This peer reviewed article claims to force SR into a contradiction. can anyone here stop that? http://article.sapub.org/10.5923.j.ijtmp.20160602.02.html
What is the basis for expecting LT and LP to be uncontradictory? Do you have a recommended link that describes the Light Postulate in layman terms?
I think it's easy to stop this one. The submitted paper has obviously not been peer-reviewed. Not by any physicists at least. It's full of basic errors, it contradicts my peer-reviewed textbooks, and finally it looks very much like the kind of posts chinglu has submitted here and on other forums, over the years, claiming he's found a problem with SR. The link is to what looks superficially like a genuine scientific journal, but isn't. It's a vanity publishing site trying to attract real papers to give itself an aura of authenticity. In short, it's buffalo chips on toast.
Please Register or Log in to view the hidden image! Buffalo chips on toast! First I've heard of that one! ps: Notice how those that seem to have a problem or making some unsupported claim re any mainstream theory, particularly SR and/or GR, are always into demands, arrogance and childish petulance and delusions of grandeur. I near pissed myself laughing at his reply in the other thread..... Please Register or Log in to view the hidden image!
Yeah, I'm pretty sure that chinglu has posted this before. It's sadly wrong. It asks, "how[sic] do we know E’ is actually the correct answer in the primed frame?" The answer is that we know this by definition of frame of reference and the choice of light as a means of synchronization. That the author thinks that SR is not amenable to mathematical introspection means that the author has some form of cognitive disability. Like many SR cranks, the author simply cannot do the mathematics and consider the relativity of simultaneity and the change of spatial coordinates.
Agreed. SAP appears on Beall's List, which, as some readers may know, is a guide to supposedly scientific publications that apply questionable quality control: https://scholarlyoa.com/publishers/ So the OP's claim that it has been "peer-reviewed" cannot be taken at face value.
Also, Andrew Banks is a well-known crank. http://vixra.org/abs/1303.0144 chinglu has been fooled by Andrew Banks before. http://www.sciforums.com/threads/special-relativity-is-refuted.109190/ http://www.sciforums.com/threads/sr-is-dead.110037/ http://www.sciforums.com/threads/new-article-shows-a-fatal-math-error-in-sr.135674/ http://www.sciforums.com/threads/apparant-2-directions.135721/ chinglu has been very wrong on simple SR topics before: http://www.sciforums.com/threads/time-dilation.105498/ http://www.sciforums.com/threads/sr-issue.141840/ http://www.sciforums.com/threads/prove-sr-is-consistent.142427/
No, because some things which are true are frame dependent. c.f. Relativity of Simultaneity, Barn and Pole paradox, Twin paradox, etc. 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. A fucking lie. 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 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) 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. 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. \(t(P) = t(K) = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}, x(P) = \sqrt{1 - \frac{v^2}{c^2}}, t'(P) = \left(1 - \frac{v}{c}\right) \frac{1}{c} , x'(P) = 1 - \frac{v}{c}, y(P) =z(P) = y'(P) =z'(P) = 0\) \(t'(Q) = t'(K) = \frac{1}{c}, x'(Q) = 1, t(Q) = \frac{ 1 + \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} \frac{1}{c}, x(Q) = \frac{ 1 + \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}}, y(Q) =z(Q) = y'(Q) =z'(Q) = 0\) 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.
