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The age of the Universe in a denser region of TimeSpace

  1. Aug 31, 2015 #1
    According to Einsteinian Relativity, is the Universe considerably younger if one was to view it from within a greater distortion of the gravitational field? Is the value we assign for flat or non-flat universal age dependant on our own point of reference for time?
     
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  3. Aug 31, 2015 #2

    PeterDonis

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    Meaning, if, for example, a person had somehow been born deep inside the gravity well of a black hole, so deep that, for them, only a few years passed while billions of years passed for an observer on Earth? Yes, to such a person, the universe would appear much younger.

    The numbers quoted for "the age of the universe" assume a "comoving" observer, i.e., one to whom the universe looks homogeneous and isotropic--the same density of matter and energy everywhere and in all directions. Someone deep in the gravity well of a black hole clearly doesn't see that; they see a huge mass close to them and the rest of the universe spread out.

    We on Earth are also not, strictly speaking, "commoving" observers, because we are in the gravity well of the Earth. But the Earth's gravity well is so weak that the difference between the age of the universe we see and the age that an idealized comoving observer in our vicinity would see is negligible--a difference of a few days [edit: years; see correction below] in 13.7 billion years. (The Earth is also moving relative to "comoving" observers in our vicinity, but we can leave that out for this discussion.)
     
    Last edited: Aug 31, 2015
  4. Aug 31, 2015 #3

    Chalnoth

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    PeterDonis addressed the time issue, so I'll address the curvature issue. The answer, it turns out, is very similar: yes, whether or not the universe appears flat depends upon the observer. Our universe appears spatially-flat for observers who see the universe as homogeneous and isotropic. Other observers will see spatial curvature, depending upon their reference frame.

    That said, the fact that our universe has any observers at all who see an (approximately) homogeneous and isotropic universe is peculiar. It didn't have to be that way. There are many more universes that you could write down equations for that do not appear homogeneous, isotropic, or flat to any observer.
     
  5. Aug 31, 2015 #4

    Bandersnatch

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    Peter, can you provide an estimate for what is the influence of the gravitational well of the Milky Way w/r to the comoving observer? Is it not more significant than Earth's gravity?
     
  6. Aug 31, 2015 #5

    PeterDonis

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    It's more significant than Earth's, yes, but that's a very low bar. It's not significant in any sense that matters for this discussion.

    The general formula we're interested in is the deviation of the metric coefficient ##g_{tt}## from 1. A good estimate of that for objects with weak fields (we'll come back to how we know which fields are weak in a moment) is simply ##GM / c^2 R##, where ##M## is the mass of the object and ##R## is its radius. We have ##G / c^2 \approx 10^{-27}## meters per kilogram, so we can do some quick order of magnitude estimates:

    Earth: ##M \approx 6 \times 10^{24}##, ##R \approx 6 \times 10^6##, so ##GM / c^2 R \approx 10^{-9}##.

    Sun: ##M \approx 10^{30}##, ##R \approx 10^9##, so ##GM / c^2 R \approx 10^{-6}##.

    Sun's field at Earth's orbit: ##M \approx 10^{30}##, ##R \approx 10^{11}##, so ##GM / c^2 R \approx 10^{-8}##.

    Milky Way: ##M \approx 10^{42}##, ##R \approx 10^{22}##, so ##GM / c^2 R \approx 10^{-7}##.

    All of these are obviously weak fields, since the correction is much less than 1; a strong field would be one in which this correction, calculated as above, was significant compared to 1. (In the strong field regime the formula I gave above, which is an approximation, breaks down, and we need to use the exact metric coefficient formula.) As you can see, the correction due to the Milky Way's field, at Earth's position, is the largest of the three; but at the surface of the Sun, the correction due to the Sun's field is larger than that due to the Milky Way's field. At Earth's orbit, the correction due to the Milky Way makes a difference of about 1000 years in the observed age of the universe (i.e., we here on Earth see a universe with an apparent age about 1000 years younger than a comoving observer far out in intergalactic space).
     
    Last edited: Aug 31, 2015
  7. Aug 31, 2015 #6

    PeterDonis

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    Actually, this should be a few years, based on the calculations in my previous post.
     
  8. Aug 31, 2015 #7

    Bandersnatch

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    Would it be correct to say that it's not the depth of the gravity well that matters here, but its slope?
     
  9. Aug 31, 2015 #8

    PeterDonis

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    Not for time dilation, no; for that the depth is what matters (and the numbers I calculated are for depth--basically they are how far below the "infinity" value of 1 an observer at radius ##R## from mass ##M## is in terms of depth in the gravity well). The slope (at least in the weak field approximation) gives the proper acceleration required to "hover" at a constant altitude.
     
  10. Aug 31, 2015 #9

    Bandersnatch

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    Cheers, Peter.
     
  11. Aug 31, 2015 #10

    marcus

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    I agree with Peter--- it's depth, that governs time dilation.
    Always seen it dealt with that way, and a handy way to think of it is as the square of the circular orbit speed, as a fraction of c.

    Our orbit speed around galaxy center is about 250 km/s and Earth orbit speed around sun is about 30 km/s.

    So about 10-3 and 10-4 of the speed of light, respectively. So squaring and adding, one gets about 10-6.

    Peter got 10-7. My calculation was very rough so I wouldn't worry about the difference.

    The galaxy has more mass outside our orbit so I think I've under estimated the depth actually. The depth should be proportional to the square of the escape velocity, so roughly proportional to the square of the circular orbit velocity when the mass is concentrated at the center, but MORE than the square of the circular orbit speed when the mass (like that of our galaxy) is spread out. So I'd guess AT LEAST 10-6, but its still pretty small potatoes.
     
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