That doesn’t make it incorrect.
The other frame sees the same image at the same time, thus obviously since it is reading its own clocks, both frames are actually shown.
And in your diagram, you only have one clock.
And in your diagram, you only have one clock at one setting.
No it doesn’t, but we’ll get back to this. First, let’s pick one set of variables to stipulate, so that I can get through the rest of your math.
Lets say, for the sake of moving things forward, that we are given (x,t), the coordinates in the train’s frame, for all the clocks. The center clocks are at C=T=(0,0), the flash clocks are at B=(-10,0) and F=(10,0) respectively. Our unknowns are in the (x’,t’) frame. We know C’=(0,0), but we don’t know B’ or F’.
Are these acceptable starting conditions for you?
Yes and no. No clock is labeled, but the time for every point on the rod is shown for both frames.
According to the pictorial it does. That is why I ask for you to tell me what might be wrong with the pictorial. The pictorial shows how and why it does.
Emm… no.
Label the other clock and set it at the other end of your rod and set that clock to the same time reading as the first clock. Then you’ll have 2 clocks.
Carl, you have this mystical notion involving t’ that causes you to misuse the Lorentz.
t’ ONLY MEANS “what the train clock reads according to BOTH frames”. It is what both frame’s observers will see as they look upon the train’s clocks.
But you have in your head that there is this magical “train frame” wherein time is “different” such that the clocks don’t reflect (literally) what time the train observer sees. That is why I drew a pictorial of a “reflection” of the train’s clocks. But time isn’t different, the clocks merely turn slower, literally. The station sees the clocks turning slower, but the train observer is internally moving slower as well, so he perceives that his clocks are just fine. BOTH parties see the same reading on his clocks. The only difference is that the station sees his own clocks as fine and the train’s clocks as slower (which they are). The train would see the station’s clocks as faster (which they are).
So until you understand that, I don’t want you to be using t’ and merely creating another argument. You were about to make the same mistake again.
In the Lorentz t’ is ALWAYS a ∆t’ = t’ - t0’. But when t0 = 0 (for a single item being analyzed), t’ is also merely the coordinate. Note that you assigned the initial conditions for both flashers as t = 0, and that would lead you into the same error of using only the t’ coordinate axes to calculate their time. But note that both of your rod ends aren’t at t = 0. That is why in my diagrams, the train was always perpendicular to the t axis, so each clock showed that it started at t = 0. Every clock has to have its own t and t’ axis and since you believe in length dilation, they each have to have their own x and x’ axes too. Each clock has its own diagram or you have to put many primary axes pairs on one diagram. Merely 2 clocks yields 8 primary axes and you cannot lay one on top of another else you hide the offsets involved.
If you will use 2 completely separate diagrams for each train timer and have each start of t = 0, you will find that t’ is the same for both of them and thus they will be identical at the station’s 4:00-t for both frames. In one diagram x = -10 to 100 and in the other x = 0 to 110. Both are equal ∆x. In both diagrams, t = 0 to 4:00-t (call it 400). Both are equal ∆t. With equal ∆x and ∆t, you cannot have anything but equal t’ or ∆t’. The pictorial I presented already shows this.
What is wrong with the logic of those pictorials? I’m guessing that you can’t find any or you would have said so by now. The reason you can’t is because there isn’t anything wrong with them. In stead, you have the belief that the mystical “train frame” must be “different” than what is shown and seen by the station when it looks at the train’s clocks.
First of all, no it doesn’t. Second of all, if you’d like to define some other idiosyncratic way to distinguish the time observed by the station observer from the time observed by the train observer, please give us more than “Emm… no.”
If you assume that there is no such thing as train time and station time, you are definitely not using Special Relativity.
They are both at t’=0.
Your diagrams simply ignore the t’ axis, but if you drew it in it would be obvious that both the flashers can only be on t=0 or t’=0.
This seems to be a recurring problem. A basic contention of Special Relativity, which you simply reject, is relativity of simultaneity. You aren’t showing that there’s no such relativity, you’re assuming it, and the relativity of simultaneity is what resolves this problem. You aren’t creating a paradox for SR, your creating a paradox in Jamesian Relativity, in which a pair of clocks synchronized in one frame is synchronized in all others. By assuming that while assuming that light travels at the same speed in all frames, you reach a paradox. Jamesian Relativity is therefore flawed, and should be rejected.
Yes. It Does.
And if you assume that the station cannot see the train’s time from looking at its clocks, then neither are you.
Only if you use 2 axes, one t’ for each end. And if you were doing that with the train, there wouldn’t be a problem now, because you would always get equal t’ for each timer.
It doesn’t matter what the mystical t’ might be. The OP is about the flash (image) getting to the center.
