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I Hubble red shifts could be gravitational red shifts rather than space-expansion red shifts?

  1. Nov 26, 2017 #1
    Several questioners ask if the Hubble red shifts could be gravitational red shifts rather than space-expansion red shifts. I understand why the answer has generally been "no". However, can I try this variation of the question.....

    Red-shifting is apparent mostly for distant galaxies, which are thus being seen at a much earlier age when the universe (I am assuming) was much smaller than currently. We (the observer) are located at a point which is beyond the part of space which was occupied at the time the red-shifted light began its journey. Thus, the red-shifted light will have needed to rise out of the gravitational potential well of the entire older universe in order to reach our detector today. And the further away the observed galaxy, the smaller the universe was when the light began its journey, and hence the longer this light would need to fight its way through the universe's potential well to reach our detector today.......and hence the greater the red shift.

    Cheers, Tom McFarland
     
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  3. Nov 26, 2017 #2

    phinds

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    You seem to be assuming that there is a center to the universe, else how to interpret your concept of "rise out of the gravitational potential well". This makes no sense and since your question is based on it, your question makes no sense.
     
  4. Nov 26, 2017 #3
    The question, please, makes no assumption about the "center of the universe". The only assumption is that the older universe was smaller, and the observer today must therefore be outside the boundary of that older universe. No specific location of anything is assumed. All locations are merely relative to other locations.

    Cheers, Tom McFarland
     
  5. Nov 26, 2017 #4

    phinds

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    I don't get how that makes any sense. Of COURSE we are further along in time, but so what? You have not defended your concept of "rise out of the gravitational potential well", which, again, does not make sense to me.
     
  6. Nov 26, 2017 #5
    I am not a physicist, but let me try to "defend" the quoted concept above.

    At any time t, the universe has a size or diameter d(t). We do not know what d(t) is equal to for any time t, though I have seen various estimates for small values of t. That said, we probably agree that at least for the early universe, d(t) was some number, and d(t) is monotone increasing.
    At any time t, all gravity-generating "stuff" (matter and energy) lay inside some sphere of diameter d(t). We do not assume any location for this sphere.

    Alright, pick a specific time t=t₀. Choose any point A within the universe at time t₀, and then another point B for which the distance between A and B is at least 3d(t₀)......B must be outside the boundary of the universe at time t₀ but inside the universe at some later time. For anything to travel from A to B, including light, that thing must rise out of the "potential well of the entire universe" as it existed at time t₀, in the same way that I would need to rise out of the Earth's gravitational potential well to travel to the moon.

    The only hitch I see in this discussion is that I may not be allowed to even pick a point B above??

    Does this satisfy you?

    Tom McFarland
     
  7. Nov 26, 2017 #6

    phinds

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    Only in that it clarifies your mistake. You are now explicitly stating what you previously denied stating which is that there is a preferred physical direction in the universe, which is equivalent to saying there is a center. You misunderstand the early universe. It was not a sphere, and it was of indeterminate size but was either infinite or was finite but unbounded. In neither case does your argument hold water.
     
  8. Nov 26, 2017 #7
    Frankly, I am finding this discussion unhelpful, but I will respond once more....

    I do not see any implication of a preferred direction or location. If you see it, show me where it is.

    I did not claim the universe was a sphere or ball, but I did claim that at each time, t, the gravity-generating stuff in the universe is bounded, that is, there is a sphere S(t) which contains all of it, outside of which there is no gravity-generating stuff. If you feel this assumption is false, justify your claim.

    Another assumption I made was that at any point in time, the sphere S(t) is imbedded in an unbounded space like E³ , most of which contains no gravity-generating stuff. As t increases, S(t) enlarges to include previous excluded points, such as B. Again, S(t) is not the universe and its center could be any point in the universe.

