliding.
My expectation in this case is that when the moving observer B halts to greet observer A - both observers will agree the B-pair clocks are synchronized and the A-pair clocks are synchronized (though all four clocks are not necessarily synchronized with eachother, and is not at issue in this question).
Bottom line: what was considered "synchronicity" by the speedy individual is accepted by the stationary individual. This seems to contradict special relativity.
Now if my scenario were modified slightly, then I believe special relativity would apply. If observer A were to hold the A-pair clocks with one held out before herself and the other held out behind herself (instead of the original left and right perpendicular angles), then finally I would expect there to be disagreement between the two observers when they stop to meet eachother.
All this suggests that position is an unaccounted for primary component of relativity, but from what little I know of SR, it doesn't hardly factor in at all. Can someone explain what I'm missing?

If you have 2 flashlights, one facing North and one facing South, how fast are the photons (or lightbeams) from both flashlights moving away from one another?
Just adding speeds would yield 2C, but that's not possible as far as I know.
The reference frame here would be the place where the flashlights are and/or .The beams relative to one another.