No I don’t care of it in this problem and no it doesn’t solve anything. It ignores obvious logic so as to propose a magical fix. “Yeah, but I’m God, so 2+2=3 if I want it too. Problem solved. See, I told you there wasn’t a problem.” And the only reason it even comes up is because of a misuse of Lorentz.
I showed you pictorials. Your only complaint is that they don’t include your mysticism. But they cover the OP issue completely.
But now since you have confirmed that they are correct, but merely missing your extraneous obfuscations, I’ll argue with you once again about your misuse of the diagrams and Lorentz, if and only if, you make the diagrams and calculations for one clock at a time and show every step of your calculations so I can AGAIN point out exactly what you are doing wrong, maybe with enough clarity that you can see it.
Again, this is the scenario you are to be duplicating in your diagrams;
t’ here refers to the time coordinate in Q’s reference frame. It does not refer to both frames or images or whatever else.
I am not, so I pass that test. The station sees the train’s clocks running slowly.
t’=0 is the the x axis in the primed frame. It is sloped relative to the line t=0. In the diagram I provided, the rod lays along that axis.
You’ve edited the OP 34 times since posting it. The OP is about whatever you need it to be about.
My “mysticism” (also known, widely, as relativity of simultaneity) is a fundamental part of Special Relativity. I challenge you to find an overview of Special Relativity that doesn’t mention the phenomenon.
Your pictorials don’t cover the OP issue. They obfuscate it by assuming that t=t’ for all values of t. But since you’re asking a problem of simultaneity between moving observers, you’re asking about different accounts of local time, i.e. t and t’.
So what makes you think the train observer doesn’t see the same time on his clocks as the station?
Oh you aren’t aware of special relativity. You are only aware of some mysticism that you think is special relativity.
YOU are the one wanting to inject extraneous issues into it. I merely gave you the rules for my participation in your confusion.
A) One train item per diagram
B) Everyone starts at t = 0
C) Fully detailed equations.
Go for it.
Again, this is the scenario;
I’ve done that. Here, here, and here are diagrams. Here is the clearest use of the equations.
What are you using for your equations? What are you solving for, and what are your givens?
The first 2 of your links attempts 2 flashers on the train leading to your confusion.
The third of your post links has no items at all.
The 4th of your post links have newly edited equations that seem to be nothing but proving my point.
I was asking of what evidence you want to provide that support YOUR point, not mine. We have plenty for mine.
A) One train item per diagram
B) Everyone starts at t = 0
C) Fully detailed equations.
Go for it.
Again, this is the scenario;
Stop posting the same picture in every post, it clutters up an otherwise quick discussion. If you have nothing new to contribute, leave it at that.
You have yet to identify your A) givens, B) unknowns, and C) equations.
The time dilation for one clock is negative, and for the other it is positive. I.E., the timers are out of sync. If that’s your point, then we’re in agreement that this paradox is resolved by a vanilla application of Special Relativity.
Avoiding something?
I offered to let you present your case for your use of Lorentz (again).
No my point was that you were misusing Lorentz. A negative time dilation means that one of your train clocks sped up while the other slowed down. It turned out that way because you didn’t use a delta ∆x but instead used an x coordinate. That was exactly my argument of your misuse for several pages. And no doubt, you would do it again. But I offered to follow along so I could point it out again.
So yeah, IF we agreed that one clock on the train was speeding up while the other was slowing down and that is why they could never be in sync, I guess I could agree that the paradox is resolved by such.
But we don’t.
You need to tell me where to start (givens) and what to find (unknowns) in order for me to do that. And I’m curious to know what equations you are using in place of those transformations.
You’ve said that, but you’ve offered no case for it. You’ve offered no source that uses the equations as you would suggest (while I’ve offered many that use the equations as I suggest), and every argument you’ve attempted to make has shown that the equations are consistent under my use of them.
That’s why you started the other topic to attempt to show that there was something wrong with the equations themselves: you ran out of plausible arguments to say that I’m using the equations wrong pages ago (or at least, you’ve provided no new ones since then).
At least identify a standard of evidence by which we can determine how the equations should be used. You’ve rejected any outside source as evidence, you’ve not accepted the result of any test of the equations. By what standard of evidence do you conclude that the equations should be used as you’re advocating? (And I don’t mean “By the standard of logic”; what are the intervening steps?)
Staying on the positive side of x will help with coordinate issues.
Unknowns?
Also, I’m not doing the math again until you satisfactorily answer this question:
I’ve done the math over and over, I’m sure you could do it out the way that I use the equations.
