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Phalcone42 t1_itxr3al wrote

Well, there are neutron stars that are functionally one giant unstable element held together by gravity instead of the strong force. The only reason they don't collapse further is because of neutron degeneracy pressure. Considering the relationship between half life and atomic size, coupled with the longevity of neutron stars, i'd say so, but it's a bit of a stretch.

Edit. For a less extreme condition where the nucleus is not so massive as to be stabilized by gravity, it seems that certain types of decay, like electron capture decay, vary with pressure, while others like ß emission does not.

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west_of_everywhere t1_itxzn0i wrote

Replying to [deleted] (#250,917)

>However, at the center of the Sun, the gravitational force is close to zero. So I'm pretty sure there would be no effect.

Isn't time dilation a function of potential energy, not force? Thanks!

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mfb- t1_ity0i9m wrote

Replying to [deleted] (#250,917)

Gravitational time dilation depends on the potential, not the gravitational force. It's maximal in the center of objects, including the Sun. But it's not a large effect overall.

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mfb- t1_ity0qbe wrote

If you think of gravitational time dilation: No. It's a tiny effect for stars, just a few parts in a million. Instead of 1 second the lifetime (as seen from far away) is now 1.000005 seconds or something like that (with the exact number depending on the star) - not a big deal.

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mfb- t1_ityc918 wrote

Replying to [deleted] (#252,789)

Gravitational time dilation is based on the potential, not based on a force. The potential has its minimum (i.e. the largest time dilation) at the center.

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DaRealKryall t1_itycuw0 wrote

Replying to mfb- (#252,831)

Oh, I didn't know that, I stand corrected then!

Just graduated high school, but our last physics unit was special relativity; we were only taught it to be based on the velocity of an object relative to the stationary observer.

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MrMurchison t1_ityp99s wrote

Replying to tobojijo (#253,994)

Sure, but that only seems like a significant difference because of human time frames. Those 20,000 years describe the difference between being at exactly half of the original mass, and functionally half of the original mass.

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Implausibilibuddy t1_ityzueg wrote

Replying to mfb- (#251,516)

What about gravity affecting ability for a particle to break the weak force, or is that negligible? Or pulling atoms closer together making them more likely to be hit by particle emissions from other atoms?

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ackillesBAC t1_itzf5ny wrote

Replying to mfb- (#251,516)

Doesnt every frame always observe thier time ticking at 1s/s?

So from the point of view of a particle in a star it lives a normal time span regardless.

However an outside observer would see thier time flowing at 1.000005s/s

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mfb- t1_itzgc2s wrote

Replying to [deleted] (#257,243)

Temperature in the Sun's core is just ~15 MK or ~1 keV, which leads to a typical gamma factor of ~1+10^(-6) for protons and ~1+5*10^(-9) for very heavy nuclei like uranium. That's a smaller effect, especially for heavy nuclei.

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rootofallworlds t1_itzgwfj wrote

On the contrary, in a star undergoing fusion the conditions may well cause nuclear reactions such as induced fission, making the effective half-life of unstable nuclides shorter than their half-lives when isolated. In the same way that uranium and plutonium in a nuclear reactor are fissioned much faster than their natural half-lives of millions or billions of years. I predict that would mask the effect of gravitational time dilation which has been calculated to be tiny by other answers.

While it's not a heavy nuclide, Lithium-7 is destroyed in fusing stars, and its presence or absence helps distinguish red dwarfs from brown dwarfs.

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mfb- t1_itzimnc wrote

Replying to Flatworldnotearth (#257,517)

2.7 GK / 15 MK = 200, so we get 1+10^(-6) for uranium. Still smaller than the gravitational effect, especially if we consider that we are now looking at the core of a far more massive star.

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Flatworldnotearth t1_itzkpwd wrote

Replying to mfb- (#257,435)

An uranium nucleus in solar core travels at around 36km/s and in a exploding star at 3GK is around 500km/s and the Lorentz factor is around 1.0000015. Reaching the effects of gravitational time dilation as you say but the heavier the star is the more gravitational time dilation it gets. Thanks for your explanation.

