Questions about spaghettification and gas cloud G2

  • #1
George Keeling
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Summary:

I want to check my formula for spaghettification and then apply it to the gas cloud G2 which approached Sagittarius A* in 2014 with disappointing results.

Main Question or Discussion Point

I did an exercise about beacons falling radially into black holes from Carroll's book and got a formula for the proper velocity$$
\frac{dr}{d\tau}=-\sqrt{\frac{R_S}{r_\ast}}\sqrt{\frac{r_\ast-r}{r}}
$$It's in natural units (##c=1)##, ##r_\ast## is where the beacon is dropped from and ##R_s=2GM## is the Schwarzschild radius. By inverting the equation and integrating it, one can calculate the proper time between two radii and it gives the correct answer for time from event horizon to the centre (##\pi GM## or 28 hours in M87*).

If I differentiate that I can get the proper acceleration ##d^2r/d\tau^2=-R_S/2r^2## and that turns out to be ##9.8\ ms^{-2}## at the surface of the earth. So that seems to work.

I can now differentiate that with respect to the ##r## to get the radial rate of change of the acceleration which will give me a measure of spaghettification. Call that ##S## and$$
S=\frac{2GM}{r^3}\ \ s^{-2}
$$The units are a bit odd, but make sense if you think of them as ##\left(ms^{-2}\right)m^{-1}## or even Newtons per kilogram per metre. Wikipedia gives the equivalent tensile force on a rod with mass ##m## length ##l## as $$
F=\frac{GMlm}{4r^3}
$$Using my formula, I calculate that for a male human at ##70 kg## & ##2 m##, ##S=0.1## would give 14 Newtons, that's how much 1.4 kg weighs. That sounds bearable - the opposite of putting on a bit of weight. Women and children would have an advantage!

Is that all correct? Would a brave black hole explorer be able to rely on me?

Now we come to the gas cloud known as G2, which was discovered heading towards Sagittarius A*, the black hole at the centre of our galaxy, in 2011 by some folk at the Max Planck Institute. G2 was destined to come closest to Sagittarius A* in Spring 2014 "with a predicted closest approach of only 3000 times the radius of the event horizon". There was great excitement because spaghettification and great fireworks were expected. However nothing much happened and G2 continues on its way, orbiting Sagittarius A*.

I re-read about it in a 2014 paper from the American Astronomical Society.

One thing continues to puzzle me: At ##r=## 3000 times the radius of the event horizon, that's ##r=4\times{10}^{13}\ m## the ##g## force is only ##0.4\ ms^{-2}## and the ##S## is a very feeble or even undetectable ##2.2\times{10}^{-14}##. Why was there such great excitement or expectation of fireworks?

It might be because they estimate the size of G2 as ##r_{G2}=3au\approx5\times{10}^{11}\ m## which is pretty big. But at closest approach where ##S=2.2\times{10}^{-14}\ ms^{-2}m^{-1}## g-forces would change across the diameter by ##r_{G2}\times S=0.001\ ms^{-2}##. Still tiny compared to ##0.4##.
 

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  • #2
Ibix
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I think your maths is correct - you just aren't thinking on a long enough timescale. This thing's orbital period is decades (one of the references in the paper you linked, Burkert et al puts it at apocenter in 1944, implying a period of around 140 years), so it'll spend months in the regime you are considering. An acceleration difference of ##0.001\mathrm{ms^{-2}}## applying for ##10^7\mathrm{s}## (three or four months) builds up a position difference on the order of ##at^2\approx 10^{11}\mathrm{m}##, comparable to the size of the dust cloud.

That's very much back-of-the-envelope, obviously.
 
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  • #3
PeterDonis
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If I differentiate that I can get the proper acceleration
This is not correctly called "proper acceleration", since it is for a freely falling object, which has zero proper acceleration.

