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Principle of equivalence

  1. May 24, 2008 #1
    Two clocks in a gravitational field separated by an altitude of x, exhibit constant time dilation.
    Two clocks accelerating in free space separated by the length x of their ship, exhibit non-constant time dilation.
    How is the principle of equivalence sustained when this fundamental measure of acceleration differs?
     
  2. jcsd
  3. May 24, 2008 #2
    Two clocks in a gravitational field, separated by any height - will not experience the same acceleration - nor the same dilation.
     
  4. May 25, 2008 #3
    That's true, but not my question.
    1) A clock at a height x above the Earth's surface will run faster than an identical clock directly below it on the Earth's surface as measured by the latter.

    2) A clock at the front of an accelerating frame will run faster than an identical clock directly behind it (on the axis of motion) in the same frame as measured by the latter.

    The time dilation in 1) is constant. i.e. the difference in "rate" of time measured between the clocks is the same no matter when it is measured.

    The time dilation in 2) is non-constant. i.e. the difference in the "rate" of time measured between the clocks "increases" over time.
     
  5. May 25, 2008 #4

    Dale

    Staff: Mentor

    I don't think this is correct. I am pretty sure that the time dilation in 2) is constant over time if the distance between the clocks is not changing as measured in the accelerating frame.
     
  6. May 25, 2008 #5
    I don't think so.
    The distance between the front clock at the time of emitting a light signal and the position of the back clock at the time the signal is observed, must continue to decrease in an accelerating frame. This results in the frequency of light signals increasing as measured by the clock at the back of the frame.
    The clock at the back thus measures the clock in the front is running fast, but it also finds it is running faster with each measurement.
     
  7. May 25, 2008 #6

    Dale

    Staff: Mentor

    Well, if the distance changed in the gravitational field then the time dilation would also be time varying in the gravitational field. If you want an equivalent situation then you need to have the distance as a function of time be the same in both the accelerating and gravitational cases. If the distance is changing in one and not in the other then obviously the results will be different.
     
  8. May 25, 2008 #7
    The distance between the two clocks does not change in 1) or 2).
    The clocks in 1) and 2) can be thought of as in the front and back of a rocket-ship, let the distance between them be x,(the length of the ship).
    In 1) the rocket-ship is standing upright on the Earth's surface.
    In 2) it is accelerating in free space.
     
  9. May 25, 2008 #8

    Dale

    Staff: Mentor

    As measured in the accelerating/gravitating frame.
    Then I am pretty sure that the time dilation is not time varying in either 1) or 2). If you believe it is could you post a derivation?
     
  10. May 25, 2008 #9
    Your assertion above, i.e. Two clocks accelerating in free space separated by the length x of their ship, exhibit non-constant time dilation is not correct for a uniformly accelerating frame of reference. In such a frame the clock rate is a constant for both clocks as measured by an observer at rest in the accelerating frame. A clock at the back of the ship will not detect changes in the wavelength of a beam of light as measured by by a clock at the front of the shift. There is no increase in frequency as you claim there is. Why do you there would be such a change?

    Also there are two forms of the equivalence principle. The weak form states
    The strong equivalence principle states
    Einstein never claimed that an arbitrary gravitational field (e.g. one corresponding to a curved spacetime) is equivalent to an arbitrarily accelerating frame of reference, in fact it they aren't.

    Pete
     
  11. May 25, 2008 #10

    Dale

    Staff: Mentor

    Are you sure about that? If you consider the momentarily co-moving inertial frame when a photon leaves the front then, by the time it reaches the back, the back clock has accelerated and there is some Doppler shift. My understanding is that this was the essence of the derivation for the gravitational blueshift.
     
  12. May 25, 2008 #11
    Those are measurements taken by different observers. No one observer in a static g-field (or a uniformly accelerating frame) will ever measure the frequency of light to change as or propagates through the spacetime. In any case, the same would happen for observers in a gravitational field. If an observer initially measures the frequency of light comming from a steady source (a source which is at rest in the g-field and which produces light with the same freuquency indpendant of time) then he'd measure that frequency to change if he moved to a place lower in the field. Same with the accelerating observer. No experiment can determine whether you're in a uniformly accelerating frame of reference or in a uniform gravitational field. That is what the principle states. If measurement shows otherwise then the equivalence principle would be wrong.

    Pete
     
    Last edited: May 25, 2008
  13. May 25, 2008 #12

    Dale

    Staff: Mentor

    Oh, I misunderstood your point at first. I agree, I cannot think of any reason for the blueshift to change over time either.
     
  14. May 26, 2008 #13
    We are talking about the time of light signals as the frequency of emission and detection, not the wavelength of light.
    If the rate of the clock at the front of the ship is measured to increase at all relative to the clock at the back of the ship, it is because the distance traveled by the light signal is less than the length of the ship. The distance traveled by the light signal continues to decrease with the increased instantaneous velocity of acceleration therefore the rate of the clock at the front of the ship will continue to increase relative to the clock at the back of the ship. As the ship approaches c the distance traveled by the light signal approaches 0.

