Relation between black hole mass and radius, and our universe's

  • Is there a graph of known black holes, with their estimated mass in the X axis and their estimated radius in the Y axis? If so, where can we find it? I would like to know if a black whole with the whole estimated mass of our universe would have the estimated radius of our universe (which means our universe could be a black hole, that's why light can't escape it and it looks "finite").


    As your basic idea is that the universe is inside a black hole I think that this question and answer explains why we're not in a black hole. There is also a link there to a question on Physics SE that goes into this as well.

    @StephenG How can my question be a duplicate, if it was asked before? I'll look at the links you provided, thank you.

    @StephenG I said that maybe our Universe IS a black hole, not "is inside one", I don't think it's the same thing. And the results are all in the same order of magnitude. Assuming all those numbers are at best good approaches, and that our Universe as a black hole don't need to follow the same rules black holes inside our Universe follow, and also considering that dark energy was discovered "just yesterday", I don't think this hypothesis should yet be discarded.

  • Gerald

    Gerald Correct answer

    9 years ago

    The Schwarzschild radius of a black hole is probably the closest we can get to your question.



    $$
    r_s = (2G/c^2) \cdot m \mbox{, with }\ 2G/c^2 = 2.95\ \mbox{km}/\mbox{solar mass}.
    $$
    This means, that the Schwarzschild radius for a given mass is proportional to that mass.
    The radius shouldn't be taken too literal in the physical sense, because space is highly non-euclidean close to a black hole.



    Present (light-travel) radius of the visible universe, as seen from the earth:
    $$13.81 \cdot 10^9\ \mbox{lightyears} = 13.81 \cdot 10^9 * 9.4607 * 10^{12}\ \mbox{km} = 1.3065\cdot 10^{23}\ \mbox{km}.$$
    So we need $$1.3065\cdot 10^{23}\ \mbox{km} / 2.95\ \mbox{km} =4.429 \cdot 10^{22}$$
    solar masses to get a black hole of the light-travel Schwarzschild radius of the visible universe, pretty close (by order of magnitude) to the number of stars estimated for the visible universe.



    The Wikipedia author(s) gets a similar result: "The observable universe's mass has a Schwarzschild radius of approximately 10 billion light years".


    WOW! Thank you! That's really amazing! It looks like we live inside a black hole, perhaps...

    I was a little surprised myself. First I thought, a black hole should be much smaller, but it isn't. But there may be solutions of the field equation of general relativity leading to a comparable "radius" besides that of a black hole.

    @Gerald So the average density of a black hole with the diameter of the universe, would be about the same as the density of our universe, almost vacuum?

    Yes. If we think of the black hole being replaced by space filled with the same mass with Euclidean metric, to get a reasonable definition of density.

    That's not as surprising as it looks at the first instant after thinking a bit: A snapshot of the universe would make light travel roughly a circle with the diameter of the observable universe, much the same as a black hole with Schwarzschild "diameter" 2/3 of the diameter of the circle.

    Yes, independent of the diameters involved. Maybe inside each new black hole some random variables are shuffled, and new rules arise in a fractal of beauty. As if black-holes are universes and universes are black-holes...

    In the first answer, (above), there is an ERROR: 1 light-year is 9.46 x 10^15 meters, not 9.46 x 10^12 ... this error skewers the resulting mass-calculation, and brings the conclusion into question ...

    9.46 x 10^15 meters is 9.46 x 10^12 kilometers. Nevertheless thanks for checking!

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Content dated before 7/24/2021 11:53 AM