4 ms·

Hypervelocity Stars Approach Speed of Light

39 points by givan 12y ago

dcre 12y ago
> (Black hole interactions also give rise to hypervelocity stars within the Milky Way, but here there’s just a single black hole, and thus a lot less energy available.) Pretty sure this is wrong. The abstract of the real article says that their conclusion depends on assumptions about the eccentricities of orbits of merging black holes, not the absolute amount of mass involved.
- Dylan16807 12y ago
  What? How is that wrong? It doesn't say anything about mass, did you read it correctly?
- throwaway_yy2Di 12y ago
  The point isn't the absolute amount of mass, but how deep the gravitational wells go -- how rapidly objects can orbit. It's not only about massive objects; it's about compact ones. If you have only one supermassive black hole, then the 2nd and 3rd objects are a binary star, and the interaction between them is very weak. Stars can't orbit each other faster than ~100's of km/s (~1,000's for white dwarfs): they're limited by their low density -- they can't get very close or they'll collide. If, instead, you have a star orbiting a black hole -- that's a very dense object, it allows much closer approaches, much deeper into the potential well, with orbit speeds approaching the speed of light. Don't forget this is a three body interaction. You want a star kicked out of a black hole, with much more kinetic energy than it had going in. This kinetic energy comes from the potential energy of some other object falling into the black hole. How strongly the hypervelocity star gets kicked out, depends on how strongly it can interact with the infalling object, to transfer its energy. (This idea is I think implicit in the paper (page 3): the "kick" velocity of a binary object falling into a black hole (v_kick) is at most a small constant times their mutual escape velocity (v_23)). http://arxiv.org/abs/1411.5022 http://arxiv.org/abs/1411.5022
JumpCrisscross 12y ago
> Superfast stars would be another sort of “particles,” albeit huge, shining ones Have I misunderstood the difference between stars and neutrinos my whole life or is the journalist getting creative?
- andrewflnr 12y ago
  I think the journalist is just being creative. However, I'm not sure "neutrino" means what you think it means, unless you know neutrinos are only one specific kind of particle and are just being creative with your phrasing. :)
  - deleted 12y ago
    [deleted]
- venomsnake 12y ago
  I think that this strip about particles should be added to some of the obligatory XKCD-s http://www.smbc-comics.com/index.php?id=3554 http://www.smbc-comics.com/index.php?id=3554 And yeah ... he is getting creative.
  - gus_massa 12y ago
    If you assume that electrons are made from "smashed electron paste", then the magnetic moment of the electron is related to the angular moment of the electron. If you divide this by the charge of the electron and some universal constants like c an h the result is the g-factor. For the "smashed particles paste" model you get that g=1. (If you have a spinning rock with constant mass density and charge density, you get g=1.) If you use a theoretical model for fermions particles, the calculation is more complicated but finally you get that g=2. If you do the experiment with actual electrons, the result is g=2.00231930436153 (yes, with a lot of decimal places). The difference is dew to the appearance of virtual particles near the electron, and can be calculated theoretically and experimentally, and the results agree up to the current precision!!! You can do the same with other particles, for example for muons g=2.0023318414, because the corrections are slightly different. But for protons g= 5.585694713 because they are not elementary particles, they are composed by quarks. (They are not elemental particles, but are neither smashed paste particles.) More details: http://en.wikipedia.org/wiki/G-factor_(physics) http://en.wikipedia.org/wiki/G-factor_(physics) I think that it's a little optimistic to say that we have the definite elementary particles, but we have hit a very strange wall.
deleted 12y ago
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throwaway_yy2Di 12y ago
"For the special case of the ejection of binary stars, described in Section 6, we employ octuple floating-point precision (~64 digits)." http://arxiv.org/abs/1411.5022 http://arxiv.org/abs/1411.5022
- HCIdivision17 12y ago
  I don't think I've ever even heard of something needing that level of precision. That's about 18 decimal places, right? I'm adding the paper to my read-later pile just to understand that need. (And I don't doubt it. But man, you can describe the height of man or the plank length in one number!) (And my apologies if it's explained in the paper - I haven't read it yet.)
  - throwaway_yy2Di 12y ago
    No, it's 64 decimal digits ~ 213 bits of precision! Octuple precision should have 8 * 32 = 256 bit word size. IEEE 754 single, double, and quadruple precision FP have 32, 64, and 128-bit word sizes, with significand precisions of 24, 53, and 113 bits. https://en.wikipedia.org/wiki/Floating_point#Internal_representation https://en.wikipedia.org/wiki/Floating_point#Internal_repres...
    - HCIdivision17 12y ago
      Even more astounding. In retrospect, the word 'digits' is pretty unambiguous, and I shouldn't have gotten confused. But man, that is some hard-core numerical work. I'll need to bring the paper to the top of my dinner reading stack. The project just seems fascinating.
  - Someone 12y ago
    "For the special case of the ejection of binary stars, described in Section 6, we employ octuple floating-point precision (~64 digits)" "That's about 18 decimal places, right?" I would 'guess' it's about 64 decimal places. At 3 digits in 10 bits, that would require 214 bits. Double precision has 53 bits for the mantissa, so that is in the right range.
deleted 12y ago
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jimmcslim 12y ago
Great article on Centauri Dreams speculating about what a civilisation living on a planet orbiting such a star might experience: http://www.centauri-dreams.org/?p=12671 http://www.centauri-dreams.org/?p=12671
bjornsing 12y ago
> “We calculate that there should be more than a trillion stars in the observable universe moving at velocities of more than a tenth the speed of light,” says Loeb. But wouldn't they be extremely easy to detect experimentally (through red/blue shift a.k.a. doppler)? I'm a bit puzzled why this sounds purely theoretical...
- thisjepisje 12y ago
  Trillion: 10^12 Number of stars in observable universe: 10^22 to 10^24
  - rajacombinator 12y ago
    That is pretty mind blowing.
- cecilpl 12y ago
  Because at these distances we can't actually detect their photons with our most powerful telescopes - we can barely see galaxies.
yummybear 12y ago
I wonder what this would do in case of a collision with another object.
qwerta 12y ago
There was scifi novel describing something like that in 70ties. Anyway, I seriously doubt this is possible. Escape velocity from center of star is tiny fraction of speed of light. We might have better luck with neutron start or black hole going at near speed of light.

