Schwarzschild radius

[huh?]

I was bored and decided to do some calculations with the schwarzschild radius. I made up a black hole with a arbitrary and ridiculous solar mass of 100 trillion and i ended up with the weirdest numbers from a size of 15 to 185,000 up to 1.9x10 to the power of 39 parsecs, lol.

 

Can anyone simplify the schwarzschild radius for my feeble brain or do the calculations for me, i'm now at the point of just wanting to know the answer.

 

Thanks

[timo]

I don't get your question. What kind of calculations are you doing with the Schwarzschild radius? The formula for the radius itself, R=2Gm/c², already is very simple - there's little to simplify or to get wrong, there.

[Martin]

I was bored and decided to do some calculations with the schwarzschild radius. I made up a black hole with a arbitrary and ridiculous solar mass of 100 trillion and i ended up with the weirdest numbers from a size of 15 to 185,000 up to 1.9x10 to the power of 39 parsecs, lol.

 

Can anyone simplify the schwarzschild radius for my feeble brain or do the calculations for me, i'm now at the point of just wanting to know the answer.

 

Thanks

 

Atheist is right, there's little that can go wrong if you understand how to compute the Schwarzschild radius.

 

It is proportional to the mass. for one solar mass it is about 3 km or roughly 2 miles.

 

for 10 solar mass it is about 30 km, or 20 miles.

 

for a trillion solar mass it is about 3 trillion km or 2 trillion miles.

 

You want it for 100 trillion solar mass, so that should be easy.

=================

 

the main thing is to know how to calculate it for the mass of the sun----the rest is just scaling up

 

just type this into Google window, and press "search"

2*G* mass of sun/c^2

 

It doesnt do a search, it actually calculates it. The google calculator knows the mass of the sun, and Newton G, and speed of light c, so it makes physics calcuations very easy. It will calculate what you put in the window, and it will come back with this:

 

2.95343236 kilometers

 

that is why I said the Schw. radius for sun mass is about 3 kilometers

[huh?]

Thanks for the explanation Martin. I have zero ability when to algebraic formulas and have no knowledge of general relativity. I was only looking at some article on some site about black holes and figured since i was bored might as well do some random calculation on a fictional black hole. Ended up just looking at black holes that were already calculated and upscaled them and did some stuff that i'm not sure why i did and got some weird numbers.

 

Thanks again guys, now i understand the Schwarzschild Radius.

[Martin]

Thanks for the explanation Martin. I have zero ability when to algebraic formulas ...

 

Indulge me, what happens when you type "mass of sun" into Google

and it hit return?

 

I mean "mass of sun" without the quote marks, just those three words.

 

Look at the very top of the list, before all the hits google gives you.

 

I am curious to know if you get the same thing I do.

 

==================

EDIT also as an afterthought. What happens when you just type "c" into Google and press return?

Just the single letter, no quote marks.

[Rakdos]

mass of sun

 

c

[Norman Albers]

Now figure out the Schwarzschild radius of an electron. . . . . ALSO, do you realize that at whatever density of 'a mass distribution', there is such a radius? If the density of the universe is such-and-such, then there is a size where it's a BH.

[Martin]

mass of sun

 

c

 

Rakdos, thanks. I was hoping "huh?" could learn to use Google's calculator to find the Schw. radius of the solar mass

 

But "huh?" did not reply, and you did instead.

 

Well how about doing the calculation----it may not be new for you Rakdos but anyway

 

Put this into the Google window

 

2*G*(mass of sun)/c^2

 

 

If you want the Schw. radius of an electron, or of the Earth, you just tell Google that, and for example type in

 

2*G*(mass of earth)/c^2

 

What is the Schw. radius of the Earth's mass?

 

Now figure out the Schwarzschild radius of an electron. . . . . ALSO, do you realize that at whatever density of 'a mass distribution', there is such a radius? If the density of the universe is such-and-such, then there is a size where it's a BH.

 

I think you are mistaken Norman.

the Schw. radius refers to a certain solution of GR called the Schw. solution, which is static. the space of the Schw. solution is not expanding.

In our universe i can easily show you a sphere within which there is the Schw. BH density for that radius and the sphere is obviously NOT a Schw. BH horizon.

 

that is, anybody out at that sphere (call it a FAKE "horizon") can travel freely back and forth as much as they want across the fake "horizon". Obviously they could not if it really was.

