Alternative Twins Paradox

Discussion in 'Pseudoscience Archive' started by Jack_, Feb 20, 2010.

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  1. Jack_ Banned Banned

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    Are you talking about this experiment or the failure of you to know when in the stationary system the moving system experiences simultaneity for the rigid body experiment.

    I think it is on you to say what that time is since you claimed the LT space is logically complete.

    If you cannot, I can do it.

    I see by the above you do not know how to calculate that. You would simply answer to prove your "superiority".

    If you need me to prove the answer I will. It has already been battle tested several times.
    [/QUOTE]

    OK, it is clear you cannot answer.

    Here we go.
    Assume two rigid body spheres of equal rest radius in relative motion v. O is called the stationary frame and O' the moving frame. O' has a light source at the origin. When the origins of the two rigid body spheres are co-located, O' emits a light pulse.

    I'll just start with the proof.

    Assume z = 0.

    Let r be the rest radius for both rigid body spheres.



    Let x = vrλ/(c(1+λ)).

    Using LT, x' = -x.

    The proof is left to the reader.

    Next, is is easily shown |x| < r. Hence, this point is forced on the rigid body sphere of O as follows.

    y = √( r² - x²)

    for O'

    y' = √( r² - x'²)
    since x' = -x, then

    y' = √( r² - (-x)²) = y

    Therefore, the point (x,y) is on the rigid body sphere of O and (x'y')=(-x,y) is on the rigid body sphere of O'.

    Now, in the stationary frame, let time elapse by t=r/c since the time of the light pulse. Since (x,y) is on the rigid body sphere of O, then this point is met by the light because of the simultaneous strike of all points.

    Next, using, t' = ( t - vx/c² )λ with x = vrλ/(c(1+λ)), it is easily shown t' = t.

    Hence, in the moving frame O', the specific point (x',y') is hit by the light if (ct')² = x'² + y'²

    By design y' = √( r² - x'²) such that (x',y') is on the rigid body sphere of O'

    So, we have x'² + y'² = x'² + ( r² - x'²) = r²

    Since r/c = t = t', then (ct')² = r²

    Hence,
    (ct')² = x'² + y'²

    Therefore, the point (x',y') which is on the rigid body sphere O' is struck by the light sphere in O' when the time r/c elapses in O.

    The following conditions are met:
    1) The time t'=r/c elapsed as required for simultaneity in the O' frame.
    2) The point (x',y') is struck when r'=t/c.

    Therefore, it is concluded this point meets the conditions for simultaneity in the O' frame.

    Two more things.
    1) Time dilation is missing for clocks at O and O'.
    2) Since the origin of the moving rigid body sphere is located at vt after any time t, then the origin of the moving rigid body sphere is located at v(r/c) when O sees the simultaneity of its sphere points.
    But, since O' sees simultaneity in its frame when the time coords of O reads r/c, then the origin of the O' sphere is located at (v(r/c), 0) in the coords of O when O' sees simultaneity. Hence, there are two light spheres in the one space of O.

    This is a contradiction.
     
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  3. AlphaNumeric Fully ionized Registered Senior Member

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    The derivation of the consistency of Lorentz transformations follows from their group structure. Time and again I've tried to engage you in this discussion and time and again you've refused. One can't help but feel that you want people to think you understand this stuff but you don't.

    Given what follows in your post I wouldn't agree.

    Yes, because I really feel I need to prove that I know more about this than you by doing a homework problem. I suppose I might need to do that to prove my superiority if I hadn't already pointed out many times where you misuse terminology and how you are ignoring a great many other mathematical results, which are covered in any undergraduate mathematical physics course.

    The fact you are refusing to respond to my direct questions and you keep saying to say that I think I'm the bees knees would suggest you have a chip on your shoulder. In your other thread you specifically asked for Rpenner and myself, which does suggest you have some axe to grind. The fact I've been fairly dismissive of you seems to nark you.

    Did you copy and paste this from somewhere? It's just that you seem to be referring to something which isn't provided in your post, like you didn't include a picture, which defines x' and λ. This suggests that you're just blindly copying from somewhere and you don't realise that without the picture your post is just a jumble of undefined expressions.

    Perhaps its just a coincidence but this is the kind of behaviour I've seen in certain other posters....

    Actually you don't need to consider y at all, since you are basically doing Lorentz transformations in 1 dimension, which you could take to be x. All other directions are irrelevant.

