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Radar versus Ruler Length

  1. Sep 2, 2013 #1
    Ideal rulers in relativity are generally assumed to be consistent with radar measurements. The modern definition of the unit of length is defined in terms of the speed of light. However, here is an a simple example where a ruler measurement does no coincide with the radar measurement.

    To paraphrase Einstein's Carousel thought experiment, Einstein states that the rulers on the circumference of the spinning carousel length contract while the rulers placed parallel to the radius do not, so that according to measurements by these ruler's, the circumference is no longer equal to the Euclidean expectation that the circumference is equal to 2*PI*R. In fact the circumference is equal to 2*PI*R*gamma according to an accelerated observer on the carousel.

    Now if the accelerated observer riding on the edge of the carousel makes measurements using radar instead of rulers, he discovers that the ruler measurement of the radius is gamma times larger than the radar measurement and that the circumference is equal to 2*PI*R*gamma^2.

    Does this mean we have to be careful to always specify ruler or radar measurement in relativity, such as when defining Born rigidity?

    Where does this leave the modern definition of the unit of length if the ruler measurement is not always consistent with the speed of light relationship using radar?

    Interestingly, in the spinning carousel example, if we recalibrate the rulers parallel to the radius, individually using radar measurements, we find we can fit more of these recalibrated rulers along the radius and might even recover the Euclidean geometry, whereby the circumference is once again equal to 2*PI*R. However, the spatial dimensions of the disc as measured by the observer riding on the spinning carousel, will still appear larger than the measurements made by a non rotating inertial observer that is at rest with the centre of the carousel.
     
    Last edited: Sep 2, 2013
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  3. Sep 2, 2013 #2

    Jano L.

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    This does not seem right.

    Let the carousel with rest radius ##R_0## rotate with angular velocity ##\omega##.

    For stationary observer, let the measured radius be ##r(\omega)##, circumference ##c(\omega)##. For the observer sitting on the carousel, let the radius be ##r'(\omega)##, circumference ##c'(\omega)##.

    The rulers on the circumference contract according to stationary observer. Since the number of rulers does not change, the circumference of the rotating carousel as measured by the stationary observer ##c## is smaller than the circumference of the same carousel when it stands still ##C_0 = 2\pi R_0##.

    Since the carousel is plane disk irrespective of whether it rotates or not, the formula ##c = 2\pi r## is maintained irrespective of the motion of the carousel. Therefore the radius of the rotating carousel ##r## is smaller than the radius of the steady carousel ##R_0##.

    In the frame of the carousel, if the centrifugal forces are low enough to not deform the carousel (that can be accomplished by suitable choice of material, angular velocity and radius), the co-moving observer on the carousel will measure the same radius ##r'## and circumference ##c'## irrespective of the motion of the carousel, so ##c' = C_0##, ##r' = R_0##and the formula ##c' = 2\pi r'## is valid also in the rotating frame.
     
  4. Sep 3, 2013 #3
    Let's assume ##\omega## is measured by an inertial observer that is at rest wrt the centre of he carousel.
    Fair enough.

    This is not correct. Please read Einstein's original description of the carousel thought experiment. The point of the carousel is that in the accelerating reference frame, the geometry is no longer Euclidean and the ratio (circumference/radius) according to the accelerated observer riding on the outer part of the disc is greater than ##2\pi##.
     
  5. Sep 3, 2013 #4

    Jano L.

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    I have read the chapter "General theory of relativity" in the book Meaning of relativity. In this text, Einstein says

    If the ratio U / D = π in the first case, and U / D > π in the second case, then either the number U or D of rods necessary to cover the circumference and radius or both have to change as a result of the rotation. However, since the measuring rods are at rest with respect to the disk, this can only happen if the disk deforms in the frame of the disk but the rulers do not deform in that frame.

    Is that the case you consider? If so, the disk is not rigid in either frame and its diameter in inertial frame is not the same as when at rest, but will depend on the elastic properties of the disk.

    If we consider disk that is rigid in the rotating frame, the geometry in this frame is the same irrespective of the rotation.
     
  6. Sep 4, 2013 #5
    Imagine that each rod is only nailed at one end. The carousel is spun up, in such a way that the radius remains constant from the point of view of an inertial observer at rest wrt to the centre of the carousel. The rods contract relative to the perimeter of the carousel and gaps open up between the rods. If the angular velocity is sufficient, another set of rods can be fitted into the gaps. Alternatively you can imagine a chain that is fixed at only one end and extends all the way around the perimeter in a channel, so that the two ends of the chain touch. The individual links of the chain can be thought of as measuring rods. When the disc is spun up a gap appears between the ends of the chain and this gap increases with increasing angular velocity of the carousel. This happens because the chain retains its proper length while the perimeter of the carousel is effectively stretched relative to the chain in the rotating frame.
     
  7. Sep 4, 2013 #6

    Jano L.

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    I do not think it likely that the radius will remain the same in the inertial frame. We should expect it will shrink or expand depending on the material of the disk. If the disk does not deform in its own rotating frame, then its circumference ##c## in the inertial frame will decrease (as you assume). But if the circumference of circle decrease, so does its radius ##r## - the relation ##c = 2 π r## holds true in the inertial frame.
     
  8. Sep 4, 2013 #7

    PeterDonis

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    Yes, at least if you want to make sure you are being clear about what particular measurements you're talking about, so that the results can be appropriately matched to the theory's predictions.

    The difference between the two measurements is not a matter of units; you can always set up a situation where the two types of measurement agree, and use measurements made in those situations to define units of distance.

    What the difference between the two measurements tells you that there is not a single unique notion of "distance" that works in all cases. In other words, it's just another way in which the universe doesn't work the way our intuitions think it ought to work.
     
  9. Sep 5, 2013 #8

    PAllen

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    Even in as simple a case as uniform acceleration you have a difference. First, I think radar simultaneity and radar distance are well understood. By ruler simultaneity I mean the spacelike 3-surface of all spacelike geodesics 4-orthogonal to a timelike vector (tangent to a world line at some event). By ruler distance, I mean invariant interval along a geodesic in such 3-surface using the induced metric; in SR, this amounts to Euclidean distance in such a 3-surface (which will be a Euclidean 3-space).

    Then, for uniform acceleration, the two concepts of simultaneity match (for arbitrary motion, they do not match); however, distances computed by the two conventions do not agree, even for this simplest deviation from inertial motion.
     
  10. Sep 5, 2013 #9

    pervect

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    It seems to me that by the time one makes the defintion of distance precise enough to be unambiguous, one tends to loose the target audience :(.

    I try to prod people like yiop into providing their own precise defintion of distance so I can talk about it in their terms (at least somewhat), but so far I haven't had a lot of luck.
     
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