back to article Remember that blurry first-ever photo of a black hole? Turns out snaps like that can tell us a lot about these matter-gobbling voids

Scientists may be able to calculate a black hole’s mass and rotation from photographs alone one day, according to research published in Science Advances. The first-ever snap taken of a black hole, located at the center of galaxy Messier 87, was unveiled by a team of physicists working with the Earth-based Event Horizon …

  1. Anonymous Coward
    Anonymous Coward

    Perhaps someone can explain something to dumb old me.

    This black hole distorts space time.

    Is it the gravity that propagates across the event horizon?

    If it's the gravity, how fast does gravity propagate and why does it propagate at that speed?

    Or is it the mass that reaches out somehow and distorts the space outside that then creates the distortion that is the gravity effect?

    If its the mass, what is it about the mass that reaches out to distort space? How fast does that effect propagate and why and what is it?

    And this accretion disc, I notice that the pictures are face on. Is there a black hole picture with an accretion disc across or partially across the black hole? What is the odds of one black-hole being face on to us, what is the odds of two, three and so on?

    1. Paul Kinsler

      Re: how fast does gravity propagate

      Gravity propagates at what we call "the speed of light", which is really the fastest possible speed allowed by the spacetime metric (i.e. 1, in dimensionless units)

      All such questions can of course be referred to the authoritative text "Gravitation", by Misner, Thorne, and Wheeler; but it's a bit of a tricky read for the layperson. Still, if you happen for some reason to be stuck at home with too much time on your hands ... :-) [1]

      [1] Or should I say "stuck at home with too much spacetime on your hands"? [2]

      [2] In practice, I suspect it will be "too much time, not enough space".

      1. Anonymous Coward
        Anonymous Coward

        Re: how fast does gravity propagate

        So presumably gravity isn't the thing that escapes the black hole if light cannot, but then what *is* it about matter that escapes the black hole?

        1. steelpillow Silver badge

          Re: how fast does gravity propagate

          Nothing escapes. All the gravity was already out there already, which is how the black hole managed to form in the first place; it just pulled the focus of the spacetime warp all together in one place.

        2. Paul Hovnanian Silver badge

          Re: how fast does gravity propagate

          "but then what *is* it about matter that escapes the black hole?"

          The matter inside the black hole distorts the space around it. This distortion isn't dependent on the motion of a particle, so it isn't constrained by the speed of light or an event horizon.

    2. This post has been deleted by its author

    3. erst

      Re: Perhaps someone can explain something to dumb old me.

      This helped me:

      1. Steve K

        Re: Perhaps someone can explain something to dumb old me.

        I think this post orbited the black hole twice....

      2. John Mangan

        Re: Perhaps someone can explain something to dumb old me.

        @erst I've seen that video before and I think it is an excellently clear explanation of what that blurry photo is revealing.

    4. Anonymous Coward

      Re: Perhaps someone can explain something to dumb old me.

      Nothing propagates across the event horizon. In particular gravity does not propagate across it. Rather the gravitational effect of a BH is essentially a sort of 'scar' left as the star collapsed: it's a record that there is some mass there – perhaps, better would be to say that the gravitational field is the mass, although that's not quite correct.

      However this isn't really right. One important thing to understand is that, with slight caveats, the gravitational field (and hence the spacetime) outside some suitably-symmetric massive object is, to use a mathematical term unique. What that means is that, no matter what is going on with the object the gravitational field outside it is unchanged. The slight caveat is that the spacetime should be 'asymptotically flat', which means that, far from the object, everything settles down to Minkowski space: this isn't true on cosmic distances (there are, for instance, other stars!), but it's a good enough approximation in practice.

      What that means, is that, if you imagine some star collapsing to a black hole, then the gravitational field outside where the star was is unchanged: if we used our giant star-crushers to crush the Sun down into a black hole, then nothing would change gravitationally (well, so long as our star-crushers were rather light).

      So, in fact, nothing needs to propagate across the horizon: the field outside the horizon is unchanged by the collapse, so long as its suitably symmetric. In particular there's no information that has to be somehow continually emitted from the thing to tell spacetime how to behave: it just does what it always did. The situation is what's called 'stationary' in GR.

      I have not explained this very well, I think: sorry.

      To answer two other questions.

