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Question about Proper Time and Rest Frame selection.

  1. May 3, 2005 #1
    An interesting time dilation experiment I thought of was to send a radio signal with two pulses a certain time apart to a receiver moving toward the emitter.

    I believe that the effect is that the accelerated reciever undergoes time dilation and would time the two pulses of light as closer together than the emitter would. I know that if I performed this experiment today, that is the result I would get but there are things I don't understand surrounding this fact.

    However, in this experiment I'm assuming earth is the proper rest frame. This could easily be refuted by looking at things differently. Say for instance my traveller is moving westward against the earth's spin at 100mph at the eaquator. This is equal to the speed of the earth and now my emitter is the one travelling and the reciever is still.

    Extend this further to the speed around the sun and say I'm travelling at 33k mph or whatever. My point is that how do I know that the sun isn't the proper rest frame? For example, in stellar aberration, the sun is the rest frame we use to calculate the expected aberration.

    Now that I mention stellar aberration being sun based I have to ask why it is that large sagnac devices like the ring laser project in new zealand (and michelson-gale) can detect the earths rotation on its axis which is very small in relation to the earth's movement around the sun but the device does not pick up the movement around the sun! :confused:

    What makes proper time proper? What makes a specific third reference point be the preferred one? How do I predict the right rest frame without arbitrarily picking the one that happens to work?

    I know that one explanation for my first problem is proposed to be because one of the objects has to accellerate isn't that explanation deprecated now? Unfortunately I don't know what the preferred explanation is but that older one does nothing for my other question anyway.

    I'm not looking for the equation for proper time; I want to understand why and how it works.

    Thanks in advance for your patience. :biggrin:
  2. jcsd
  3. May 3, 2005 #2


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    Before I write a long response, are you looking for someone to explain your misconceptions about relativity or not?

    There is no such thing as a "proper rest frame" in relativity - it does not matter which reference frame one adopts as far as observations go. Several of your assumptions in the scenario that you give are incorrect (according to relativity) but unless you are interested in what relativity has to say, I don't see much point in going into the details. (Since your chosen handle is "anti-relative", I have a nagging suspicion that writing a long detailed response about what relativity has to say might just be a waste of my time).
  4. May 3, 2005 #3


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    while the reciever would time the pulses as closer together than the the emitter, this is due to Doppler effect and not time dilation. If you factor out the Doppler effect, the reciever would determine that the pulses were further apart than the emmiter would as long as the reciever maintains a constant velocity towards the emitter..
    Small in what way? the magnitude of the Earth's rotation in angular velocity: 0.000073 rad/sec is much larger than the Earth's angular velocity due to its orbit around the sun: 0.0000002 rad/sec ( 365 times larger, in fact).
    Proper time is simply time as measured in the frame you decided to work from. The 'rest' frame is whatever frame you want, usually it is chosen as such because it is the one that it is easiest to work from. It doesn't matter which frame you choose, as long as you take all relativistic effects into account you won't go wrong.
  5. May 4, 2005 #4
    I just got the nickname from asking hard questions in class...
    Still something I'm not getting my head around. With the doppler effect combined with the constancy of lightspeed, don't you have to assume that the same number of waves fit in a smaller distance in regard to the detectors frame of reference? Say for instance that 100 waves fit in 100 inches from the emitter's frame and the duration of the beam of light happens to be equal to 100 waves. Say that the listener traverses 50 inches (from the emitter frame) from the time it receives the first wave till it recieves the last. This results in double the frequncy. So in the listeners frame the total beam length is actually only 50 inches instead of the 100 that the emitter observes. The total duration of the light beam would be half what the emitter believes it to be. Isn't that time dilation at work?

    The way in which an optical gyroscope works allows it to measure speed through real space from what I understand though. If I were to put one on my finger and move my finger around in a little circle, the rad/sec would be immense (far far greater than the earth's revolution) but would not be detected by the sagnac device any greater than a straightline path at the same speed as my finger.

    The speed of the surface of earth when considering only its spin is somewhere around 1000 mph at the equator. The speed of the earth's surface in regard to its orbit around the sun is around 33ish times that.

    Does a predominant gravitational field somehow generate a preferred frame of reference or something like that? There has to be some rule about this that I don't know. The michelson gale and the ring laser in new zealand pick up that surface speed component in relation to the earths center instead of the surface speed in regard to the sun or any other object. Therein lies my curiosity.

    Ahh, so each frame of reference you might examine has it's own proper time?
  6. May 5, 2005 #5
    No, time dilation is an effect additional to that. Here is a good link about it. Check out the other excellent relativity links there too.

    Yes. Scatter deep space with clocks moving randomly relative to one another. Each clock runs at a 100% rate in its own frame; e.g. as observed by someone at rest with respect to the clock. From the perspective of any given clock's frame, the other clocks run at various rates between above 0% and 100%, which can be measured directly to discount the Doppler effect, as many laymen's books explain.
    Last edited: May 5, 2005
  7. May 5, 2005 #6


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    There is a fairly easy way to derive time dilation (and the other consequences) of relativity from doppler shift.

