A Test of Special Relativity

Discussion in 'Physics & Math' started by MacM, Jun 14, 2005.

  1. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    also by funkstar in a previous post:
    "The standard second aboard the GPS satellites is not 9192631770 beats of the particular radiation, but slightly longer in order to cancel out the predicted time contraction aboard the satellites."
    ==========================================================

    I explained that 9192631770 'beats' of the cesium ALWAYS defines a second
    and cannot be changed. You cannot make the atom beat slower. The quartz
    oscillator will synchronize the orbital clock with the Earth clock by counting
    more than 9192631770 beats of the cesium atom to match an Earth second, not by making the cesium atom 'beat slower'. This corrects for gravitational BLUESHIFT, not redshift.
    The transmitted signal is blueshifted as it falls in a gravity well.

    I said the light clock must agree with an unadjusted cesium clock in the orbital frame of reference. The cesium clock in high orbit beats relatively faster than a cesium clock in low Earth orbit. In case you have forgotten again, my question was how do you make a light clock in high orbit beat relatively faster than a light clock in low orbit, to slightly rephrase the question so you will not misunderstand. Is the path travelled by the light in the light clock shorter in high orbit, or does light travel faster in high orbit relative to another light clock in low orbit? The second is shorter in high orbit, so distance must also contract or light must travel faster to make the light clock run faster.
     
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  3. funkstar ratsknuf Valued Senior Member

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    That's what I'm saying!

    To clarify, if we were to bring the orbital and earth clocks together, the orbital clock would tick slightly slower, because each tick on the orbital clock is actually set to be more than the standard second. Agree?
    True. I was using "gravitational redshift" as the generic term for the effect.
    I think I understand what you're trying to ask. I'll try to explain, but it is a bit difficult.

    Imagine overlaying space with a grid corredponding to the higher orbit clocks frame of reference. Now, point a laser from the orbit clock at the earth clock. The laser is set to a definite frequency, say, 1Hz, in the orbit clock frame, so we can walk along it, and it will fall perfectly "on" the orbit clock's reference frame. The length of a period on this laser wave is exactly one lightsecond in the orbit clock frame, everywhere along the wave. Now, due to the equivalence principle, it will be blueshifted when falling into a gravity well, but this doesn't change that one period corresponds to exactly one lightsecond in the orbit clock frame! Now, from the earth frame, because the light is blueshifted by gravity, the length between the periods will be less than one lightsecond, i.e. the orbit clock is ticking faster than the earth clock. Conversely, a 1Hz pulse from the earth will be redshifted when viewed from orbit, but the same reasoning holds - the period length is exactly one lightsecond in the earth frame.

    How do we interpret this wrt. the light clock? Well, the length between the periods corresponds to exactly one lightsecond in the orbit frame, but is too short in the earth frame - so the earth observer sees the orbiter as slighty contracted. Hence, from the earth's point of view, the distance between the mirrors in the orbiters light clock is too short. Conversely, the distance between the mirrors of the earth clock will appear to be too long from the orbiters point of view*.

    So, to answer your question: No, light doesn't move faster than lightspeed, it's a problem of differing frames. Yes, it comes out as length contraction. In this case there is no symmetry, as there was in the relative velocity frame.

    *You might ask whether this is dependent of the orientation of the mirrors with respect to the gravitational field, which it is, but we're assuming that locally, spacetime is flat, that is, the difference in gravitational potential across the expanse of the lightclocks is negligible - this assumption correspond to saying that light isn't gravitationally shifted across the length of the clocks. This is a reasonable assumption for atomic clocks on the surface of, and GPS satellites orbiting, the earth. For light clocks with mirrors a lightsecond apart it wouldn't be, for obvious reasons.
     
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  5. 2inquisitive The Devil is in the details Registered Senior Member

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    Thanks, funkstar, now we are getting somewhere! Now, by what mechanism does this
    length contraction occur? I ask because I have never seen it explained. Of course, I
    realize that it may not be explained in detail or that you may not know the details either. Part of my confusion comes from the Lorentz transformations of Special Relativity. Yes, I know we are speaking General Relativity in this example, but it is
    Einsteinian Relativity also. In STR, distance is observed to contract in the direction of motion only, and when looking at a frame of reference with a relatively SLOW ticking clock. In this GTR example, you seem to believe distance contracts in all directions
    when looking at a frame of reference with a relatively FAST beating clock. See what
    I mean? In GTR, distance would EXPAND when viewing a frame with a slow ticking clock. GTR makes sense to me, STR doesn't. In STR, it is stated that a relative motion
    between frames of .866c leads to a gamma of .5. The observer at rest 'sees' the clock in the moving frame as ticking at 1/2 the rate of his rest frame's clock and distance contracted by 1/2 the meter in his rest frame. That is like saying a car
    going 60 miles per hour in the rest frame would (assuming gamma) be seen to travel
    30 miles in 1/2 hour in the moving frame. There would be no net difference, the car would still be traveling the same distance in the same amount of time, neither time nor distance would have a net loss in the moving frame.
     
