Relativistic parallel rods

Discussion in 'Physics & Math' started by Pete, Apr 30, 2013.

  1. Pete It's not rocket surgery Moderator

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    This thread is for Tach, who wants to argue that parallelism is frame invariant:
    Tach maintains that violating this rule would violate the principle of relativity.

    I acknowledge that Tach and I have [thread=111112]argued about invariance of zero-angles[/thread] before in a different context (where the angle was between a surface and a velocity).
    I also acknowledge that Fedris48 [post=3064880]posted a counterproof[/post] to Tach's claim in a concurrent thread which Tach has essentially ignored.

    In this thread I present a very simple graphical demonstration that parallelism is not frame invariant.
    Two boxes of the same proper size pass each other. Each box contains a diagonal rod.

    Here is the rest frame of Box 1:

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    Here is the rest frame of Box 2:

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    Note that the each box is length contracted in the other box's rest frame, so the rods are not parallel.

    Now, here is the frame in which the boxes have equal and opposite velocity:

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    In this reference frame, The boxes are equally length contracted, and the rods are parallel.

    Tach, and anyone else interested, does this not adequately demonstrate that parallelism is relative?
     
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  3. Tach Banned Banned

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    Pictures prove nothing, you should know better. Math, on the other hand, does.

    The rods "touch first" at one end as viewed from the blue frame and at the opposite end as viewed from the red frame. Same scenario, opposite conclusions.
    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"? This means that the conclusion that you are trying to get from your scenario is unphysical, something that I have been telling you for a long time. The only picture that you got right is the third picture, from a frame that the two rods move with equal and opposite velocities. This should give you some pause.


    BS, of course I did NOT ignore, I answered it immediately but you were too self-absorbed in your trolling to notice it.
     
    Last edited: Apr 30, 2013
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  5. renislaj Registered Member

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    Ladies and gentlemen, you don't get any lamer than this.
     
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  7. RJBeery Natural Philosopher Valued Senior Member

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    Pete, why you didn't make this an "alpha rules" thread I have no idea. Your pictures are clear enough though. Tach's response is literally breaking the definition of frame invariance by demanding that the rods only be measured for parallelism in a particular frame. It's cringe-worthy that he's standing his ground; I almost feel sorry for him.
     
  8. Tach Banned Banned

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    Quite the opposite, it points out the pitfalls of doing "proofs" via "pictures".
    Given your total ineptitude in terms of science (math, physics), it is no wonder that you did not understand an iota and you rushed to "answer" before Pete. Pete is still "chewing" on it, this should have clued you in not to rush and get a taste of your own feet. I think Pete realizes that he's dug himself into a very deep hole and he's struggling to see how can he extricate himself. This will take a few hours, or days. Or maybe he'll just admit that he's wrong.
     
  9. przyk squishy Valued Senior Member

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    3,171
    So work it out mathematically. Consider two lines moving with equal speed but in opposite directions, defined by the equations

    \(x \,=\, \alpha y \,+\, vt \qquad \text{(line 1)}\)​

    and

    \(x \,=\, \alpha y \,-\, vt \qquad \text{(line 2)} \,.\)​

    These lines are parallel since the \(\Delta x/\Delta y\) gradient, \(\alpha\), is the same for both lines.

    You can work out what these lines get boosted to by substituting an inverse Lorentz boost:

    \( \begin{eqnarray} t &=& \gamma (t' \,+\, vx') \\ x &=& \gamma(x' \,+\, vt') \\ y &=& y' \end{eqnarray} \)​

    (in units where \(c \,=\, 1\)). Do that for both lines above and rearrange the terms, and the end result is

    \(x' \,=\, \gamma \alpha y' \qquad \text{(line 1)}\)​

    and

    \(x' \,=\, \frac{\alpha}{\gamma(1 \,+\, v^{2})} y' \,-\, \frac{2v}{1 \,+\, v^{2}} t' \qquad \text{(line 2)} \,.\)​

    The boosted \(\Delta x'/\Delta y'\) gradients are respectively \(\gamma \alpha\) and \(\alpha / (\gamma(1 \,+\, v^{2}))\). These aren't the same, so the lines are no longer parallel in the boosted frame.


    That is expected due to relativity of simultaneity. In the third picture both ends of the rods touch simultaneously. Those are spacelike separated events, so both ends are generally not going to touch simultaneously in other reference frames.
     
  10. Tach Banned Banned

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    I did. Click on the link.

    You are making the same mistake as Pete, you are transforming trajectories, not line segments. . Pete and I have already spent a lot of time on this subject.
     
