I don't like Tach. He's an abusive troll. But I'm sorry guys, I think he might be right about this. Look at Pete's pictures: Here is the rest frame of Box 1: Please Register or Log in to view the hidden image! Here is the rest frame of Box 2: Please Register or Log in to view the hidden image! I think there might be a problem with them, because length contraction doesn't work the way that you've been taught. I'll try to explain: Imagine you're in a gedanken spaceship shaped like a thin parallelogram, like this / . I'm in an identical spaceship nearby. We both look at the CMBR. We don't detect any CMBR blueshift in front or redshift behind, and we deem ourselves to be at rest with respect to the universe, and to each other. You wave goodbye through your window, and then cruise away to a suitable distance. Then you turn round, and you step on the gas. You accelerate to considerable speed, such that you can see significant CMBR blueshift and redshift, and then you pass me by. As you do you notice that my ship looks length-contracted, totally in line with the Lorentz factor. But note this: my ship hasn't moved an inch, and it hasn't changed an inch. To you the distance to the nearest star appears to have reduced, but it hasn't changed a jot. I didn't change, my ship didn't change, and I along with everybody on Earth says that star is still four light-years away. Instead you changed. You see everything length-contracted because you got longer. You're "smeared out" in what I deem to be the space of the universe. Your ship is smeared out to a wider parallelogram. But you can't see this. You don't look down at your own body and say Oh, I'm smeared out. You see yourself exactly as you always did: Lorentz invariance holds good. So does the symmetry of our relative motion. Regardless of that CMBR, I see you like you see me. When we're separated by space we both say you look smaller than me and we don't shout paradox. When we're separated by relative motion we both say you look shorter than me and we still don't shout paradox. It's because of the wave nature of matter, wherein the world is painted in light. But if our parallelogram-ships changed angle, when we collided we'd have a contradiction. I'd say the tops collided first whilst you said the bottom. Then we would shout paradox. And we'd be wrong, because collisions are collisions. They occur at the same place at the same time, whatever you think that time or place is. Your ship is like this / and mine is like this / , and whichever one you say is smeared out, it's a full-on collision for both of us. It isn't quite the same for the horizontal bar. If it's falling like this ‾ and you're moving like this ←. It can hit you on the head. It doesn't have to poke you in the eye. Now you've got orthogonal motion, and aberration. I can't find what I said about it, but it might have been wrong. Maybe we should look at it again after this.
Let me ask you Tach, before you take a dump do you think about it? No. That's what you're doing right now. Not thinking and just crapping all over the place.
Twice you failed simple algebraic substitution: (Mindless duplication is not the same as confirmation!) Again, this is a demonstrated flaw in your algebra. In your example, world sheet 1 is \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'\) and world sheet 2 is \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'+x'_0+Vt'\) and at time \(t' = \frac{- x'_0}{V}\) the two equations reduce to \(x'=\frac{\alpha}{\gamma}y'+\gamma x'_0\) and \(x'=\frac{\alpha}{\gamma}y'+\gamma x'_0 +x'_0 - x'_0 = \frac{\alpha}{\gamma}y'+\gamma x'_0 \) and so at that time the two lines coincide. Then you accuse me of "surreptitiously" editing a post to remove a purported "error" when my edits are time-stamped two minutes after the initial post and long before you began your reply, and you identify an equation which remains in the original. As you can see, I neglected to cancel a factor of gamma in one instance. Naturally you look hypocritical when you concentrate on a term which is not at issue. Later expressions are correct, so this was not an algebraic error so much as a transcription error. But the correct intermediate should have been: \(V t' + \not\gamma x' - \frac{\alpha}{\gamma} y = \frac{1}{\gamma} x_0\) Parallel lines which are are co-moving in one frame are parallel lines which are co-moving in another frame. This is a consequence of the linearity of the Lorentz transform, so is too unsurprising to evoke comment. When you say Pete got it his third animation (my double-primed coordinates) right in post #2, you can't meaningfully examine his scenario of parallel entities in relative motion by only looking at co-moving parallel entities. Where is the evidence that I went "away" ? If you thought I was acting like you, I'm sorry. I should leave the wild claims and hate speech to you.
