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Time dialtion tick tick skeptick

  1. Nov 14, 2007 #1
    Please read the attached document and images. The other images can be found in the next post. Please feel free to critisise what I propose.
     

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  2. jcsd
  3. Nov 14, 2007 #2
    next images

    here are the rest of the images for my original post.
     

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  4. Nov 15, 2007 #3
    revised document

    Sorry folks I have made some minor amendments to the document as it contained some grammatical errors, it is it now easier to read and understand. The images all remian the same.
     

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  5. Nov 15, 2007 #4
    e_Freddo

    The first person to find a flaw in my reasoning wins a zero calorie e_freddo :)
     
  6. Nov 15, 2007 #5

    JesseM

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    The flaw is that you seem to be unaware of the relativity of simultaneity, which says that there is no frame-independent definition of what it means for clocks to be "synchronized"--if two clocks are synchronized in their own rest frame, they will be out-of-sync in a frame where they are moving. This is a natural consequence of the fact that each frame assumes light moves at c in that frame, so that clocks at rest in this frame can be synchronized using this assumption...if I am on board a rocket ship and I want to synchronize two clocks at the front and back of the ship, I can just set off a light flash at the midpoint of the two clocks, and then set each clock to read the same time at the moment the light from the flash reaches it. But if in your frame the rocket is moving forward, then naturally if you assume the light from the flash moves at c in both directions in your own frame, then you must say the light reaches the back clock before the front one, since the back clock is moving towards the point where the flash was set off, while the front clock is moving away from that point. So, my synchronization procedure is going to leave the two clocks out-of-sync in your frame.

    A useful formula in SR is that if two clocks are synchronized in their mutual rest frame, and the distance between them in their rest frame is L, then in a frame where the clocks are moving at speed v along the axis between them, the time on the back clock will be ahead of the time on the front clock by vL/c^2. So although you're right that in fig. 2, the time as measured in this frame for the light to go from the green end to the red end will be L^/(v + c), where L^ is the Lorentz-contracted distance between the clocks (L^ would be equal to [tex]L * \sqrt{1 - v^2/c^2}[/tex]) and v is the speed of the clocks in this frame, this does not mean that the time on the green clock when the light leaves it will differ from the time on the red clock when the light hits it by L^/(v + c)...instead you must take into account both the fact that the red clock was ahead of the green clock by vL/c^2 when the light departed from the green clock, and the fact that both of these clocks were ticking more slowly in this frame by a factor of [tex]\sqrt{1 - v^2/c^2}[/tex] due to time dilation, so that even though a time of L^/(v + c) passed in this frame the two clocks only ticked forward by [tex]\sqrt{1 - v^2/c^2}[/tex] * L^ /(v + c), and plugging in L^ = [tex]L * \sqrt{1 - v^2/c^2}[/tex] this means they ticked forward by L * (1 - v^2/c^2) / (v + c). Simplifying a little:

    L * (1 - v^2/c^2) / (v + c) =
    L * (1/c^2) * (c^2 - v^2) / (v + c) =
    L * (1/c^2) * (c + v)*(c - v) / (c + v) =
    L * (c - v) / c^2.

    Now, if both clocks ticked forward by L * (c - v) / c^2 in the time between the moment the light left the green end and the moment it reached the red end, and the clock on the red end was ahead of the clock on the green end by vL/c^2, this means the time on the clock on the red end when the light reaches it will be greater than the time on the clock on the green end when the light departed it by:

    [L * (c - v) / c^2] + [vL/c^2] =
    [(cL - vL)/c^2] + [vL/c^2] =
    cL/c^2 =
    L/c.

    Of course, this is exactly the same difference you'd expect if you calculated things from the perspective of the two clocks' rest frame, where the distance between them was L and the light moved at c between the green end and the red end.