It is also not quite the exact formula for "acceleration due to gravity", i.e., the coordinate acceleration that an object free-falling downward would have in the momentarily comoving inertial frame of an observer at rest at radial coordinate ##r##. That formula is

$$
\frac{R_S}{2 r^2 \sqrt{1 - R_S / r}}
$$

The extra factor in the denominator makes this quantity diverge as the horizon ##r = R_S## is approached.

I can now differentiate that with respect to the to get the radial rate of change of the acceleration
What you are doing happens in this case to be equivalent to computing the appropriate component of the Riemann curvature tensor (the one describing radial geodesic deviation) in an orthonormal basis. More generally, shortcuts like this don't always work, but the Riemann tensor can always be computed and is the thing that describes what you are looking for.

The units are a bit odd
First, you need to include a factor of ##c^2## in the denominator, since ##R_S = 2 G M / c^2##, not ##2 G M##.

Second, the units (inverse meters squared) are the correct geometric units for curvature, and you've been using geometric units.

The units of inverse square seconds are what you would get if you did not use geometric units. So you've basically switched units at this point.

the equivalent tensile force on a rod
Is obtained, as the Wikipedia article notes, by integrating your quantity ##S## from the center of the rod to the ends. That will of course give a smaller answer than just multiplying your ##S## by the length of the rod; but the smaller answer is the correct one, since not all points on the rod are separated by its total length.

I calculate that for a male human at 70 kg & 2 m, S = 0.1 would give 14 Newtons
Where are you getting ##S = 0.1## from?
 
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  • #4
George Keeling
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Where are you getting S=0.1 from?
S is also Newtons per kilogram per metre. Set S=0.1 then 0.1×70×2=14
I am struggling to work out how you get proper acceleration =$$
\frac{R_S}{2 r^2 \sqrt{1 - R_S / r}}
$$As you say it diverges at the horizon. Wouldn't that mean S diverges too?
 
  • #5
Ibix
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I am struggling to work out how you get proper acceleration
That's the proper acceleration of an observer hovering at constant altitude. You can write down the ##t## component in Schwarzschild coordinates easily because ##g_{\mu\nu}U^\mu U^\nu=1## (give or take sign convention and assuming ##c=1##) and the spatial components of their four velocity are zero. Then use ##U^\mu\nabla_\mu U^\nu## to get the four acceleration. The formula @PeterDonis gives is the magnitude of the spatial component. It's the closest GR equivalent to a Newtonian g-force - the acceleration your accelerometer will measure while you remain at constant ##r##. It is divergent at the event horizon, as would be its derivative, yes. I don't think Peter is saying that this is at all relevant to your problem, just that your ##d^2r/d\tau^2## isn't interpretable as a proper acceleration. The formula he gives is the only really physically intetesting proper acceleration, and it's irrelevant to your free-falling gas cloud.
 
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  • #6
PeterDonis
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S is also Newtons per kilogram per metre. Set S=0.1
Why are you setting ##S = 0.1##? Where is that number coming from?

It's no answer to say "well, that's what you need to set ##S## to to get 14 Newtons", because that just raises the question of where you are getting 14 Newtons from.
 
  • #7
Ibix
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Why are you setting ##S = 0.1##? Where is that number coming from?
I had the impression that @George Keeling was merely establishing that ##S=0.1\mathrm{s^{-2}}## was not a stupidly large number, in that it seems perfectly survivable to a human. That goes some way to establish that the ##S## value he gets for the cloud really is small. Of course, the cloud is much larger than a person, not bound (at least that was the hypothesis - the reality appears to be different), and is applied for a rather lengthy period.
 
  • #8
PeterDonis
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I had the impression that @George Keeling was merely establishing that was not a stupidly large number, in that it seems perfectly survivable to a human.
That might be true, but it doesn't tell us anything about where in a person's fall into the black hole they will see ##S = 0.1##. Seeing ##S = 0.1## way inside the horizon, almost to the singularity, is very different from seeing ##S = 0.1## well before you even reach the horizon. In the former case, you know you won't be spaghettified until you're almost to the singularity (and hence the end of your worldline) anyway. In the latter case, you have no such guarantee.
 
