This is "[thread=95873]Special Relativistic time dilation and length contraction derived[/thread]" which is a closed thread. You also might like my repost of my 2008 argument from symmetry and a 1859 measurement: "[post=2826191]What's this Theory of Relativity? (post #66)[/post]"
Glad you are back. Still have this post outstanding. http://www.sciforums.com/showthread.php?141840-SR-Issue&p=3199693&viewfull=1#post3199693 I would be surprised if you attempted to encounter it. But, hey, so be it.
Special Relativity/Mathematical transformations http://en.wikibooks.org/wiki/Special_Relativity/Mathematical_transformations
rpenner already has. You may have noticed, that people knowing your bad record, here and elsewhere, [where you have been banned] are not replying to your goading type posts, and the incessant claim you keep making re "prove it" I'm not as bright as those other folk that have chosen to ignore your trash, and I give you what you crave for in replying. But I'll think I'll now desist in replying, as you have shown your hand in this thread by refusing to answer my questions. So you have some fum chingy baby and carry on with the nonsensical ridiculous claims.
Exactly who are you to speak for rpenner? Further, you confess you do not even understand basic math. It is amusing you post in this thread. Rpenner has a post waiting from me that he has refused to address. Now, can you address to support your accusations?
Honestly, I do not know. I will even confess why. There is an earth rotational sagnac but there is no orbital sagnac or milky way sagnac. This is all verified by GPS. But. simultaneity is decidable in the rotating earth frame since GPS works.
More on straight lines Previously, I have written out physical descriptions of lines in terms of x (or x') in terms of t (or t', respectively) or falling back to constant expressions for t or t' if the line no such dependence. The same lines may be written in a form which does not treat horizontal and vertical lines as separate cases: And by explicit Lorentz transformation, we can calculate what these lines are in the other coordinate system: But a third mechanism to describe these straight lines exists, description in terms of a dimensionless parameter: \(\begin{array}{c|cc|cc} \textrm{Line} & x & t & x' & t' \\ \hline \\ f & 0 + 0 \lambda_f & 0 + \frac{d'}{c} \lambda_f & 0 - \beta \gamma d' \lambda_f & 0 + \gamma \frac{d'}{c} \lambda_f \\ g & 0 + \beta \gamma d' \lambda_g & 0 + \gamma \frac{d'}{c} \lambda_g & 0 + 0 \lambda_g & 0 + \frac{d'}{c} \lambda_g \\ h & - \beta \gamma^{-1} \, d' + \beta \gamma d' \lambda_h & 0 + \gamma \frac{d'}{c} \lambda_h & - \beta d' + 0 \lambda_h & \beta^2 \, \frac{d'}{c} + \frac{d'}{c} \lambda_h \\ j & 0 - \gamma^{-1} d' \lambda_j & \gamma^{-1} \frac{d'}{c} + 0 \lambda_j & -\beta d' - d' \lambda_j & \frac{d'}{c} + \beta \frac{d'}{c} \lambda_j \\ k & 0 + d' \lambda_k & \gamma^{-1} \frac{d'}{c} + \beta \frac{d'}{c} \lambda_k & - \beta d' + \gamma^{-1} d' \lambda_k & \frac{d'}{c} + 0 \lambda_k \\ \ell & 0 + \frac{1 + \gamma + \beta \gamma}{1 + \gamma} d' \lambda_{\ell} & 0 + \frac{1 + \gamma + \beta \gamma}{1 + \gamma} \frac{d'}{c} \lambda_{\ell} & 0 + \frac{1 + \gamma - \beta \gamma}{1 + \gamma} d' \lambda_{\ell} & 0 + \frac{1 + \gamma - \beta \gamma}{1 + \gamma} \frac{d'}{c} \lambda_{\ell} \end{array}\) Note that in every case above the Lorentz transform of \(\left( x(0), \; t(0) \right)\) gives \(\left( x'(0), \; t'(0) \right)\) and the Lorentz transformation of \(\left( x(1) - x(0), \; t(1) - t(0) \right)\) gives \(\left( x'(1) - x'(0), \; t'(1) - t'(0) \right)\), which is what is meant by affine transformation. Spend some time with these three representations of lines in the x-t plane of frame Σ and show that they are all exactly equivalent to each other. Likewise, examine the lines in the x'-t' plane of frame Σ'. Finally prove that for any particular value of the relevant parameter, the point given in the x'-t' plane is the Lorentz transformed event given by the same value of the parameter in the x-t plane. Thus this work with lines in both frames Σ and Σ' is based entirely on the Lorentz transformation. Intersections are just particular values on the