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noldig t1_itzm5gt wrote

Replying to Ahandgesture (#257,956)

Not completely, there is still neutron decay going on, but it is heavily surpressed because of electron degeneracy pressure, and balanced by electron capture

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Ahandgesture t1_itzmhnb wrote

Replying to noldig (#258,376)

Interesting, thank you for the answer! I wonder what the neutron lifetime in this case is. There's already the discrepancy between beamline and "jar" lifetime experiments. If we could measure neutron decay in a neutron star, we'd probably have another point of discrepancy 😆

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noldig t1_itzmxzz wrote

Replying to Ahandgesture (#258,423)

The lifetime will mostly depend on phase space and temperature, so two quantities a free neutron doesn't care about. But these processes are responsible for cooling neutron stars down so we try to measure it. I have computed the rate a few times but never converted the units haha I will check

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unkilbeeg t1_itzny0y wrote

At the core of the star there would be very little effect from gravity, because the gravity is close to zero there. The pressure from all the surrounding mass (which has gravity working on it), on the other hand, would be huge.

Inside a body, gravity decreases as you go towards the center. The maximum gravity is right at the surface.

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pinocola t1_itzub5m wrote

Replying to mfb- (#257,169)

Also the shell theorem shows that there is no net gravitational force at the center of a sphere. Lots of pressure down there, but the mass on all sides of the core cancels the gravity out.

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mfb- t1_iu0bfwx wrote

Replying to alien_from_earth_14 (#261,300)

No, it can't be stable at the singularity, nothing (that we know) can. A nucleus will get ripped apart by tidal forces before it reaches the singularity (or whatever might be there - but for sure not atoms).

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8spd t1_iu0pfw6 wrote

Replying to mfb- (#257,169)

I mostly like the term "well inside a black hole", because it implies the existence of "barely inside a black hole".

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fore4runner t1_iu1rs2f wrote

Replying to pinocola (#259,480)

Does that cancel out the time dilation as well? (It seems like it should logically, but intuitively it 'feels' like the time dilation should be at a maximum at the very centre of the body).

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pinocola t1_iu23ksq wrote

Replying to fore4runner (#269,500)

No, because light and information have to cross out of the gravity well to get to an observer, and time dilation is always measured in reference to some other point, in this case presumably outside of the core of the star.

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Emuuuuuuu t1_iu23luc wrote

Replying to fore4runner (#269,500)

Force can be applied in many directions such that the net is zero. The force is still being applied though. Time only changes in one direction (slows down) and can't be cancelled by other directions.

It could maybe be sped back up with a very dense clump of anti-matter? Can a general relativitist chime in here?

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fore4runner t1_iu2aue3 wrote

Replying to pinocola (#271,417)

But, if a person, in the centre of the earth shout out a laser every 1ms to a relatively stationary observer in space they would get the pulse every 1ms? But if someone on the earths surface did that, they wouldn’t, right?

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pinocola t1_iu2m0r6 wrote

Replying to fore4runner (#272,580)

The pulse from the surface would be a bit slower than once every ms due to time dilation, and the pulse from the center of the earth would be even more time dilated than that.

Time dilation comes from a speed difference or the gravity gradient you travel/observe across. The pulses of light don't observe any gravity at the core itself, but once they start traveling out from the earth, they feel the gravity from whatever portion of the earth they've traveled out from, and would appear proportionally time dilated.

If the earth were a big hollow shell like an inflated ball, two observers anywhere inside the shell would feel no gravity due to the mass of the earth, and would not observe any time dilation when observing one another (even if one was at the geometric center and another was hanging onto the inside of the shell). But an observer outside the shell would feel the entire gravity of the earth and would see both people inside the shell as time dilated by the same amount.

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mfb- t1_iu2wbsw wrote

Replying to 8spd (#263,751)

The event horizon is usually seen as boundary. You can be just behind the event horizon (you can't stay there, of course).

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mfb- t1_iu2wlzo wrote

Replying to Ituzzip (#262,005)

Also irrelevant for radioactivity. Electron degeneracy pressure can have an influence on electron capture and beta decays, neutron degeneracy pressure (neutron stars) can have an effect on the stability of nuclei.

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RobusEtCeleritas t1_iu3t52l wrote

Replying to ToolFO (#256,250)

There's not really any meaningful sense in calling a neutron star a giant nucleus. A nucleus is bound by the residual strong force, and the heaviest nuclei have radii on the order of tens of femtometers or so.

A neutron star is bound by gravity, and has a radius of around a few kilometers.

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