    You don't seem to have proven anything here other than your inability to describe things. Given you have missed out a picture or some other set of definitions I question if you even know what you're posting since without further information your post is nonsense.

    It's much easier to just look at the system from two different frames. Frame 1 is the stationary one. The sphere satisfies \(x^{2}+y^{2}+z^{2} = (\tau c)^{2}\) but we can reduce this to just \(x^{2} = (c\tau)^{2}\) without a problem. The moving sphere is \(\gamma^{2}(x-vt)^{2} +y^{2} + z^{2} = (c\tau)^{2}\) and again we reduce to \(\gamma^{2}(x-vt)^{2} = (\tau c)^{2}\).

    At t=0 a photon sphere is formed and will have the form \(x^{2}+y^{2}+z^{2} = (ct)^{2}\). Reducing again gives \(x^{2} = (ct)^{2}\). This intersects the stationary sphere when \(t=\tau\). It intersects the moving sphere in such a way that it hits a single point first, which will be that nearest to the origin of Frame 1. ie when x=-ct and \(\gamma^{2}(x-vt)^{2} = (c\tau)^{2}\), so \(\gamma(c+v)t = c\tau\) so \(t = \sqrt{\frac{c-v}{c+v}}\tau\). This makes sense since if the moving sphere is moving at v=c then it'll Lorentz contract to a disk and so will hit the photon sphere immediately, t=0.

    Now do it from Frame 2's point of view. The second sphere is at \(x'^{2}+y'^{2} + z'^{2} = (c\tau)^{2}\) and the first sphere is at \(\gamma^{2}(x'+vt')^{2} + y'^{2} + z'^{2} = (c\tau)^{2}\). Reduce both to 1 dimension and you've got \(x'^{2} = (c\tau)^{2}\) and tex]\gamma^{2}(x'+vt')^{2} = (c\tau)^{2}[/tex]. Photon sphere forms about origin at t'=0 and so is of the reduced form \(x'^{2} = (ct')^{2}\). Photon sphere hits second sphere at \(t'=\tau\) and first sphere when \(\gamma(c+v)t' = c\tau\) so \(t' = \sqrt{\frac{c-v}{c+v}}\tau\). Precisely as you'd expect given the system is symmetric, the Lorentz boost by v and a parity flip leaves the system invariant.

    A lot more coherent and a lot simpler than the rambling ill defined crap you're peddling.
     
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  5. Jack_ Banned Banned

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    I am engaging you. You just do not understand that I am.

    Uhmm, this x point I created is mine. Can you find this point anywhere else while you are giving your opinions?

    No, I am simply interested in demonstrating my point. Though I do have an attitude with rpenner.

    It is my property this point and I have proven t' = t. I was waiting for you to say you could not prove it.

    Wrong, this point requires a rather large y. This implies you do not understand LT.



    I have a picture, why do you need it?

    This stupidity above indicates you do not realize this point is not in the domain of the rigid body sphere without a y component.

    You should have read my proof more closely.



    The point exists between O and O' and a light beam could not be there unless O' travels faster than light.

    You just made an ass of yourself.
     
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  7. Jack_ Banned Banned

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  8. RJBeery Natural Philosopher Valued Senior Member

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    Hi Jack! I could digress about the demeanor of this thread (and frequently this forum) when someone has challenging ideas, but I won't. Your LT conclusions are presuming that Lambda does not equal 1, which I believe it would since O and O' are in the same inertial frame. That is the only solution that satisfies both the clock-sync and LT analysis.:bagpuss:
     
  9. Jack_ Banned Banned

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    If you are saying what I think you mean then I agree.

    The only solution is that Lamda = 1 for the entire experiment.

    It is impossible that Lambda > 1 for a valid clock sync, ie in the same frame.

    The only way LT and the clock sync are satisifed is if v never existed which is a contradition to the experiment.
     
  10. RJBeery Natural Philosopher Valued Senior Member

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    Wait, what? What do you mean by "if v never existed"? V does not exist in the inertial frame, does it? My point is that both the LT and clock-sync methods would show that O and O' are the same age. Despite the "sausage making" in the first few steps, your experiment would make t = t' and lambda = 1.

    Please Register or Log in to view the hidden image!

     
  11. Jack_ Banned Banned

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    Well, technically you are correct.

    However, SR proves time dilation is reciprocal.

    As such, O believes O' is younger and O' believes O is younger.