    A similar analysis could show that in fig. 3, where the green end is now positioned on the back end instead of the front end (relative to the direction of motion), although it will indeed take longer for the light to go from one end to the other in the frame where the clocks are moving, this frame will nevertheless predict that the difference between the time on the green clock when the light leaves it and the time on the red clock when the light arrives will still be L/c, because of the way that both clocks are ticking slowly and the clock on the green end is ahead of the clock on the red end.
     
    Last edited: Nov 15, 2007
  7. Nov 15, 2007 #6

    Dale

    Staff: Mentor

    Hi Skeptick,

    Many people will not download .doc files or any other type of file that could contain a macro. I would recommend simply posting your comments directly rather than through a word file if you want a wider review.

    That said the mistake or flaw is almost always neglecting the relativity of simultaneity.
     
  8. Nov 15, 2007 #7

    robphy

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    I wonder... are attachments uploaded to PF scanned for virii or other malware?
     
  9. Nov 15, 2007 #8

    JesseM

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    I've already downloaded everything so I'll post the text myself, although if you want to see the diagrams you'll have to download them:
    (edit: I uploaded the diagrams to imageshack so you could also see them that way--fig. 1, fig. 2, fig. 3, fig. 4, and fig. 5)
     
    Last edited: Nov 15, 2007
  10. Nov 15, 2007 #9

    Dale

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    Probably. I'm just paranoid.
     
  11. Nov 15, 2007 #10

    robphy

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    I hope so.
    I'm paranoid too... but I opened the attachment anyway... [after making sure my anti-virus program is up to date].

    Too bad something like http://view.samurajdata.se/ didn't work on that attachment.
     
  12. Nov 15, 2007 #11
    virus

    FYI

    You cannot catch a virus from a bitmap file .bmp

    You can catch a virus from a word doc but only if you have macros enabled. This word doc does not use a macro so you can view it with macros disabled.

    Also you can catch a virus from carney so be wary of them as well
     
  13. Nov 15, 2007 #12

    robphy

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    It's better to use standard web formats like .gif or .png or .jpg... which have better support for compression. 100kb is a lot for very simple graphics.

    I'm thinking of use either antiword, Open Office, or some online-service to read .doc files from the internet.
     
  14. Nov 15, 2007 #13

    jtbell

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    I took "fig 1.bmp" from the first post and converted it to a GIF. Let's see how big it turns out to be after I attach it here... wow, only 1.9 KB! :bugeye:

    The difference between 1.9 KB and 139.1 KB may not be important for people with broadband connections, but for someone like me who uses a dialup connection at home, it makes a real difference.
     

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    Last edited: Nov 15, 2007
  15. Nov 16, 2007 #14

    Chris Hillman

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    A Degree of Paranoia is Healthy (on the web)

    As we all should be, unfortunately.

    My understanding is that at PF, user-uploaded images (e.g. gif files to illustrate a post) are not made "live" until they can be checked for content and scanned for viruses by a moderator, but I've seen a number of practices I consider troubling, such as hiding the url of a link using VB code (some browsers can be configured to state the url and even do an ARIN lookup if one mouses over a live link) to a possibly unsafe external website. I am not sure whether all uploaded files are scanned at PF before being made live but they certainly should be. There was a recent discussion about security issues raised by a proposal to loosen restrictions for regular posters at the Homework Help forum in the Feedback forum.

    I also caution PF users from believing non-authoritative (or even seemingly authoritative) statements of the form "it is not possible to catch a virus from..."). Unfortunately, closer examination often reveals that someone making such statements simply hasn't enough knowledge of the amazing variety of "exploits" used by criminal phishing gangs. As I wrote in the discussion just mentioned, it is also possible to go too far in the other direction; as always, its a matter of finding the right balance, which comes with knowledge and experience.
     