lines where the values of the parameters conspire to put the event at the same place at the same time: \( \begin{array}{c|c|c|c} \textrm{Event} & \textrm{Frame} \; \Sigma & \textrm{Frame} \; \Sigma' & \textrm{Parameters} \\ \hline \\ O=f \cap g \cap \ell & \left( x=0, \; t=0\right) & \left(x'=0, \; t'=0 \right) & \lambda_f = 0, \; \lambda_g = 0, \; \lambda_{\ell} = 0 \\ P = f \cap h \cap j \cap k & \left( x = 0, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right) & \left( x' = - \frac{d'}{c} v, \; t' = \frac{d'}{c} \right) & \lambda_f = \gamma^{-1}, \; \lambda_h = \gamma^{-2}, \; \lambda_j = 0, \; \lambda_k = 0 \\ Q = j \cap \ell & \left( x = d' \, \sqrt{1 - \frac{v^2}{c^2}}, \; t = \frac{d'}{c} \sqrt{1 - \frac{v^2}{c^2}} \right) & \left( x' = d' \, \left(1 - \frac{v}{c} \right) , \; t' = \frac{d'}{c} \left(1 - \frac{v}{c} \right) \right) & \lambda_j = -1, \; \lambda_{\ell} = \frac{( 1 - \beta ) ( 1 + \gamma )}{1 + \gamma - \beta \gamma} \\ R = k \cap \ell & \left( x = \frac{c d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v}, \; t = \frac{d' \sqrt{1 - \frac{v^2}{c^2}}}{c - v} \right) &\left( x' = d' , \; t' = \frac{d'}{c} \right) & \lambda_k = \gamma ( 1 + \beta ), \; \lambda_{\ell} = \frac{1 + \gamma}{1 + \gamma - \beta \gamma} \end{\array}\) Note in all three cases, it doesn't matter if you Lorentz transform the lines and then find the intersections or find the intersections and Lorentz transform the events. That is what is meant by the self-consistency of the Lorentz transformations. Likewise since in frame Σ, \(Q \neq R\) means Q and R are events that don't happen in the same position or same time or both, in frame Σ' self-consistency requires that Q and R are events that don't happen in the same position or same time or both. This turns out to be the case, and so the OP is a confused demonstration of relativity of simultaneity rather than the purported demonstration of physical inconsistency.
In the OP, chinglu proposes that we consider objects M and C' moving inertially with non-zero relative velocity \(v = \beta c\) and colliding at event P. Chinglu also asks us to consider a unidirectional flash of light originating at event O (in the past light cone of P) and asks for a description of the light "at the same time" as event P. Naturally, this does not give rise to a single description because "at the same time" is a function of which coordinate frame we use to determine when distant points are simultaneous. Specifically, if we use frame Σ, the frame where object M is at rest, then the definition of simultaneity in frame Σ finds only one event, Q, which is both in the future of the flash of light and "at the same time" as event P. And if we use the definition of simultaneity in frame Σ', the frame where object C' is at rest, then the definition of simultaneity in frame Σ finds only one event, R, which is both in the future of the flash of light and "at the same time" as event P. Because the direction of relative velocity between M and C' is not orthogonal to the direction of the propagation of the flash of light, it follows that \(Q \neq R\) and necessarily, the time and space coordinates of Q will not match those of R in either frame. There is no great mystery here unless one mistakenly assumes "at the same time" means something physical for events that don't happen in the same place. For those that haven't seen any other post in this thread, object M travels on line f from event O to event P and is at rest in frame Σ. Object M' travels through event O and is at rest in frame Σ'. Object C' travels through event P and is at rest in frame Σ'. Line j is all events in the x-t plane of frame Σ which are considered simultaneous to P in that frame. Line k is all events in the x'-t' plane of frame Σ' which are considered simultaneous to P in that frame. Line ℓ is a light-like line which passes through O and intersects both lines j and k to the right of P. This intersections we call events Q and R, respectively. You can't "agree" with anyone else in this thread, because both Special Relativity and myself have said that "when M and C' are co-located" has no physical meaning except for objects whose world-lines pass through