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In this section explaining relativity

  1. Aug 22, 2005 #1
    In this section explaining relativity,

    " If a mirror clock and a conventional clock in the spacecraft agree with each other on the ground but not when in flight, the disagreement between them could be used to find the speed of the spacecraft independently of anyh outside frame of reference - which contradicts the principle that all motion is relative."

    i don't understand what is meant by why the principle is contradicting...
    can somebody explain why?
  2. jcsd
  3. Aug 22, 2005 #2


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    Let some outside inertial observer be in communication with the spacecraft and assume, contrary to actuality, that there is some velocity dependent difference between the mirror clock and the conventional clock. The observe uses radar to determine thespeed of the spacecraft relative to him, and queries the spacecraft crew as to the clock difference. On learning what it is, he can compute (by hypothesis) what the "absolute" speed of the spacecraft is, and be adding the relative speed from the radar, he can get his own absolute speed. This contradicts the principle that all inertial observers are equal, each can regard himself as at rest, or assume he has any particular velocity and work out his physics just as well. So there is no absolute velocity for inertial observers. This makes the contradiction.
  4. Aug 22, 2005 #3
    This sounds like your bringing both clocks with you in the spacecraft so that you can continue to directly compare them to figure speed. That's wrong from the start, a clock is a clock, mirror or conventional they read the same time. The point is they will both see the same time and display the same time, they will not disagree with each other unless one remains in some other referance frame.

    Where are you quoting "this section explaining relativity" from?
  5. Aug 22, 2005 #4


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    I believe the point of the quote is this:
    A light (mirror) clock is used to deduce and formulate the effects of special relativity.
    One then asks if these effects are somehow due to choice of a "light clock" as a clock, or is this true of all clocks?
    As you say, the two different types of clocks carried by the same observer will read the same.
    Otherwise, as selfAjoint explained, if there was a discrepancy when the observer is in inertial motion, one could distinguish that from being at rest [where they, by construction, read the same]... which violates the principle of relativity [that all inertial motion is relative].
  6. Aug 22, 2005 #5


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    I read that as a grammatical error: that one clock would remain on the ground.
  7. Aug 22, 2005 #6
    Except that the original quote is claiming the use of the two clocks to find:
    "disagreement between them could be used to find the speed of the spacecraft independently of any outside frame of reference"

    No outside reference - means both in the same reference.

    If you could bring along a clock that would stay on the Earth time reference I guess you could figure your speed from earth without looking outside but clocks don't work that way.

    edit) If asfd1 could tell us where the quote comes from it would help.
    Last edited: Aug 22, 2005
  8. Aug 22, 2005 #7


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    Ehh, you may be right. I'm not sure.
  9. Aug 23, 2005 #8
    You can find your velocity relative to the ground using the clock rate of the spacecraft - e.g., a GPS satellite clock will run slower than a clock on the earth's surface if it has not been preadjusted to account for its velocity relative thereto. That does not counterdict SR. (In actuality, of course, if there is no preadjustment, the GPS satellite clock would run faster because of the lower gravitational potential, but that is a GR correction).
  10. Aug 23, 2005 #9
    Can somebody describe how the calculations are made?
  11. Aug 26, 2005 #10
    the quote is from "concepts of modern physics" by arthur beiser pg7~
    thanks! :)
  12. Aug 26, 2005 #11


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    I'm slightly troubled by how the quote is written. The essential principle of relativity is that any physical experiments yields identical results in any inertial frame (all things being equal except the frame that you're in).

    Two clocks agreeing in one frame, and disagreeing in another... or any physical experiment yielding different results in two different frames... is an immediate violation of the principle of relativity. It seems to me like this would be a more straightforward way of saying it... but the context of the quote is important.
  13. Aug 26, 2005 #12


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    The calculations of what? What type of problem are you trying to find the answer to?
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