... LT translation matches what the target frame agrees is true.
No, because some things which are true are frame dependent. c.f. Relativity of Simultaneity, Barn and Pole paradox, Twin paradox, etc.
So for example, given LT translates light event E to E’, other than by blind acceptance. how do we know E’ is actually the correct answer in the primed frame?
This pair of sentences proves that the SAP not only didn't peer-review the article, they didn't review it for proper use of English punctuation. The term "correct" has no definition here; a more proper phrasing in math is "self-consistent", while the correct term in physics is "consistent with observation." Banks uses neither definition.
Also, the passive Lorentz transform does not translate event E to event E', but rather converts the coordinates of event E to equally admissible coordinates given a different standard of rest. It's a change of the description of E, not a change of E itself.
SR does not lend itself well to such introspection,
A fucking lie.
Now, it is true that the translated light event E’ does measure $$c$$ and that can proven mathematically. However, there are many space-time coordinates that measure $$c$$ such as $$\left( x', y', z', \sqrt{x'^2+ y'^2+ + z'^2} / c \right)$$. (1)
So, the fact that E’ measures $$c$$ is not a sufficient condition to prove that the primed frame agrees E’ is the correct specific space-time coordinate for the circumstances.
Bullshit.
To prove LT is consistent with LP, you need to look at transformed
pairs of events. Let $$\mathcal{M}$$ be the set of all events, and $$\mathcal{L}$$ be the set of all admissible Lorentz transforms, then $$\mathcal{M}^2$$ is the set of all ordered pairs of events and let
$$\mathcal{C} = \left\{ m \, | \, m \in \mathcal{M}^2 \wedge 0 \leq t(m_2) - t(m_1) \wedge c^2 \left( t(m_2) - t(m_1) \right)^2 = \left( x(m_2) - x(m_1) \right)^2 + \left( y(m_2) - y(m_1) \right)^2 + \left( z(m_2) - z(m_1) \right)^2 \right\}$$
be the set of all ortho-chronous light-like pairs of events, where $$t,x,y,z$$ are functions that give the coordinates for an event.
Then for a given Lorentz transform $$\Lambda \in \mathcal{L}$$, we have $$t', x', y', z'$$ which are non-singular
linear functions of $$t,x,y,z$$. Thus the inverse Lorentz transform exists is also a Lorentz transform, $$\Lambda^{-1} \in \mathcal{L}$$. And it follows from linearity that
$$\Delta t'(m) = t'(m_2) - t'(m_1) = \left( \Lambda\left( t(m_2), x(m_2) , y(m_2) , z(m_2) \right) \right)_t - \left( \Lambda\left( t(m_1), x(m_1) , y(m_1) , z(m_1) \right) \right)_t
\\ = \left( \Lambda\left( t(m_2) - t(m_1), x(m_2) - x(m_1), y(m_2) -y(m_1), z(m_2) -z(m_1) \right) \right)_t = \left( \Lambda\left( \Delta t(m), \Delta x(m) , \Delta y(m) , \Delta z(m) \right) \right)_t $$.
So only because the Lorentz transform is linear that we can we, without loss of generality, observe that the conditions which renders a pair of events as ortho-chronous, $$ 0 \leq t(m_2) - t(m_1) $$, and light-like, $$ c^2 \left( t(m_2) - t(m_1) \right)^2 = \left( x(m_2) - x(m_1) \right)^2 + \left( y(m_2) - y(m_1) \right)^2 + \left( z(m_2) - z(m_1) \right)^2$$, are expressible in solely in terms of coordinate differences for coordinate pairs, $$ 0 \leq \Delta t(m)$$ and $$c^2 \left( \Delta t(m) \right)^2 = \left( \Delta x(m) \right)^2 + \left( \Delta y(m) \right)^2 + \left( \Delta z(m) \right)^2 $$,
and that the Lorentz transform operates equally well on coordinate differences as coordinates. From these two conditions, it is possible to prove that LP and LT are consistent by working with fixed event O, such that $$t(O) = x(O) = y(O) =z(O) = 0$$ as the first of the two events. This induces a subset of $$\mathcal{M}^2$$ is naturally isomorphic to $$\mathcal{M}$$ and is identified by E. $$\Delta \mathcal{M}^2 \sim \mathcal{M}, \; \Delta m \sim E, \; \Delta t(m) \sim t(E), \; $$ etc.
$$\Delta \mathcal{C} \sim \left\{ E \, | \, E \in \mathcal{M} \wedge 0 \leq t(E) \wedge c^2 \left( t(E) \right)^2 = \left( x(E) \right)^2 + \left( y(E) \right)^2 + \left( z(E) \right)^2 \right\} $$.
