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What are the four clock readings?

  1. Sep 30, 2012 #1
    The fact that two rocket ships passing each other see the other ships clock run slow fascinates me. But what about total elapsed time?

    Suppose a person on earth and a rocket ship on Mars 40,000,000 miles away are at rest relative to each other. They synchronize their clocks so that at t= 0 seconds the Martian ship accelerates to 95% of the speed of light and heads toward Eartha on Earth. The Martian ship flies very close to the Earth clock(about 1 meter) where Eartha and the Martian pilot can see each others clocks.
    At the moment they are side-by-side(with no Doppler Effect), they look at each others and their own clock, so that at closest approach there are four clock readings:
    1 Eartha looks at her own clock.
    2 Eartha looks at the Martian pilot's clock
    3 The Martian pilot looks at his own clock
    4 The Martian pilot looks at Eartha's clock

    The only answer that I am sure of is 1) Eartha will see a reading of 226 seconds on her clock, the time it takes the ship to travel 40,000,000 miles at 0.95c.
    What are the other clock readings for 2, 3, and 4.
     
  2. jcsd
  3. Oct 1, 2012 #2

    ghwellsjr

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    Since gamma at 0.95c is 3.2, we divide that into 226 seconds to get a reading of 71 on the Martian's clock. They both will see 226 on Eartha's clock and they both will see 71 on the Martian's clock.

    Why do you find this challenging?
     
  4. Oct 1, 2012 #3
    Little side note: In principle there always is a Doppler effect. What has been emitted perpendicularly wrt a moving source, is propagated at an angle and also arrives at that angle at a stationary detector. However, Doppler is irrelevant for answering your question if one can neglect the distance (but if not understood then it's bound to come back another day!).
     
  5. Oct 1, 2012 #4

    mfb

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    I think the unintuitive thing is 4 (rocket looks at clock on earth).
    It must be the same as earth's value (226s). But due to time dilation, only 71/3.2=22.2 seconds passed on earth's clock during the journey. The solution: In the frame of the moving rocket, earth's clock already started with 203.8 seconds on its clock. In the frame of mars, viewed from the same point in spacetime (just with a different velocity), the clock is still at 0 seconds.
     
  6. Oct 1, 2012 #5
    Mfb is right that #4 is unintuitive. The martian pilot reasoned that Eartha's clock runs slower than his own clock during his journey. The pilot is surprised that Eartha's clock didn't have a lower reading than his own clock reading of 71 seconds.
    But if the earth clock starts out with 203.8 seconds, that would solve the problem.
     
  7. Oct 3, 2012 #6
    starfish99, special relativity is only correct in uniform motion, get familiar with that concept. You have to use GTR for more correct calculations. Basically, the one who accelerates notices time dilation. In STR everyones clock is the right one, but that doesn't match physical reality very well.
     
  8. Oct 3, 2012 #7
    There is no need to bring in GR if you want to calculate the effects of acceleration; as long as you stick to inertial frames you will be fine and starfish apparently understands that. GR is required for the effects of gravitation.
     
  9. Oct 3, 2012 #8

    ghwellsjr

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    Special Relativity is also correct with accelerated motion of any kind. Where did you learn that SR cannot handle accelerations?
    Not if it is only motion that you are concerned with.
    Nobody ever notices time dilation. It's a calculation based on an arbitrarily selected inertial Frame of Reference.
    Yes, in Special Relativity, every clock keeps track of what is called its own Proper Time which can be calculated based on its speed in a Frame of Reference. All you have to do is calculate the reciprocal of gamma from its speed to determine how much slower than normal it is advancing. That matches physical reality very well as long as you can ignore the effects of gravity.
     
  10. Oct 3, 2012 #9

    They synchronize the clocks which is possible if we idealize mars and earth being at rest in the same inertial frame of reference.
    But let's look at what acceleration does to the Martian.

    Acceleration switches the Martian's inertial frame of reference.

    Now you have to map all the events you know of in the earth/mars rest frame into this new inertial reference frame the Martian stepped in and is moving at vrel 0.95c relative to earth/mar's rest frame.
    Let me add that along the path of mars to earth, a huge amount of clocks have been placed, which were also synced.

    Mapping the events from the earth/mars inertial reference frame, into the inertial frame of reference the Martian has accelerated into and is now moving at v= 0.95c relative to earth/mars, properly, according to SR, you will notice that every clock in the direction the Martian has accelerated towards and which resides in the earth/mars rest frame, has it's counter shifted up. The further away the clock, the higher the counter is up.

    Pretty much what mfb said, just a bit more elaboration on it.
     
    Last edited: Oct 3, 2012
  11. Oct 3, 2012 #10

    Erland

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    Again, it is the relativity of simultaneity that causes trouble. To understand this, we should also imagine a clock on Mars (considered at rest wrt Earth), synchronized with the clock on Earth, and a clock in another space ship going with 0.95 c behind the first one and going in the same direction, passing Mars when the clock on Mars reads 226 s. The clocks in the space ships should be synchronized with each other.

    I have said before that this a little difficult to imagine, and that it is easier to see what is happening if we think of a train, which is long with many cars, instead of a space ship with a quite small extension.

    This is done here, in a situation analogous to yours, in the following way (with the difference that the velocity is 0.87 c, not 0.95 c, and that there is no accelerartion up to 0.95, but that is not necessary, it would be the same if your space ship just passed Mars at t=0 with speed 0.95 c):

    West end of platform = Mars
    East end of platform = Earth
    Locomotive = space ship
    Last car of train = extra space ship mentioned above
    Train driver = pilot on space ship
    Person at east end of platform (there was none in the example but we can easily add one, agreeing with the station master about time) = Eartha
    Station master = person stationary on Mars (can be added to your example)
     
    Last edited: Oct 3, 2012
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