event P. Instead, I wrote in [post=3199686]post 65[/post]: The only way that the phrase "when M and C' are co-located" has any meaning when one's world line does not pass through event P is by the artificial establishment of an inertial coordinate frame. Thus line j is a line of simultaneity that applies to just frame Σ (or by the reckoning of people at rest in frame Σ). Likewise, line k is a line of simultaneity that applies to just frame Σ'. So you agree that lines j and k are distinct lines that only meet at event P? And you agree that any two points on line j have the same coordinate time in frame Σ? And you agree that any two points on line k have the same coordinate time in frame Σ'? Because understanding that about all of lines j and k except at event P is a general feature of special relativity. It's very simple, I agree. But if you had any knowledge of linear algebra you wouldn't have to prove it for the coordinates of a single event P. Given \(\Lambda\) is a homomorphism between coordinates in frame Σ and frame Σ', then it follows that \(\Lambda^{-1}\) exists and \(\Lambda^{-1} \Lambda = \Lambda \Lambda^{-1} = I\) where \(I\) is the identity transform. Thus this proves \(\begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix} = \Lambda^{-1} \Lambda \begin{pmatrix} x \\ y \\ z \\ t \end{pmatrix}, \quad \quad \quad \begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix} = \Lambda \Lambda^{-1} \begin{pmatrix} x' \\ y' \\ z' \\ t' \end{pmatrix}\) for the coordinates of all events in space-time. Given \(x' = \gamma \left( x - \beta c t \right), t' = \gamma \left( t - \beta c^{-1} x \right)\) it is easy to prove \(x = \gamma \left( x' - (-\beta) c t' \right), t = \gamma \left( t' - (-\beta) c^{-1} x' \right)\) thus demonstrating that Lorentz transform has an inverse which is a different Lorentz transform. Thus you've labored to prove a trivial result while relying on a stronger theorem. No. I guess you are admitting you didn't understand a single one of my posts, including the very short [post=3198606]post #2[/post]. To misunderstand my post to this extent looks like trolling. If it is not required that Q=R, then why are you arguing that Q and R can't be different? And you demonstrate that you are delusional in that you think I'm scared of your bad English and worse math and that you think you are entitled to immediate responses to misunderstandings that were addressed in post #2. Is it wrong to hope that one day you might finally understand trivial affine mappings between coordinate systems? Perhaps. But the only way you can demonstrate that I am totally wrong is by dying before learning better. What questions? If you mean the sentences that end with the question mark, they appear to be predicated on false assumptions to the point where you are merely trolling. No. An expert has a subject matter that defines the boundaries of purported expertise. An expert's opinion in the area of expertise is revered. That I make pronouncements on your physics simply indicates I have walked the path that all physics experts walk, far enough to have a better idea where the line between layman and experts exists. For example, I have read sections one and two of Einstein's 1905 paper that end with: http://www.fourmilab.ch/etexts/einstein/specrel/www/ I really don't care who you think "they" are, but this purported thing you hope they will understand and my fear seem to be your baseless delusions. M and C' are only co-located at one event in all of space and time: event P. What do you mean "valid for the frame" ? I'll tell you what you mean. You mean that the t-value of event P is equal the the t-value of event Q and that Q is part of line ℓ. So a shorter and more mathematical way to say the same thing is that event Q is the intersection of lines j and ℓ. Or \(Q = j \cap \ell\). Likewise \(R = k \cap \ell\). See post #2. Presumably LP is the speed of light postulate which says both O to Q and O to R are paths taken by something moving at the speed of light in every frame. But that was already true about all events on line ℓ. \( O \in \ell, \quad Q \in \ell, \quad R \in \ell\) are true independent of any choice of coordinate frame. The Lorentz