  • #9
PeterDonis
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I re-read about it in a 2014 paper
Where does this paper say it expects the gas cloud to be spaghettified? It says it expects the cloud to be "tidally disrupted", but that's a much weaker statement.

(More precisely, the paper said it expects G2 to be tidally disrupted if it's a gas cloud; so since it was not observed to be tidally disrupted, that is evidence against it being a gas cloud.)
 
  • #10
George Keeling
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Where does this paper say it expects the gas cloud to be spaghettified
It doesn't. But I read this in 2012: "But the black hole has a tremendous gravitational force, and so the gas cloud will fall into the direction of the black hole, be elongated and stretched and look like spaghetti, said Stefan Gillessen, astrophysicist at the Max Planck Institute for Extraterrestrial Physics in Munich, Germany, who has been observing our galaxy’s black hole, known as Sagittarius A* (or Sgr A*), for 20 years." here. I was suspicious at the time ...
 
  • #11
George Keeling
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I had the impression that @George Keeling was merely establishing that
##S=0.1\mathrm{s^{-2}}## was not a stupidly large number, in that it seems perfectly survivable to a human.
That is indeed what I was trying to establish.
 
  • #12
PeterDonis
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I read this in 2012
The actual distance of closest approach might not have been known then. But it's also quite possible that the physicist quoted was making an overstatement of the kind physicists (unfortunately) often make in informal contexts when they're not talking to other experts.
 
  • #13
PeterDonis
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That is indeed what I was trying to establish.
But establishing that ##S = 0.1## is survivable for a human is irrelevant unless you know that ##S = 0.1## is a reasonable value for ##S## at some point of interest on the human's trajectory. So far I have seen nothing whatever to show why one would think that is the case. The value for ##S## calculated in the OP for the distance of closest approach to the hole of the object G2 is many orders of magnitude smaller.
 
  • #14
George Keeling
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survivable for a human is irrelevant
I am guilty! But it may be useful another day.
 
  • #15
George Keeling
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proper acceleration of an observer hovering at constant altitude
Something terrible happens when I use ##U^\mu\nabla_\mu U^\nu## for proper acceleration! Here is what I calculate for the hovering observer with ##U^i=0##
\begin{align}
U^\mu\nabla_\mu U^\nu=U^\mu\left(\mathrm{\partial}_\mu U^\nu+\Gamma_{\mu\lambda}^\nu U^\lambda\right)=U^0\frac{\partial U^0}{\partial t}+U^0\Gamma_{00}^\nu U^0&\phantom {10000}(2)\nonumber\\
=U^0\frac{\partial U^0}{\partial t}+U^0U^0\frac{\left(r-2GM\right)GM}{r^3}&\phantom {10000}(3)\nonumber
\end{align}Now using ##g_{\mu\nu}U^\mu U^\nu=-1## at a fixed position we get
\begin{align}
g_{00}U^0U^0=-1&\phantom {10000}(4)\nonumber\\
\left(1-\frac{2GM}{r}\right)U^0U^0=1&\phantom {10000}(5)\nonumber\\
U^0=\sqrt{\frac{r}{r-2GM}}&\phantom {10000}(6)\nonumber
\end{align}Put that into (3) and we get
\begin{align}
U^\mu\nabla_\mu U^\nu=\frac{GM}{r^2}&\phantom {10000}(7)\nonumber
\end{align}That's the result that is 'wrong'.
 
  • #16
Ibix
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That's the result that is 'wrong'
Assuming that's ##U^1## in eq. 7, you're fine. The proper acceleration (now I've looked it up instead of relying on memory) is ##\sqrt{g_{\mu\nu}A^\mu A^\nu}##, i.e. the modulus of the four acceleration, not just its spatial part. ## g_{11}## supplies the missing factor.

Edit: 0s replaced by 1s, with thanks to Peter.
 