      Changes in gravity propagate at the speed of light: so gravitational waves propagate at the speed of light. Indeed it's really better to think of the speed of light as being the speed of causality: it's the speed at which information about things happening far from you gets to you (or the maximum speed at which it gets to you). That means that nothing propagates across the event horizon because the speed you need to travel to stay still (to not get closer to the centre of the BH) at the event horizon is the speed of light: no information, of any kind, can get out.

      (Caveat: what I'm describing is the classical GR picture: quantum gravity may be different, but quantum gravity needs to reduce to classical gravity in suitable limits so it won't be very different in those limits.)

      What accretion disks look like is complicated. I think that we are looking at the M87 BH (the one the EHT image is of) from somewhere near one of its poles – so we're looking down onto the disk. And since that's the BH we've got images like this of, well, it is what it is. However even if you are looking at a BH from its equator, what you see is not what you'd expect – a thin line which is the disk. Instead you see something much more complicated, because you can see right round a BH, as light can orbit the BH. So, for instance, you can see the part of the accretion disk which is directly behind the object. And indeed you can see lots of copies of it, depending on how many times the light from it orbited the BH before escaping. And this, of course, is what the people this article is about are interested in being able to see.

      What you actually see then, is complicated by this huge distortion of the paths light takes. The images of the BH from Interstellar are quite good: they were actually computed by building numerical models of what the thing should look like (Kip Thorne was involved in this, and he knows what he's about), and in them you can see this weird thing where the accretion disk seems to rise up over the central object in a very strange way.

      So in a strange sense, you're always looking 'down' on the accretion disk (at least the inner regions of it).

      1. Paul Kinsler

        Re: Nothing propagates across the event horizon.

        The statement "Nothing propagates across the event horizon" is not true; although it is true that for a distant observer nothing is seen to cross the event horizon, ... things just approach ever closer and become more and more red-shifted. But from the perspective of an in-falling object (photon, rock, whatever), crossing the event horizon is easy enough.

        Perhaps it is more-or-less implicit from your description that you mean "propagate *outwards* across the event horizon", but I think it worth the clarification.

        1. Anonymous Coward
          Anonymous Coward

          Re: Nothing propagates across the event horizon.

          Yes, sorry, that is what I meant: nothing gets from inside it to outside it. Some earlier version of the comment even had that text but I botched an edit.

          1. JCitizen

            Re: Nothing propagates across the event horizon.

            I think I know just enough to make a short observation - I think we laypeople forget that gravity has a relatively short range - just looking at the equation shows that up - so this is why science has been in a quandary about dark matter, because the mass of the universe doesn't somehow track with what gravitational effects should be in that elusive "theory of everything" that physicists are endlessly chasing after. Or at least the movement of the galaxy doesn't track with gravity itself, it doesn't explain what is holding this whirlpool together.

            Mine's the one with the .32 ACP pocket pistol please!

            1. Paul Kinsler

              Re: gravity has a relatively short range

              Note that gravity is 1/r^2 in the Newtonian limit ... so not short range in the sense a physicist would use (i.e. not a faster fall off such as e.g. a decaying exponential). It is *weak*, though.

    5. steelpillow Silver badge

      Re: Perhaps someone can explain something to dumb old me.

      Einstein's theory of General Relativity is essentially a theory of geometry. It is this geometrical model of spacetime which predicts the formation of black holes.

      The mass warps spacetime before it ever collects into a black hole. That warp is known as gravity (a gravity wave is literally a ripple in the fabric of spacetime) and propagates at the speed of light.

      When the black hole forms, its event horizon bounds the region from which nothing can escape - not even more gravity. But nothing else really changes. If more matter falls in, it brings its gravitational distortion with it but most of that was already way out there and simply merges with the rest.

      Black holes do come with their accretion disks at all angles. The thing about this particular black hole is that it is not far off face-on, that was one reason it was chosen for the photograph. ISTR it is about 15-20 deg off being exactly face-on, so a little spherical geometry shows that maybe one black hole in 30 will be striking a suitable pose for the camera.

    6. Rich 11 Silver badge

      Re: Perhaps someone can explain something to dumb old me.

      And this accretion disc

      It's not an accretion disc. The animation shows you how the appearance of a disc is formed.

    7. Anonymous Coward

      Re: Perhaps someone can explain something to dumb old me.

      The other answers addressed your gravity questions.

