7 ms·

According to Einstein there is no gravity that pulls us to the center of Earth. Instead mass curves spacetime in such a way to make you experience traveling alo

by WhoBeI 9y ago

According to Einstein there is no gravity that pulls us to the center of Earth. Instead mass curves spacetime in such a way to make you experience traveling along the curvature as gravity. If the mass is rotating the curvature will also drag on the surrounding spacetime. Imagine something like a whirlpool and then consider what happens if two of those merge. While merging some of their energy will cause waves in the surrounding water that eventually fades away with range. Those waves would be the gravitational waves - ripples in spacetime.

lomnakkus 9y ago
Incidentally, I also find the "geometrical" approach a great way to look at the whole "light vs. black holes" thing: It's not that black holes directly "trap" light or "prevent light from escaping" per se, it's more that spacetime gets curved so much that regardless of which direction a ray of light is traveling it'll always find itself back inside the event horizon. (Not sure where I first heard this explanation, but I can definitely say that I didn't originate it!)
- nerfhammer 9y ago
  Man, that made me realize that black holes really are literally "holes".
  - lomnakkus 9y ago
    Yup. A "hole" in space in some sense, though I must admit I haven't thought of it such extremely concrete terms[1]. I imagine that realization went into the naming :). [1] I mean, we usually imagine we can retrieve things when they drop into a hole, but that's not really a thing in this case. (Excepting Hawking radiation and the absurd amount of computation/time you would need to reconstruct the original state information from that. If you thought the time-of-evaporation for a Black Hole was long, you've got another thing coming when it comes to reconstructing the state from that evaporated radiation! Not that this has has any practical bearing, but it's fun to think about, right?)
raattgift 9y ago
Spherically symmetrical masses (even ones that are rotating and even ones that aren't exactly spherical as a result) do not shed gravitational radiation. If you put two such masses on a parallel course through space, although they'll eventually collide, they won't release gravitational radiation. Barbell-like configurations of matter emit gravitational radiation when they spin about an axis perpendicular to a line through both bodies' centres of mass. A binary star system is barbell-like, with two heavy masses at opposite ends of an so-infinitesimally-thin-it's-not-really-there bar. This "barbell" rotates around the barycentre, which for all practical purposes lies along a line through the two stars' centres-of-mass. Even more massive binary objects can have a rotating-barbell type of arrangement. More detail here if you want it: https://www.wikiwand.com/en/Gravitational_wave#/Sources https://www.wikiwand.com/en/Gravitational_wave#/Sources Rotation of each black hole is relevant in black hole mergers in the late stages of the inspiral, and there have been several numerical relativity studies of BHs with extreme rotation rates (each BH rotating around its own axis, and possessing only axisymmetry rather than approximately spherical symmetry). The results are interesting, but don't really have an impact on the shedding of gravitational radiation; that comes down to the barbell-like configuration. In order for there to be frame-dragging one must first fix a frame of reference. If you fix Cartesian coordinates with the origin at the middle of a bucket of water, and then you spin that bucket around on a rope, you're dragging those coordinates around with the rope. If the coordinates serve as the basis of a frame of reference then presto, you have frame-dragging. But if you have a suitably heavy bucket of water and spin it hard enough, you'll notice the barycentre shifts outside of your body along the rope towards the bucket. Congratulations, you're now shedding gravitational radiation! That gravitational radiation is there even if you only ever calculate in coordinates where the origin is on the floor of the corner of the room in which you and the bucket are spinning; that ("laboratory") frame of reference is not being dragged by you.[1] In the lab frame and in the bucket frame, your precession as you struggle with your footing are pretty different: the rope remains the same length so you are at a constant position in bucket-basis coordinates, but you are moving around relative to the lab coordinates. You can even fix (0,0,0,t) on some part of your own body; you are dragging those coordinates around as you spin the bucket, and the bucket eventually settles at some fixed point in your coordinates. The walls of the lab move around in those coordinates though, and might move inwards and outwards from you as you spin around. The Lense-Thirring effect is similar to how the walls move around in your coordinate basis as you spin around. It arises when one uses an exact solution of the Einstein Field Equations (typically the Kerr metric) and a set of coordinates suitable for a static observer. A static observer is one who sees [a] no change to the gravitational or matter fields over time and [b] can slice spacetime into space+time in a particular way. Static observers are not realistic observers in a universe full of moving matter. The Kerr metric is unsuitable for the real Earth, since it's lumpy inside. These approximation choices are mathematically convenient, but come with the side effect of needing to import fictitious forces to explain orbital precessions. The extra mathematics of Lense-Thirring precession however are much much more convenient than dealing with a more realistic metric and set of coordinates for Earth. Moreover, it is perfectly reasonable to do physics in preferred frames of reference (Kerr + Boyer-Lindquist, in this case) as long as you admit to yourself that that is what you are doing. There exists a Bogoliubov transformation from this preferred frame of reference to any other reasonable frame of reference, so you're not "stuck" with Lense-Thirring. You are however stuck with the physical result that something orbiting such that 24.something orbits "should" give it the same view below (and above!) but doesn't. We've shown this around various artificial satellites put into polar orbits around various bodies in the solar system. The underlying source of this precession is a combination of gravitational and special-relativistic time dilation. These are not "forces", though. - -- [1] It is being dragged around by the Earth's movements though; if you treat it as an exact inertial frame and do extremely sensitive experiments you'll see deviations in motions of things that you may be tempted to describe using e.g. coriolis forces. If you use exquisitely precise atomic clocks and insist on Cartesian coordinates, you will eventually discover that you are not in an inertial frame after all, and will have to take the local curvature of spacetime into account in the basis of your frame of reference, or shrug and use fictitious forces as corrections to Newtonian or special relativistic geometry. The magnitudes of the fictitious forces will be small.