    Hence, the only way to make all agree, ie LT and the clock sync, is if the relative motion period did not exist. That is the v I meant.
     
  12. RJBeery Natural Philosopher Valued Senior Member

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    lambda = 1 IS reciprocal. The relative motion period is irrelevant. Why do you think that the earlier relative v requires that lambda not equal to 1 under the inertial frame?
     
  13. Jack_ Banned Banned

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    We are not communicating.

    lambda = 1 IS reciprocal.

    Please write the math for this.
     
  14. RJBeery Natural Philosopher Valued Senior Member

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    Err, I don't understand what you mean I guess. You said that SR time dilation is reciprocal and I don't dispute this; I said that a time dilation factor of 1 (meaning, NO time dilation factor exists because there is no relative movement) conforms to this reciprocality. What math are you asking for?

    Also it would help if I understood why you think that the earlier relative velocities require that lambda is not equal to 1 under the inertial frame, and whether you believe that t and t' are necessarily unequal.

    Put simply, why don't you plug in true numbers to your original experiment and maybe we can discover the source of confusion?
     
  15. AlphaNumeric Fully ionized Registered Senior Member

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    The motion is in the x direction therefore the Lorentz transformation between the frames of O and O' will only affect the (x,t) and (x',t') coordinates. Directions transverse to the boost do not transform so you have y'=y and z'=z. Further more since the first points to be hit by the photon sphere in particular frames are those on the y=z=0 part of the sphere, precisely due to the motion in the x direction, the first moment at which the spheres intersect can be found by simply ignoring the y and z directions. Just draw a picture and its obvious.

    The fact transverse directions don't transform is seen in any book on special relativity. A Lorentz boost in a particular direction in N dimensional space reduces to a Lorentz boost in 1 dimensional space because you can choose your coordinates (or use Lorentz rotations) to make the direction the boost is applied be the x direction. All other directions are left unchanged.

    Hence its all the most humorous you keep saying I don't understand Lorentz transformations when I have to keep walking you through the stuff covered in a 1st year introduction to SR course. Further more, why is it you're the only person in 100 years to see such an error in such a simple experiment? Initially relativity was not popular and a great many people tried to come up with contradictions in it and they didn't succeed. You give the distinct impression you've never even done special relativity which would mean you believe you grasp something you've never studied better than people who spent 50 years studying it.

    Further more you never replied to my comment that the consistency of special relativity is equivalent to that of standard Euclidean geometry. Was it a little over your head?

    If you haven't copied and pasted it from somewhere then you have done an extremely poor job explaining yourself because you use expressions you don't define. If you can't explain yourself properly you haven't supported your case. And as my bit of algebra shows the claim t'=t is not true.

    A sphere centred on the origin is defined by \(\sum_{n} x_{n}^{2} = R^{2}\). A deformed sphere is obtained by rescaling each direction to get \(\sum_{n} \lambda_{n}^{2}x_{n}^{2} = R^{2}\). A deformed sphere centred on (y_{N}) is \(\sum_{n} \lambda_{n}^{2}(x_{n}-y_{n})^{2} = R^{2}\).

    In Frame 1 the sphere of O is centred on the origin and is not deformed. The sphere of O' is centred on \((vt,0,0)\) and is Lorentz contracted in the x direction. The photon sphere is centred on the origin, is not deformed and has radius ct. In Frame 2 the sphere of O' is centred on the origin. The O sphere is centred on (-vt',0,0) and Lorentz contracted in the x direction. All that remains is the photon sphere and it follows that you need only consider the sphere's x' coordinates as the \((\pm ct',0,0)\) point is on the sphere and will hit the spheres at the same time as or before the rest of the sphere.

    Hence you can collapse the situation down into considering the x'/x coordinates and set y=z=y'=z'=0 because the point on the sphere with the largest |x| or |x'| value is the one of interest since they are the ones which feel the Lorentz contractions.

    Its basic geometry.

    You have not justified this claim in any way. You set up the system and then you just wave your arms with undefined and unexplained mathematics and even then you don't come down to showing 1=2, you just state an expression and make the jump to "And thus it moves faster than light". Surely if SR implied that you could end up with an expression of the form "V = c + (some positive quantity)"? Because that's what you're claiming but its not what you've shown.