    The procedure is called Bondi "k-calculus", though it doesn't actually involve any calculus, just basic math. Google finds http://www.geocities.com/ResearchTriangle/System/8956/Bondi/intro.htm
    as one online reference, but I'll go into some details myself. There are a few books on relativity which also use this appraoch ("Relativity and Common Sense by Bondi is one of them). This is not necessarily the most powerful approach to special relativity, but it's one of the simplest, requiring only high school algebra.

    The base assumption is this - there is a constant ratio of interval of reception to interval of transmission for two moving observers. Let's say the two observers are located at the same point at t=0, and they are therefore ncessarily moving away from each other for t>0. This implies a redshift, which implies that the ratio of reception to transmission is greater than 1. (You can work out what happens when t<0 using the same approach, it's just a little more work, and I'm just going to sketch out the math as fast as I can - which is still going to take a lot of typing).

    For an example for concretenss, I will use a value of k=3. When I emit a signal at 1 second intervals, the moving receiver recieves the signals at an interval of k=3 seconds. When the other person emits signals at intervals of 1 second, I recieve them every 3 seconds. The situation is perfectly symmetrical, it does not matter who is considered to be moving or who is considered to be stationary.

    So far so good, now what we need to argue is that if I emit a signal at every t seconds, and that signal either bounces or is re-transmitted by the moving observer, it will arive back to me at an interval of k^2*t seconds. In our simple example, I send a signal every 1 second, and k=3, so I receive a return signal once every 9 seconds.

    The argument is fairly simple, if I transmit a signal every second, it gets received every 3 seconds. We've already noted that if the moving observer transmits a signal every second, I receive it every three seconds. In this case, we simply have the moving obsever send a signal every 3 seconds - I must by the laws of the doppler shift receive one every 9 seconds.

    Now comes the math - if I send a signal at 1 second, at it returns at 9 seconds, and the speed of light is constant, I can use radar timing to compute the position of the moving object as a function of time.

    The speed of light is constant, so the signal arrives at the target at the midpoint of the time interval, which is (1+9)/2 seconds, or 5 seconds.

    The amount of distance the signal travelled during that time (from t=1 seconds to t=5 seconds) is equal to the amount of distance the signal travels during its return trip (from t=5 seconds to t=9 seconds). So twice the distance is equal to c*(9 seconds - 1 second), and the distance at t=5 seconds is 4 light seconds. This means that with k=3, the velocity of the object is 4/5 of the speed of light.

    Now here is where time dilation comes in. My time of 5 seconds at a distance of 4 light seconds corresponds to the moving observers time of only 3 seconds.

    I conclude that his clock is running slow. This conclusion is known as time dilation.

    Note that the other moving observer concludes my clock is running slow. Time dilation is symmetrical.

    You are right in that distances are also affecteed, but this post has run on long enough. The full story can be found in the Lorent transformation, which gives one time dilation, length contraction, and the relativity of simultaneity. All three of these effects are necessary for a consistent mental picture of relativity.

    The important result of the final results (which would take too long to derive, see Bondi's book if you want the details of how to use k-calculus to derive the following results) is that there is a consistent set of transformations from my coordiante system (x,t) to the moving observer's coordinate system (x',t'). These coordinate transformations play a central role in relativity, and are known as the Lorentz transformations.

    The Lorentz transformations are
    x' = gamma*(x-v*t)
    t' = gamma*(t-v*x/c^2)

    The fact that t' = gamma*t when x=0 is the mathematical expression of time dilation.

    That fact that t' depends on x as well as t is the mathematical expressionof the relativity of simultaneity.

    Length contraction takes a little more work to dervive from these formulas, but it is also a logical consequence of them.

    I'm not going to get into the rotating observer issues in the same level of detail, I'm only going to to note that a laser gyroscope cannot and does not measure absolute velocity. It can and does measure absolute rotation.

    A laser gysroscople used to measure the rotation of the earth will detect it's sidereal rotation rate, which is different from the solar day due to the Earth's orbital motion. So it's wrong to say that the Earth's orbital motion does not affect the rotation rate of the Earth - the total angular frequency of the Earth is equal to the angular frequency due to it's daily rotation plus the angular frequency due to its yearly orbit.

    It's also quite wrong to conclude that a laser gyroscope can measure velocities - it cannot measure velocities, only rotation.
    Last edited: May 5, 2005
  8. May 5, 2005 #7


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    Side comment: I find it interesting that Bondi's k-calculus isn't used as much as I feel it should be. I have noticed that British authors seem to use it... probably influenced by Bondi's BBC presentation.

    There's a good reason why consequences in relativity can be derived from the Doppler Effect and the k-calculus. To see this, find the eigenvectors and eigenvalues of the Lorentz Transformation matrix.
  9. May 5, 2005 #8


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    The standard approach to relativity emphasises Lorentz invariance. I think this better prepares students for General Relativity than the k-calculus approach. I still suspect that the k-calculus approach would work very well for a high-school course in special relativity. I think that a high school course in special relativity would be a Good Thing.

