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I am teaching myself astronomy and I'm working on a basic problem regarding time dilation around black holes.

The question is basically this: if a neutron were ejected from a nucleus at a distance of 3km from a 1 solar mass black hole, how long would it appear to take for the neutron to decay (the assumption here is that the neutron would decay in 15 minutes... right on the half-life) to an outside observer?

My first thought was that we'd never see it decay, because 3km is the Schwarzschild radius of a BH of that mass, so the frequency would basically be $0$ and we'd never observe the decay event.

The text notes the time dilation

$$ \frac{\Delta t'}{\Delta t_\text{obs}} = \frac{\nu_f}{\nu_i}$$ where $\Delta t'$ is the the time difference between two events at the source and $\Delta t_{obs}$ at the observer.

We can use the relativistic red shift equation to get the wavelength difference (9.2 in my case using 1 solar mass and a radius of 3km) so that wouldn't be infinite, just big. For a 15 min $\Delta t'$ then, the $\Delta t_{obs}$ would be 15 min * 9.2, yeah?

So my two answers obviously contradict each other. What am I missing in my approach here?

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  • $\begingroup$ BTW, the free neutron half-life is around 10.2 minutes. The mean lifetime of a neutron is just under 15 minutes. OTOH, the mean lifetime is probably more appropriate for this problem anyway. $\endgroup$ – PM 2Ring Jan 27 at 9:10
  • $\begingroup$ Can you give a reference to the text? $\endgroup$ – John Duffield Jan 28 at 19:17
  • $\begingroup$ The text is Zeilik and Gregory Introduction to Astronomy and Astrophysics (1998). $\endgroup$ – PSR-1937-21 Jan 28 at 20:53
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    $\begingroup$ The Schwarzchild radius of a 1 solar mass black hole is 2.95km, so your neutron is just outside the event horizon. A redshift factor of 9.2 seems plausible for that. $\endgroup$ – Steve Linton Feb 2 at 11:45
  • $\begingroup$ @SteveLinton, right. So I need to adjust the gravitational red shift a wee bit b/c got that number using a radius of 3km, not 2.95 km. So it will be similar, but slightly different. However, your clarification of the size of the Schwarzschild radius does dissolve my paradox, so thanks for that. $\endgroup$ – PSR-1937-21 Feb 3 at 1:07
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If you use the equation of the Schwarzschild metric. $$ds^{2}=-(1-\frac{2M}{r})dt^{2}+(1-\frac{2M}{r})^{-1}dr^{2}+r^{2}d\Omega^{2}$$ (equation 11.1 from Schutz's book A first course in GR).

If we think the neutron stay at rest in $r=r_{S}=2M$ then between the born and dead of the neutron $dr=0$ and $d\Omega=0$, and $ds$ will be the proper time, which I will call $\tau$ (I'm folowing the convention $c=1=G$).

Putting this on the metric we have

$$ d\tau^{2}=-(1-\frac{2M}{r})dt_{obs}^{2}$$

$$ \frac{d\tau^{2}}{dt_{obs}^{2}}=-(1-\frac{2M}{r})$$

and if we put $r=r_{s}$ (where the neutros was), we have

$$ \frac{d\tau^{2}}{dt_{obs}^{2}}=0$$

Then $\Delta t_{obs}=\infty$ .

I couldn't write a comment that's why I answered, but I think you had some mistake using the red shift equation to get the wavelength difference.

Can you show your calculus? May be I can help.

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    $\begingroup$ No calculus here. This is for an introductory astronomy text, so I'm using using the gravitational red shift equation given in the text: $\frac {\lambda_f}{\lambda_i} = [1 - 2GM/Rc^2]^{-1/2}$ $\endgroup$ – PSR-1937-21 Feb 3 at 1:09
  • $\begingroup$ If you put $R=R_{s}=\frac{2GM}{c^{2}}$ everything simplifies and you have a $0^-1/2=\infty$ as you wished. You musn't put $R=3Km$. May be you replaced the true values for $G$, $M$ and $c^2$, it can take you to a bit error because of the $3Km$ is not the exact Schwarzschild radio for the earth if you wanted to be really acurate. $\endgroup$ – Gabriel Palau Feb 3 at 2:09

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