transform cannot change anything physically fundamental. It can't change inertial motion to non-inertial motion. It can't reverse cause and effect. It can't make something moving at the speed of light move at a different speed. So you are incorrect to assert that LT(Q) is a thing. The Lorentz transform changes coordinates, not events. \(Q = j \cap \ell\) and \(R = k \cap \ell\) are statements about subsets of space-time, not coordinates. But since \(Q \in j, \quad R \not \in j\) then of course R can never take the special role in frame Σ of being simultaneous with P that Q has. Likewise, \(Q \not \in k\) so Q and P are not simultaneous in frame Σ'. So it is untrue that the Σ' time coordinate of Q is the same as the Σ' time coordinate of P, even though the Σ time coordinate of Q equals the Σ time coordinate of P. That's as simple as noticing line j is different than line k. Again, you seem confused. Coordinate systems are just systematic ways of naming events. But the name Q or \(j \cap \ell\) is just as good a name from the field of geometry. There is no reason to suspect a connection between the coordinates in one frame of R and the coordinates in another frame of Q. For one, in every frame R happens later than Q. By this point you haven't done any logic. You have no definitions of what you mean, and you are just making empty assertions and confusing yourself. You don't even know what universal property connects lines f and j ( or h and k). Take any two events on line f, say P and O. Take any two events on line f, say Q and P. Then in every case, in every inertial coordinate system: \(c^2 \Delta t_f \Delta t_j - \Delta x_f \Delta x_j - \Delta y_f \Delta y_j - \Delta z_f \Delta z_j = 0\) It works in frame Σ: \(c^2 \Delta t_f \Delta t_j - \Delta x_f \Delta x_j - \Delta y_f \Delta y_j - \Delta z_f \Delta z_j =c^2 \left( \frac{d'}{\gamma c} - 0\right) \left( \frac{d'}{\gamma c} - \frac{d'}{\gamma c} \right) - \left( 0 - 0 \right) \left( \frac{d'}{\gamma} - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) = c^2 \times \frac{d'}{\gamma c} \times 0 - 0 \times \frac{d'}{\gamma} - 0 \times 0 - 0 \times 0 = 0\) It works in frame Σ': \(c^2 \Delta t'_f \Delta t'_j - \Delta x'_f \Delta x'_j - \Delta y'_f \Delta y'_j - \Delta z'_f \Delta z'_j =c^2 \left( \frac{d'}{c} - 0\right) \left( \frac{d}{c} \left( 1 - \frac{v}{c} \right) - \frac{d'}{c} \right) - \left( - \frac{v d'}{c} - 0 \right) \left( d' \left( 1 - \frac{v}{c} \right) - \left( - \frac{v d'}{c} \right) \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) - \left( 0 - 0 \right) \left( 0 - 0 \right) = c^2 \times \frac{d'}{c} \times \left( - \frac{v d}{c^2} \right) - \left( -\frac{v d'}{c} \right) \times d' - 0 \times 0 - 0 \times 0 = 0\) That's because this type of inner product is Lorentz invariant and represents a feature of the geometry of space-time, not merely a statement about coordinates.
I read through your posts but you did not answer this specific question in the OP. So, could you please answer this simple question? And you did? You agreed based on the light postulate in the M frame, when C' and M are co-located the lightning is located at \((d'/\gamma,0,0,\frac{d'}{c\gamma})\). You agreed based on the light postulate in the M' frame, when C' and M are co-located the lightning is located at \((d',0,0,\frac{d'}{c})\). Now, I don't want to assign meaning from you unless you agree with the above. Is the above what you intend? Please a simple yes if you agree and a no with why you disagree.
Wellwisher, You are misusing words there. But your primary problems are: You can't use GR to explain the finite speed of massless particles when that's built-into GR as an assumption. You can't explain why the universe is one way without a model of how the universe can be two ways and somehow picked this one way, which you did not do. The finite speed of massless particles is what allows electromagnetism to work (by which I mean, carry momentum and energy from place to place so that light from the sun gets to Earth, radios work, etc.) so it seems you need at a minimum have to explain the origin of electromagnetism before you can have any faith in your claims about the origin of special relativity.