    Last edited: Jun 24, 2005
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  7. wesmorris Nerd Overlord - we(s):1 of N Valued Senior Member

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    Now let me ask of the ridiculous.

    Let's for a moment, accept GR and SRT as reasonable models..

    Do they not imply directly that each atom of your body is traveling in its own frame? Does the interaction of the atoms somehow lock them into a frame? Are your toes aging slightly faster than your brain?
     
  8. c3dlc Registered Member

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    spam deleted
     
    Last edited by a moderator: Jun 25, 2005
  9. funkstar ratsknuf Valued Senior Member

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    1,390
    Well, like the footnote was meant to illustrate, there is a dependency on the direction and strength of the gravity field. However, the gravitational redshift effect is not only in the direction of the field and that can be visualized in the following way: Imagine that you could overlay the orbiter's reference frame with a rubber sheet with coordinate markings, the idea being that it's marking coincide with how the orbiter's reference frame is - light in the orbiter frame can and will move from one coordinate point to an adjacent one in exactly one second, say. This rubber sheet will deform under the influence of gravity in exactly the same way as light will. Note that this analogy has a flaw in that while a normal rubber sheet would stretch under the influence of a ball, this rubber sheet is actually compressed in a gravitational field, because light is blueshifted. So adjacent coordinate points on the sheet are pressed together deeper in the gravity well, including those that aren't directly in the line with the orbiter.

    The contraction is not quite uniform in all directions, but if we are viewing small enough patches of it, this doesn't really matter that much. Of course, in huge gravitational fields, and considering wide expanses, it becomes quite a bit more involved.
    Well, there is the caveat in gtr that you have to be two different places in a gravitational field for that statement to make sense. It is not generally true.
    Ah, no. The length contraction which is, quite right, described by the gamma factor (which would usually be given as 2, in this case), is not on the meter as the rest observer sees it, but on how long the rest observer measure the the moving observer (in a car, say), to be. The half length car is still moving at the velocity 0.866c in the observers rest frame.
    Of course. The relative velocity is something both observers agree on. It is only once you compare the passage of time in both frames, that they disagree. Or am I failing to understand what you mean, here?
     
  10. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    OK, this is what I have noticed before. For an observer at rest, the coordinates of his
    frame of reference extend to infinity in all directions. But from the rest observer's viewpoint, the moving observer's coordinates do not extend to infinity, only to the measurements of his craft, correct? From the rest observer's viewpoint, the frame of reference of the moving observer is a box. Only when an observer considers himself at rest do the coordinates of his frame extend to infinity, correct? From the rest observer's view of the moving observer, the moving observer's meter stick will be shorter in the direction of travel compared to orthogonal to travel, and his clock will
    beat slower in the direction of travel and normal if turned orthogonal to direction of travel. The weird physics always happens to the 'other guy' in motion, while everything stays neat and tidy in his own frame of reference. Plus, how do you explain the SRT observation that even though speed of light is constant in both frames, it will travel a shorter distance with a relatively slower clock, more time between ticks for light to move? If the speed of light is constant, it should travel a greater distance given the relatively slower ticking clock of the moving frame when compared to the rest frame. Or do you believe the speed of light is relative to frame of reference, slower in a moving frame? If so, which is the moving frame?
     
  11. MacM Registered Senior Member

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    10,104
    Thank you. However, the problem is that I have (while personally in disagreement with the "Perception" issue) have stated I would, temporarily at least, not debate the perception issue and James R and others continue to insist that the perception is physical reality and refuse to acknowledge that of the trillions of perceptions of each clock that exist in the universe the clock only records one accumulated time.

    They completely disregard the fact that 0.866c relative velocity does not compute a proper time dilation except in gendankins. In reality it is very much a matter of actual composite velocities which comprise the total relative velocity. i.e. in a symmetrical -0.433c and +0.433c situation there is no dilation between A and B even though the relative velocity 0.866c (excluding veloicty addition) is still there.
     
