# Swallowed by a Black Hole? **URL:** \ **Category:** Cecil's Columns/Staff Reports\ **Created:** [November 9, 2003, 12:48pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221 "2003-11-09T12:48:04Z")\ **Posts on this page:** 7\ **Page:** 1
**Author:** ![1010011010](https://avatars.discourse-cdn.com/v4/letter/1/ea666f/32.png) [@1010011010](https://boards.straightdope.com/u/1010011010)\ **Post date:** [November 9, 2003, 12:48pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/1 "2003-11-09T12:48:04Z")
[What would happen if you were swallowed by a black hole?](http://www.straightdope.com/classics/a1_200.html) Cecil says: "For reasons that I confess are not entirely clear to me, when a black hole grows to enormous mass, it becomes less dense. " Basically, the radius of a black hole is proportional to it’s mass, while the volume of a black hole is proportional to the cube of its radius. I.E. the volume of a black hole increases faster than its mass, ergo its density decreases. The radius of a black hole can be given using the escape velocity equation ( v=sqrt(2GM/R) ) with the understanding that the event horizon of a black hole is defined as the distance from the center of a black hole at which the escape velocity equals the speed of light. Thus, the raduis of a black hole is given by R=(2GM)/c^2, which, by stuffing all the constants into a single figure, becomes R=kM with k=1.48485e-27 The volume of a sphere is given by V=(4/3)piR^3, substituting our last equation, and again compressing our constants we get: V=kM^3 with k=1.37132e-80 ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 10, 2003, 9:25pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/2 "2003-11-10T21:25:21Z")
> [@](#): > > Thus, the raduis of a black hole is given by R=(2GM)/c^2, which, by stuffing all the constants into a single figure, becomes R=kM with k=1.48485e-27 Not quite. What you mean is that k = 1.48494276 × 10[sup]-27[/sup] m / kg . You need to include the units in your constant, unless you’re using a system of units where the fundamental constants used are dimensionless. But in such a system of units, k = 2. Similarly for your second example. ---
**Author:** ![The\_Bad\_Astronomer](https://avatars.discourse-cdn.com/v4/letter/t/f05b48/32.png) [@The\_Bad\_Astronomer](https://boards.straightdope.com/u/The_Bad_Astronomer)\ **Post date:** [November 12, 2003, 5:18am UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/3 "2003-11-12T05:18:44Z")
I actually wonder if this is correct. I don’t think you can simply take the Schwarzschild radius, plop it into the equation for the volume of a sphere, and calculate the “volume” of a black hole. Space is warped infinitely near a black hole, and the volume becomes a meaningless term. In fact, I think radius is meaningless too. You can’t go across a black hole! You can go _around_ one, though. So I believe that a circumference has meaning. Putting Euclidean equations to work near a black hole is only going to buy you trouble. Caveat: I am not a relativity astrophysicist, and I may be wrong. Anyone take GR and can say what’s what? ---
**Author:** ![MC\_Master\_of\_Ceremonies](https://avatars.discourse-cdn.com/v4/letter/m/bb73d2/32.png) [@MC\_Master\_of\_Ceremonies](https://boards.straightdope.com/u/MC_Master_of_Ceremonies)\ **Post date:** [November 12, 2003, 1:25pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/4 "2003-11-12T13:25:18Z")
I’m no GR maven, but the Scwarzchild solution describes space-time outside any spherically symmetric mass, like the sun (roughly) and you can defintely assign a meanigful voulme to the sun, though the equation only describes the vacuum outside the sun, not the interior. The equation describes all points of a black hole except the singularity and the quantity r (where r[sub]BH[/sub] = 2GM/c[sup]2[/sup]) is in the equation so it is certainly meaningful to talk about the radius of a black hole. I imagine though that something like volume would depend on your rest frame. ---
**Author:** ![bonzer](https://avatars.discourse-cdn.com/v4/letter/b/45deac/32.png) [@bonzer](https://boards.straightdope.com/u/bonzer)\ **Post date:** [November 12, 2003, 8:20pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/5 "2003-11-12T20:20:39Z")
In the case of something like the sun, the Schwarzschild solution technically only applies outside the surface of the mass. Inside the star, another (non-singular) solution “takes over”. As for **Bad As …**'s question, the key notions are _proper_ length, area and time. Instrumentally, proper length is just what you measure by laying out metre sticks. Proper area is marginally more delicate. Locally, it’s just what you get using metre sticks. Imagine laying out lots of little squares and adding them up. Proper volume is the volume equivalent. A couple of simple rules apply in Schwarzschild geometry. In it the radius coordinate is conventionally _r_, as in **MC** ’s post. The most familiar rule is that if we take the sphere of radius _r_, then this has a proper area _A®_ = 4 pi _r[sup]2[/sup]_. In fact, this pretty much follows from the spherical symmetry of the situation. In talking about volumes, we want to avoid getting tangled up in issues of the horizon, never mind the central singularity. So consider the volume of spherical shells about central mass/horizon. The neat result is that a shell of width _dProper Radius_ has a proper volume of _V®_ = _A® dProper Radius_ However, _dProper Radius_ =/= _dr_. And that screws things up. So if you do start trying to measure areas and volumes of spherical shells about a black hole, the relationships in terms of their measured differences in radius are not what you’d expect in a Euclidean world. So the objection is reasonable, though it may not make that much difference in practice. You presumably still can unambiguously define a volume for a black hole; there’s a nasty integration, so I haven’t worked it through. ---
**Author:** ![bonzer](https://avatars.discourse-cdn.com/v4/letter/b/45deac/32.png) [@bonzer](https://boards.straightdope.com/u/bonzer)\ **Post date:** [November 12, 2003, 8:26pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/6 "2003-11-12T20:26:27Z")
Incidentally, that first comment reads like I’m correcting some nuance in **MC** ’s post. I’m not - I merely overlooked his statement of the same qualification. ---
**Author:** ![Chronos](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/chronos/32/134_2.png) [@Chronos](https://boards.straightdope.com/u/Chronos)\ **Post date:** [November 13, 2003, 7:29pm UTC](https://boards.straightdope.com/t/swallowed-by-a-black-hole/212221/7 "2003-11-13T19:29:53Z")
> [@](#): > > The most familiar rule is that if we take the sphere of radius r, then this has a proper area > > A® = 4 pi r[sup]2[/sup]. > > In fact, this pretty much follows from the spherical symmetry of the situation. While this is true, it doesn’t really follow from the spherical symmetry. We have to make a choice for what to call our “r” coordinate. We could define r such that dr = d(proper radius). In that case, we would still have the spherical symmetry, but we would not have A = 4 pi r[sup]2[/sup]. As it happens, we usually do define r such that A = 4 pi r[sup]2[/sup], but this is a matter of convention, not necessity. **BadAs** is right that the volume of a black hole won’t simply be 4/3 pi \* R[sub]s[/sub][sup]3[/sup], but it’s still going to have to be proportional to R[sub]s[/sub][sup]3[/sup], and therefore to m[sup]3[/sup]. I don’t know the constant of proportionality offhand, and like **bonzer** , I’d rather not work it out. But the scaling follows from dimensional analysis, without doing the integral.