Scientists and scholars use "read" to mean "read and understood your communicated thoughts" which does not appear to be the case. I think it is obvious to any informed readers that it is not a simple question. Instead you have asked a complex question (複合問題, see also 既定觀點問題) which contains the preconception that the concept of "when C' and M are co-located" has the same meaning in both frames. This is a false assumption in any purported thought experiment examining special relativity, because it directly contradicts the relativity of simultaneity. By including it in the question you have been intellectually dishonest in your study of special relativity. Therefore, your entire intellectual posture is one of empty-headed denialism. Until you confront the flaws in your own preconceptions, you will never actually understand the topic. So the most correct answer to your complex question is to split it into two fair questions about separate coordinate frames, each of which can and has been answered: According to frame Σ, which uses x to describe position and t to describe when, the flash of light at "the same time as P" is described as event Q, so the answer to "For coordinate system Σ, where is the “lightning” along the positive x-axis when C' and M are co-located is answered with \(x = \gamma^{-1} d'\), as was stated in post #2. According to frame Σ', which uses x' to describe position and t' to describe when, the flash of light at "the same time as P" is described as event R, so the answer to "For coordinate system Σ', where is the “lightning” along the positive x'-axis when C' and M are co-located is answered with \(x' = d' \), as was stated in post #2. Event Q is not the same event as event R, even though both Q and R are along the same world-line of the same ray of light, ℓ. Instead, event Q is not the same event as event R because your question is phrased in terms of a frame-dependent definition of simultaneity but you are talking about two frames, not one. Therefore instead of one universal concept of simultaneity, like existed before 1905, we have two separate definitions: line j for frame Σ and line k for frame Σ'. How is it that you have never looked at the meaning of line j and its definition in frame Σ as \(t = \gamma^{-1} \frac{d'}{c}\) or line k and its definition in frame Σ' as \(t' = \frac{d'}{c}\) and not figure out that these were two different definitions of simultaneity? I did in [post=3198606]post #2[/post] for the whole world to see. How is it that you haven't asked any questions about post #2 and still don't understand the shortest of my posts to you? You make three mistakes: This isn't about just the hypothesis that light in a vacuum always moves at speed c in every frame. That's only important because of the way you specified the problem is that world-line ℓ just happens to correspond to movement at the speed of light. The exact location of Q and R depend on the nature of line ℓ, but it could easily move at any constant velocity that was not at rest in frame Σ and you would get the same qualitative answer that event Q is not the same as event R. The real reason that event Q is not the same as event R is because line j is all things simultaneous with event P as defined by frame Σ, while line k is all things simultaneous with event P as defined by frame Σ'. When you list the coordinates in frame Σ, you make the mistake of not labeling them clearly by event and frame. A coordinate is a function of what event in space-time we are talking about and what coordinate frame we are using to describe its position. Failure to label things clearly will get you in trouble because x is not the same thing as x' and t is not the same thing as t'. So I say, that in frame Σ, event Q and event P are simultaneous and the coordinates of event Q are in frame Σ: \(x_Q = \gamma^{-1} d', \quad y_Q = 0, \quad z_Q = 0, \quad t_Q = \gamma^{-1} c^{-1} d'\) So many conceptual errors are repeated here. But I say, that in frame Σ', event R and event P are simultaneous and the coordinates of event R are in frame Σ': \(x_R = d', \quad y_R = 0, \quad z_R = 0, \quad t_R = c^{-1} d'\) My meaning has been clear since [post=3198606]post #2[/post], you have ignored relativity of simultaneity and therefore you are confused as to why event Q is not the same as event R, even though that is the simplest of predictions of special relativity made at the bottom of section 2 of Einstein's 1905 paper introducing the concept. My intention is to demand that if you ask questions you read and understand the replies. You need to start reading [post=3198606]post #2[/post]. It's very short, but clearly you don't yet understand the words being used. As [post=3198606]post #2[/post] shows, my disagreement with your OP isn't based on your algebra, but on your space-time geometry and your assumption that you have two frames but only one definition of simultaneous. The algebra disagrees with you and clearly shows that each frame has its own different definition of simultaneous. Thus \(j \neq k\) thus \(Q \neq R\). Further, this is not an unexpected result to anyone who understands special relativity.