  11. rpenner Fully Wired Staff Member

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    4,833
    Let A,B,C,D be inertial world lines such that A and B are co-moving and C and D are co-moving. Let neither of them have motion in the y or z directions.
    \(x_A = u t \\ x_B = L \cos \psi + u t \\ y_A = 0 \\ y_B = L \sin \psi \\ x'_C = v' t' \\ x'_D = M' \cos \theta + v' t' \\ y'_C = 0 \\ y'_D = M' \sin \theta\)
    Let the primed frame be the twice-transformed unprimed-frame where each Lorentz transform is parameterized by speed w in the x direction.
    \(\begin{pmatrix} ct'' \\ x'' \\ y'' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \end{pmatrix} \\ \begin{pmatrix} ct' \\ x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct'' \\ x'' \\ y'' \end{pmatrix} \\ \begin{pmatrix} ct' \\ x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ \frac{w}{c \sqrt{1 - \frac{w^2}{c^2}}} & \frac{1}{\sqrt{1 - \frac{w^2}{c^2}}} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \end{pmatrix} = \begin{pmatrix} \\ \frac{c^2 + w^2}{c^2 - w^2} & \frac{2 c w}{c^2 - w^2} & 0 \\ \frac{2 c w}{c^2 - w^2} & \frac{c^2 + w^2}{c^2 - w^2} & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} ct \\ x \\ y \end{pmatrix}\)
    Then in the unprimed frame we have:
    \( v = \frac{c^2 v' - 2 c^2 w + v' w^2 }{c^2 - 2 v' w + w^2} \\ x_A = u t \\ x_B = L \cos \psi + u t \\ x_C = v t \\ x_D = \frac{c^2 - w^2}{c^2 - 2 w v' + w^2} M' \cos \theta + v t \\ y_A = 0 \\ y_B = L \sin \psi \\ y_C = 0 \\ y_D = M' \sin \theta\)

    So in the unprimed frame, the angle between AB and CD at any time is:
    \(\phi = \tan^{\tiny -1} \frac{y_B - y_A}{x_B - x_A} - \tan^{\tiny -1} \frac{y_D - y_C}{x_D - x_C} = \tan^{\tiny -1} \frac{L \sin \psi}{L \cos \psi} - \tan^{\tiny -1} \frac{M' \sin \theta}{\frac{c^2 - w^2}{c^2 - 2 w v' + w^2} M' \cos \theta} = \tan^{\tiny -1} \tan \psi - \tan^{\tiny -1} \left( \frac{c^2 - 2 w v' + w^2}{c^2 - w^2} \tan \theta \right)\)

    In the correctly illustrated example by Pete \(u = v' = 0, \; w \neq 0, \; \psi = \theta = \frac{\pi}{4} \) and so \( v = \frac{- 2 c^2 w }{c^2 + w^2}\) and \(\phi = \tan^{\tiny -1} \tan \frac{\pi}{4} - \tan^{\tiny -1} \left( \frac{c^2 + w^2}{c^2 - w^2} \tan \frac{\pi}{4} \right) = \tan^{\tiny -1} 1 - \tan^{\tiny -1} \frac{c^2 + w^2}{c^2 - w^2} = - \tan^{\tiny -1} \frac{w^2}{c^2} \neq 0\)
    So Pete has made a valid proof without words.
     
    Last edited: Apr 30, 2013
  12. przyk squishy Valued Senior Member

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    3,171
    Lorentz transformations don't act on Euclidean vectors and there is no unique well-defined answer as to what the Lorentz transformation of a Euclidean vector is. The closest thing you can apply a Lorentz transformation to is a four-vector and the result you get depends on what you put as the timelike component of that four-vector. Generally if you apply a Lorentz transformation to a spacelike vector of the form \(a^{\mu} \,=\, (0,\, a_{x},\, a_{y},\, a_{z})\) then the result will have a nonzero timelike component and as such cannot be interpreted as a Euclidean vector. So your whole manipulation is meaningless.


    No, the equations I gave you define moving line segments: for fixed t each equation fixes a constraint between x and y, but doesn't tell you what x and y are. So for any fixed t my equations define lines and not points.
     
  13. Fednis48 Registered Senior Member

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    725
    Ok, we really need to get this particular claim ironed out. Your position, as you explained it to me, is that we should take a "snapshot" of the rod with fixed \(t\) in some reference frame, then apply only the spatial component of the Lorentz transform when we boost it to other frames. This is what you do in the notes you keep linking to prove that parallelism is invariant. Since the Lorentz transform is generally expressible as a \(4\times 4\) matrix, the burden of proof is on you to show that it's correct to throw out the first row and column in this situation. As I said in the other thread, I would love to see any source that says your "snapshots" are even physically meaningful, let alone a correct solution to this problem.
     
  14. Tach Banned Banned

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    well, look at the derivation for the aberration angle, it uses 3 vectors, not 4 vectors and it gets a result consistent to what I have told you, zero aberration angle in one frame translates into zero aberration angle in all other frames in inertial motion wrt the first frame.





