I am perfectly capable of understanding Pete's picture and relativity. I do not begin to discuss the specifics of Pete's picture until I get to the line that starts: Which makes \(\psi\) the angle AB makes with the x-axis in the frame where AB is at rest, \(u = 0\), and \(\theta\) the angle CD makes with the x-axis in the frame where CD is at rest, \(v' = 0\). I then can compare the angles AB makes with CD in the frames where AB is at rest and CD is moving, \(t,x,y\), where AB is moving and CD is at rest, \(t',x',y'\) and the frame where AB and CD are moving with equal and opposite velocities, \(t'',x'',y''\). Pete started the thread, so he has priority in defining what the discussion is about -- you need to stop carrying baggage around from thread to thread. Pete's examples are parallel in the frame where both are equally Lorentz-contracted -- my AB and CD are both parallel \(\phi = 0\) in the case where \(\psi = \theta = \frac{\pi}{4}\) AND \(u = v' = 0\) but only in the frame where \(\left| u'' \right| = \left| v'' \right|\). Thus two relatively moving extended objects in space-time which are parallel in one particular frame need not be parallel in another frame. The math is self-consistent so you can start in the double-primed coordinates where the lines are parallel and equally well calculate the angle between AB and CD in the frame where only one of them is moving. Either way, objects which are parallel in one frame need not be parallel in another. Inertially moving objects are always stationary with respect to themselves, so a discussion about two objects in different states of inertial motion is ALWAYS a discussion about two objects stationary wrt two different reference frames. This is horrible notation. First of all, these are not the equations of lines since you have introduced space-time frame S which at the minimum has coordinates t,x,y and so the equation \(x=\alpha y\) defines a locus of points (t,x,y) that is more than 1-dimensional as can be seen in the equivalent \(0 t + x - \alpha y = 0\). Moreover you don't label the locus, so the next locus: \(0 t + x - \alpha y = x_0\) doesn't have a name to distinguish it. Because these locii are more than 1 dimensional, they don't have a well-defined velocity associated with them. For example both \(x = 0, y = 0\) and \(x = \alpha c t, y = c t\) define world-lines which are contained entirely within the first sheet. In addition to the family of inertial and sometimes superluminal motion, there are countless families of non-inertial motion contained in these locii. This latter expression is incorrect. if \(t = \gamma(t'+Vx'/c^2)\) and \(x = \gamma(x'+Vt')\) and \(y' = y\) then \(x = \alpha y' + x_0\) is equivalent to \(\gamma(x'+Vt')=\alpha y'+ x_0\). Another equivalent expression is \(0 t + x - \alpha y = \gamma V t' + \gamma x' - \alpha y = x_0\) which correctly leads to \(V t' + \gamma x' - \frac{\alpha}{\gamma} y = \frac{1}{\gamma} x_0 = x'_0\) So \(x_0 = \gamma x'_0\) is the correct substitution for the last term of your expression. Again, the last expression is \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'+x'_0\). As you wrote it, non-intersecting locii ( I have assumed \(x_0 \neq 0\) ) are incorrectly transformed to locii that intersect at \(t' = \frac{- x'_0}{V}\). But you have said nothing about the case where two objects which don't share the same state of inertial motion are judged parallel in one frame, because your locii don't have well-defined motions -- just no relative motion in the x-direction. In the concrete example with AB and CD having different inertial motions they need not be parallel in all frames.
In post 2, you point out that the tops of the rods touch first in the red frame, while the bottoms of the rods touch first in the blue frame. I agree with this. You then say that this is an unphysical contradiction, which implies that you think the order in which the points touch should not depend on what frame we're in. If that isn't what you meant, please clarify for me. If it is what you meant, please choose from the following three (I made an edit) options, or introduce a fourth, to explain why you think the order in which the points touch should be frame-invariant: 1. The order in which events occur should not depend on reference frame. or 2. These two events are both connected to the same physical object, so they obey a different set of rules from physically unconnected events. or 3. Events can be spatially extensive, and all of rod A touching all of rod B is in fact a single event.