Limiting the form of the Lorentz transform makes this easy. If we have \(t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}}, \; x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}}, \; y' = y, \; z' = z\) with \( -c \lt v \lt c\) then it follows that \(c^2 t'^2 - x'^2 - y'^2 - z'^2 \\ \quad \quad \quad \quad = c^2 \frac{ \left( t - \frac{v}{c^2} x \right)^2 }{1 - \frac{v^2}{c^2}} - \frac{ \left( x - v t \right)^2 }{1 - \frac{v^2}{c^2}} - y^2 - z^2 \\ \quad \quad \quad \quad = \frac{ \left( c t - \frac{v}{c} x \right)^2 }{1 - \frac{v^2}{c^2}} - \frac{ \left( x - v t \right)^2 }{1 - \frac{v^2}{c^2}} - y^2 - z^2 \\ \quad \quad \quad \quad = \frac{ c^2 t^2 - 2 v t x + \frac{v^2}{c^2} x^2 }{1 - \frac{v^2}{c^2}} - \frac{ x^2 - 2 v t x + v^2 t^2 }{1 - \frac{v^2}{c^2}} - y^2 - z^2 \\ \quad \quad \quad \quad = \frac{ c^2 t^2 + \frac{v^2}{c^2} x^2 }{1 - \frac{v^2}{c^2}} - \frac{ x^2 + v^2 t^2 }{1 - \frac{v^2}{c^2}} - y^2 - z^2 \\ \quad \quad \quad \quad = \frac{ c^2 t^2 - v^2 t^2 }{1 - \frac{v^2}{c^2}} - \frac{ x^2 - \frac{v^2}{c^2} x^2 }{1 - \frac{v^2}{c^2}} - y^2 - z^2 \\ \quad \quad \quad \quad = \frac{ 1 - \frac{v^2}{c^2} }{1 - \frac{v^2}{c^2}} c^2 t^2 - \frac{ 1 - \frac{v^2}{c^2} }{1 - \frac{v^2}{c^2}} x^2 - y^2 - z^2 \\ \quad \quad \quad \quad = c^2 t^2 - x^2 - y^2 - z^2 \) which proves Lemma 1. An obvious corollary is \(c^2 t'^2 - x'^2 - y'^2 - z'^2 \geq 0\) iff \(c^2 t^2 - x^2 - y^2 - z^2 \geq 0\). Thus if \(c^2 t^2 \geq x^2 + y^2 + z^2 \) and \(t = 0\) then it follows that \( x^2 + y^2 + z^2 = 0\) thus \(x=y=z=0\) and by linearity \(t' = x' = y' = z' = 0\). Likewise, if \(c^2 t^2 \geq x^2 + y^2 + z^2 \) and \(t' = 0\) then it follows that \( x'^2 + y'^2 + z'^2 = 0\) thus \(x'=y'=z'=0\) and by linearity \(t = x = y = z = 0\). Thus Lemma 3 is proven. Next if \(c^2 t^2 \geq x^2 + y^2 + z^2 \) it follows that \(|c t| \geq | x |\). By assumption \(0 \leq |v /c| \lt 1\). Thus \( |c t| \geq |x v / c| \), thus \( | t | \geq \left| \frac{v}{c^2} x \right|\), thus \(t \times \left( t - \frac{v}{c^2} x \right) \geq 0\). Thus \( t \times t' \geq 0\) and Lemma 4 is proven. Lemma 2 follows from Lemmas 3 & 4 trivially by definition of sgn.
Lemma 1 is nearly equally straightforward to prove with \(0 < | \vec{v} | < c \) (while the case \(\vec{v} = 0 \) is necessarily trivial) with the Lorentz transformation in the form: \( t' = \frac{1}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}} \left( t - \frac{1}{c^2} \vec{v} \cdot \vec{x} \right) \\ \vec{x}' = \left[ \vec{x} - \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } \vec{v} \right] + \frac{1}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}} \left[ \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } - t \right] \vec{v} \) First, since \( \left[ \vec{x} - \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } \vec{v} \right] \cdot \vec{v} = \vec{x} \cdot \vec{v} - \vec{v} \cdot \vec{x} = 0\) we see the expression for \( \vec{x}' \) is a partition into components orthogonal and parallel to \(\vec{v}\) which makes the squaring simpler. Thus: \(c^2 t'^2 - \vec{x}'^2 \\ \quad \quad = \frac{ \left( c t - \frac{1}{c} \vec{v} \cdot \vec{x} \right)^2 }{ 1 - \frac{\vec{v}^2}{c^2} } - \left( \left[ \vec{x} - \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } \vec{v} \right] + \frac{1}{\sqrt{1 - \frac{\vec{v}^2}{c^2}}} \left[ \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } - t \right] \vec{v} \right)^2 \\ \quad \quad = \frac{ \left( c t - \frac{1}{c} \vec{v} \cdot \vec{x} \right)^2 }{ 1 - \frac{\vec{v}^2}{c^2} } - \left[ \vec{x} - \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } \vec{v} \right]^2 - \frac{1}{1 - \frac{\vec{v}^2}{c^2}} \left[ \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } - t \right]^2 \vec{v}^2 \) \( \\ \quad \quad = \frac{ c^2 t^2 - 2 t \vec{v} \cdot \vec{x} + \frac{1}{c^2} \left(\vec{v} \cdot \vec{x} \right)^2 }{ 1 - \frac{\vec{v}^2}{c^2} } - \left[ \vec{x}^2 - 2 \frac{ \vec{v} \cdot \vec{x} }{ \vec{v}^2 } \vec{x} \cdot \vec{v} + \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^4 } \vec{v}^2 \right] - \frac{ \vec{v}^2 }{1 - \frac{\vec{v}^2}{c^2}} \left[ \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^4 } - \frac{ 2 t \vec{v} \cdot \vec{x} }{ \vec{v}^2 } + t^2 \right] \) \( \\ \quad \quad = \frac{ c^2 t^2 - 2 t \vec{v} \cdot \vec{x} + \frac{\vec{v}^2}{c^2} \frac{\left(\vec{v} \cdot \vec{x} \right)^2}{\vec{v}^2} - \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } + 2 t \vec{v} \cdot \vec{x} - \frac{\vec{v}^2}{c^2} c^2 t^2 }{ 1 - \frac{\vec{v}^2}{c^2} } - \vec{x}^2 + 2 \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } - \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } \) \( \\ \quad \quad = \frac{ c^2 t^2 + \frac{\vec{v}^2}{c^2} \frac{\left(\vec{v} \cdot \vec{x} \right)^2}{\vec{v}^2} - \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } - \frac{\vec{v}^2}{c^2} c^2 t^2 }{ 1 - \frac{\vec{v}^2}{c^2} } - \vec{x}^2 + \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } \) \( \\ \quad \quad = \frac{ c^2 t^2 - \frac{\vec{v}^2}{c^2} c^2 t^2 + \frac{\vec{v}^2}{c^2} \frac{\left(\vec{v} \cdot \vec{x} \right)^2}{\vec{v}^2} - \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } }{ 1 - \frac{\vec{v}^2}{c^2} } - \vec{x}^2 + \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } \) \( \\ \quad \quad = c^2 t^2 - \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } - \vec{x}^2 + \frac{ \left( \vec{v} \cdot \vec{x} \right)^2 }{ \vec{v}^2 } \\ \quad \quad = c^2 t^2 - \vec{x}^2 \) An obvious corollary is \(c^2 t'^2 - \vec{x}'^2 \geq 0\) iff \(c^2 t^2 - \vec{x}^2 \geq 0\). Thus if \(c^2 t^2 \geq \vec{x}^2 \) and \(t = 0\) then it follows that \( \vec{x}^2 = 0\) thus \(\vec{x}=0\) and by linearity \(t' = 0, \vec{x}' = 0\). Likewise, if \(c^2 t^2 \geq \vec{x}^2 \) and \(t' = 0\) then it follows that \( \vec{x}'^2 = 0\) thus \(\vec{x}'=0\) and by linearity \(t = 0, \vec{x} = 0\). Thus Lemma 3 is proven in the more general case. Next if \(c^2 t^2 \geq \vec{x}^2 \) it follows that \(|c t| \geq | \vec{x} |\). By assumption \(0 \lt | \vec{v} /c| \lt 1\). Thus \( |c t| \geq |\vec{v} \cdot \vec{x} / c| \), thus \( | t | \geq \left| \frac{ \vec{v} \cdot \vec{x}}{c^2} \right|\), thus \(t \times \left( t - \frac{\vec{v} \cdot \vec{x}}{c^2} \right) \geq 0\). Thus \( t \times t' \geq 0\) and Lemma 4 is proven in the more general case. As before, lemma 2 follows from Lemmas 3 & 4 trivially by definition of sgn.
