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Preferred and Unique Coordinate System

  1. Oct 8, 2012 #1
    I am confused about one of the basic findings of relativity, that all coordinate systems are equal and there is no preferred coordinate system.

    A simple thought experiment is to consider three spacecraft called left, middle, and right. Left speeds off at half the speed of light in the left direction relative to middle, middle stays put, and right speeds off at half the speed of light in the other direction. Also on each spacecraft is an equal pile of material which radioactively decays. The classic conclusion would be that middle's clock would tick faster and the material on spacecraft middle would have decayed more than the material on either left or right and that the material on left and right would have decayed an equal amount. However why is that the case? In left’s opinion if there is no unique velocity, left would be perfectly justified in concluding that it wasn’t moving and that middle would be speeding away and right would be speeding away even faster. With that assumption left would conclude that its pile of material would have decayed the most and that right’s would have decayed the least. Right could conclude the exact opposite. An independent observer with relative motion could conclude something else. In reality each pile of radioactive material decayed a specific unique amount. In the classic sci-fi scene were an astronaut speeds away in an advanced spaceship and returns to meet his great great grandkids, who is to say that the astronaut could not spend years on the spaceship and once they returned only seconds went by on Earth? To insist that Earth's time ticked slower implies that Earth is a more relavent or special coordinate system.

    Basically I think the rate of time's passage must be unique or else most of physics falls apart. Therefore there must be a unique coordinate system which energy levels must be measured against to get the "proper" rate of time's passage.

    I do not think that any simple coordinate system applies to any large region of space, but instead each bit of space has a unique coordinate system which may be twisted and deformed relative to other locations.

    Am I missing something basic here?
    Thanks
     
  2. jcsd
  3. Oct 8, 2012 #2
    Nope. Each spacecraft's description of what happens on the other ships is correct.

    In this case, the Earth is on a free-fall trajectory while the spacecraft has a proper acceleration. A free-fall trajectory between two points always is either the longest or shortest path. In this case, it is the shortest. This does mean that the spacecraft must experience less time according to its own clock than Earth does according to its own clock.
     
  4. Oct 8, 2012 #3

    zonde

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    Let me modify this example a little bit so that the astronaut is the one who experiences more time.
    And example is like this. Some time after astronaut have left Earth another astronaut leaves Earth and chases after him. When both astronauts meet second astronaut has aged less.

    Do you see how this example is similar to yours?
    Imagine triangle - then the person who is taking two sides of triangle is the one who experiences less time (other person takes one side).
     
  5. Oct 9, 2012 #4

    ghwellsjr

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    Ok, let's stipulate that your thought and conclusion are correct. How do you identify this unique coordinate system?
    Ok, let's stipulate that your thought and conclusion are correct. How do you identify these unique coordinate systems in each bit of space?
     
  6. Oct 9, 2012 #5
    Yes, you are missing the global view, still there is something to what you are saying in the local case, there are static solutions of the EFE that are limited to a certain region of the unverse, that give very good experimental predictions for solar system physics (think of the 4 classical GR tests). This regional static (and asymptotically flat) solution (not the general spacetime solution) has a coordinate system (schwarzschild coordinates) by definition.

    Even for cosmological solutions you do have "preferred" coordinates (like the FRW coordinates) that are identified as those that equate rest to lack of dipole anisotropy wrt the CMB, and are the only ones with wich one mathematically can clearly infer the universe expansion.
    All of these are however not "unique" according to GR's mainstream view.
    See above caveats.
     
  7. Oct 9, 2012 #6
    Do the same thought experiment, but before the rockets accelerate, place a huge amount of clocks from the middle towards the right side, and from the middle towards the left side. The clocks are synced before the acceleration.

    Assume the rockets with each having an observer at the back of the rocket, accelerate instantaneously. The back of the rockets are at the same place as where the atomic clock in middle resides. The back of the rockets is also where the atomic clocks inside the rockets reside.

    Acceleration is responsible for forcing the rockets left and right to change which inertial frame of reference they are at rest in.

    You have to map all of the spacetime events into the new reference system, left is now at rest in for example.
    When you do this, according to SR, you will see that locally, shortly after the acceleration, the atomic clocks are at zero still.

    However, the clocks in the direction the rocket is moving towards(seen from middle's perspective), are now showing a higher counter (seen from left/right's perspective). Higher, the further away they are.

    All the clocks put along the way are indeed ticking slower (seen from left/right's perspective). There is no doubt about this.
    An observer inside the left rocket comparing clocks which are moving towards him at vrel (remember he considers himself at rest), will notice that all clocks passing by his atomic clock will always show a higher counter, even though they tick slower. This is because they move too fast towards him, not giving his atomic clock at the back of the rocket enough time to make up for the initial shift of the counters after the acceleration.

    Just like when he accelerated initially, if he accelerates again and changes the frame he is at rest with to the frame middle is at rest in, locally there won't be any changes in the time clocks show. Again, in front and behind him, clocks will shift the counters. The direction he accelerated towards the clock counters will go up, while the counters of clocks behind will go down.

    Since he never managed to get any clock moving towards him when being at rest in the other frame to display a lower value than his clock (edit: actually, the longer he is in that frame at rest, the higher the difference between his atomic clock and the incoming clocks will become, when comparing clocks moving towards him, right when they are next to his atomic clock).
    When he accelerates and switches the frame he is at rest, to the same middle is at rest with, he will see all clocks along the line display the same time, as it should be in middle's frame, with his clock showing a lower counter. Lower, the longer he was in the former frame at rest.

    He and his clock therefore aged less.
     
    Last edited: Oct 9, 2012
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