    That's a cool point about the eigenvalues/vectors of the Lorentz transform - the eigenvectors are null vectors (i.e. x=ct, x=-ct), but I never realized that the eigenvalues were the Bondi k values (the doppler shifts), though it's somewhat obvious in retrospect.
  10. May 6, 2005 #9
    Thanks for the info on Bondi's stuff. Looks like a valuable simplification that conveys a lot more understanding mopre quickly than other mothods of looking at things.

    I need to argue with you on one point though

    They are used in airplanes and missle technologies to measure distance travelled over time (velocity?). They are now made in sizes similar to a thick postage stamp. Many modern laser "gyroscopes" are not gyroscopic at all. They are completely unmoving fiber-optic ring interferometers.
    These devices most assuredly pick up absolute motion. This includes straight up and down movement if aligned to detect that movement.

    The problem still remains that when one of these devices is on the "fast side" of the earth away from the sun there is a momentwhere it is completely analogous to it simply orbiting the sun by itself. It is also analagous kinematically to it sitting on the surface of a gigantic planet the size of the earth orbit.(ignoring gravity)
    Again. My confusion still remains that it does not pick up the rotation of this larger planet as 33,000 MPH but instead still detects movement through absolute space as 1,000ish MPH (dependant on latitude) as the rotation of the earth.

    It still seems like gravitational dominance sets up a preferential reference frame with sagnac devices. To put it bluntly, it's freakin wierd. :uhh:
  11. May 6, 2005 #10


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    Gyroscopes do not measure linear velocity, and neither do laser gyroscopes. - they measure the rate of rotation (angular velocity) or can be used to create a gyro-stabilized platform that does not rotate.

    Combined with an accelerometer, one can do "inertial guidance" on a missile with a combination of plain gyroscopes (or laser gyros), and accelerometers.

    See for instance


    Conceptually, the old-fashioned gyro-stabilized platform illustrates the principle. The gyroscopes keep your accelerometers always pointing in the same direction. A set of three accelerometers (one for each direction, x,y,z) measure your acceleration, determining your change in velocity. Given that you know that your missile (or vehicle) started out "at rest", you can determine it's velocity by measuring and integrating the accleration from accelerometers (a different device than a ring laser gyroscope) - and you can determine the position by integrating the velocity.

    Laser gyroscopes DO NOT measure linear velocity. Like any other gyroscope, they only measure rotation (angular velocity).

    For a bit more detail try the wikipedia again, compare and contrast the entries on ring laser gyros and on accelerometers

  12. May 6, 2005 #11
    If they can only measure their own internal rotation rate then how is it that the michelson-gale experiment measured the rotation of the earth without building the device all the way round the world? I could maybe understand if it was vertically aligned but it wasn't.

    How does the ring-laser project in new zealand continue to monitor the earth's rotation? It's also not vertically aligned.
    http://www.phys.canterbury.ac.nz/research/laser/ring_2000.shtml [Broken]

    I'm still confused :frown:
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  13. May 6, 2005 #12


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    I'm not quite sure I totally understand the source of your confusion

    http://www.phys.canterbury.ac.nz/research/laser/ring_2000.shtml [Broken]

    seems fairly clear on the topic, one sample quote is

    which makes it very explicit that a ring laser measures rotation, and not absolute motion. It's basically a very sophisticated and compact Foucault pendulum.

    Perhaps the Foucault pendulum will answer your question. Much like a ring laser gyroscope, the pendlum is sensitive to rotation around one specific axis, and not sensitive to rotation around the other two axes. A Foucault pendulum at the north or south pole will measure the correct total rotation of the Earth, approximately once a day (more precisely, once per sidereal day). A Foucault pendulum at the equator will not be able to measure the rotation of the Earth at all. However, at intermediate lattitudes, the Foucault pendulum will precess at a rate of 23.93 hours / sin(latitude)

    see for example

    http://www.phys.unsw.edu.au/PHYSICS_!/FOUCAULT_PENDULUM/foucault_pendulum.html [Broken]

    or the more detailed mathematical derivation linked to this page at


    At 51 degrees Lattitude, Canterbury is not on the equator - you can calculate that the rotation pariod will be increased by a factor of 1/.77 (that's 1/sin(51 degrees) - which can be alternatively described as saying that the rotation rate/angular velocity is multiplied by .77.

    A factor of .77 has to be taken into account in interpreting the results, but clearly it will not significantly degrade the sensitivity of the instrument.

    The factor of .77 occurs because angular velocities represented by vectors add using vector rules, and decompose into components using vector rules (that's why the standard vector convention is used). This is a very basic and general statement about how angular velocities behave, which is independent of the exact technology (Focault pendulum, laser interferometer) which is used to measure said angular velocities.
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