Yes. I am saying you can't use eqn 1.6 with also using eqn 1.3 which require that you have both temporal and spatial information.
...which you edited quietly, after I flagged it. I do not understand why you persist in lying about it. I guess that this is the closest you can come to admitting that I was right all along. I use a different approach at looking at Pete's scenario, see post 2. Pay attention.
I highly doubt anyone disputes that point, but it does not apply to the rod and floor scenario of the other threads because they are not co-moving in any frame; they are in relative motion, as I've pointed out to you many times before. What part of "they are not co-moving in any frame" do you fail to understand? You're making colossal, classic amateur blunders... (sorry...started digressing into Tachese at the end there)
The above is not a very good description. Let's start with the bottom (3-rd) image: Please Register or Log in to view the hidden image! You have two boxes , representing the train car floor and the rod moving towards each other at speed \(\frac{u}{2}\) in a frame where they have equal and opposite velocities. Let's call this frame \(F\). When the two boxes overlap, i.e. the rod hits the train car floor, the rod and the floor have zero relative velocity, i.e. they stop, unlike your unphysical drawing where the two continue their motion passing through each other. The electromagnetic force precludes that from happening. But I guess, you knew that. Since the two segments come to rest wrt each other in \(F\), their Lorentz transformation in any other frame is a single, common segment. So, images 1 and 2 in your above scenario are unphysical. It is just another way to point out that you can continue opening as many threads , the fact that you don't understand the physics of the scenario will be repeated in all of them. You have opened only two new threads , each one made up of hundreds of posts, maybe it is time you stopped and thought about the physics? The biggest irony of all these exchanges with you is that this puzzle has a very simple solution in GR, you don't have to strip it off gravitational fields, you don't have to progressively dumb it down from thread to thread.
Put a clock on each end of each black line in each respective box, and synchronize all of them in F such that they all read the same time when the black lines make contact. Do you believe all of those clocks will remain synced from the red or blue frames of reference?? In fact, do the same thing but string each black line with an infinite number of clocks synced in F. Now if you analyze when the string of clocks are equal (and touching) from the red or blue frame you can visualize that the rods "zip together" with a bend.
Shh, adults are talking. One day, when you go back to finish 9-th grade in high school , you may learn a lot of the things that you obviously don't know right now. Very funny, too bad you can't back it up with anything.
You're screaming at the tsunami of Truth, hoping to change it. Doesn't it strike you as odd that EVERYONE, from Physicists to amateurs to cranks, is pointing out that you're wrong on this issue? Not even your usual sycophants like brucep and Syne are backing you up, they're just letting you drown in your own stubborn ignorance.
Yes, RJBeery, you don't need to scream any more Please Register or Log in to view the hidden image! Now, if you don't mind , please stop spamming the thread, I am waiting for an answer to a point I made to rpenner, so please let adults speak without your annoying droning.
Yes, please continue playing the martyr, fighting for truth and justice. I'll go back to lurking and feeling slightly embarrassed for you.
Illustration of linear transformation of parallel vectors in two and three dimensions Let A,B,C,D be four points in the plane such that vector AB is parallel to vector CD. What does that mean? It means \( \left\| \vec{AB} \, + \, \vec{CD} \right\| \; = \; \left\| vec{AB} \right\| \, + \, \left\| \vec{CD} \right\| \) which is a bit stricter than the definition of parallel lines in that AB and CD have to point in the same direction. Squaring both sides and subtracting common terms we get \(2 \vec{AB} \, \cdot \, \vec{CD} = 2 \, \left\| vec{AB} \right\| \, \left\| \vec{CD} \right\| \). From this we conclude that a vector of zero magnitude is parallel to any vector.
In this in place of not answering my question from post 89? Is this non-sequitur the best you could do? Certainly, you can do a lot better than that.