    I am not talking about arbitrary physical equivalence, I am talking about the fact that the principle of equivalence ("we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system." (Einstein 1907)) is statement about the "instantaneous" physical equivalence of the laws.
     
  15. May 26, 2008 #14

    Dale

    Staff: Mentor

    I don't understand this at all. The ship doesn't approach c in the accelerating reference frame, it is always at rest in the accelerating reference frame. That is the whole point of the accelerating reference frame.
     
  16. May 26, 2008 #15
    It approaches c in the frame of any observer that measures it in constant acceleration. In the ship the distance x, (length of the ship) and the constancy of the speed of light are as (x/c) what determines the clock at the front to be running fast relative to the clock at the back.
     
  17. May 26, 2008 #16
    Thanks for clarifying. If this is the case then your statement This results in the frequency of light signals increasing as measured by the clock at the back of the frame. was confusing.
    No. It is because the clock is actually ticking faster as reckoned by the accelerating observers.
    You're making coordinate dependant statements. As such please specify which observer
    you are referring to. As measured by an observer at rest in the ship the distance is constant.
    Then you are mistaken. You are assuming that "we [...] assume the complete physical equivalence of a gravitational field and a corresponding acceleration of the reference system. is the equivalence principle. It is not in that neither the weak or strong equivalence principle make such an assertion.

    Pete
     
  18. May 26, 2008 #17

    Dale

    Staff: Mentor

    Sure, but that frame is an inertial frame so it has nothing whatsoever to do with the equivalence principle.

    The equivalence principle relates an accelerating reference frame to a gravitational field, not an inertial reference frame to a gravitational field. Of course you don't get equivalence in your comparison.
     
  19. May 26, 2008 #18
    pmb_phy, DaleSpam, all of your comments make it appear you are not familiar with this thought experiment.
    Let me make it clear.
    A rocket-ship stands upright on the Earth. We will call the front of the ship A and the back B.
    We synchronize clocks, call them Ac and Bc and then place them at A and B respectively.
    A clock "Ac" at the front of the ship "A" is designed to emit a flash of light at regular intervals toward B (let's say once per second).
    An observer at the back of the ship (B) is marking the frequency of the signals (i.e. the time intervals between each flash)
    against their clock Bc.
    In a gravitational field, the observer at B will measure the light signals to arrive at a greater frequency than once per second according
    to their clock Bc. This is gravitational time dilation.
    When the ship is accelerating in free space, (let's say relative to the Earth) the observer at B will again measure an increased frequency
    of the signals which they can only reconcile as follows:
    The speed of light is constant, the length of the ship (x) is constant, therefore x/c is constant, therefore the clock at A is running faster than
    the clock at B. This is the equivalence of gravitational time dilation
    In a gravitational field, such as when it is standing on Earth, the clock at A, will remain at a constant rate of time that is faster than the
    clock at B.
    When accelerating, the clock at A will continue to increase its rate of time with respect to B as the ship's instantaneous velocity
    relative to the Earth, increases with its acceleration. Why, because relative to the Earth and for the same reason Ac is fast in the
    first place, the distance between the emission at A and detection at B decreases over time with the constant acceleration of A,B.
    The constancy of the speed of light (i.e. independent of A and B) will result in a constantly decreasing time of detection at B between each signal.
     
  20. May 26, 2008 #19
    Ya never know! :tongue:
    Yup. That is exactly how I understand gravitational time dilation to be.
    Woa there! Where did you get the idea that the speed of light is constant in a gravitational field. In fact it isn't. It is a function of position.
    Okay so far (except for the speed of light in a g-field).
    I disagree. The rate at which the clocks tick will remain constant as measured in the ship's frame of reference. Why do you think that the rate will increase. Please state with what frame of reference you are referring to when you use the term "increase".
    With what frame of reference are you referring to when you say that the distance increases. Also recall that the speed of light is not constant in a g-field but varies with position.

    Pete
     
  21. May 26, 2008 #20
    You're right the "speed" of light is not the "same" in all positions, but it is constant at each position. It's constancy at B is all that's needed to calculate the rate of signals from A.
    The rate at which A ticks will remain constant as measured by an observer "at" A and the rate at which B ticks will remain constant as measured by an observer "at" B.
    That is not the issue here.
    The rate at which A ticks according to the measured rate of signals detected at B must increase with the increased velocity of the ship with respect to the Earth.

    You seem to be agreeing with the experimental mechanics in principle and then questioning the mechanics you agree with.
    If you agree with the principle of equivalence of acceleration and gravitation what do you think is the means by which a clock at A is detected to run faster than a clock at B in an accelerating frame?
     
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