    Cheers, Tom McFarland
     
  9. Nov 26, 2017 #8

    phinds

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    This is incorrect. There IS no "outside"

    And by the way

     
  10. Nov 26, 2017 #9
    Phinds:

    I appreciate your interaction, but it is time to say goodbye.

    The statement "There IS no "outside" is completely unverifiable, unless you define "universe" to make it true. I proposed a different definition of "universe", perhaps implying a different metric, and a different interpretation of red shift. You are unwilling to think this way. I will work on the math, and come back another day......but not tomorrow.

    Best wishes, Tom
     
  11. Nov 27, 2017 #10
    22 / 2

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    Phinds:

    I appreciate your interaction, but it is time to say goodbye.

    The statement "There IS no "outside" is completely unverifiable, unless you define "universe" to make it true. I proposed a different definition of "universe", perhaps implying a different metric, and a different interpretation of red shift. You are unwilling to think this way. I will work on the math, and come back another day......but not tomorrow.

    Best wishes, Tom
     
  12. Nov 27, 2017 #11

    Bandersnatch

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    @Tom Mcfarland the theory describing the expansion of the universe assumes homogeneity and isotropy throughout the universe (aka 'cosmological principle'). It is this assumption that allows the Einstein field equations to be solved in order to obtain the Friedmann equations. Seeing how our locally observable patch of the universe looks homogeneous and isotropic on large enough scales, and also about just like the theory predicts in other ways (i.e. it does look like it evolves according to Friedmann equations), the cosmological principle seems like a good descriptor.

    So, if you want to talk about what modern cosmology has to say about the expansion of the universe, you have to assume the cosmological principle. If you do that, then there can be no part of the universe that is outside all the matter in it. The only case where all the matter can be within a bounded radius d is when the universe is a closed hypersurface (such as a hypersphere), as only this case allows for a finite amount of matter to be distributed evenly across the entire space.
    So far, attempts at detecting intrinsic spatial curvature have failed to find any, which points to a flat universe. Cosmological principle then dictates that matter should be spread evenly across infinite space, with no large (in cosmological sense) empty areas, unbounded.
    Whether the universe is bounded or not*, there can be no global gravity wells when matter is distributed evenly.
    The only gravity wells that can develop are due to local overdensities of matter, and these don't evolve much in the expanding universe (if you're interested in the effect these have on redshifts - look up the Sachs-Wolfe effect).

    However, if you want to talk about alternatives to cosmological models discussed in textbooks and literature, e.g. by developing new solutions to EFE, then this forum is not the place to do it.

    *note to @phinds, mark the difference between bounded and having boundary - even after all this time we've been talking about it, you're still using it incorrectly
     
  13. Nov 27, 2017 #12

    phinds

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    Hm ... guess I'm still confused. I intended my statement to mean that the universe is without bound, meaning that if it is finite, there is no edge, you would just keep going forever. I did not mean to imply that it has a boundary.
     
  14. Nov 27, 2017 #13

    phinds

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    @Bandersnatch, I have been corrected on this bounded/boundary thing a number of times but it seems to have just eluded me. In an attempt to straighten myself out, I have gone back through some older posts and I THINK I have it now. I'm going cross-eyed from having read those words so many times in the last 5 or 10 minutes that I may STILL be getting it wrong, but I would appreciate it if you would verify that I have the following correct:

    if the universe is finite then it is "bounded" because all parts are within some specific distance of each other

    If the universe is infinite then it IS "unbounded" because it is not true that any two parts are within a specifiable distance of each other.

    I have been mistakenly using the term "bounded" (or more specifically "UNbounded") to describe whether or not you would go on forever in the same direction. You would, but this apparently is described by saying that it has no bound, NOT that it is "unbounded".

    A sphere is bounded, not unbounded as I have been saying it is, but it has no boundary.