    Given that the photon sphere doesn't behave ballistically the situation is symmetric between the two spheres. Each one sees a photon sphere appear at their coordinate origin and remain centred on it. Hence why they see reciprocal times. A symmetric system, a symmetric answer. This is nothing more than just a system where two observers do symmetric motions. It only seems like it isn't if you think the sphere of photons doesn't obey special relativity and behaves ballistically because then you'd have one of the spheres seeing it's centre move.

    You make an extraordinary claim and no matter how many people discuss it with you you still tap dance around your supposed evidence. If you are right then you can either work through clearly explained special relativity algebra and get 1=2 or you can, so you claim, get a velocity larger than light for something physical. You have done neither, the only algebra you've provided has been poorly explained and several expression pulls from nowhere.

    I'm also wondering why you have a beef with myself and Rpenner? Perhaps I'm too suspicious but given your "You're lecturing me on mathematics?!", your unwillingness to say what maths you actually know, your dislike for the two of us (even though I don't recall ever cross paths with you in other threads before this week), your willingness to proclaim enormous mathematical achievement but then provide very poorly written explainations which don't define the expressions you use you act a lot like Reiku/NeoNo.1/gluon/Photino/Gareth Lee.

    But hey, cranks often share so many attributes they all blur into one.... :shrug:
     
  16. RJBeery Natural Philosopher Valued Senior Member

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    AlphaNumeric: I know you're involved in a personal exchange but, if you care, I posted a possible resolution to the OP in post #25.
     
  17. Jack_ Banned Banned

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    There is no reason to plug in numbers.

    O believes O' is younger.

    O' believes O is younger.

    The clock sync produces one answer.

    There is no confusion on my part since this is a contradiction.
     
  18. Jack_ Banned Banned

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    He and I are on a different subject.

    No, you did not resolve it.

    There is a relative motion period like the normal twins paradox.
     
  19. RJBeery Natural Philosopher Valued Senior Member

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    What you write is not true when lambda = 1. Lambda = sqrt(1-(v^2/c^2))^-1, and when v = 0 (which it would because they are in an inertial state with no relative movement), then you get Lambda = 1/sqrt(1). SR being reciprocal, as you say, does not necessitate that each observer believes the other is younger - they can both believe they are the same age and the reciprocality is not broken.
     
  20. Jack_ Banned Banned

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    You need first to mmake sure they are in the domain of the light sphere.

    In this case x' = -x.
    Hence,

    So, this co-located point between the two frames is located left of the origin of O' and right of the origin of O.

    Therefore, since O and O' were co-located at light emission, it is impossible for light to be behind the O' frame, ie left of it, when y'=0 and z'=0.

    That is why the thoight experiment was careful to define a y ≠ 0 in order to satisify the properties of beiing on the light sphere.

    I did give a direct answer. We are exploring that right now.

    As I said, this is my work, but I guess you are calling me a liar.

    Proof. It is ugly and I hope I have all the parens correct.

    t' = ( t - vx/c² )λ

    set x = (rλv)/(c(1+λ)) and t = r/c

    t' = ( r/c - (rλv²)/(c³(1+λ)) )λ
    t' = rλ/c( 1 - (λv²)/(c²(1+λ)) )

    t' = r/c( λ - (λ²v²)/(c²(1+λ)) )

    t' = r/c( (λc² + λ²c² - (λ²v²))/(c²(1+λ)) )

    t' = r/c( (λc² + λ²(c² - v²))/(c²(1+λ)) )

    since
    λ = 1/√( 1 - (v/c)²)
    then
    λ² = c²/( c² - v²)
    substitute for λ²
    t' = r/c( (λc² + c²/( c² - v²))(c² - v²))/(c²(1+λ)) )
    t' = r/c( (λc² + c²))/(c²(1+λ)) )
    t' = r/c( c²(λ + 1))/(c²(1+λ)) )
    Hence,
    t' = r/c = t


    y' and y cannot be set to zero for the light circle in this case. This circle however, can be rotated about the x-axis to generate the sphere.

    But, if you want to work just with LT, it is OK to ignore the y and z for length contraction.

    As you can see in the picture I posted, the moving sphere will be an ellipsoid. So, this part we are not in disagreement.

    So, I used LT to show x' = -x and t' = t.

    Then, I included the entire light circle with these values and since r is the radius, the y is forced.



    You are now reduced to calling me names.

    Now, I proved the point is on the light sphere in both frames.

    Also, I provided an extremely specific point x = (rλv)/(c(1+λ)) for all to examine and plug into LT.