So from the existence of inverses, and linearity, we can prove that LT and LP are consistent by proving:
Lemma 1: $$\vdash \forall \Lambda \in \mathcal{L} \; \forall E \in \mathcal{M} \; c^2 \left( t'(E) \right)^2 - \left( x'(E) \right)^2 - \left( y'(E) \right)^2 - \left( z'(E) \right)^2 = c^2 \left( t(E) \right)^2 - \left( x(E) \right)^2 - \left( y(E) \right)^2 - \left( z(E) \right)^2$$
Lemma 2: $$\vdash \forall \Lambda \in \mathcal{L} \; \forall E \in \mathcal{M} \; c^2 \left( t(E) \right)^2 \geq \left( x(E) \right)^2 + \left( y(E) \right)^2 + \left( z(E) \right)^2 \rightarrow \textrm{sgn} \left( t(E) \right) = \textrm{sgn} \left( t'(E) \right)$$
the latter of which can be decomposed into:
Lemma 3: $$\vdash \forall \Lambda \in \mathcal{L} \; \forall E \in \mathcal{M} \; c^2 \left( t(E) \right)^2 \geq \left( x(E) \right)^2 + \left( y(E) \right)^2 + \left( z(E) \right)^2 \rightarrow \left( t(E) = 0 \Leftrightarrow t'(E) = 0 \right) $$
Lemma 4: $$\vdash \forall \Lambda \in \mathcal{L} \; \forall E \in \mathcal{M} \; c^2 \left( t(E) \right)^2 \geq \left( x(E) \right)^2 + \left( y(E) \right)^2 + \left( z(E) \right)^2 \rightarrow t(E) \times t'(E) \geq 0 $$
This is not the approach Banks uses, because he is not a scholar of SR or mathematics.
This entire analysis depends on the ability to create common agreed upon circumstances for both frames. This can be achieved by considering configurations of the two coordinate systems. So, an experiment is designed below such that the start of the experiment is the configuration where the two origins of the frames are common and a light pulse is emitted from the common origins in the positive direction of the common x-axis. The end of the experiment occurs when the primed coordinate $$\left( -v/c, 0, 0\right) $$ is co-located with the unprimed frame origin.
This is ridiculous language. That the coordinate of $$O$$ are given as $$t(O) = x(O) = y(O) =z(O) = 0$$ implies via linearity of the Lorentz transform that $$t'(O) = x'(O) = y'(O) =z'(O) = 0$$.
Instead of working with all Lorentz transforms, Banks limits us to the case $$0 \lt v \lt c$$, $$t' = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} t - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \frac{v}{c^2} x$$, $$ x' = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} x - \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} v t$$, $$y' = y'$$, $$z' = z$$. Instead of working with 4 space-time coordinates, Banks describes the "motion" of spatial "points". So it's left to the reader to puzzle out what does he mean the "when" of the end of the experiment, because he doesn't describe a time, but an event K with coordinates
$$ x(K) = y(K) =z(K) = y'(K) =z'(K) = 0, x'(K) = -v/c$$ the last of which is not even a measure of length. (This is more evidence that this article was not peer reviewed.)
This is consistent only with $$t(K) = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}, t'(K) = \frac{1}{c}$$. (because the position was not given in units of length, the time is not given in units of distance)
At the end of the experiment,
As calculated, that's TWO DIFFERENT TIMES. The events where t(H) = t(K) is not the same set of events where t'(H) = t'(K). That's basic relativity of simultaneity.
the light pulse is located at some unique position on the positive x-axis of the primed frame coordinates.
Uniqueness is not possible as the x-axis is parallel to the motion characterizing the Lorentz transform. Thus even in section 1, all we have in this paper is are deceptions built on misunderstandings and willful ignorance.
So we have two events, P and Q, where Banks falsely assumes there is one.
$$t(P) = t(K) = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}, x(P) = \sqrt{1 - \frac{v^2}{c^2}}, t'(P) = \left(1 - \frac{v}{c}\right) \frac{1}{c} , x'(P) = 1 - \frac{v}{c}, y(P) =z(P) = y'(P) =z'(P) = 0$$
$$t'(Q) = t'(K) = \frac{1}{c}, x'(Q) = 1, t(Q) = \frac{ 1 + \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} \frac{1}{c}, x(Q) = \frac{ 1 + \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}, y(Q) =z(Q) = y'(Q) =z'(Q) = 0$$
In the primed frame, the light’s position is determined by LP. In the unprimed frame, the light’s position is first determined by LP and then translated to primed frame coordinates by LT. If the result of LT does not match the primed frame LP result then LT got the answer wrong.
No, that's just Banks confusing relativity of simultaneity for talking about the light postulate.
In the unprimed coordinates K and P happen at the same time. In the primed coordinated, K and Q happen at the same time. That's because the definition of "at the same time" is coordinate-sensitive when events don't happen both at the same place and same time.
In Section 2, Banks calculates $$Q = O_1$$ and $$P = O_2$$ but never notices the effect of relativity of simultaneity.
Buffalo chips on toast! First I've heard of that one!
