4 ms·

Shouldn't there be a big lump of dark matter at the center of every planet, moon, and star? If we were to hollow out an asteroid, would there still be a blob

Shouldn't there be a big lump of dark matter at the center of every planet, moon, and star? If we were to hollow out an asteroid, would there still be a blob of intangible dark matter inside of it?

manigandham 10y ago
We don't know what dark matter is (including whether or not it's even matter).
dogma1138 10y ago
Which type of dark matter would that be?
wyager 10y ago
No, not if it's weakly interacting. Matter forms big balls because it collides with other matter and turns kinetic energy into heat. If dark matter is weakly interacting, it won't collide, so after falling into a star it will just shoot right back out the other side. It might form a stable orbit, but it also might just fly away and fall through some other gravity well. You can probably expect some bulk gravitational effects (e.g. Dark matter might tend to cluster along large matter distributions like galaxies), but in general it wouldn't be amenable to staying in one place.
saboot 10y ago
In fact, the answer for planets is Yes! http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA20176 http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA20176
platz 10y ago
I like to look through the lens of quantum field theory instead of 'dark matter particles'. Dark matter may be some field or interaction of existing fields that we don't understand yet.
- nialv7 10y ago
  Particle is an excited state of a field. What you are saying is the exact same thing.
  - platz 10y ago
    Yes, this misses the point i'm making however - I think fields connote more than just the "excited state" part (the particle). Particles aren't exactly the same idea because they don't explain the coupling between fields, or things like 'virtual particles' which don't "count" as real particles but have important real behaviors of the field itself.
    - raattgift 10y ago
      The problem is that your "lens" statement doesn't convey much information; if it goes deep, unfortunately it's not obvious. For example, it doesn't tell anyone whether you think that CDM is sparticles or a quantization of a function on the Ricci scalar or of a second metric or whatever. The latter two are awfully hard to distinguish from relativistic approaches to MOND; indeed you could quantize most of the approaches in Chapter 7 of https://arxiv.org/abs/1112.3960 https://arxiv.org/abs/1112.3960 . More generally, you can put a scalar or other degree of freedom on either side of the usual write-down of the Einstein Field Equations - on the R side it's "modified gravity", but on the T side it's "modified matter". There's no strong reason to choose one side over the other (Einstein and Schrödinger had a couple of letters to each other about "which side" as early as 1918![1]). Either way you're adding a field to General Relativity -- and you'd see that in a Langrangian formulation like an expansion of the Einstein-Hilbert action[2]. There is no real reason that you couldn't quantize a field on either side of G = T (consider how g is quantized in perturbative quantum gravity, for example), and if they couple at all non-gravitationally to standard model particles, you'd very much want to. Finally, historically, particle CDM was strongly motivated by QFT considerations in the first place, with the expectation that they would be lightest MSSM sparticles or mass-explaining sterile neutrinos on the WIMP hand, or alternatively strong-CP resolving axions. So your "lens" is standard, so I agree with your parent comment. Lastly, you wrote essentially the same comment several days ago https://news.ycombinator.com/item?id=13600085 https://news.ycombinator.com/item?id=13600085 . [1] https://arxiv.org/abs/1211.6338 https://arxiv.org/abs/1211.6338 [2] e.g. S = \int \left[ 1/2 R + \mathcal{L}_{baryons} + \mathcal{L}_{photons} + \mathcal{L}_{neutrinos} + \mathcal{L}_{CDM} + ... \right] \sqrt{-g} {d}^{4}x.

Shouldn't there be a big lump of dark matter at the center of every planet, moon, and star? If we were to hollow out an asteroid, would there still be a blob of intangible dark matter inside of it?

manigandham 10y ago

We don't know what dark matter is (including whether or not it's even matter).

dogma1138 10y ago

Which type of dark matter would that be?

wyager 10y ago

No, not if it's weakly interacting. Matter forms big balls because it collides with other matter and turns kinetic energy into heat. If dark matter is weakly interacting, it won't collide, so after falling into a star it will just shoot right back out the other side. It might form a stable orbit, but it also might just fly away and fall through some other gravity well. You can probably expect some bulk gravitational effects (e.g. Dark matter might tend to cluster along large matter distributions like galaxies), but in general it wouldn't be amenable to staying in one place.

saboot 10y ago

In fact, the answer for planets is Yes! http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA20176 http://www.jpl.nasa.gov/spaceimages/details.php?id=PIA20176

platz 10y ago

I like to look through the lens of quantum field theory instead of 'dark matter particles'. Dark matter may be some field or interaction of existing fields that we don't understand yet.

nialv7 10y ago
Particle is an excited state of a field. What you are saying is the exact same thing.
- platz 10y ago
  Yes, this misses the point i'm making however - I think fields connote more than just the "excited state" part (the particle). Particles aren't exactly the same idea because they don't explain the coupling between fields, or things like 'virtual particles' which don't "count" as real particles but have important real behaviors of the field itself.
  - raattgift 10y ago
    The problem is that your "lens" statement doesn't convey much information; if it goes deep, unfortunately it's not obvious. For example, it doesn't tell anyone whether you think that CDM is sparticles or a quantization of a function on the Ricci scalar or of a second metric or whatever. The latter two are awfully hard to distinguish from relativistic approaches to MOND; indeed you could quantize most of the approaches in Chapter 7 of https://arxiv.org/abs/1112.3960 https://arxiv.org/abs/1112.3960 . More generally, you can put a scalar or other degree of freedom on either side of the usual write-down of the Einstein Field Equations - on the R side it's "modified gravity", but on the T side it's "modified matter". There's no strong reason to choose one side over the other (Einstein and Schrödinger had a couple of letters to each other about "which side" as early as 1918![1]). Either way you're adding a field to General Relativity -- and you'd see that in a Langrangian formulation like an expansion of the Einstein-Hilbert action[2]. There is no real reason that you couldn't quantize a field on either side of G = T (consider how g is quantized in perturbative quantum gravity, for example), and if they couple at all non-gravitationally to standard model particles, you'd very much want to. Finally, historically, particle CDM was strongly motivated by QFT considerations in the first place, with the expectation that they would be lightest MSSM sparticles or mass-explaining sterile neutrinos on the WIMP hand, or alternatively strong-CP resolving axions. So your "lens" is standard, so I agree with your parent comment. Lastly, you wrote essentially the same comment several days ago https://news.ycombinator.com/item?id=13600085 https://news.ycombinator.com/item?id=13600085 . [1] https://arxiv.org/abs/1211.6338 https://arxiv.org/abs/1211.6338 [2] e.g. S = \int \left[ 1/2 R + \mathcal{L}_{baryons} + \mathcal{L}_{photons} + \mathcal{L}_{neutrinos} + \mathcal{L}_{CDM} + ... \right] \sqrt{-g} {d}^{4}x.