[/QUOTE] Rpenner, you do not understand how to prove SR is consistent. LT claims when it maps a light position to the other frame that mapping produces what the other frame says is the true position for the light given circumstances. Where you have gotten yourself confused is you accept LT as producing that truth without proving it. From there, you produced lots of irrelevant equations. Now, the paper demonstrates an experiment that shows LT’s calculated position for the light is different from what the LP in the target frame says is true. NO THEORY IS ALLOWED TO PRODUCE ANYTHING THAT SAYS ITS POSTULATES ARE FALSE. But, the paper shows LT does claim the LP is wrong in the target frame. That is a contradiction. Now, you mentioned the relativity of simultaneity (ROS) as an antiseptic for this contradiction in SR. ROS’s claims cannot override the fact that SR contradicts itself. So ROS is completely irrelevant. So, your task is to prove that LT gives the correct answer for the light position in the other frame given C’ and M are co-located. It is a simple math problem. Can you perform that easy math yes or no?
[Mod note: chinglu has now been twice warned about asserting untruths in math and physica on the main science forums. Continuing to do so, even in replies, will generate more warning points and moderator action.] I'm not claiming to prove that Special Relativity (SR) is entirely self-consistent but merely that the Lorentz transform (LT) is entirely consistent with the postulate that if one frame describes light moving from event A to event B, then so do all other frames (LP) where an admissible homogeneous ( linear) Lorentz transform describes the relation between how events are assigned coordinates in one frame and how events are assigned coordinates in another. Lemma 1 (combined with the observations that the Lorentz transform is homogeneous and that the definition of light-like depends only on differences of Cartesian coordinates) says that if one frame says the space-time relationship between events A and B are light-like, then all do. Lemma 2 says similarly that if one frame says the space-time relationship between events A and B are light-like OR time-like then all frames agree on the order in which A & B happen. The LT makes no such claim. The LT is a transformation of space-time coordinates for all events. SR is what is saying that if the space and time coordinates of any event are valid to do physics with, then so are the transformed coordinates. Since it is a transform not only of position, but of time as well, Banks fails to account for this and produces two results by neglecting to take into account Relativity of Simultaneity (RoS). Because Banks fails to deal with the ambiguity of saying distant events happen "at the same time" it is he who produces two separate true results and falsely tries to equate them. The first is permissible by ambiguity of the problem setup (Banks' fault) and the second is simply wrong (also Banks' fault). Wrong. Banks' claim was that the results of the LT (aka assuming LT is admissible) produces results incompatible with LP. Lorentz gave the LT first, then in 1905 Einstein assumed the LP and the Principle of Relativity (PoR) and Maxwell's equations to conclude that the LT was admissible transformation of coordinates. Banks hoped to be able to proved that if you started (assumed to be true) the LT that one could conclude a result incompatible with LP. In post 8, Banks was proven to be wrong in a silly way which easily demonstrates the journal has no meaningful peer review. Post #8 introduced four lemmas which demonstrate that the LT was always going to predict results compatible with the LP. Since the equations in posts #9 & 10 prove those lemmas, how can they be irrelevant? Wrong. The problem arises from neglect of RoS, which is implicit in use of the LR. t=0 is not the same thing as t'=0. Likewise, \(t = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) is not the same thing as \(t' = \frac{1}{c}\). Why? Because t & t' don't have a proportionality since neither event P or Q happen where x=0 (position M in Banks' paper) . The relationship is \(t' = \frac{t - v x /c^2}{\sqrt{1 - \frac{v^2}{c^2}}}\) and x is NOT zero. Banks is OK with x=x(M)=0 (position M) and x'=x'(M')=0 (position M') being distinguishable. Why would he object to t=t(K) and t'=t'(K) being distinguishable, unless he rejected RoS? That's not a principle of physics. Physics approximates the behavior of reality with mathematical models and has no fundamental problem if a model has a limited domain of applicability. In math, when one encounters a contradiction, one knows one has started from an incompatible collection of assumptions. But the fault may be in any of the axioms, definitions or givens. Here Banks introduces an elaborate setup and thus has many givens. Incorrect. It shows that if one assumes both LT and absolute time (i.e. neglecting PoR) that one runs into contradictions. Since both events P and Q are in fact compatible with the LP, it is unclear why you think the LP is shown to be wrong. A contradiction which lies entirely at the feet of the responsible author: Andrew Banks. RoS is part of SR and trivially derivable from the LT. You have not shown that Banks did not introduce into the problem setup the assumption that t = t(K) and t' = t'(K) meant the same thing so you have not addressed my argument in post #8 that Banks introduced the contradiction rather than discovering it in SR. Thus your claims are baseless. Also, Wrong. C' and M are not "co-located" but rather only meet at event K. Since in all frames M and M' meet at the event O where light is produced, and in all frames M moves slower than light, and in all frames Event K happens after event O, it follows that t = t(K) and t' = t'(K) don't mean the same thing when x ≠ x(M) = x(K) so the assumption that there is one correct answer for the position of the light is incompatible with the LP when you and Banks assert that there are two different defintions of time in use.
