Einstein, in his derivation of the relativistic aberration formula would certainly disagree. This is why I added the paragraph at the end of the proof, in the hope you'll take the hint. But you couldn't..... Sigh
The Lorentz length contraction factor is a very narrow result. It compares the distance between two parallel worldlines (two trajectories with the same velocity) in a given frame with the rest distance between the trajectories. Simply dividing the longitudinal distance by \(\gamma\) is only correct if 1) the two reference points used to define the distance are moving with the same velocity, and 2) both reference points are at rest in the initial frame considered before the boost is applied. It is nonsensical otherwise. In particular in the exercise Pete defined in the OP, there is generally no reference frame in which both line segments are at rest and parallel.
Yes. The second equation in that proof shows how we are transforming the time coordinate. Meaning we are dealing with four-vectors, not three-vectors. The proof goes on to note that we need to mark both ends of the rod simultaneously in the \(S'\) frame, and that we therefore need to take different ends of the rod from different times in the \(S\) frame. Specifically, equation 1.3 shows exactly how far apart in time the two ends of the rod are in the \(S\) frame. In your "snapshot" argument, you try to do the opposite: take two ends of the rod at the same time in \(S\), then transform them into \(S'\) while ignoring the fact that their times are now different. This proof contradicts your reasoning by the middle of the first page.
What? The rod and the floor are not both stationary in any single reference frame, and that problem is what spawned this thread.
You could use the same logic with velocity or length. Make the proclamation that length is frame invariant, then show that SR proves otherwise...then declare there is a contradiction and that our setup is somehow "unphysical"...rather than doubting our proclamation. I swear to GOD you're sounding a lot like a stubborn crank who never actually took a Physics course but is sure that this whole Relativity thing is hooey.
Trout never has been a physicist or philosopher -- just a bully who thinks he's the elected defender of the hegemony.
But being a bully implies being in a position of power. There's no power in being wrong. Being this wrong about something and maintaining arrogance and stubbornness just makes a person look like a rude dunderhead.
I disagree. I believe bullies are self-appointed illegitimate tyrants who impose their will through uncontested belligerency. Bullies don't stop just because they are asked or encounter social criticism. Certainly they won't stop just because their math or physics has been shown to be wrong. Consequently, a bully's reign ends when the assumption of non-contested belligerence is contradicted by those in legitimate authority taking decisive action or likewise, someone taking equally illegitimate decisive action. Consequently, effective reduction in bullying can closely resemble the acts of tyranny or anarchy -- someone has to take extreme action for the greater good.
Well, he's certainly a bully to newcomers here but that's because it's easy for them to mistake bold assertions, condescension and insults as coming from a legitimate authority. I figured his game out quickly, though, and have "contested" him ever since. He's a tool and it actually makes me upset that I'm taking the time to write about him. Attention is what he wants; I suspect his mommy left him on a doorstep...
So, you did not read the derivation I provided for Fednis48. Figures. ...because the scenario that Pete defined is not the scenario we started with. As such, it has a different set of deficiencies than the original one, deficiencies outlined in post 2.
Marking the ends of a vector at the same time in a given frame is the standard way of measuring length in that frame <strike one> . ...because the rod is moving, so, you need an instance of that rod, not its trajectory <strike two> Not. The proof is simply a basis of the proof of the invariance of zero angles. You know, the one that relativistic aberration fulfills as well.
Let's see: 1. You surreptitiously removed the basic algebra error by quietly editing your post, never acknowledging the error. 2. You never admitted that your calculations, flawed as they are, confirm that the two parallel lines are parallel in all frames. 3. This is not the first time I pointed out errors in your arguments just to watch you respond with an ad-hominem followed by you quietly slithering away. I don't know why you feel the need of acting like I jerk whenever you are being proven wrong. I think you can do much better than that.
Good. So at least we agree that marking the ends of a vector at the same time in a given frame is a correct way to measure length in that frame. Why isn't it also a correct way to measure angle in that frame? The rod is moving in your proof, too. If your argument about "instances" was true, the proof would have to start by setting \(t=t_0\) in the \(S\) frame, then transforming the spatial coordinates into \(S''\) while ignoring time. It doesn't do this, because "instances" are in fact just cross-sections of trajectories, and have no physical meaning outside the reference frame in which they are seen.
Before there is more BS piled in this thread, let's clarify a few things: 1. The scenario that you are presenting is different than the scenario we were talking about. 2. Nevertheless, it suffers from a serious flaw, as explained in post 2. 3. It turns out that this exact scenario is analyzed correctly in a published paper. I have been holding off mentioning this in the hope that you will answer the challenges to your conclusions. 4. Since you are not, you can see that the authors "explain away" the disagreement between the red and blue observers via introducing the notion of "extended present" (top of page 7). In other words, the relative "angle" between the rods is not a physically measurable entity, the "relative angle" is no less a figment of your imagination in the red frame than it is in the blue frame since it cannot be detected through any measurement.
I did not say that, the mathematical proof I gave in this thread you shows the opposite, fixing \(t'\) simply fixes the rod in S'. I am not using \(t'=t^'_0\) as a line of simultaneity in S', I am simply taking an instance of the rod. You have a knack of twisting what I am saying. Since we know the angle to be zero in frame S' (the previous thread), it follows that the angle is zero in frame S", exactly like the case of the relativistic aberration angle.
:bugeye: How can you be smart enough to catch algebra errors and track down an excellent paper on the subject at hand, yet dumb enough not to realize that the paper contradicts everything you've been saying? From the paper itself: In other words, the two ends of the rods do not have to collide simultaneously in all reference frames, unlike you've been claiming. (Emphasis added. Immediately after this line, the paper goes on to calculate \(t_k\) for the collision of B and D, showing that it is different from \(t_k\) for the collision of A and C.) In other words, it's important to re-match the temporal coordinates in the new reference frame, rather than just taking a constant-time slice from the original frame and transforming its spatial part. In other words, Janus58's solution way back in the original thread, based on maximum speed of signal propagation, is qualitatively right.
So far, the dumb person has been you. Actually the authors try (in vain) to teach you the exact opposite: "This paradox is NOT explained by stress propagation because the direction of propagation of the stress in the rods is reversed in both frames" (bottom of page 3, top of page 4). Give it rest, you posted enough stupidities.
Then why didn't you make that objection from the very beginning? If you sincerely believed that Pete was giving the correct solution to an irrelevant problem, then the correct and sane way for you to respond to that would have been to say exactly that. If you think that the problem Pete presented in the OP is irrelevant to what you may have been discussing in earlier threads, then posting calculations relating to a different problem doesn't help your case. It just makes you look obtuse. All I gathered from your post #2, and in particular this bit: is that you have apparently forgotten or never heard about the relativity of simultaneity effect. The "vector analysis" you keep trying to shoehorn into this problem is very suggestive of the same thing, since Lorentz transformations generally doesn't map four-vectors with a null timelike component to other four-vectors with a null timelike component (only the rotation subgroup does that). Whether two spacelike separated events happen simultaneously or not, and the order in which they occur, does depend on the reference frame under consideration.
