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I Expansion of universe

  1. Jun 27, 2016 #1
    i have a question about the motion of galaxies and the speed limit of light.
    using Hubble's law, it is theoretically found that after a certain distance(about 4200 mega parsec or something) from a observer, the galaxies are moving faster than the speed of light.
    consider the following scenario:
    let galaxy A be in any part of the universe. let galaxy B be at a distance greater than 4200 mega parsec from A. as per Hubble's law, B is moving away from A at speed faster than light and to B , A is moving away from A at speed faster than light . the movement of these two galaxies is observed by a observer in a absolute frame of reference F (Newtons's idea; assume such a frame exists)
    here are the question:
    1)for the observer in F , is the galaxy B and A moving by themselves , or are they moving due to the expanding space ?. because if galaxies were to move due to expansion of space, then aren't they violating the speed limit of light due to the following fact. let V be the velocity of the galaxy B and v be the velocity of the expanding space in which the galaxy is present . hence the total velocity of the galaxy B measured by the observer in F would be V+v ;where both V and v are comparable to the speed of light .
    2) are the galaxies actually embedded in space and moving with space or are they moving in space and not embedded in the space itself?
     
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  3. Jun 27, 2016 #2

    Drakkith

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    The speed of light is a local speed limit, not a global one. That means that two galaxies can recede from each other at a velocity greater than c thanks to the expansion of space, yet you will never be able to outrace a light-beam.

    There's no observable difference between the two that I know of. Keep in mind that we can't observe what space does. We can only observe what objects within space do. So there's no way to tell if space is "actually moving" or if objects are merely moving "through space". As far as I know at least.
     
  4. Jun 27, 2016 #3
    then, would that mean its possible that galaxies have their own motion and the space itself has its own motion. and for both their motion the speed limit is c. hence the maximum possible observable velocity of a galaxy is 2c for a observer in a absolute frame of reference F (the frame F is at absolute rest - Newton proposed the idea of such a frame of reference) . hence the speed of light is not violated by the components in motion individually(space ; galaxy) but the speed of the galaxy is greater than the c . is this possible?
     
  5. Jun 27, 2016 #4

    Drakkith

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    Keep in mind there is no limit to recession velocity, not merely 2c.

    Nope. A preferred, absolute reference frame is ruled out by observations as far as I understand.
     
  6. Jun 27, 2016 #5
    isn't a stationary inertial frame of reference similar to the absolute frame of reference? also if it is proved that something in the universe is at absolute rest, than would a absolute frame of reference be accepted?
     
  7. Jun 27, 2016 #6

    Drakkith

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    I don't believe so. If an absolute reference frame existed then an inertial frame of reference which was in motion with respect to the absolute frame should have differences in experiments run in that inertial frame. In other words, if you perform two identical experiments in each frame, there should be a difference in the results, and those differences should increase as the inertial frame moves with a higher and higher velocity relative to the absolute frame. This is not what we observe. All inertial frames behave identically, no matter which direction they are moving in nor how fast they are moving.

    Certainly. Absolute rest necessarily requires that an absolute reference frame exist.
     
  8. Jun 27, 2016 #7
    how is it possible that measurement made from a inertial frame is correct irrespective of the velocity of the inertial frame, which itself is not possible to know without an absolute frame . hence any measurement made from a inertial frame is relative , isn't it? then how is the measured value same in all inertial frames of reference? [ I seem to have deviated from the original topic :-) ]
     
  9. Jun 27, 2016 #8

    Drakkith

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    It's not that they are "correct", it's that they are the same. And to clarify, I'm talking about experiments you can perform in a single reference frame, not observations of external events outside of your reference frame.
     
  10. Jun 27, 2016 #9
    so is it impossible to determine whether any measurement like velocity made by an observer(in a inertial frame) of an event occurring in the same frame of the observer or any other frame(inertial or non inertial) is the correct value of the quantity , since the velocity of the frame of the observer is unknown ?
     