 

You can take the sphere of radius c/H centered at the Milkyway galaxy, so that we (or our galaxy) is the center of the sphere.

 

and you can show that this sphere contains the same mass as a Schw. BH of that radius.

So if things were STATIC, we would be at the singularity of a Schw. BH.

 

but we are obviously not.

 

This statement is false:

If the density of the universe is such-and-such, then there is a size where it's a BH.

 

It is also stated rather vaguely.

 

If you would like to be more precise, the density of the universe is, as best we can tell, around 0.8 or 0.9 joules per cubic kilometer.

 

What radius sphere, centered at the Milky Way, would you claim encloses a BH?

[Norman Albers]

Thanks for the explanation, Martin. I must have been reading on stationary models.

[Rakdos]

Rakdos

 

What is the Schw. radius of the Earth's mass?

[hide] 8.87134507 mm[/hide]

[Norman Albers]

What radius sphere, centered at the Milky Way, would you claim encloses a BH?

Well, I calculate part of a light-year, maybe 2/3, at 0.9 Joules/km^3.[bIGTEETH] (I will rerun the numbers.) This is interesting fun, as its tells us the scale at which expansion becomes notable. Can we not calculate a certain higher density at which a 'dust cloud' becomes critical? I shall read in my cosmology chapters. . . In mks units, G is 6.67E-11.

[Martin]

[hide] 8.87134507 mm[/hide]

 

this is great! I think Rakdos just used the Google calculator (I love that calculator because it knows many basic natural constants, like Planck's hbar and Boltzmann's k so you can do really neat stuff easily!)

 

Let's see if we put this in Google window and press search do we get what Rakdos says?

2*G*(mass of earth)/c^2

 

Yes! we get exactly what Rakdos says. So very likely he was using the same calculator (at least the same values of the fundamental constants). Yay!

[Norman Albers]

Pardon me if I didn't speak to whichever specific: I figured Schw. radius as 4.4 meters for Earth, without the factor of 2. [math]m=\frac{GM}{c^2}[/math] and checking terminology I see I am speaking of the geometric mass while the Schwarzschild radius is twice this or 8.8 m.

[Martin]

Pardon me if I didn't speak to whichever specific: I figured Schw. radius as 4.4 meters for Earth, without the factor of 2. [math]m=\frac{GM}{c^2}[/math] and checking terminology I see I am speaking of the geometric mass while the Schwarzschild radius is twice this or 8.8 m.

 

I think you are off by a factor of 1000.

 

Well, I calculate part of a light-year, maybe 2/3, at 0.9 Joules/km^3.[bIGTEETH] (I will rerun the numbers.) This is interesting fun, as its tells us the scale at which expansion becomes notable. Can we not calculate a certain higher density at which a 'dust cloud' becomes critical? I shall read in my cosmology chapters. . . In mks units, G is 6.67E-11.

 

You are off by many many orders of magnitude. I can't take the time now to see what you are doing wrong. Maybe Rakdos will help to get you straightened out. Or one of the others.

[Norman Albers]

Yes, indeed, now I get about a centimeter for the Sch. rad. of Earth, and a very different answer for the geometric radius of an energy density of about one Joule per cubic kilometer. Putting that energy density over [math]c^2[/math], and figuring total mass of a uniform sphere I get ten billion light-years.

[Martin]

Putting that energy density over [math]c^2[/math], and figuring total mass of a uniform sphere I get ten billion light-years.

 

that is approximately right!

it is certainly the right order of magnitude

 

earlier I gave an estimate of the density of the universe to be 0.8 to 0.9

joules per cubic kilometer

 

when I calculate out the best estimate I know I get 0.85

 

you are using 0.9. If it is not too much trouble, I'm curious what you would get if you used 0.85. If you don't want to bother that's fine---just mildly curious.

[Norman Albers]

[math]R^2=\frac3 {4\pi} \frac {c^4 }{GN}, [/math] where N is energy density in Joules per cubic meter. I used N at 0.9E-9, yah? Is my trusty sliderule serving? This is good, clean fun. Yes calculators won't make mistakes adding and subtracting magnitudes but you're gonna bury me, like Max Born, with my rule. I got exactly 1E26 m , and light-years are 8.6E15m. It comes out about 12 billion light-years I think. Sort of rings a bell, huh?

[Martin]

[math]R^2=\frac3 {4\pi} \frac {c^4 }{GN}, [/math] where N is energy density in Joules per cubic meter...