And now that I think of it, once you give unknowns, you can show how you find those unknowns, too. So be ready to do that. This should really be the last time we need to do this math out, once we’ve both solved for the unknowns and stated our equations.
YOU are the one questioning. Pick whatever unknowns you think you need to prove that the clocks CAN’T read the same in both frames at 4:00-t. If you think you need to find the engine’s oil pressure go for it. The paradox doesn’t need anything that hasn’t already been stated time and time again.
That doesn’t even make sense. What evidence causes me to conclude that they should be used as I use them?? Emm… logic, arithmetic, common sense,…? … NOT getting asynchronous clocks on the same train and NOT getting different ∆t’ for 2 clocks on the same train as seen by the station.
One at a time would be best.
I’m sorry, that’s not good enough.
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“Common sense” dictates that when an equation contains a variable written as x, it is not intended to be understood to say ∆x. A source which gives the equation with ∆x instead of x would bolster this point.
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“Arithmetic” doesn’t seem to have anything to do with our points of contention. We both agree what +, –, etc. mean, but we disagree over whether x refers to x or to ∆x. Again, a source that supports the use of ∆x would bolster your position.
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“Logic” would fall squarely under the proviso I included: “I don’t mean “By the standard of logic”; what are the intervening steps?” Providing those steps would go a long way towards offering a justification of your contention that we should use ∆x where the equations say x.
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As I show here, we do not get “different ∆t’ for 2 clocks on the same train” using the equations as I am suggesting, so this cannot be a valid justification for your belief.
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And lastly, it seems that the entire question question of this thread, the crux of the supposed paradox, is whether the clocks on the train are asynchronous in either the train’s or the station’s frame. I say yes, you say no. If you are rejecting my use of the Lorentz equations simply because it solves your paradox, it seems to be a paradigm case of begging the question. If this is a part of your justification for why there is a paradox, the paradox is easy to unravel.
Not good enough for what?? What are you trying to dig up?
Entirely false. Almost every variable of measurement is a ∆ between itself and its zero state. When you have 2 clocks, you have 2 different x = 0 for the clocks.
I pointed out your misuse of Lorentz time and time again exactly.
Are you now stalling?
I went point by point through why the paradox exists. You state, “no it won’t be seen that way”. But you haven’t provided anything at all that justifies that response other than pages ago when I pointed out your mistake in using Lorentz.
This is the scenario;
Now what exact equation do you propose to justify saying that this pictorial is impossible or irrelevant? It shows synchronicity. It is very obvious.
Vague, “it doesn’t work like that”, “the train won’t see it synchronously”, and “I don’t know what t’ is for that” phrases don’t cut it.
The obvious questions;
A) Can the first portion of that pictorial be done?
B) Can the second portion?
C) Can the third portion?
Saying, “I don’t know” or demanding to calculate t’ for each is silly, but if you want to, go for it.
The only thing relevant to the paradox is that 2 flashers flash at the same moment. The pictorial shows that they can do exactly that and show the exact same time both on and off the train.
And you are also claiming that this picture from the train’s perspective simply cannot be physically accomplished;
But you know that it can be.
Not good enough to change my opinion that you have no reasonable standard by which you believe the claim that x should be read as ∆x. Indeed, your assertion that you believe it to be true because to believe otherwise would confirm relativity of simultaneity, and thus defeat your paradox, confirms my suspicion that your belief is ad hoc and simply held so as not to yield the point.
Almost every? Indeed, for any x, ∆x from x to zero is x. But that doesn’t mean that x always equals ∆x. Some paths never intersect the origin. Some are stationary at x. Sometimes we are measuring ∆x from x1>0 to x2>0. In these situations where x≠∆x, what leads you to believe we should use ∆x in an equations that says x?
If the standard by which you are replacing x with ∆x is that in a limited set of circumstances, x=∆x, why not replace x with y, because certainly in some situations x=y?
So, when you write x, you mean to say ∆x? Because they may have different ∆xs, and there may be different times at which ∆x=0, or something like that. But the distinction between x and ∆x is mathematically important, as this discussion clearly demonstrates. x is a coordinate, ∆x is the change in x between two coordinates (one of which need not be zero). In a coordinate plane, there is only one line x=0.
Claimed a mistake, which I am asking you to justify. If you cannot justify the claim, it is a non sequitur with no logical force.
Can you demonstrate that what this claim follows as a mathematical consequence of anything I’ve said?
I’ve never claimed that the “pictorial is impossible or irrelevant”, but that it is incomplete. It shows one frame, and we are concerned with two. Indeed, where is the paradox in your picture? In one of the many versions of your OP, you relied entirely on the existence of multiple frames to argue that you had created a paradox. It doesn’t seem like anything in your picture is remotely contradictory.