Let's get this out of the way first. You claimed, Now, the relativity of simultaneity (ROS) concerns itself with 2 events in each frame. We are talking about one event where each frame agrees on the times of each other's clocks. Can you explain why ROS applies for only one event? I mean, does ROS apply when the origins are co-located? You need to answer this or you demonstrate you do not understand SR. You also need to come to the conclusion that ROS does not apply to one event. I expect you to admit you are in error.
For the sake of young ones and newbies. Chinglu claims against all evidence over 100 years that SR is false. He contradicts all the experts, and refutes all those that show he is wrong. He will argue the toss repeatedly and never recede from his claim that SR is false. He has had most of his material moved to pseudoscience and alternative hypothesis sections. He has been banned here more then once for his continued unsupported claims. He has been totally banned elsewhere including "Cosmoquest" [never to be allowed to post again] for the same sort of nonsensical claims. He will continue to ask for proof [when proof is not the goal in scientific theories] while supplying nothing himself. Take care in accepting ANYTHING he says.
Where did you get this idea from? It's wrong. Can you explain why it wouldn't apply to one event, along with a pair of observers in relative motion? The single event is observed by two independent observers, right? So the observers and the event form a pair of simultaneous spaces which are only congruent when both observers are moving with the same velocity. They don't know what the time is on the other's clock; they would have to communicate this information between them. Do you understand that "simultaneous" is only relevant to a single observer? Two people walking slowly past each other have different simultaneous spaces, because they have low relative velocities there is no noticeable effect locally, but for events a large distance away, there's a big difference. However, the large distance means the events there won't have any effect on the two observers, there's a big time delay as well so no "paradox" or anything. What SR says is that simultaneous events for one observer are not simultaneous for another observer moving relative to the first, they each have their own simultaneous space. Simultaneity is not a difficult concept, nor is the notion that every observer has their own space in which they observe simultaneous events.
Sure, the relativity of simultaneity says if two events with different x coordinates y > 0, z = 0, are simultaneous to one frame, then will not be simultaneous to another. One event does not qualify under this definition. And, if you are claiming simultaneous spaces are created on one event, you are claiming absolute time. Now, what we have on the P predicate is C' and M are co-located. We then apply LT to get the space-time coordinate of the other observer. Note, this is not an issue of simultaneity, but an issue of coordinate transformation. Assume said observer is located at (0,0,0) and a light pulse is emitted from there. (0,0,0) claims (-d,0,0) and (d,0,0) are simultaneous events. Also, all y of the nature (0,y,0) and in the same frame will agree those events are simultaneous visually. Any other observer in the same frame will calculate the events are simultaneous. So, you are wrong.
Well we have at least two now disputing your maths, and your Interpretations, and your understandings of what SR entails. But you remain certain you are still correct. If you are correct, does this not invalidate SR? If you still think you are correct, the consequences are rather enormous. Why not get it peer reviewed then? I mean, if as you say, you are correct, then get it 100% verified. Don't be afraid of peer review.
\ There have been two that have gone off ST yo dispute me. But, I brought them back into the SR fold. Note their silence. So, you post is meaningless.
To demonstrate the relativity of simultaneity, you need two observers who don't share the same simultaneous space, and you need at least one event. With two events, the two observers can have different views of which event occurs before the other, but you don't need two events, a single event will still mean the two observers have different views of "when" it occurs. So far all you've managed, as always, is to tell everyone how poor your understanding of Einstein's ideas is. You can't even present a "thought experiment" that isn't so vaguely described as to be close to meaningless (viz. rpenners' comments). Oh, please.
OK, prove with a thought experiment that ROS is valid with 2 observers each in different frames. Make a commitment and I will prove to you why you are wrong. I am not interested in your vague posts. Oh, here is one. prove ROS with C' and M being co-located. That is rpenner's contention. Prove it. I am waiting on rpenner to prove which I hope he attempts. I will demonstrate he is wrong. If you cannot accept this challenge, then you have absolutely no idea what you are talking about and the reader's will laugh at you.