39 points by givan 12y ago

dcre 12y ago

> (Black hole interactions also give rise to hypervelocity stars within the Milky Way, but here there’s just a single black hole, and thus a lot less energy available.) Pretty sure this is wrong. The abstract of the real article says that their conclusion depends on assumptions about the eccentricities of orbits of merging black holes, not the absolute amount of mass involved.

Dylan16807 12y ago
What? How is that wrong? It doesn't say anything about mass, did you read it correctly?
throwaway_yy2Di 12y ago
The point isn't the absolute amount of mass, but how deep the gravitational wells go -- how rapidly objects can orbit. It's not only about massive objects; it's about compact ones. If you have only one supermassive black hole, then the 2nd and 3rd objects are a binary star, and the interaction between them is very weak. Stars can't orbit each other faster than ~100's of km/s (~1,000's for white dwarfs): they're limited by their low density -- they can't get very close or they'll collide. If, instead, you have a star orbiting a black hole -- that's a very dense object, it allows much closer approaches, much deeper into the potential well, with orbit speeds approaching the speed of light. Don't forget this is a three body interaction. You want a star kicked out of a black hole, with much more kinetic energy than it had going in. This kinetic energy comes from the potential energy of some other object falling into the black hole. How strongly the hypervelocity star gets kicked out, depends on how strongly it can interact with the infalling object, to transfer its energy. (This idea is I think implicit in the paper (page 3): the "kick" velocity of a binary object falling into a black hole (v_kick) is at most a small constant times their mutual escape velocity (v_23)). http://arxiv.org/abs/1411.5022 http://arxiv.org/abs/1411.5022

JumpCrisscross 12y ago

> Superfast stars would be another sort of “particles,” albeit huge, shining ones Have I misunderstood the difference between stars and neutrinos my whole life or is the journalist getting creative?