  12. funkstar ratsknuf Valued Senior Member

    Messages:
    1,390
    Both frames conceptually cover all of space.
    Well, all inertial observers consider themselves at rest. The moving frame also covers all of space.
    You're right about the meter stick, but not about the clock. Think about it: What happens to length in the direction of motion? Why do you think gammma is the same in length contraction as in time dilation?
    From your frame his physics seem weird, but from his frame they don't...
    I linked this page earlier in the thread. It might clear up some of your problems...
     
  13. 2inquisitive The Devil is in the details Registered Senior Member

    Messages:
    3,181
    Sure, I have seen many such examples before. OK, Jasper views Zoe's clock as beating
    slower because the the photon follows a longer path because her vehicle is moving
    forward, orthogonal to the photons travel, causing the photon to follow a V-shaped
    trajectory. Now, keep everything in the example the same, EXCEPT let the photon
    move front to back in Zoe's vehicle. Jasper views the scene from the same position
    as before. According to Jasper, Zoe's vehicle contracts in length along her direction of travel. The photon will then have a shorter path to travel because of the contraction.
    Zoe's clock will beat faster than Jasper's when the photon travels front to back. That is what I meant when I said the tick rate of the traveling clock was dependent upon
    orientation of the light clock.
     
  14. 2inquisitive The Devil is in the details Registered Senior Member

    Messages:
    3,181
    by funkstar:
    "Imagine overlaying space with a grid corredponding to the higher orbit clocks frame of reference. Now, point a laser from the orbit clock at the earth clock. The laser is set to a definite frequency, say, 1Hz, in the orbit clock frame, so we can walk along it, and it will fall perfectly "on" the orbit clock's reference frame. The length of a period on this laser wave is exactly one lightsecond in the orbit clock frame, everywhere along the wave. Now, due to the equivalence principle, it will be blueshifted when falling into a gravity well, but this doesn't change that one period corresponds to exactly one lightsecond in the orbit clock frame! Now, from the earth frame, because the light is blueshifted by gravity, the length between the periods will be less than one lightsecond, i.e. the orbit clock is ticking faster than the earth clock. Conversely, a 1Hz pulse from the earth will be redshifted when viewed from orbit, but the same reasoning holds - the period length is exactly one lightsecond in the earth frame."
    also:

    "Well, like the footnote was meant to illustrate, there is a dependency on the direction and strength of the gravity field. However, the gravitational redshift effect is not only in the direction of the field and that can be visualized in the following way: Imagine that you could overlay the orbiter's reference frame with a rubber sheet with coordinate markings, the idea being that it's marking coincide with how the orbiter's reference frame is - light in the orbiter frame can and will move from one coordinate point to an adjacent one in exactly one second, say. This rubber sheet will deform under the influence of gravity in exactly the same way as light will. Note that this analogy has a flaw in that while a normal rubber sheet would stretch under the influence of a ball, this rubber sheet is actually compressed in a gravitational field, because light is blueshifted. So adjacent coordinate points on the sheet are pressed together deeper in the gravity well, including those that aren't directly in the line with the orbiter."
    =============================================================

    Do you see the discrepancy? In the first example above, the tick of the light clock is relatively faster because the meter must be relatively shorter than an Earth meter.
    Remember, a meter is define as the distance light will travel in 1/299,792,458 seconds.
    Assume a hypothetical, but very precise, light clock that operates by bouncing a photon off a reflector one meter from the emission and recording point. Each photon will travel two meters for one 'beat', beating exactly 149,896,229 times per second.
    To have this light clock beat relatively faster than an identical clock deeper in the gravity well, the length of the meter must be relatively shorter or the speed of light must be relatively faster. Contracted meters, faster light clocks.

    Now, in your second example above, you have the rubber sheet 'grid' as contracting
    between coordinate points as it 'falls' in a gravity well. That would correspond to shorter meters deep in the gravity well and longer meters higher in the gravity well.
    Remember from the light clock example that the meter must do the OPPOSITE for the light clock to beat faster higher in the well. See what I mean?
     
  15. 2inquisitive The Devil is in the details Registered Senior Member

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    3,181
    When Einstein formulated the Special Theory of Relativity in 1905, he used postulates
    that were assumed true, then used gedankens to support his postulates. But the postulates had to be true for the gedankens to be true.

    Let me make a false statement and formulate a gedanken to support it.