  11. Jun 27, 2016 #10

    Drakkith

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    The velocity of the observer isn't unknown, it's relative. A spaceship flying by the Earth at 100 km/s is viewed by an observer on the Earth as traveling at 100 km/s, yet an observer on the ship views the Earth as passing by at 100 km/s. And both are correct.
     
  12. Jun 27, 2016 #11

    Dale

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    Newtonian gravity works on the scale of the earth and moon, but it does not work at the scale of the universe.
     
  13. Jun 27, 2016 #12
    For the most part it does:

    newton,cosmology.png
    Also see this, this, this and this reference on the subject. For the gravity in cosmology Newton does most of the job since it is rather weak; the relativistic equations give the same result like Newtons, with the little difference that the energy of light is inversely proportional to the scalefactor. But you don't need hardcore relativity in standard cosmology like you would in black hole astronomy, the formalism is completely newtonian for the most part.

    That's right. The first is the local and the second the comoving recessional velocity. We observe a combination of both.

    Only for the local velocity. The recessional velocity has no limit.

    UDFy-38135539 as one of the farthest observable galaxies with a redshift of 8.55 was receding with 2.08 c when it emitted its light and is receding with 3.62 c today.The CMB surface had 63.12 c relative to us when it emitted the light which we receive today 400 million years after the big bang, and is receding with 3.11 c today.

    Later he rejected it.

    newton+einstein.png , st13.png
     
  14. Jun 27, 2016 #13
    All comoving observers, even when having a relative recessional velocity to each other, are at rest relative to the CMB. If we want to translate the measurements made from earth to that frame we first have to substract the CMB dipole and then add the proper distance times the Hubble constant to the equation if we want to transform into the system of another Robertson-Walker observer. The kinetic energy of an object in that frame is the energy it would have relative to an observer at rest to the CMB when passing by right next to him.
     
  15. Jun 27, 2016 #14

    PeterDonis

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    He said the "equations of an expanding universe" are "Newton's equations", but he does not mean the fundamental equations of Newtonian gravity. He only means the equations from one particular solution in Newtonian gravity. The fact that one particular GR solution happens to yield the same equations as one particular solution in Newtonian gravity does not mean Newtonian gravity explains the expansion of the universe. It just means the same equations happen to come out of GR in this particular case. You certainly can't use the fundamental equations of Newtonian gravity to derive this particular solution for the universe.
     
  16. Jun 27, 2016 #15
    but if both of the observers (on the earth and space ship) are already in motion such that both earth and spaceship are moving at 5 km/s , and 100km/s is the measured relative value. but the actual value is 105 km/s . hence the 5 km/s value is not known unless a absolute frame of reference is defined. isn't this correct?
     
  17. Jun 27, 2016 #16
    but that does not have anything to do with the absolute frame of reference.
     
  18. Jun 27, 2016 #17

    PeterDonis

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    Relative to what?

    Relative to what?

    No, it's not known unless you specify what it is relative to. There is no absolute frame of reference.
     
  19. Jun 27, 2016 #18
    so if the galaxy is moving with the resulting speed of 3.11 c , then aren't the particles in that galaxy violating the local speed limit. (because the recession velocity is added to the local velocity of the particle.).

    also in calculating the recession velocity using Hubble law, is the recession of the observer taken into account?
     
  20. Jun 27, 2016 #19
    not all motion is relative. only the measured value is relative. whether the motion is observed or not , the particle or whatever being observed is in motion because of the expansion of the universe and also due to its own energy. hence it would be impossible to find the correct value of the observed value without defining a absolute frame of reference.
     
  21. Jun 27, 2016 #20

    PeterDonis

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    No, it isn't. The recession velocity is a coordinate velocity in FRW coordinates. The local velocity of the particle is the relative velocity that would be measured by a comoving observer at the same spatial location. You can't add these two things; the result is physically meaningless.
     
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