    So my previous statements that "the universe is either infinite or finite but unbounded" should be "the universe is either infinite or finite without a boundary"
     
  15. Nov 27, 2017 #14

    Bandersnatch

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    'Bounded' means that there is such a sphere of finite radius R that you could 'draw' in the universe, which would encompass all of its 3D space. You can draw such a sphere in a closed hypersphere, in the same way as you can draw a circle of finite radius R on the surface of a sphere, so as to encompass all of its surface.

    As such, 'bounded' means pretty much the same as finite. 'Unbouded' then means the same as infinite, because in infinite universe you'd have to draw a sphere/circle of infinite radius to encompass of all the 3D/2D space under consideration.

    So when you say
    it's an oxymoron. It cannot be both.

    Just say what you intended to say, i.e. 'without boundary' whenever you want to say unbounded.

    (Just now reading your latest post I see you got it right)
    Yes. Which incidentally is the same as saying 'the universe is either unbounded or bounded without a boundary'
     
  16. Nov 27, 2017 #15
    @phinds Do u have any ideas about how to solve the problem of conservation of energy on gravitational fields according to GR?
     
  17. Nov 27, 2017 #16

    phinds

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    Sorry, but I don't even understand the question. I don't know the math of GR, which is likely a prerequisite to answering your question in a helpful way.
     
  18. Nov 27, 2017 #17

    phinds

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    Good. Thanks. Just to be SURE I have all this right, is EVERYTHING I said in that post correct? If so, I'm going to try to remember to look at it before ever using those terms again.
     
  19. Nov 27, 2017 #18
    I understand.

    @phinds , I imagine you know that in general relativity the curvature of space-time does
    What happens is that General relativity doesn't seem to take conservation of energy into consideration. I hope the following extract from "
    Is Energy Conserved in General Relativity?", of the maths departament of the university of California.

    "The differential form says, loosely speaking, that no energy is created in any infinitesimal piece of spacetime. The integral form says the same for a finite-sized piece. (This may remind you of the "divergence" and "flux" forms of Gauss's law in electrostatics, or the equation of continuity in fluid dynamics. Hold on to that thought!)

    An infinitesimal piece of spacetime "looks flat", while the effects of curvature become evident in a finite piece. (The same holds for curved surfaces in space, of course). GR relates curvature to gravity. Now, even in newtonian physics, you must include gravitational potential energy to get energy conservation. And GR introduces the new phenomenon of gravitational waves; perhaps these carry energy as well? Perhaps we need to include gravitational energy in some fashion, to arrive at a law of energy conservation for finite pieces of spacetime?

    Casting about for a mathematical expression of these ideas, physicists came up with something called an energy pseudo-tensor. (In fact, several of 'em!) Now, GR takes pride in treating all coordinate systems equally. Mathematicians invented tensors precisely to meet this sort of demand — if a tensor equation holds in one coordinate system, it holds in all. Pseudo-tensors are not tensors (surprise!), and this alone raises eyebrows in some circles. In GR, one must always guard against mistaking artifacts of a particular coordinate system for real physical effects. (See the FAQ entry on black holes for some examples.)

    These pseudo-tensors have some rather strange properties. If you choose the "wrong" coordinates, they are non-zero even in flat empty spacetime. By another choice of coordinates, they can be made zero at any chosen point, even in a spacetime full of gravitational radiation. For these reasons, most physicists who work in general relativity do not believe the pseudo-tensors give a good local definition of energy density, although their integrals are sometimes useful as a measure of total energy".
     
  20. Nov 27, 2017 #19

    Bandersnatch

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    Pretty much. Maybe apart from some minor details. E.g.
    In an unbounded, i.e. infinite, universe you can easily find any two parts that are within a specifiable distance (like between me an my coffee cup, or us and the surface of last scattering). But you can't specify a finite distance that will encompass ALL of its parts.
     
  21. Nov 27, 2017 #20

    phinds

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    Point taken. Thanks. I think in all of this in my past posts, I had the concepts correct, but was using the wrong terminology (which DID of course make my statements incorrect). Thanks for the help.
     
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