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A Relativity Experiment I'd Like to See

  1. Nov 29, 2011 #1
    So imagine that time is very slow, like 1 meter per second. Yes I know this is impossible, but I'm tired of space ships, OK?

    So you and another cowpoke named Pedro want to have a mosey contest. You face each other ten meters apart. You each have a clock and a mirror. Each cowpoke can see his own clock in the other guy's mirror. You fiddle with your clock until the clock in Pedro's mirror and Pedro's clock match. Pedro of course sees his clock in your mirror matching your clock.

    After this synchronization you then commence your mosey towards Pedro. Your clock in Pedro's mirror will end up behind Pedro's clock. I have trouble visualizing the result because the slow light, acceleration, and dilation get all confounded so it is hard to say exactly what happens when.
     
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  3. Nov 29, 2011 #2

    Dale

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    This is a self contradictory setup. If you see your mirrored clock match Pedro's, then Pedro sees his mirrored clock behind yours.
     
  4. Nov 29, 2011 #3

    Suppose your clock and Pedro's are already synchronized at zero. It takes ten seconds for the light from your clock to get to Pedro's mirror. During that time his clock ticks ten seconds. So his clock and your clock's reflection are synchronized. It takes ten more seconds for you to see this, so you see two synchronized clocks that are both twenty seconds behind your clock.

    It is exactly the same for Pedro. So everything is synchronized until you commence your mosey.
     
  5. Nov 29, 2011 #4

    PAllen

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    No, this doesn't work. Each would see the reflection of their clock 10 seconds behind the image of the other's clock (with the description above). At t=20, each would see a reflection of their clock reading t=0, and an image of the other clock reading t=10.

    As Dalespam said, you can achieve your goal for either observer, but not for both.
     
  6. Nov 29, 2011 #5

    Dale

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    No. So your clock's reflection is ten seconds behind. The image of your clock at t=0 arrives at Pedro when his clock reads t=10.
     
  7. Nov 29, 2011 #6
    By gum you are right. You have to compare the two Pedro clocks with your own clock and split the difference.
     
  8. Nov 29, 2011 #7

    ghwellsjr

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    The whole point of synchronizing clocks is so that they have the same time on them even when they don't look like it because of the delay caused by the speed of light. So if the two clocks are synchronized to zero while ten meters apart, then when you look at Pedro's clock, you won't see it reading zero until ten seconds later. Meanwhile your clock sent an image of zero time to Pedro's mirror which will take the same ten seconds and another ten seconds for the image to get back to you. So you won't see your own clock in the mirror reading zero until twenty seconds after it actually read zero.

    The process of synchronizing your clock to a remote clock is to divide the round-trip time delay of sending a signal from your clock to the remote clock in half. So what you do is look directly at the time on your own clock and subtract the time you see on the reflected image of your own clock. This gives you the round-trip time delay between your clock and the remote clock. You divide this time in half. We'll call this the half-time. Then you adjust the time on your clock so that it reads the same as the remote clock [STRIKE]minus[/STRIKE] plus the half-time.

    You can always verify that your two clocks are synchronized by making sure that the time on the remote clock is half way between the actual time on your clock and the time on the reflected image of your clock.

    If there is no time dilation in your fictitious physics, then when you mosey towards Pedro, your clocks will remain synchronized and when you get together and all clocks and reflections read the same time.

    EDIT: got the arithmetic backwards.
     
    Last edited: Nov 29, 2011
  9. Nov 29, 2011 #8

    ghwellsjr

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    You've got this backwards. You have to compare your two clocks--the actual with the reflected and and adjust yours to split the difference with Pedro's clock.
     
  10. Nov 29, 2011 #9

    PAllen

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    The way I think about this is that if Pedro's clock (as you see it) remains half way between your clock (now) and the reflected image of your past clock, then Pedro's clock is synchronized with yours.
     
  11. Nov 29, 2011 #10

    ghwellsjr

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    OK, but if it's not, then what do you do? Send a message to Pedro to get his act together or do you take action yourself and change the time on your clock? Whatever you do, the two of you need to agree on which clock is the master and which is the slave. Only the slave should be changed or otherwise you'll never get them synchronized.
     
  12. Nov 29, 2011 #11

    PAllen

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    You could change neither. Each adopts a correction to the other's clock. Then each can treat other clock readings+correction as synchronized. No master/slave needed.
     
  13. Nov 29, 2011 #12

    ghwellsjr

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    Does that work with more than two clocks?
     
  14. Nov 29, 2011 #13

    PAllen

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    Yes, but does get messy - everyone needs a correction for every other clock. In that sense, prior agreement on a master is simpler. Then each non-master adjusts theirs by the difference between halfway between current time and reflected image of past clock versus image of master clock.
     
  15. Nov 29, 2011 #14

    ghwellsjr

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    I don't see how this can work. The whole point of synchronizing clocks is to establish a Frame of Reference in which every clock at every location has the same time on it. So let's work out an example for your way of doing things:

    Let's say that three clocks are 10 meters apart from each other forming an equilateral triangle. Let's say that A reads 0, B reads 10 and C reads 20. 20 seconds later, they will be able to do the calculation and A will say that the sync time is either 30 or 40, B will say it is 20 or 40, and C will say it is 20 or 30. There isn't even an option that produces the same time for all of them.

    So I say let's agree with Einstein's 1905 paper:
    If the clock at A synchronizes with the clock at B and also with the clock at C, the clocks at B and C also synchronize with each other.​
     
  16. Nov 29, 2011 #15

    PAllen

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    For 3 clocks with no prior synch, that are at mutual relative rest, you would simply have 3 variants of the same frame.

    A, B and C each have their own version of synchronized time. Each is as good as the other. Each can call themselves the master. (I assume you are pretending c = 1 meter / second from you statement about 20 seconds later).

    A's version: what I literally see on B clock is what I call now at B's location; what I see on C clock - 10 is now at C.

    B's version: what I see on A clock + 20 is now at A; what I see on C is now at C.

    C's version: what I see on A + 30 is now at A; what I see on B + 20 is now at B.

    Each one synchronizes all other clocks 'virtually' by having a correction for each other clock.

    It is the same synchronization convention, except that the synchronization is done by per clock correction factor by whoever is reading the clocks.
     
  17. Nov 29, 2011 #16

    ghwellsjr

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    OK, if you are describing three individual Frames of Reference, then yes, I see that it can work. Thanks for clarifying.
     
  18. Nov 30, 2011 #17
    Anyway, now that we've taken care of synchronization, what do you see on Pedro's clock and in the mirror as you walk toward Pedro? Any takers?
     
  19. Nov 30, 2011 #18

    ghwellsjr

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    I already answered your question in post #7:
     
  20. Nov 30, 2011 #19

    ghwellsjr

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    If you do want to consider time dilation similar to what actually happens, then you can use Einstein's formula (found near the end of section 4 of his 1905 paper) for calculating the time difference between two previously synchronized clocks when one of them is moved to the location of the other one:

    ½tv2/c2

    So if you want your mosey to be done at one-half the speed of light so that it takes you 20 seconds to get there, then your clocks will differ by ½(20)(½)2 = 10/4 or 2.5 seconds.
     
  21. Nov 30, 2011 #20
    Yes, I can get that far. Also you won't see any dilation for the first fifteen seconds or so of the mosey because of delay due to the slow light. Instead you will see Pedro's clock and the mirror clock running fast at the same rate, then the dilation comes in. The question is, does the dilation happen linearly or does much of it occur during the acceleration?
     
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