  16. Nov 16, 2007 #15
    Forget relavtivity this exeriment is taking place inside a snail

    Sorry but I have thought of the synchronisation of the clocks. I would synchronise them thus:
    I would place the two clocks as close together as possible. Depending on the physical construction of the clocks this means the actual timing mechnism could be only atoms apart. I would then synchronise the clocks thus ensuring the time difference between the two clocks was virtually zero, or the distnace between atoms divided by C. I would then move the clocks apart to thier final positions. As they were synchronised before they were moved apart they will still be synchronised afterwards. All mopvements are as slow as you like say 0.00000000001 m/ 1 billion years.

    In my scenario the speed of the MFR 10000 m/sec which is too slow for the effects of relativity to make a difference. Make the MFR's velocity as slow as you like 1 kilomteer per 1 million years if you like the maths remians the same. As the speed of the MFR is not relativisitic the effects of simutenaiety can be ignored the time diffrenece of the two clocks would be negligable.

    Secondly if you look at fig 1 I have a clock at the green end and a clock at the red end. As explained the clock at the green end stops when a phton leaves the green end.

    If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is travelling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end simutenaiety effects can be ignored.

    Same for the red end the clock at the red end stops when a ptohon arrives at the red end. The clock is as close as you like to the red end. Again negating simultenaiety.


    .........next
     
  17. Nov 16, 2007 #16
    this should read ...............

    If the clock at the green end stopped when a photon from the red end struck it then simultenatoiety may be a problen assuming this whole gizmo is travelling very fast but as the clock at the green end stops when a photon leaves the green end and the clock at the green end is as close as you like to the green end and the whole gizmo is travelling at the pace of a snail simutenaiety effects can be ignored.
     
  18. Nov 16, 2007 #17

    JesseM

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    No, this method won't synchronize them in any universal sense. Are you familiar with the concept of "limits" in calculus? In the limit as the speed of the two clocks relative to one another approaches zero, there will be some frame in which the velocity of both clocks is approaching zero as well (the rest frame of whichever clock is not being changed in the limit), and in this frame and this frame only your method will result in the clocks being arbitrarily close to being perfectly synchronized in the limit as their velocity relative to one another approaches zero. But in some other frame--say, a frame where the first clock is moving at 0.8c and the second clock is moved away from it at 0.800000000000000000000000000000000000000001c, they will not remain close to synchronized if they are given enough time to move some significant distance apart...in fact, if you move them a distance of L apart from one another, in this frame your method will result in them being out-of-sync by an amount that's arbitrarily close to the amount they'd be out-of-sync if you use the conventional Einstein synchronization method involving light-signals. I gave a proof of something basically identical in post #41 of this thread, where someone suggested that you could synchronize two clocks C1 and C2 at a fixed distance apart by moving a third clock C3 very slowly between them, with C1 and C2 set to read the same time as C3 at the moment C3 passed each one. Here's what I said there:
    But if the velocity v of the clocks in the MFR is negligible, then the difference of L^ /(v+c) from L/c will be negligible too (because the Lorentz contraction will be negligible, and the difference between c and v+c will be negligible). To the degree there is any small difference, it will be exactly compensated for by the small Lorentz contraction, time dilation and lack of simultaneity between the two clocks, in exactly the right way to ensure that, if the green clock stops when the light leaves it and the red clock stops when the light arrives at it, then the difference in readings between the two stopped clocks will end up being precisely L/c.
    The difference in simultaneity means that in the MCR, if the clock at the green end reads a time of T at the exact moment the light departs from the green end (the moment when the clock is stopped), then at that precise moment in the MCR the clock at the red end will read a time of T + vL/c^2, where L is the distance between the clocks in their rest frame and v is the speed of the two clocks in the MCR (this is assuming the clocks have been synchronized using a method that gives the same definition of 'synchronized' as the Einstein clock synchronization convention, which would include your proposal of synchronizing them at a common spot and moving them apart with an arbitrarily small relative velocity).