    I proved when t=r/c this point is hit in the O frame because of the y which puts it on the rigid body sphere.
    All can examine this.

    Also, I proved when t=r/c in O, then t'=r/c for light hitting this point in the O' frame.
    All can examine this.
    So, the mapping when t = r/c, is specifically,
    (x,y,r/c) is mapped to ( x',y',r/c)

    Each piece of proof can be examined and scrutinized.
    It is one thing to say this is all wrong.
    Since it is based on math proof, it is another thing to prove it is wrong.

    Say, are you in disagreement with the twins paradox above?

    Math and proof only please.
     
  21. Jack_ Banned Banned

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    You have left of the history of the clocks because they were brought back into the same frame.

    Here is a partial solution to the normal twins paradox.
    It is only from the view of the stationary observer.
    http://en.wikipedia.org/wiki/Twin_paradox

    If you look at phase 2 and 5, you will note time dilation is part of the clock's history and counted when the clocks are brought together in the same frame.
     
  22. AlphaNumeric Fully ionized Registered Senior Member

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    No, x' = -x is a parity transformation. It is neither a boost nor a rotation. If you knew about Lorentz transformations you'd know that they have determinant 1 in the GL(4,R) representation and the transformation of coordinates you've just written down has -1 determinant. If you expand your considerations to the multiply connected set of SO(3,1) matrices you can include such a transformation but it is not the correct one.

    The parity transformation will not lead you to the correct results. By expanding your Lorentz group from the orthochronous case, where (in the GL(4,R) rep) \(\Lambda_{0}^{0} > 0\) and \(|\Lambda|=1\), to include the cases of \(\Lambda_{0}^{0} < 0\) and \(|\Lambda|=\pm 1\) you end up with non-physical results. Swapping the axes as you've done will get your left and right (or the sign on your x) the wrong way around and that forms the entirity of your argument.

    Oh and if you're so mathematically adept surely you know that \(x^{2} = R^{2}\) is a sphere? It's an \(S^{0}\) submanifold of \(S^{n}\). All spheres \(S^{n}\) contain \(S^{m}\) for m<n as submanifolds, done by seeing some of the Cartesian coordinates to zero. So the point (R,0,0) is on the sphere \(x^{2}+y^{2}+z^{2} = R^{2}\). Don't you remember your basic geometry?

    Yes, the result when they measure the photon sphere hitting their sphere is the same value. But that's not the same as saying t=t' as that equated coordinates.

    They can. I explained why transverse coordinates are not relevant.

    Once again you are sloppy with your explainations. What you mean to say is that the event happens in the (x,t) coordinates at (X,T) and in the (x',t') coordinates at (-X,T). Saying x=-x and t'=t is equating coordinates and obviously they do not relate to one another in that manner.

    So what? I mentioned in a previous post the system is symmetric, since the motion of the photon sphere's emitter is irrelevant, hence by doing a boost and a parity flip you get the same system. This isn't contradictory in any way.

    You haven't proven any paradox. Of course for symmetric setups you get symmetric timings. You don't seem to like the fact they each say the other is younger but this is not a problem since they are different points in space and in relative motion. In order to properly compare one another's clocks (so as to avoid the additional complication of having to account for message relay times) they must undergo acceleration and meet up. If they do it symmetrically they find each is the same age, a result which follows from how one another sees time past for the other. If they do it asymmetrically they'll get a difference.

    This kind of warping of aging is a standard textbook example. You find that as they accelerate to approach one another they see the other age much much faster than normal and suddenly their clocks tend towards agreement, which is obtained when they stop next to one another. You can see this by considering what would happen if they shine torches at one another all the way through. While moving away they see red shifted light, which equates to seeing one another age more slowly. When moving towards one another they see blue shifted light, which equates to seeing one another age quicker. Should there be any asymmetry in their accelerations there's an imbalance and one ages less than the other. If its all symmetric then they age the same amount.

    No contradiction, just a bit of counter intuitive results and annoying algebra. You have done the 'away' trip but not the return trip and unless they are at the same point in space to compare clocks you aren't immediately say its a contradiction. They move symmetrically and both find that T amount of time has passed before the photon sphere hits their sphere. No contradiction. They then come back to compare clocks and this blue shifting sorts everything out.
     
  23. RJBeery Natural Philosopher Valued Senior Member

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    The history of the clocks does not affect the time dilation calculation for lambda. Do you think it does?
     
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