by WhoBeI 9y ago

lomnakkus 9y ago

Incidentally, I also find the "geometrical" approach a great way to look at the whole "light vs. black holes" thing: It's not that black holes directly "trap" light or "prevent light from escaping" per se, it's more that spacetime gets curved so much that regardless of which direction a ray of light is traveling it'll always find itself back inside the event horizon. (Not sure where I first heard this explanation, but I can definitely say that I didn't originate it!)

nerfhammer 9y ago
Man, that made me realize that black holes really are literally "holes".
- lomnakkus 9y ago
  Yup. A "hole" in space in some sense, though I must admit I haven't thought of it such extremely concrete terms[1]. I imagine that realization went into the naming :). [1] I mean, we usually imagine we can retrieve things when they drop into a hole, but that's not really a thing in this case. (Excepting Hawking radiation and the absurd amount of computation/time you would need to reconstruct the original state information from that. If you thought the time-of-evaporation for a Black Hole was long, you've got another thing coming when it comes to reconstructing the state from that evaporated radiation! Not that this has has any practical bearing, but it's fun to think about, right?)

raattgift 9y ago

Spherically symmetrical masses (even ones that are rotating and even ones that aren't exactly spherical as a result) do not shed gravitational radiation. If you put two such masses on a parallel course through space, although they'll eventually collide, they won't release gravitational radiation. Barbell-like configurations of matter emit gravitational radiation when they spin about an axis perpendicular to a line through both bodies' centres of mass. A binary star system is barbell-like, with two heavy masses at opposite ends of an so-infinitesimally-thin-it's-not-really-there bar. This "barbell" rotates around the barycentre, which for all practical purposes lies along a line through the two stars' centres-of-mass. Even more massive binary objects can have a rotating-barbell type of arrangement. More detail here if you want it: https://www.wikiwand.com/en/Gravitational_wave#/Sources https://www.wikiwand.com/en/Gravitational_wave#/Sources Rotation of each black hole is relevant in black hole mergers in the late stages of the inspiral, and there have been several numerical relativity studies of BHs with extreme rotation rates (each BH rotating around its own axis, and possessing only axisymmetry rather than approximately spherical symmetry). The results are interesting, but don't really have an impact on the shedding of gravitational radiation; that comes down to the barbell-like configuration. In order for there to be frame-dragging one must first fix a frame of reference. If you fix Cartesian coordinates with the origin at the middle of a bucket of water, and then you spin that bucket around on a rope, you're dragging those coordinates around with the rope. If the coordinates serve as the basis of a frame of reference then presto, you have frame-dragging. But if you have a suitably heavy bucket of water and spin it hard enough, you'll notice the barycentre shifts outside of your body along the rope towards the bucket. Congratulations, you're now shedding gravitational radiation! That gravitational radiation is there even if you only ever calculate in coordinates where the origin is on the floor of the corner of the room in which you and the bucket are spinning; that ("laboratory") frame of reference is not being dragged by you.[1] In the lab frame and in the bucket frame, your precession as you struggle with your footing are pretty different: the rope remains the same length so you are at a constant position in bucket-basis coordinates, but you are moving around relative to the lab coordinates. You can even fix (0,0,0,t) on some part of your own body; you are dragging those coordinates around as you spin the bucket, and the bucket eventually settles at some fixed point in your coordinates. The walls of the lab move around in those coordinates though, and might move inwards and outwards from you as you spin around. The Lense-Thirring effect is similar to how the walls move around in your coordinate basis as you spin around. It arises when one uses an exact solution of the Einstein Field Equations (typically the Kerr metric) and a set of coordinates suitable for a static observer. A static observer is one who sees [a] no change to the gravitational or matter fields over time and [b] can slice spacetime into space+time in a particular way. Static observers are not realistic observers in a universe full of moving matter. The Kerr metric is unsuitable for the real Earth, since it's lumpy inside. These approximation choices are mathematically convenient, but come with the side effect of needing to import fictitious forces to explain orbital precessions. The extra mathematics of Lense-Thirring precession however are much much more convenient than dealing with a more realistic metric and set of coordinates for Earth. Moreover, it is perfectly reasonable to do physics in preferred frames of reference (Kerr + Boyer-Lindquist, in this case) as long as you admit to yourself that that is what you are doing. There exists a Bogoliubov transformation from this preferred frame of reference to any other reasonable frame of reference, so you're not "stuck" with Lense-Thirring. You are however stuck with the physical result that something orbiting such that 24.something orbits "should" give it the same view below (and above!) but doesn't. We've shown this around various artificial satellites put into polar orbits around various bodies in the solar system. The underlying source of this precession is a combination of gravitational and special-relativistic time dilation. These are not "forces", though. - -- [1] It is being dragged around by the Earth's movements though; if you treat it as an exact inertial frame and do extremely sensitive experiments you'll see deviations in motions of things that you may be tempted to describe using e.g. coriolis forces. If you use exquisitely precise atomic clocks and insist on Cartesian coordinates, you will eventually discover that you are not in an inertial frame after all, and will have to take the local curvature of spacetime into account in the basis of your frame of reference, or shrug and use fictitious forces as corrections to Newtonian or special relativistic geometry. The magnitudes of the fictitious forces will be small.