    My statement is that Einstein believed the Earth to be flat, a Flat-Earther in 1905. He used
    STR to support this belief.
    How, do you ask? In STR, there is no gravity, only mass and weight. For instance, an
    object dropped from a height above the surface of the Earth will fall straight down, not toward the center of the Earth due to gravity. Only mass and weight affects the object. An observer standing on the surface of the Earth will see the sun orbiting around the flat Earth, he assumes the position of rest with the sun in motion. The observer cannot assume HE is in motion and cannot prove it. According to this observer, if the Earth were rotating, objects would roll downhill as the sun was seen to be closer to the horizon, no gravity directed toward the center of the Earth to cause them to do otherwise. Therefore, the
    Earth must be stationary, at rest, and the sun orbiting around the Earth. Within the
    confines of this gedanken and Special Theory, can you prove otherwise? Einstein used
    many such gedankens to support his postulates. Gedankens are not proofs and can lead to false conclusions.
     
  16. geistkiesel Valued Senior Member

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    2,471
    Your post makes me think that Funkistar is now feeling that somebody cheated him out of something, but he's not quite sure what it was that he got cheated out of, or even who cheated him out of it.
    Geistkiesel
     
  17. funkstar ratsknuf Valued Senior Member

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    1,390
    No, I don't.

    The observer higher up in the gravity well sees his own meter as shorter than the meter of the observer deeper in the gravity well - hence, his own clock is ticking faster than the observer deeper in the well's clock.

    The observer deeper in the gravity well sees his own meter as longer than the meter of the observer higher up in the gravity well - hence, his own clock is ticking slower than the observer high in the well's clock.

    There's anti-symmetry in gravitational redshift. They both agree on who's clock is ticking faster than the others, because they both agree that the meter of the observer deeper in the gravity well is longer than the meter of the observer higher in the gravity.
     
  18. 2inquisitive The Devil is in the details Registered Senior Member

    Messages:
    3,181
    quote also by funkstar:

    , this rubber sheet is actually compressed in a gravitational field, because light is blueshifted. So adjacent coordinate points on the sheet are pressed together deeper in the gravity well, including those that aren't directly in the line with the orbiter.
    =========================================================

    I agree with you that meters are relatively shorter for an observer higher in a gravity well. But read again your post using the rubber sheet analogy. In it,
    you state coordinate points are pressed CLOSER deeper in the gravity well,
    a shorter meter deeper in the gravity well. Did you really not see this? I also
    know the rubber sheet analogy is used often at physics sites and universities.
    You are just repeating someone else's mistake. The light clock shows this is a mistake. This is the discrepancy I spoke of.
     
  19. funkstar ratsknuf Valued Senior Member

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    1,390
    You're interpreting it wrong.

    Well, perhaps I framed the analogy poorly: When I say that the coordinate points are pressed closer, I meant the orbiter frame relative to the earth frame. The distance between the points is still a meter in the orbiter frame! But due to the gravitational redshift, the earth and orbiter observer disagree on how long a meter is, as described above.
     
  20. 2inquisitive The Devil is in the details Registered Senior Member

    Messages:
    3,181
    What orbiter? There was no 'orbiter' in the rubber sheet analogy. I never disagreed that each observer would not measure a meter as a meter in his own frame of reference, I have stated often that he would. I am not confused, sorry, but you are.
    In GR, IT DOES NOT MATTER from which frame of reference an observer views the other frame. It has been empirically tested that clocks run relatively faster higher up
    in the gravity well, from both frames of reference. To beat faster, the meter must relatively shorten in a light clock. But GR states coordinate points, meters, are compressed
    DEEPER in the gravity well. SR states meters are relatively shorter (in direction of travel) along with a slow beating clock in frames moving away with relativistic relative
    velocity. Light clocks beat FASTER, not slower, with relatively shorter meters.
     
  21. funkstar ratsknuf Valued Senior Member

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    1,390
    Let's try to calculate the, from the rest frame, how far the photon travels from one mirror to the other and back again. We'll use lightseconds as the unit of length, and assume that the mirrors are placed length l proper lightseconds apart.