 

Your trusty sliderule may be serving OK but you forgot a factor of 2 in the equation you were using. So your answer will be off by approximately the squareroot of two on that account. But that may not be the only trouble.

Instead of wading thru this, why don't I just state the answer.

 

the problem is find R such that a sphere centered at our galaxy with that radius would contain enough mass so that R would be the Schwarzschild radius for that mass. That doesnt mean that we are in a Schwarzschild black hole because the universe is expanding.

anyway the solution is

R = c/H

where H is the hubble parameter at the present time.

[Norman Albers]

Yes, gotcha. Lo, I'm fading and eating dinner. [bIGTEETH]. . . . Thus fortified, I can see that my problem should be stated, "with a density filling a Schwarzschild radius.." so yes, we take the cube of (2m) for the mass statement, but then also twice the result, so the answer comes out less by the factor of \sqrt(2) , or like 8-9 billion lt-yrs. Is this what you see, Martin?

 

In detail, I'm asking what is the Schwarzschild radius of an energy density which fills that sphere. [math]r_s=\frac{2GM}{c^2} [/math] and we say the total mass [math] M=\frac4 3 \pi\rho (r_s)^3[/math]. Mass density [math]\rho[/math] is taken as energy density (in kg/m^3) divided by c^2.

[Martin]

 

In detail, I'm asking what is the Schwarzschild radius of an energy density which fills that sphere. [math]r_s=\frac{2GM}{c^2} [/math] and we say the total mass [math] M=\frac4 3 \pi\rho (r_s)^3[/math]. Mass density [math]\rho[/math] is taken as energy density (in kg/m^3) divided by c^2.

 

Norman IMO it's time for you to dig a little deeper. Instead of just taking the estimate of the density that I told you

 

(namely 0.85 joules per cubic km-------or 0.8-0.9 E-9 joule/m^3---or whatever)

 

why don't you CALCULATE IT in the usual way cosmologists do.

 

then your whole thing will be more stand-alone self-contained.

 

you have a choice of calculating the mass or the energy density, and what you need for your purpose is the MASS density, so I will give you the formula for that

 

[math]\rho_{crit} = \frac{3 H^2}{8 \pi G}[/math]

 

current observational measurements show the real mass density is very close to mass rho_crit

I recently saw an errorbar based on a lot of different data which was a 68 percent errorbar that the real rho was [1.01, 1.04] times the crit.

Ned Wright gave a "best fit" figure of 1.011.

so there is uncertainty but it looks like the formula I gave you here for crit is good to within one or two percent.

Good enough for government work. So go with it. And use THAT estimate of the mass density to find your radius.

 

 

Remember we are not in a black hole, the radius is only a pretend Schw. radius, because the universe is expanding (if it were static and matched the Schw. picture then we would be at the singularity which we aint).

 

But you can still calculate the what-if pretend Schw. radius and it should come out equal c/H

[Norman Albers]

Thanks, I am trying to put all these perspectives together, and one just has to mess around to learn a territory. Is my approach not also clear? . . . . Lo, it is! At least once I lay it out correctly. Getting the factors of two straight (??) I come up with [math] r_s=1.2E26 meters[/math]. Since a light-year is 8.6E15 meters, this gives 14 billion. Ah, now this is quite near 13.8.

[Martin]

Thanks, I am trying to put all these perspectives together, and one just has to mess around to learn a territory. Is my approach not also clear?

 

blindingly clear:D

you may find other people to interest in

I want to move on

Like a lot of other people, I am mainly motivated by pleasure and curiosity (not a sense of responsibility or moral duty). If you want to keep me interested, calculate the density of the universe the way I showed you and see what you get.

[Norman Albers]

You just missed it. Actually I did not yet put in the 0.85J/km^3, so one over the square root of that adds another 8%, so this is yielding 15 billion years. <Sorry, now the root's right>

[scalbers]

Interesting to consider this Schwarzchild radius calculation can come up with the radius of the observable universe. Almost makes me think we are living in a black hole (neglecting the expansion). On the other hand, we have the actual universe being much larger than the observable part.

[Norman Albers]

Yah, scalbers, I'm not up on the cosmology yet; you can see I'm just approaching, and I have worked in the small with geometric mass in the context of angular momentum and Kerr metric. I figure there's something to be learned by how much this figure differs from what is our "horizon".