andrewflnr 12y ago
I think the journalist is just being creative. However, I'm not sure "neutrino" means what you think it means, unless you know neutrinos are only one specific kind of particle and are just being creative with your phrasing. :)
- deleted 12y ago
  [deleted]
venomsnake 12y ago
I think that this strip about particles should be added to some of the obligatory XKCD-s http://www.smbc-comics.com/index.php?id=3554 http://www.smbc-comics.com/index.php?id=3554 And yeah ... he is getting creative.
- gus_massa 12y ago
  If you assume that electrons are made from "smashed electron paste", then the magnetic moment of the electron is related to the angular moment of the electron. If you divide this by the charge of the electron and some universal constants like c an h the result is the g-factor. For the "smashed particles paste" model you get that g=1. (If you have a spinning rock with constant mass density and charge density, you get g=1.) If you use a theoretical model for fermions particles, the calculation is more complicated but finally you get that g=2. If you do the experiment with actual electrons, the result is g=2.00231930436153 (yes, with a lot of decimal places). The difference is dew to the appearance of virtual particles near the electron, and can be calculated theoretically and experimentally, and the results agree up to the current precision!!! You can do the same with other particles, for example for muons g=2.0023318414, because the corrections are slightly different. But for protons g= 5.585694713 because they are not elementary particles, they are composed by quarks. (They are not elemental particles, but are neither smashed paste particles.) More details: http://en.wikipedia.org/wiki/G-factor_(physics) http://en.wikipedia.org/wiki/G-factor_(physics) I think that it's a little optimistic to say that we have the definite elementary particles, but we have hit a very strange wall.

deleted 12y ago

[deleted]

throwaway_yy2Di 12y ago

"For the special case of the ejection of binary stars, described in Section 6, we employ octuple floating-point precision (~64 digits)." http://arxiv.org/abs/1411.5022 http://arxiv.org/abs/1411.5022

HCIdivision17 12y ago
I don't think I've ever even heard of something needing that level of precision. That's about 18 decimal places, right? I'm adding the paper to my read-later pile just to understand that need. (And I don't doubt it. But man, you can describe the height of man or the plank length in one number!) (And my apologies if it's explained in the paper - I haven't read it yet.)
- throwaway_yy2Di 12y ago
  No, it's 64 decimal digits ~ 213 bits of precision! Octuple precision should have 8 * 32 = 256 bit word size. IEEE 754 single, double, and quadruple precision FP have 32, 64, and 128-bit word sizes, with significand precisions of 24, 53, and 113 bits. https://en.wikipedia.org/wiki/Floating_point#Internal_representation https://en.wikipedia.org/wiki/Floating_point#Internal_repres...
  - HCIdivision17 12y ago
    Even more astounding. In retrospect, the word 'digits' is pretty unambiguous, and I shouldn't have gotten confused. But man, that is some hard-core numerical work. I'll need to bring the paper to the top of my dinner reading stack. The project just seems fascinating.
- Someone 12y ago
  "For the special case of the ejection of binary stars, described in Section 6, we employ octuple floating-point precision (~64 digits)" "That's about 18 decimal places, right?" I would 'guess' it's about 64 decimal places. At 3 digits in 10 bits, that would require 214 bits. Double precision has 53 bits for the mantissa, so that is in the right range.

deleted 12y ago

[deleted]

jimmcslim 12y ago

Great article on Centauri Dreams speculating about what a civilisation living on a planet orbiting such a star might experience: http://www.centauri-dreams.org/?p=12671 http://www.centauri-dreams.org/?p=12671

bjornsing 12y ago

> “We calculate that there should be more than a trillion stars in the observable universe moving at velocities of more than a tenth the speed of light,” says Loeb. But wouldn't they be extremely easy to detect experimentally (through red/blue shift a.k.a. doppler)? I'm a bit puzzled why this sounds purely theoretical...

thisjepisje 12y ago
Trillion: 10^12 Number of stars in observable universe: 10^22 to 10^24
- rajacombinator 12y ago
  That is pretty mind blowing.
cecilpl 12y ago
Because at these distances we can't actually detect their photons with our most powerful telescopes - we can barely see galaxies.

yummybear 12y ago

I wonder what this would do in case of a collision with another object.

qwerta 12y ago

There was scifi novel describing something like that in 70ties. Anyway, I seriously doubt this is possible. Escape velocity from center of star is tiny fraction of speed of light. We might have better luck with neutron start or black hole going at near speed of light.