I proposed such a "thought experiment" in post #119, or perhaps an outline of one.. It's the same example as in a book written by Roger Penrose. "ROS with C' and M being co-located" is too vague to say anything much about it. I seriously doubt you can prove rpenner is wrong, you would then be proving that Einstein was wrong, you would then be a genius. That's just too funny to be anywhere near serious.
Fine if C' and M being co-located is too vague, that means we can never know such things which makes SR useless. Just a point rpenner does not dispute the co-location event. Now, prove the example from Penrose shows ROS from one event.
Rpenner, you have many defending you, all of which are simple minions that cannot follow the math. But, you have failed to prove the co-location event of C' and M is an application of ROS. When do you intend to show this claim so that I can quickly dismiss it with proof?
Relativity of simultaneity and the fact that \(P \not \in \ell\) are sufficient to require \(Q \neq R\) as a matter of self-consistency. Because \(j \neq k\), the definitions \(P = j \cap k\), \(Q = j \cap \ell\), \(R = j \cap \ell\) and the observation that \(P \not \in \ell\) lead to the conclusion that \(Q \neq R\). Line f describes the trajectory through space-time of object M. Since the inner product of the difference of any two events on line f and any two events on line j is zero, it proves that j is a line of simultaneity for any inertial system of coordinates where M is at rest. Frame Σ is such an inertial system of coordinates. Line h describes the trajectory through space-time of object C'. Since the inner product of the difference of any two events on line h and any two events on line k is zero, it proves that k is a line of simultaneity for any inertial system of coordinates where C' is at rest. Frame Σ' is such an inertial system of coordinates. Lines f and h cross at a single event P. Lines j and k cross at a single event, which is also P. Thus \(P = f \cap h \cap j \cap k = f \cap h = f \cap j = f \cap k = h \cap j = h \cap k = j \cap k\) is the only place and time where C' and M are co-located. You don't even have to look at all the mathematical content of post #2 to see that. Here, let me remove some of the distractions for you: So what did we learn by actually reading post #2? In frame Σ, look at the t-coordinate of event P and event Q. Therefore events P and Q are simultaneous in frame Σ. They have the same t-coordinate value. Look at line j, it has constant t-coordinate value. Therefore both events P and Q are part of line j. Therefore line j is a straight line connected two events simultaneous events in frame Σ. Therefore every event which is parts of line j is simultaneous with event P in event Σ. In frame Σ', look at the t'-coordinate of event P and event R. Therefore events P and R are simultaneous in frame Σ'. They have the same t'-coordinate value. Look at line k, it has constant t'-coordinate value. Therefore both events P and R are part of line k. Therefore line k is a straight line connected two events simultaneous events in frame Σ'. Therefore every event which is parts of line k is simultaneous with event P in event Σ'. In frame Σ, look at the x- and t-coordinates of events Q and R. Which comes later? It's R. What is the relation between the difference between values of the x coordinates and the differences of the t coordinates? It's \(\frac{x_R - x_Q}{t_R - t_Q} = c\). So a straight line between events Q and R represents movement at the speed of light. If you extend this movement backwards in time to t = 0 does it pass through the position of x = 0 ? Yes. So O, Q and R are all on the same light-like straight line: ℓ In frame Σ', look at the x'- and t'-coordinates of events Q and R. Which comes later? It's R. What is the relation between the difference between values of the x' coordinates and the differences of the t' coordinates? It's \(\frac{x'_R - x'_Q}{t'_R - t'_Q} = c\). So a straight line between events Q and R represents movement at the speed of light. If you extend this movement backwards in time to t' = 0 does it pass through the position of x' = 0 ? Yes. So O, Q and R are all on the same light-like straight line: ℓ Therefore the principle of relativity of simultaneity is summarized in this problem by the statement \(j \neq k\). And even though in frame Σ, P and Q are simultaneous, and in frame Σ', P and R are simultaneous, self-consistency requires that Q and R not be simultaneous in any frame. One would be mistaken to attribute that result from this example to the assumption of consistency of light speed, since all that is required is for ℓ to not be space-like and not pass through event P.
With all this noise above you did not prove the following which is the task at hand. This was in your post before. Now prove this and then we can move on. Are you going to prove your assertion or continue the distractions. I am prepared to move on once we get your statement out of the way. After all, it is your statement. We are playing chess and you have already been checkmated. I wonder if you know it and this is why you continue to get off task with distractions. Again, this is your statement and it is up to you to prove it.
Time will tell. I see you didn't notice RPenner ran from one of his own assertions. Why do you think that is?
Because frame Σ and frame Σ' are in relative motion, you need both t and x coordinates to figure out what the t' coordinate is. Thus a line of constant t-coordinate value cannot equal a line of constant t'-coordinate value. Thus the statement (in the context of [post=3198606]post #2[/post]) that \(j \neq k\) embodies the relativity of simultaneity. j is the straight line which in frame Σ happens to satisfy the equation \(t = t_P\). k is the straight line which in frame Σ' happens to satisfy the equation \(t' = t'_P\). It's obvious that event P is therefore common to both lines. Because of special relativity and the assumption that non-zero motion in the X direction differentiates the standard of rest in frame Σ from the standard of rest in frame Σ', it follows immediately that line j and line k share ONLY event P. Therefore \(j \neq k\), which is a precise example of relativity of simultaneity. Therefore for any time-like or light-like world-line that does not pass through event P has only a frame-dependent definition of what event along that world line is simultaneous with event P. Thus \(Q \neq R\).
This does not explain your statement below. Otherwise, use ROS and prove it. Maybe you need to state ROS and then prove your statement. That way all readers can see what you are saying.