      As for your question about the angle of the black hole to us, the black hole in the picture was chosen because it was oriented that way. Given the number of black holes in the universe, every orientation exists. Many are too small to image the accretion disk (the one in question is a super massive black hole at the center of a galaxy). Others are obscured by dust (like the super massive black hole at our galaxy's center). And many are just too far away for current imaging technology.

    8. Paul Hovnanian Silver badge

      Re: Perhaps someone can explain something to dumb old me.

      "how fast does gravity propagate"

      In a static situation, gravity doesn't propagate. It's just bent space. When one removes or adds some mass, the change in that bend propagates at the speed of light. Much as a wave travels along the surface of a pond.

  2. Anonymous Coward

    Photon orbiting a black hole (though not indefinetly)

    Yeah. My brain has worked on these spatial systems and logical effects of the statements here. But wow, I'm gonna need a bit of time trying to think of how and what it means for a photon to "orbit" a black hole for more than a single orbit... That's making my brain hurt... more than usual.

    A single warping of spacetime and a bit of a loop of a photon is easy to visualise. But it doing a loop, and then escaping, instead of being drawn in, has gotta be complex.

    It might have to do with the rotation of the black hole and non-circular orbits? I guess, due to the kind of system, you only ever fall in or fly away, you never "orbit forever" as such, as too much going on to throw you off course. But even then, trying to visualise the trajectory of a 2 or 3x orbit before then being thrown (wrong term, but poetic) away, is just *warping* my mind. ;)

    1. Pascal Monett Silver badge


      If you really want a mind warp, check this out (SFW).

      1. Anonymous Coward
        Anonymous Coward

        Re: Heh

        Yeah, I might have seen that lecture. There are some great ones out there too, and some visualizations as well. Just not seen the multiple orbits one for a photon just yet (or if I did, it went whizz past my head and I need to watch it again!).

    2. steelpillow Silver badge

      Re: Photon orbiting a black hole (though not indefinetly)

      Photons orbit a black hole for two reasons. The basic reason is that, close to the event horizon, spacetime is so warped that a photon which happens to be going that way ends up circling the black hole just like we circle the sun. Put on your mad scientist hat and a spinning black hole actually drags spacetime round with it as it spins, a phenomenon known to mad scientists as frame-dragging. This helps keep those durn photons from gettin' too ornery. But the outliers do manage to spiral outwards, or bump into other stuff and get bounced away; these are the ones we see.

      1. Mooseman Silver badge

        Re: Photon orbiting a black hole (though not indefinetly)

        Eddies in the space-time continuum?

    3. Anonymous Coward

      Re: Photon orbiting a black hole (though not indefinetly)

      It's worth mentioning that the 'not indefinitely' thing is because there's a radius below which there are no stable circular orbits. That is at 3R where R is the radius of the event horizon. That radius is the innermost edge of the accretion disk for a BH, because below that, well, there are no stable orbits so things either fall in or get ejected.

      However orbits do exist below 3R: it's just that they're not stable. There is an innermost radius at which orbits can exist at all which is 3R/2. At that radius you have to be moving at the speed of light to be in an (unstable) orbit. Below that radius you'd have to be moving faster than that so there are no orbits at all.

      And that's the clue: light has to move at, well, the speed of light, so if it's going to orbit the BH it must necessarily do it at 3R/2. But that's not a stable orbit so pretty quickly it either ends up falling in or getting ejected out to infinity where we can see it. That's why it can't orbit indefinitely.

    4. Peter2 Silver badge

      Re: Photon orbiting a black hole (though not indefinetly)

      In orbital terms, there is a joke that says to speed up, slow down.

      That pretty much explains what your trying to get your head around in a nutshell. Imagine your in a spacecraft that doesn't have the fuel to escape orbit. If you deliberately change your orbit to a decaying orbit that loops in closer and closer to the stellar body of your choice then you speed up. If you keep speeding up then you might get to exceed the speed required to break orbit from the gravity well or the stellar body that your sitting in.

      This is a gif of the orbit of the Galileo spacecraft which shows what i'm talking about reasonably well, although obviously we put Galileo in orbit around Jupiter deliberately whereas this wasn't deliberately steered, but it's the same general principle.

  3. WibbleMe

    I keep being sent photos of black holes but there nothing there!

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