Jacques Posted September 10, 2012 Share Posted September 10, 2012 What would be de radius of the event horizon of a blackhole having the mass of the universe ? Same question whith the barionic mass of the universe ? Just a curiosity... Link to comment Share on other sites More sharing options...
LaurieAG Posted September 11, 2012 Share Posted September 11, 2012 (edited) Hi Jacques, Regardless of what mass you use the following applies. The largest Compton wavelength (rest mass) in the visible universe would currently be 13.7 Billion ly long and this would have a reduced form (relative mass) of 13.7 B ly/(2 * Pi). The Schwarzchild radius is twice the reduced Compton wavelength for the mass so the circumference of the event horizon of a black hole with all of the visible universes mass would be 13.7 Billion ly/Pi (i.e. reduced Compton wavelength x 2). http://en.wikipedia....pton_wavelength Edited September 11, 2012 by LaurieAG -1 Link to comment Share on other sites More sharing options...
imatfaal Posted September 11, 2012 Share Posted September 11, 2012 Laurie 1. Universe isn't 13.7 Gly across 2. The schwarzchild radius equalling twice the reduced compton wavelength only occurs at the planck mass. Remember that compton wavelength is inversely proportional with mass - yet the schwazchild radius in linearly proportional. At the planck mass they are equal - higher than the planck mass then compton wavelength is shorter and continues to shorten and the s radius continues to grow. Jacques I am not sure if you can do the simple calculation of plugging the mass of the universe into the equation for the schwarzchild radius - but if you do put in 3*10^54 kg (total mass estimated by multiplying critical density by volume) you get a black hole with a schwarzchild radius of 5x10^11 light years. The mass of atoms is esitmated to be about 4.56% of that total mass - so the radius of all the atoms would be 4.56% of the above figure - ie schwarzchild radius and mass are linear with each other Link to comment Share on other sites More sharing options...
Jacques Posted September 11, 2012 Author Share Posted September 11, 2012 5x10^11 light years That is more than the visible universe, so we basically live in a black hole. Or am I missing something ? Thanks Link to comment Share on other sites More sharing options...
imatfaal Posted September 11, 2012 Share Posted September 11, 2012 That is more than the visible universe, so we basically live in a black hole. Or am I missing something ? Thanks Jacques - I will re-iterate this bit of my post: "I am not sure if you can do the simple calculation of plugging the mass of the universe into the equation for the schwarzchild radius" This is from the Wiki-page on Schwarzchild Radius Assuming constant density, the Schwarzschild radius of a body is proportional to its mass, but the radius is proportional to the cube root of the volume and hence the mass. Therefore, as one accumulates matter at normal density (103kg/m3, for example, the density of water), its Schwarzschild radius increases more quickly than its radius. At around 150,000,000 times the mass of the Sun, such an accumulation will fall inside its own Schwarzschild radius and thus it would be a supermassive black hole of 150,000,000 solar masses. So at 2*10^39 kg a mass the density of water would form a black hole? That is more than the visible universe, so we basically live in a black hole. Or am I missing something ? An object of any density can be large enough to fall within its own Schwarzschild radius, where: is the volume of the object; is its density. This strikes me as slightly batty - in terms of the entire universe; but I think we have to think in terms of viewing an isolated sphere of mass and volume the size of the universe from a huge distance. In that very weird scenario then maybe the idea of event horizon and s radius becomes more palatable Link to comment Share on other sites More sharing options...
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