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  • #17
PeterDonis
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## g_{00}## supplies the missing factor.
No, ##\sqrt{g_{11}}## supplies the missing factor. The only nonzero component of the vector ##A^\mu## is the ##1## component (the radial component). So the norm is ##|A^\mu| = \sqrt{g_{11}} A^1##.
 
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  • #18
Ibix
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No, ##\sqrt{g_{11}}## supplies the missing factor. The only nonzero component of the vector ##A^\mu## is the ##1## component (the radial component). So the norm is ##|A^\mu| = \sqrt{g_{11}} A^1##.
Thanks - corrected above.
 
  • #19
George Keeling
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Assuming that's ##U^1## in eq. 7, you're fine
I now see that my eq, 7 was awful. The ##\nu## has disappeared! It should be $$
U^\mu\nabla_\mu U^\nu=\left(0,\frac{GM}{r^2},0,0\right)
$$ as Ibix hinted. I also needed to 'guess' that ##A^\nu=U^\mu\nabla_\mu U^\nu##, which nobody had explicitly mentioned. So now I have resolved PeterDonis's first and last comments of 5. Only three more to go 😥. Thanks guys!
 
  • #20
George Keeling
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What you are doing happens in this case to be equivalent to computing the appropriate component of the Riemann curvature tensor
This was fascinating and I found a more detailed explanation at the end of the Wikipedia article on the Schwarzschild metric. So I had accidentally calculated the ##R_{\ \ \ \hat{t}\hat{r}\hat{t}}^{\hat{r}}## component of the Riemann tensor. It is $$
R_{\ \ \ \hat{t}\hat{r}\hat{t}}^{\hat{r}}=-\frac{R_S}{r^3}
$$The hats are for orthonormal coordinates. The same component in Schwarzschild coordinates is $$
R_{\ \ \ trt}^r=-\frac{R_S\left(r-R_S\right)}{r^4}
$$So I have understood a bit about the geodesic deviation equation and soon will have to work out how to get to orthonormal coordinates 😥 .

First, you need to include a factor of ##c^2## in the denominator, since ##RS=2GM/c^2##, not ##2GM##.
I find units are problematic and have developed a recipe for converting from what Carroll calls natural units where ##c=\hbar=k=1##. ##k## is Boltzmann's constant and I have not used it yet. ##\hbar##, the reduced Planck, comes into conversions to SI units, but in my limited experience, has always cancelled out. I don't know if these are what you are calling geometric units. Wikipedia says that they have ##G=1## which is not the case for Carroll, I think. At any rate, Carroll does not usually include ##c##'s in his equations and always includes ##G##'s when appropriate.

So in natural units (à la Carroll) I had ##g##-force and my ##S## factor as$$
g=\frac{d^2r}{d\tau^2}=-\frac{GM}{r^2}\ \ \ ,\ \ \ S=\frac{d}{dr}\left(\frac{d^2r}{d\tau^2}\right)=\frac{2GM}{r^3}
$$The recipe converts those to SI units using the dimensions of everything in the equation thus ##G\rightarrow G\hbar^{-1}c^{-5}\ ,\ M\rightarrow Mc^2## , ##r\rightarrow r\hbar^{-1}c^{-1}## , ##\tau\rightarrow\tau\hbar^{-1}##. (##\tau## having dimensions of time). That gives for ##S## $$
S=\frac{d}{\hbar^{-1}c^{-1}dr}\left(\frac{d}{\hbar^{-1}d\tau}\left(\frac{\hbar^{-1}c^{-1}dr}{\hbar^{-1}d\tau}\right)\right)=\frac{2G\hbar^{-1}c^{-5}Mc^2}{\hbar^{-3}c^{-3}r^3}
$$miraculously all the conversion factors cancel and we end up with the same thing:$$
\ S=\frac{2GM}{r^3}
$$and I work out that the units are ##\rm{s^{-2}}## because I know the units of ##G,M,r##. I haven't had to convert proper time before so for a day I thought that its conversion might be like distance so I had ##\tau\rightarrow\tau\hbar^{-1}c^{-1}##. This gets a ##c^2## in the denominator of the expression for ##S## but it also gets a ##c^2## in the denominator of the expression for ##g## and that gives a very wrong answer for the ##g## force at the surface of the earth!