    Zoe is travelling at speed v (in lightsecond/second, to simplify, making c = 1), so she appears length contracted by a factor of

    gamma = 1 / ((1 -v^2)^½)

    Let t_1 be the time the photon has to travel from the back mirror to the front mirror (the long way). t_1 can described by the following equation

    t_1 = v * t_1 + l/gamma

    The mirrors move v * t_1, but the photon, moving exactly t_1, also has to move the length between the mirrors, which is contracted to l/gamma, from which we get the equation. This equation solves to

    t_1 = l/gamma * 1/(1-v)

    Similarly, the way back, with the back mirror rushing to meet the photon takes the time t_2 which is described by

    t_2 = - v * t_2 + l/gamma

    which solves to

    t_2 = l/gamma * 1/(1+v)

    So the total roundtrip is

    t_1 + t_2 = l/gamma * (1/(1+v) + 1/(1-v))
    = l/gamma * 2 * 1/(1-v^2)
    = l/gamma * 2 * gamma^2
    = 2 * l * gamma.

    This means that on average, Zoe's clock will tick slower by a factor of gamma. This is exactly the same as when the clock is orthogonal to the direction of travel! That Jasper doesn't agree with Zoe that two such clocks would tick perfectly in synch is not a paradox, because simultaneity is relative.

    This shows that even with length contraction in the picture, the gamma factor is constant. So you can orient the light clock any way you want it, and the result is the same.
     
    Last edited: Jul 29, 2005
  22. funkstar ratsknuf Valued Senior Member

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    1,390
    Yes, there was. I very clearly stated that the rubber sheet corresponded to the orbiter's reference frame.
    This is wrong, and the root of your misconception. Frames are still extremely important - how could they possibly not be? There's no universal agreement on what constitutes a meter or a second between frames in GR, either, and gravitational redshift is an example of this even when there's no relative velocity between observers. Gravity distorts the frames so they don't agree.
    Yes. Now, think about what this means, from a perspective of frames. Is there any possible way for the two clocks to be in the same reference frame when this happens? Please think, before you answer.
    As compared with the local earth meter, yes. Not as compared with the orbiter meter. Which is the only way it can be, if the clocks are to beat faster in the orbiter frame! The compression of the rubber sheet corresponds to the fact that an earth observer will measure an orbiter meter to be shorter than his own meter.
    You're still not getting it, and I think you've been confused by the what the rubber sheet analogy is trying to say.

    I've written it before, but here's the qualitative example, again.: A is higher up in the gravity well than B. A measures a rod he's holding to be of length l, in standard meters. Now, B also measures the length of A's rod (by some method) to be l', also in standard meters. Because of gravitational redshift, l' < l. Why? Because what constitutes a meter in A's frame is compressed to something smaller than a meter in B's frame, and what constitutes a meter in B's frame is expanded to something longer than a meter in A's frame. Neither A nor B moves, and they have no relative velocity, so there's no need to involve SR. Also, it is important to note that if you were to move the rod from A to B, those measurement would no longer hold: A would now measure the rod to be some l'' > l, and B would measure it to be l.

    Think about it.
     
  23. 2inquisitive The Devil is in the details Registered Senior Member

    Messages:
    3,181
    by funkstar:
    "So adjacent coordinate points on the sheet are pressed together deeper in the gravity well"
    "Well, perhaps I framed the analogy poorly: When I say that the coordinate points are pressed closer, I meant the orbiter frame relative to the earth frame."
    =============================================================

    Here is where the confusion arose on my part. I know reference frames very well, all
    inertial, non-inertial, global and the ECI, ECEF and ICRF frames used in GPS. Again, a
    frame of reference is not small location, it is a set of coordinates extending to infinity
    from the observer's location. Reread your second statement. It can be interpreted to mean all coordinate points are pressed closer together in the orbiter frame relative to the Earth frame. That is still not correct. Now, what I stated without your separation of the two sentances:
    "In GR, IT DOES NOT MATTER from which frame of reference an observer views the other frame. It has been empirically tested that clocks run relatively faster higher up
    in the gravity well, from both frames of reference."
    It does not matter which frame of reference one uses, the clocks always beat relatively faster when they are higher in the gravity well, from either frame of reference. Understand now? I never said that frames of reference were not extremely important, only that in GR, the faster clock is always in the same frame, regardless of which frame one views them from.

    by funkstar:
    "Yes. Now, think about what this means, from a perspective of frames. Is there any possible way for the two clocks to be in the same reference frame when this happens? Please think, before you answer."
    ===============================================================

    Of course, both clocks can be in the same reference frame, the reference frame of the observer. The observer can only be in one reference frame at a time, but not the clocks. Do you understand this?

    Edit: A correction to the global reference frame. It is not necessary to have an observer at the center of a global reference frame. The coordinates extend
    from a point in space, not necessarily corresponding to any observer's location.
     
    Last edited: Jun 27, 2005

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