RPenner I have decided to help you out on the Relativity of simultaneity. This is from wiki. Now, note the co-location of C' and M is one event and ROS needs two. Are you understanding this now? Otherwise, you need to prove the co-location event is actually two different events for both frames. That means both frames agree are two different events are required for each frame for co-location. Good luck!!!!
How is paddoboy empowered to notice an event that didn't happen? Be specific, (1) what assertion did you mean, (2) where did I make it, and (3) where is this "running" evidenced? Here are my posts to you: [POST=3198606]Post #2[/POST] "You have ignored relativity of simultaneity and therefore ignored basic characteristic of special relativity, dating back to Einstein's original 1905 paper." [POST=3198777]Post #4[/POST] "Where you got confused is by trying to assert absolute time, even in the face of two different inertial frames and events which don't happen in the same place." [POST=3198939]Post #15[/POST] "the opposite of ROS is built into your assumptions." "If at any event which is not P you talk about the time when C' and M are co-located, then you have made an assumption about time. If you don't explicitly talk about line j or line k, then you have ignored relativity of simultaneity. If you ignore that both j and k are physically distinct concepts of when C' and M are co-located then you have made the assumption of absolute time." [POST=3199300]Post #18[/POST] "Two human-defined coordinate frames in relative motion leads to two different definitions of what events are simultaneous with event P. Lines j and k encapsulate those definitions, and lead to intersecting with two different events on line ℓ." [POST=3199356]Post #21[/POST] "You can't talk about anything being at two different positions at the same time if they are only at different positions at different times. Looking at events Q and R, they are clearly distinct events but both are part of line ℓ so if Q and R are at different positions (as they are in every coordinate system) then they must also happen at different times (which is also true in every coordinate system)." [POST=3199363]Post #23[/POST] "unless you are exactly at the position of event P, lines j and k correspond to different human-chosen conventions of what is simultaneous with event P. Therefore events Q and R happen at different times." [POST=3199618]Post #58[/POST] "Using the Minkowski inner product, line f is orthogonal to line j in any frame and line h is orthogonal to line k in any frame." [POST=3199686]Post #65[/POST] "you[r] only mistake is assuming the phrase "when M and C' are co-located" has some universal meaning independent of frames. No law of nature is violated -- the only thing that is violated is your assumption of the universality of simultaneity. Lines j and k are different lines. [POST=3200578]Post #110[/POST] "the OP is a confused demonstration of relativity of simultaneity rather than the purported demonstration of physical inconsistency." [POST=3200657]Post #111[/POST] "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." ... "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." [POST=3200974]Post #115[/POST] "you have asked a complex question ... 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. " [POST=3201642]Post #140[/POST] "Relativity of simultaneity and the fact that \(P \not \in \ell\) are sufficient to require \(Q \neq R\) as a matter of self-consistency. Because \(j \neq k\), the definitions \(P = j \cap k\), \(Q = j \cap \ell\), \(R = j \cap \ell\) and the observation that \(P \not \in \ell\) lead to the conclusion that \(Q \neq R\)." Your quote of my [POST=3200974]post #115[/POST] was exposing a fault with your question, not my assumptions. You assumed that because frame Σ says events P and Q happen at the same time and frame Σ' says events P and R happen at the same time, that all frames say events Q and R happen at the same time, when in fact, none of them do. I think I have, many times. Line j is the line where t (which is defined in frame Σ) matches the value of t for event P. Line k is the line where t' (which is defined in frame Σ') matches the value of t' for event P. Event Q is on line j. Event R is on line k. But no event which is distinct from P is on both lines j and k. Thus the concept of simultaneity is frame-dependent unless you actually pass through the event you are are trying to be simultaneous with. In short, \(j \neq k\). Does line ℓ pass through event P? No. So are events Q and R both separated in space from event P ? Yes. Therefore events P and Q are separated in space and "it is impossible to say in an absolute sense that" P and Q happen "at the same time." Likewise events P and R are separated in space and "it is impossible to say in an absolute sense that" P and R happen "at the same time." We can say relative to clocks at rest in frame in frame Σ, events P and Q happen at the same relative coordinate time. Since that's true about any two events on line j, we can say \(P\in j, \; Q \in j\). Likewise, we can say relative to clocks at rest in frame in frame Σ', events P and R happen at the same relative coordinate time. Since that's true about any two events on line k, we can say \(P\in k, \; R \in k\). But in no frame do Q and R have the same relative coordinate time. This is because line j is not the same line as line k. The only event common to both lines j and k is event P, which is the only place and time that C' and M are co-located. As a practical manner, in every inertial coordinate frame, not just the two under discussion, event R happens later than event Q. I have understood this since I wrote [POST=3198606]post #2[/POST]. Have you understood what the definitions of events Q and R are? That make no sense, especially in light of the calculations in [POST=3198606]post #2[/POST].
Clearly you are not understanding. Relativity of simultaneity "depends" on two independent observers in relative motion. It does not depend on two events, since one event is sufficient. Co-location is just where the worldlines of the two observers intersect. It's a single point in space and it makes no sense to say "when" this occurs because each observer has their own clock, so all they can do is compare their local clocks. Hence, co-location is an event defined differently for each observer, they can only agree on where co-location occurs.
Here is your post, You did not admit you are totally wrong. Now, are you going to confess your statement is false, yes or no? Otherwise, prove it. Your did neither above.
RPenner here is your statement. Now prove to the readers why you are right! If you run from this, then you lose.
Please Register or Log in to view the hidden image! Even I had alarm bells ringing when I read that absurd sentence you are referring to. And that's saying something taking into account I'm a real novice with SR maths.
Understood. Your post even to me makes 100% sense. This I suggest, reveals chinglu1998's real reason for starting this thread. Obviously saying SR is false.
The important question to all readers is that it really beggars belief what you do not know about SR and associated maths. this makes no sense...