    You are free to assume the the velocity of the two clocks v in the MCR is negligible compared to see, in which case T + vL/c^2 will differ negligibly from T, but the result will be the same--if the clock on the red end also stops when the light from the green end reaches it, and we examine the readings on both clocks after they have stopped, their readings will differ by precisely L/c.
     
    Last edited: Nov 17, 2007
  19. Nov 18, 2007 #18
    Lorentz contraction LC

    from wikipedia

    Length contraction, according to the special theory of relativity, which was formulated in the early twentieth century through the seminal work of Einstein, Poincaré and Lorentz, is the physical phenomenon of a decrease in length detected by an observer in objects that travel at any non-zero velocity relative to that observer

    In my example the observer is in the MFR there is no relative movemenet hence no LC so that can be diregarded. The clock is fig 2 would be exactly the same length as the clock in fig three they are both in the same MFR moving at the same velocity the only difference is the clock in fig 3 is 180 degree rotation to clock fig 2

    Time dialation same deal ... again wikipedia
    Time dilation is the phenomenon whereby an observer finds that another's clock which is physically identical to their own is ticking at a slower rate as measured by their own clock. This is often taken to mean that time has "slowed down" for the other clock, but that is only true in the context of the observer's frame of reference. Locally (i.e., from the perspective of any observer within the same frame of reference, without reference to another frame of reference), time always passes at the same rate. The time dilation phenomenon applies to any process that manifests change over time.

    In my example the observer is in the MFR there is no relative movement/Frame hence no Time Dialation so again that can be diregarded


    The person in the MFR synchronises the clocks, runs the experiments as shown in fig 1, 2, 3, 4, 5
    when experiment in fig 2 is run the observer in the MFR will see time speed up. When experiment 3 is run the observer in the MFR will see time slow down. explain this without time being a vector


    An observer external to the MFR would see the same as the person internal to the MFR.

    In the experiment depicted by fig 2 an external observer would percieve time to speed up
    In the experiment depicted by fig 3 an ext observer would percieve time to slow down

    Thankyou for your post it has given me food for thought
     
  20. Nov 18, 2007 #19

    JesseM

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    I thought you were using the "MFR" to refer to the frame in which the two clocks are in motion, ie the frame where you calculated the time to be L^ /(v + c). If you intend the MFR to be the frame where the two clocks are at rest, then of course if the distance between the clocks is L and light moves at c in this frame, the time for the light to go from the green end to the red end will be L/c. But my analysis shows that if you analyze the situation from the point of view of the frame where the two clocks are in motion--whatever you want to call this frame--then even though the time measured in this frame will be L^ /(v+c), the difference in readings between the two clocks after they both stop will still be predicted to be L/c, because of the way the clocks are ticking slowly and the way the distance between them is less than L and the fact that they are out-of-sync in this frame.
    Again, I'm not entirely clear on what you mean the "MFR" to be--is it the clocks' own rest frame, or the frame in which they are in motion? Your figures 2-5 seem to be drawn from the perspective of a frame where the clocks are in motion, since they show the position of the green and red end being different at the moment the light departs the green end and the moment the light arrives at the red end.
    If by "external observer" you mean the observer in the frame where the clocks are moving, I've already shown that this observer will predict the difference in the two clocks' readings will be L/c (although the time as measured in this frame will be different from L/c).

    Would you like to go through a numerical example? You can tell me the velocity v of the clocks in the external observer's frame, the distance L between the clocks in their own rest frame, and I can show that because of the way the clocks are slowed down and out-of-sync in the frame where they're moving, the difference in their readings after both stop (assuming the clock on the green end stops at the moment the light departs that end, and the clock on the red end stops at the moment the light arrives at that end) will work out to L/c. This will be true regardless of whether the clocks are moving left-to-right or right-to-left in the external observer's frame, I can do the calculation both ways.
     
  21. Nov 20, 2007 #20
    Ill attempt to clear up one problem at a time

    I am in a space shipthat is moving, can I sunchronise two clocks in the space ship Yes / No?
     
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