So ##g## and ##S## have exactly the same expression in SI units and in (Carroll's) natural units. It's not the same for ##R_S##. In natural units ##R_S=2GM## and in SI units ##R_S=2GM/c^2##. Perhaps the misunderstanding is that I had switched to SI units when I said that the units were ##\rm{s^{-2}}##.

Can you point to me a definition of your geometric units?
 
  • #21
PeterDonis
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I don't know if these are what you are calling geometric units.
Geometric units (as used, for example, in Misner, Thorne, and Wheeler) are units in which ##c = G = 1## (so mass and energy have the same units as length and time). You cannot also have ##\hbar = 1## in these units.

Units in which ##c = \hbar = 1## are often called "natural" units in quantum field theory. In these units you cannot have ##G = 1##. In these units mass and energy have the same units as inverse length and time.
 
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  • #22
pervect
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So I have understood a bit about the geodesic deviation equation and soon will have to work out how to get to orthonormal coordinates 😥 .
It's not hard compared to what you've already done. You create an orthonormal basis. Then you just do a standard change of basis.

For the Schwarzschild line element
$$-(1-\frac{r_s}{r}) dt^2 + (1-\frac{r_s}{r})^{-1} dr^2 + r^2 d\theta^2 + r^2 \sin^2 \theta d\phi^2$$

an orthonormal basis of one forms is

$$dt' = \sqrt{1-\frac{r}{r_s}} dt \quad dr' = \frac{dr}{\sqrt{1-\frac{r}{r_s}}} \quad d\theta' = r \,d\theta \quad d\phi' = r \sin\theta \,d\phi$$

If you prefer to write or think of your orthonormal basis as a vector basis rather than as it's dual, the one-form basis, you can convert. MTW uses the one-form basis a lot, and I suspect other texts do as well, as it's very easy to get from the line elment.

dt' in this context is a one-form, because it operates on a vector, and returns a scalar. This makes it a one-form, a rank 1 tensor. If it didn't operate on a vector, it'd just be a scalar. Some texts use the convention to use a boldface d to distinguish one-forms from scalars,but I was lazy and didn't do that.

You can do the change of basis with just algebra, if you like. For instance, in the basis of one-forms I just wrote, the metric is just

$$-dt' \otimes dt' + dr' \otimes dr' + d\theta' \otimes d\theta' + d\phi' \otimes d\phi'$$

Note the change in notation. The rank 2 tensor g is written as the sum of tensor products of rank 1 tensors ##dt', dr', d\theta', d\phi'## using index-free notation. The only thing you have to be a bit wary of is that tensor products don't commute, but that isn't an issue here.
 
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  • #23
PeterDonis
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You create an orthonormal basis. Then you just do a standard change of basis.
It should be noted that this is finding an "orthonormal basis", which is not the same as finding "orthonormal coordinates" (which is the term @George Keeling used). The latter cannot always be found in a general curved spacetime, except in a local patch that is small enough that curvature can be ignored (in which case you can always find a Minkowski coordinate chart on the small patch). But you can always use an orthonormal basis to do computations, even if there is no coordinate chart for which that basis is a coordinate basis.
 
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  • #24
George Keeling
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an orthonormal basis of one forms is
Is there a typo? Should it be $$
dt' = \sqrt{1-\frac{r_s}{r}} dt
$$ and similar for ##dr'##?
 
  • #25
pervect
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Yes, sorry
Is there a typo? Should it be $$
dt' = \sqrt{1-\frac{r_s}{r}} dt
$$ and similar for ##dr'##?
Yes, sorry
 

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