Jump to content

JVNY

Senior Members
  • Content Count

    52
  • Joined

  • Last visited

Community Reputation

8 Neutral

About JVNY

  • Rank
    Meson

Profile Information

  • Favorite Area of Science
    Physics
  1. I agree that it is tough, but that is part of what makes it interesting. The light emitter and the photographic film that recorded the fringe shift were fixed on Sagnac's rotating disk, so what he recorded was from the POV of the rotating disk, not the inertial frame attached to the center. Trying to understand the effect from the POV of the disk is worthy of effort. And I agree, the c+v and c-v refer to closing speeds. If the observers on the rim try to synchronize their clocks using the Einstein convention sending signals in one direction, then yes I think that each clock is ahead o
  2. A couple of thoughts. First, I think that even on the rim (not just in the lab) each clock will tick at the same rate. Say the interior of the rim is mirrored, so that a light signal can travel around the interior. If a clock sends a series of signals around the ring (in either direction), one every second of its proper time, then each other clock on the rim should receive the signals one every second of its proper time (after the initial delay for the first signal of the series to arrive). Second, you are on to something with the simultaneity shift suggestion -- some analyze the Sa
  3. I am trying to get at the point of view of the rim, not the axis. If a rim is rotating in an inertial laboratory, it makes sense that signals sent in opposite directions around the rim travel in the lab frame relative to their starting point on the rim at c+v or c-v. However, the result of +v or -v applying is that the difference in lab time between the arrival of counter- and co-rotating signals back at their starting point on the rim will depend on the speed of the signals. On the rim, however, the difference in arrival times does not depend on the speed of the signals. Any pair of
  4. I suspect that you can do this. You might not even need more than one section. Say that you have (1) a stationary, hollow cylinder with its inside surface mirrored, and (2) a single light emitter/light receiver unit. You set the unit into motion circling around the interior of the cylinder like a motorcycle in the classic "wall of death" carnival sideshow, see http://en.wikipedia.org/wiki/Wall_of_death_(carnival_sideshow). Then the unit sends a flash of light in each direction. I suspect that the same result would occur (the unit would record a fringe shift for the returning flashes when t
  5. Thanks to both swansont and phyti. The thing that confuses me about the particular quotation is the phrase "the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system . . ." Given that the emitter and detector are both on the disk, the result (fringe shift recorded on film) has nothing to do with any inertial coordinate system. If you set the interferometer into translational movement rather than rotation, the film will not record any fringe shift (swansont: "you get a fringe shift in a rotating Sagnac interfe
  6. Here is a typical explanation of the Sagnac effect: Clearly the pulse traveling in the same direction as the rotation of the loop must travel a slightly greater distance than the pulse traveling in the opposite direction, due to the angular displacement of the loop during the transit. . . [T]he expressions "c+v" and "c-v" appearing in the derivation of the phase shift . . . do not refer to the speed of light, but rather to the sum and difference of the speed of light and the speed of some other object, both with respect to a single inertial coordinate system, which can be as great as 2c a
  7. I think that they even state that the line for the actual path is straight (no aberration) because the aberration is so small that it cannot be represented in the simple drawing.
  8. It is essentially their drawing (without copying it outright and violating any copyright they might have in it). But remember that they expressly state that they are modeling a different path than the actual experiment, in order to simplify the math. They acknowledge that their answer is therefore only an approximation of the actual experiment. Their method yields 250 microseconds; xyzt recalls that the actual is around 200 microseconds. So it is clearly only an approximation.
  9. Sure. Although I do not understand them, I can post them for you. For radial path segments they use: dr/dt = 1 - (2M / r) [note that for c=1, I believe that this is the same as your post 3 formula: ] and then integrate: dt = dr / (1 - (2M / r)) They define r as the "coordinate" radius (or reduced circumference, or r-coordinate) and say that it is called a radius "despite its being no true radius" (page 2-9). They use an approximation for 2M/r: (1 + d)^n = 1 + nd And from this they get 53 microseconds delay for the radial segment Earth to Sun. Then, for tan
  10. It seems that the biggest issues are (1) whether slower coordinate speed of light has any physical meaning, (2) if one believes so whether one is mainstream or fringe, and (3) whether the coordinate speed of light (the speed as measured by a distant observer) is a valid explanation of the Shapiro delay. Taylor and Wheeler are certainly mainstream, and here are excerpts from their book "Exploring Black Holes: Introduction to General Relativity." The text answers the questions (1) yes, (2) mainstream, and (3) yes. Project (i.e., chapter) E is titled "Light Slowed Near Sun." That pretty m
  11. I agree with this. What I did in the other exercises was to proportionately adjust all of the clocks' tick rates to the same rate (the farther rearward, the more elapsed ticks show on the clock than actual ticks of time for that rocket's clock). I think that you would call this putting all of the clocks on the same coordinate time. That eliminates the time dilation effect. It leaves only the distance effect. If you calculate the speed of light using the proper distance, then you get the speed of light slowing the closer the rocket is to the rear of the accelerating row (or the deeper in a
  12. I get the same, thanks, and I apologize for getting the question numbering wrong (and for stipulating that the ships accelerate at 0 rather than 60 in their frame, making it a bit awkward). I will put together a Minkowski diagram shortly and think about your question about the time it takes for the ship to pass, but here is a summary and finishing off of question 5. The ships initially present 80 apart in the ground frame, with their clocks not synchronized there. The rear (Bert) clock is ahead by the product of velocity and proper length, hence 100 * 0.6 = 60. Events that are simultane
  13. Getting back to the initial posts, I will go out on a limb and say that yknot is not nuts. There are three events: (1) emission of light, (2) light striking O1, and (3) light striking O2. The light emission and the light strikes are all events, and serious physicists would agree. Witness Taylor and Wheeler, Spacetime Physics, page 10: "Another event is the emission of a flash of light from an atom . . . A fourth event . . . is the strike of a lightning bolt on the rudder of an airplane." See also the (less august) wikipedia entry on "Spacetime": "Events which occur to . . . a photon
  14. This site says 20% of all humans who ever lived past age 65 are alive today: http://www.trinity.edu/mkearl/ger-biol.html So I suspect you have to go a lot higher than age 50, perhaps age 90 or 100.
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.