Let's redo Banks with three concrete choices. 1) speed. Let \(v = \frac{4}{5} c\) 2) scale. Let \(x'(C') = - 234.6712 \, \textrm{km} = - \frac{v}{c} \times 293.339 \, \textrm{km} = - \frac{v}{c} \times c \times \frac{1}{1022} \, \textrm{s} = - c \times \frac{2}{2555} \, \textrm{s}\) so that we may have conventional measures. 3) Let's suppress the y and z coordinates. Then the Lorentz Transformation is: \(t' = \frac{5}{3} t - \frac{4}{3c} x, \quad x' = \frac{5}{3} x - \frac{4 c}{3} t\) (which is why my choice of v was made) So Banks identifies six lines in the x-t and x'-t' planes, but names only three of them. C' : \(x' = - c \times \frac{2}{2555} \, \textrm{s}, \quad x = \frac{4}{5} c t - c \times \frac{6}{12775} \, \textrm{s}\) M' : \(x' = 0, \quad x = \frac{4}{5} c t\) M : \(x' = - \frac{4}{5} c t', \quad x = 0\) and three more that I will give names to: L : \(x' = c t', \quad x = c t\) This is the propagation of light in the +x direction from event O T : \(t' = \frac{9}{25550} \, \textrm{s} - \frac{4}{5 c} x', \quad t = \frac{3}{5110} \, \textrm{s}\) T' : \(t' = \frac{1}{1022} \, \textrm{s}, \quad t = \frac{4}{5c} x + \frac{3}{5110} \, \textrm{s} \) These are lines where either t=t(K) or t'=t'(K). The fact that T is a different line than T' is entirely due to relativity of simultaneity. Event O lies on lines M', M and L and has coordinates \(x' = 0, \; t' = 0, \quad x = 0, \; t' = 0\) Event K lies on lines C', M, T and T' and has coordinates \(x' = - c \times \frac{2}{2555} \, \textrm{s}, \; t' = \frac{1}{1022} \, \textrm{s}, \quad x = 0, \; t = \frac{3}{5110} \, \textrm{s} \) Event P lies on lines T and L and has coordinates \(x' = c \times \frac{1}{5110} \, \textrm{s}, \; t' = \frac{1}{5110} \, \textrm{s}, \quad x = c \times \frac{3}{5110} \, \textrm{s}, \; t = \frac{3}{5110} \, \textrm{s} \) Event Q lies on lines T' and L and has coordinates \(x' = c \times \frac{1}{1022} \, \textrm{s}, \; t' = \frac{1}{1022} \, \textrm{s}, \quad x = c \times \frac{3}{1022} \, \textrm{s}, \; t = \frac{3}{1022} \, \textrm{s}\) That P is not the same event as Q has nothing to do with the LP and everything to do with the RoS requiring T ≠ T'.
this is a simple issue. Answer the question in the term of science. Given M and C' are co-located does LT give the correct position of light in the primed frame yes or no.
I invite science. prove any given mapping by LT is the same vector in the target frame using the light postulate. let's have it. http://article.sapub.org/10.5923.j.ijtmp.20160602.02.html
Maybe you could start with the science, by asking a relevant question. I know you already think you're asking relevant questions, but that's your problem.