Pete starts generating threads every time he gets in trouble, so the context needs to be carried from thread to thread. Sorry, try to stick to the original content, do not introduce different material. I am not going to bother with your pedantic nitpicking, the point is that the lines are parallel in S'. I am quite sure that you got the fact that they share the slope \(\frac{\alpha}{\gamma}\). Huh? You sure about that? You made an elementary algebraic error, I hope you realize that.
Again, you are incapable of reading, what I am saying is that depending on what frame is considered at rest, you get a different result. Hence, a contradiction. Need I remind you that motion is relative, so the outcome should not depend on which frame you consider "at rest" and which one you consider as "moving"?
I get that. Specifically, the different result that we get is that either "the tops of the rods touch before the bottoms" or "the bottoms of the rods touch before the tops" depending on our frame. Two separate events, and the differing result is which one happens first. My question is, why do you think this is a contradiction? I can only think of three reasons why it would be, and I would appreciate it if you'd pick one or introduce another I haven't thought of: 1. The order in which events occur should not depend on reference frame. or 2. These two events are both connected to the same physical object, so they obey a different set of rules from physically unconnected events. or 3. Events can be spatially extensive, and all of rod A touching all of rod B is in fact a single event.
...because motion is relative and you cannot get different results depending on which frame you "fix". You seem to have a tremendous difficult understanding this. Maybe you can now lay off the tactic of asking the same questions over and over.
I'll lay off the tactic of asking the same questions over and over as soon as you lay off the tactic of failing to answer them over and over. But your response is getting somewhere this time, so I'll run with it. You say that "you cannot get different results depending on which frame you fix." Would it be fair to translate this as "you cannot get different results of any observation depending on which frame you fix?" If not, what specific kinds of observations have to give the same results in all frames? Edit: Upon reflection, I realized it might look like I'm getting drawn off into a tangential debate again. To keep this on-topic, the supposed contradiction in Pete's scenario is the only reason Tach has given why his fixed-time "snapshots" have any physical meaning at all. By disproving the contradiction, I intend to demonstrate the the snapshots are arbitrary mathematical constructs with no basis in reality.
You have neither justified nor explained the relevance of the calculations you perform in the linked PDF. Consequently you haven't shown anything. The OP is about parallelism of lines in relative motion with respect to one another. Pete started this thread, and he, not you, defined what it was about.
I didn't fail to answer it, you fail to understand the answer. Frame S moves wrt frame S'. Frame S' moves wrt frame S. If you bring \(rod\) over \(rod'\) you should be getting the same result as if you brought \(rod'\) over \(rod\). PoR tells you that you cannot tell one frame from each other. I already explained that to you, SR is a theory of invariants, experiments concentrate on measuring invariants, this is why there is no experimental test for RoS. On the other hand, a zero angle between two line segments is such an invariant. Measuring invariant quantities is what enables us to do the measurement in the lab and generalize it to all other frames. Another good example is the zero motion of the interference fringes in MMX.
As a scientific argument, the above rates on par with "I don't want what you are saying to be true, so there".
Ok. This claim keeps coming up, and it needs to stop. przyk keeps touching on this, but I'll come right out and say it: Your pdf contains a serious, basic math error. In equation 8, you apply a Lorentz transform to a three-vector, which is not possible. As a result, all of page 2 amounts to manipulating symbols with no physical relevance, and your conclusion have no bearing on the real world. A zero angle is not invariant.
That might be true if: 1) You had defined what the vectors \(\bar{a}\) and \(\bar{b}\) were and how they relate to moving lines. 2) You had explained why it was sensical to simply divide the longitudinal components of those vectors by \(\gamma\), including why it would give the correct result even for a moving line. You haven't done either, so what I say is actually correct. You launched into a calculation that, as far as I'm concerned, looks like you pulled the whole thing out of your ass.
Err, try again. It is simply the Lorentz transform applied to an arbitrary segment that says that the extent in the direction of motion is contracted while the extent in the direction perpendicular to the direction of motion is not. I could provide you with this rather basic proof, I wasnt aware that you don't know it.
It doesn't relate to "moving lines". It relates to the invariance of zero angles. The title should give you a hint <shrug> Same answer as above, to Fednis48. I wasn't aware that you were ignorant of this very basic result. My fault.
You can't apply a Lorentz transform to a three-vector. Period. Show me a proof that does, and I'll show you a bad proof.
