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Question about the universe and a black hole

  1. Feb 2, 2013 #1
    I recently asked myself the question "If all the stars in the universe condensed together to form a black hole how big would that black hole be?" Using the information from http://en.wikipedia.org/wiki/Observable_universe I got the approximate mass to be 3x10^52kg. After that it was just a matter of finding the Schwarzschild Radius. So...

    2GM/c^2 = 2(6.67300x10^-11)(3x10^52)/8.98755x10^16 = 4.45482918x10^25m

    Is this right? I feel like this number is too large since the overall size of the observable universe is ~4.3x10^26. It seems strange that the radius of this black hole would take up this much space since the universe is mostly empty already. Am I not taking something in to account? Am I just doing the math wrong? Or am I right and that's just how the universe is?
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  3. Feb 2, 2013 #2


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    I haven't looked at your math but a black hole made up of all the matter in the observable universe would be WAY smaller than the observable universe, not just a factor of 10. I would expect MANY orders of magnitude smaller.
  4. Feb 2, 2013 #3
    That's what I assumed too which is why I came here. Especially since I was only using the mass of stars!
  5. Feb 3, 2013 #4


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  6. Feb 3, 2013 #5
    That's pretty interesting. I think there they are using the mass for the entire observable universe (though it's not clear) where I was just trying to use the mass of the stars. So it's HIGHLY probable that I simply got the wrong information about total stellar mass and used the full universal mass instead, in which case my answer seems much more plausible.
  7. Feb 3, 2013 #6


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    I used an estimate of just the ordinary (baryonic) matter in the currently observable portion of the universe---1.5 x 1053 kg.

    Does that seem right to you?

    If I put this into google calculator 1.5e53 kg/(mass of sun)
    it tells me 7.5e22

    This for the radius of the observable currently being about 46 billion LY.

    So the ordinary matter content of the observable is 7.5 x 1022 solar masses

    The bulk of that would be gas and dust, not stars. I think there are an estimated 1022 stars. So that 7.5 seems the right order of magnitude.

    The Schw. radius of a solar mass is 3 km.

    So let's put this into google calculator: "3*7.5e22 km in light years"

    It does the conversion and gives us 24 billion lightyears.

    That would be my back-of-envelope rough estimate of the Schw. radius of all the ordinary matter in the observable.

    Drakkith found an estimate of around 10 billion lightyears. I suspect that is an order-of-magnitude estimate even rougher than mine. Very round numbers. The two are the same order of magnitude.

    The more massive a BH is, the lower "density" it has if you compare its mass with the conventional volume. because the mass goes up linearly with radius and the conventional sphere volume goes up as the cube of the radius. So it is not so surprising.

    I guess if you could carve out and isolate just the observable universe region, and stop its expansion, it would start to collapse. It is dense enough compared with a BH of the same general size.

    But of course it is not isolated as far as anyone knows. As far as we know the universe as a whole has no center and no boundary. And it is expanding very rapidly. So doesn't really have a chance of collapsing. Conditions just aren't right for collapse, even though on paper the density looks sufficient.
  8. Feb 3, 2013 #7

    Jonathan Scott

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    This is just another way of looking at a well-known idea which is related to the Dirac Large Numbers Hypothesis (LNH), to Mach's Principle and to ideas relating inertia to gravitational induction (usually known as the "Sum for inertia").

    The result is known as the "Whitrow-Randall Relation" (sometimes with other names such as "Sciama" included) and states that the sum of the potential terms (Gm/rc2) for every mass in the universe gives a result of order 1. If you replace the sum with a total mass and an appropriately weighted "average" distance, this "average" distance is then of the same order as the Schwarzschild radius of the total mass of the universe.

    In General Relativity, if you accelerate local sources relative to a test mass, then this induces a tiny acceleration in the test mass in the same direction, requiring a tiny (unmeasurably small) opposing force to keep it at rest. The "Sum for inertia" idea says that perhaps you could extend this idea to considering what would happen if you consider a viewpoint where the whole universe is apparently accelerated relative to the test mass, in which case the induced force necessary to oppose that acceleration would be proportional to the mass of the test mass and the relative acceleration. This would effectively provide a neat explanation for inertia if the constant of proportionality for the whole universe happened to come out to exactly 1. Unfortunately, this cannot be the case in General Relativity, as exact equality requires G to vary slightly depending on the distribution of masses, with time and location, but experimental observations show that any variability of G with time or location seems to be too small to support this possibility.
  9. Feb 3, 2013 #8


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    I'm surprised to see I was so far off on this. I've GOT to learn to stop trying to apply my own intuition to physics. :smile: I've HAVE already gotten over that in quantum mechanics.
  10. Feb 4, 2013 #9
  11. Feb 5, 2013 #10

    Since this is my first (actual) post on this site, I will make it clear that I am not
    even close to being a math expert, fortunately Einstein was good at making very
    visual examples, and I was able to adapt quite a bit of his mathematics from rela-
    tivity to variables I could relate to, also Einstein was very capable of explaining his
    ideas in either complex or simpler forms. The reason am writing about this is that
    I am an artist and poet, so my thoughts are made up of
    flowery words and images,
    not equations, But I have worked very hard to understand the General Theory of
    relativity, I won't use a lot of math in my \proofs" but instead use logic that leans
    toward philosophy.
    As far as your question is concerned, I am kind of on the fence when it comes
    to the big bang - vs - the big bounce theory, so there might be a possibility of this
    situation actually happening. Other than that I think the question is a bit extraneous, surely the theoretical mathematical model (or philosophically in my case) is
    a good activity to keep one's mind sharp, but the question has very little practical
    Looking through the \lenses" of General relativity, and using the concept of
    a singularity. the question you had is unresolvable (in the literal sense). If the
    universe has coalesced into one object, this factor would negate the terms \mass"
    and \size" because there would be no reference body to compare the black hole to,
    then it's mass would be irrelevant to the way the universe exists today.
    Because this singularity comprises the sum of all information, there would be no
    where for the object to go, so velocity wouldn't be a factor anymore, time would
    stop, because the gravitation would be infinate, unless gravitation ceased due no
    reference bodies available. in other words, without any other objects for comparison, the Black hole would not even exist by current standard
    On the other hand, we understand only a small portion of how black holes
    react within the event horizon,or do we understand exactly and dark matter/ en-
    ergy is or what it does, other than create an apparent gravitational "field", which
    shouldn't happen because dark matter/ energy doesn't seem to have any mass, yet
    dark matter is probably the reason that galaxies formed and dark energy seems to
    cause expansion, but otherwise it can only be a wild guess as to what other function
    these forms of matter have. and that is something that needs to be factored in or
    out. I hope this made sense, it might not answer your question (especial due to my
    weakness in math) but I hope this answer will oer more questions.
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