Already answered. So the answer is YES if you don't define "correct" incorrectly. Example: "What is the position of the light pulse at the same time as determined by clocks in the unprimed frame as when C' and M are co-located?" This is a version of Banks' question that doesn't ignore Relativity of Simultaneity in some fatal way. In the unprimed frame you work out the equations in x&t for C' to be \(x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}} \) AND \( x'= -v/c\) or \(x - vt = -\frac{v}{c} \sqrt{1 - \frac{v^2}{c^2}}\) In the unprimed frame you work out the equations in x&t for M to be \(x=0\) Then you solve this system of two equations to learn what the coordinates in x&t for the intersection K are: \(t = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) AND \(x=0\) In the unprimed frame you work out the equations in x&t for T to be \(t = t(K) = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) In the unprimed frame you work out the equations in x&t for L to be \(x - c t = 0\) Then you solve this system of two equations to learn what the coordinates in x&t for the intersection P are: \(t = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) AND \( x = \sqrt{1 - \frac{v^2}{c^2}}\) Finally you use the Lorentz Transformation to learn what the coordinates in x'&t' for the intersection P are: \(t' = \frac{t - \frac{v}{c^2} x}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{c} - \frac{v}{c^2}\) AND \(x' = \frac{x - vt}{\sqrt{1 - \frac{v^2}{c^2}}} = 1 - \frac{v}{c}\) Alternately: In the primed frame you work out the equations in x'&t' for C' to be \( x'= -v/c\) In the primed frame you work out the equations in x'&t' for M to be \(x=0\) and \(x = \frac{x' + vt'}{\sqrt{1 - \frac{v^2}{c^2}}} \) or \(x' + vt' = 0\) Then you solve this system of two equations to learn what the coordinates in x'&t' for the intersection K are: \(t' = \frac{1}{c}\) AND \(x'= -v/c\) In the primed frame you work out the equations in x'&t' for T to be \(t = t(K) \) AND \(t = \frac{t' + \frac{v}{c^2} x'}{\sqrt{1 - \frac{v^2}{c^2}}} \) AND \(t(K) = \frac{t'(K) + \frac{v}{c^2} x'(K)}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{ \frac{1}{c} + \frac{v}{c^2} \left(-\frac{v}{c}\right)}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{c} \frac{1 - \frac{v^2}{c^2}}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) or \(t' + \frac{v}{c^2} x' = \frac{1}{c} - \frac{v^2}{c^3}\) In the primed frame you work out the equations in x'&t' for L to be \(x' - c t' = 0\) Then you solve this system of two equations to learn what the coordinates in x'&t' for the intersection P are: \(t' = \frac{1}{c} - \frac{v}{c^2}\) AND \( x' = 1 - \frac{v}{c}\) Finally you use the inverse Lorentz Transformation to learn what the coordinates in x&t for the intersection P are: \(t = \frac{t' + \frac{v}{c^2} x'}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{c} \sqrt{1 - \frac{v^2}{c^2}}\) AND \(x = \frac{x' + vt'}{\sqrt{1 - \frac{v^2}{c^2}}} = \sqrt{1 - \frac{v^2}{c^2}}\) There is nothing mysterious in saying T and T' are different lines, so if you don't ignore RoS in the question, then you will know which one to use. Then you get only the correct answer out. Because Banks implicitly treated t=t(K) and t'=t'(K) as the same condition, even at places not co-located at K, then he doomed himself to failure: the failure to understand SR. [Mod note: chinglu is warned a third time not to post pseudoscience in the main science forums.] As used here, the LT doesn't change geometric vectors, but provides alternate algebraic coordinates for them. As used here, vectors subject to the LT are necessarily space-time vectors, a fact Andrew Banks frequently forgets.
you did not answer. Given M and C' are co-located does LT give the correct position of light in the primed frame yes or no. put the answer right here and prove your answer under SR.
Oh, can you prove any calculation i made that is not SR. Write the proof out here step by step for everyone. and write out any math calculation made in the paper that is false. That means prove it. So show everyone any math in the paper that is wrong. Otherwise, you will confess all math in the paper is true. http://article.sapub.org/10.5923.j.ijtmp.20160602.02.html