    In other words, trajectories (or bundles of trajectories). Let's do it over, correctly this time:

    In frame \(S\) you have two lines , parallel to each other:

    \(x=\alpha y\)
    \(x=\alpha y+x_0\)



    In frame \(S'\), moving wrt \(S\) at speed \(V\) :

    \(\gamma(x'+Vt')=\alpha y'\)
    \(\gamma(x'+Vt')=\alpha y'+\gamma(x'_0+Vt')\)

    so:

    \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'\)
    \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'+x'_0+Vt'\)

    Lines are parallel in \(S'\)
     
  15. Tach Banned Banned

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    I do not throw that out, I simply fix it, Because , if you don't, you are transforming trajectories (or bundles of trajectories).


    The point is that your blind transforms are the ones that are causing all the problems, rendering the results as unphysical.
     
  16. Tach Banned Banned

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    Err, no, the line segments are fixed in their respective reference frames, please pay attention to Pete's "proof by picture". You are making up a different scenario, there is no \(u\), no \(v'\). There is just two lines parallel to each other in S (Pete has somewhat messed up the scenario by putting a line in each frame). The discussion is about frame invariance of parallel lines. Comparing two lines stationary wrt two different reference frames is a different discussion.


    In frame \(S\) you have two lines , parallel to each other:

    \(x=\alpha y\)
    \(x=\alpha y+x_0\)



    In frame \(S'\), moving wrt \(S\) at speed \(V\) :

    \(\gamma(x'+Vt')=\alpha y'\)
    \(\gamma(x'+Vt')=\alpha y'+\gamma(x'_0+Vt')\)

    so:

    \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'\)
    \(x'=\frac{\alpha}{\gamma}y'-\gamma Vt'+x'_0+Vt'\)

    Lines are parallel in \(S'\)

    Now, in your general scenario, you do get a non-zero angle but that only raises more questions than answers as you can see from post 2. The math may be correct but is the answer physical? Not so, since it leads to a contradiction.
     
    Last edited: Apr 30, 2013
  17. Fednis48 Registered Senior Member

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    725
    Ok, then, the burden of proof is on you to show that the usual \(4\times 4\) form of the Lorentz transform needs to be "fixed". Again, I would love to see a source that says your snapshots have physical meaning or are useful in the context of any problem. Because from where I stand, it looks like you're just making up a bizarre mathematical formalism for no reason.

    Since simultaneity is relative, there is nothing unphysical about two reference frames disagreeing on the order in which spacelike-separated events occur.
     
  18. Tach Banned Banned

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    The rods in Pete's drawings "touch first" at one end as viewed from the blue frame and at the opposite end as viewed from the red frame. Same scenario, opposite conclusions.
    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"? This means that the conclusion that you are trying to get from your scenario is unphysical, something that I have been telling you for a long time
     
  19. przyk squishy Valued Senior Member

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    3,171
    So? You're not calculating the aberration angle. It may be the case that in certain very specific situations you can get away with manipulating Euclidean vectors in relativity. But they're not what the Lorentz transformation applies to, so if you manipulate Euclidean vectors you have to carefully justify what you're doing with them and why such manipulation is meaningful in the context of relativity. There is no such justification in the calculation you linked to.



    Or, more simply, a moving line, which is what the problem in the OP is actually about. Why are you so reluctant to call a moving line what it is?


    This defines two lines that are both at rest in a given reference frame. This is only a very special case of the general problem considered in the OP. Why are you so insistent on pretending a moving line is the same thing as a stationary one? It's not, and as I already showed above your result does not generalise to the case of two lines in motion relative to one another.
     
  20. Tach Banned Banned

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    5,265
    I am calculating an angle. The calculations should be consistent.

    I agree. The OP case is one of those cases.


    Sure there is, it proves the invariance of zero angles. Exactly the point of this thread.




    Because the exercise is not about moving lines, it is about parallelism.



    Yep, we are talking about the invariance of the notion of parallelism: two lines parallel in S are also parallel in S'.


    Nope, this is exactly what we were talking about but Pete decided to change the scenario, thus introducing even more contradictions as explained in post 2.


    Please don't put words in my mouth, the exercise is about comparing stationary line segments, not about trajectories. Once you introduce trajectories you get a whole slew of internal contradictions, as I already pointed out in post 2.
     
  21. renislaj Registered Member

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    Tach you're wasting everyone's time.
     
  22. Fednis48 Registered Senior Member

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    Let me make sure I understand you. The top of rod A touching the top of rod B is an event. The bottom of rod A touching the bottom of rod B is another event. You say that the order in which these events occur should not depend on what frame we're in. That tells me you believe one of (edit) three things:

    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.

    edit: or

    3. Events can be spatially extensive, and all of rod A touching all of rod B is in fact a single event.

    Which of these do you believe? Or is there a fourth option I'm missing?
     
    Last edited: Apr 30, 2013
  23. Tach Banned Banned

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    shh, adults are talking
     

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