Why don't Neutron Stars form event horizon?

  • Trying to compare density of Black Holes and Neutron Stars I came up with the following:

    A typical neutron star has a mass between about 1.4 and 3.2 solar masses1[3] (see Chandrasekhar Limit), with a corresponding radius of about 12 km. (...) Neutron stars have overall densities of 3.7×10^17 to 5.9×10^17 kg/m^3 [1]


    You can use the Schwarzschild radius to calculate the "density" of the black hole - i.e., the mass divided by the volume enclosed within the Schwarzschild radius. This is roughly equal to (1.8x10^16 g/cm^3) x (Msun / M)^2 (...)

    The value of the Schwarzschild radius works out to be about (3x10^5 cm) x (M / Msun) [2]

    Let's take a neutron star from the top of the spectrum (3.2 Msun) and same mass black hole.

    Converting units:

    • Neutron star: 5.9×10^17 kg/m^3 = 5.9 × 10^14 g/cm^3

    • Black hole: 1.8x10^16 g/cm^3 x (1/5.9)^2 = 5.2 x10^14 g/cm^3

    The radius of the black hole would be (3x10^5 cm) x ( 5.2 ) = 15.6km

    The 3.2Msun Neutron Star of this density would have volume of 1.08 x 10^13 m^3 which gives radius of 13.7 kilometers

    According to Shell Theorem, spherical objects' gravity field strength at given distance is the same for spheres as for point masses so at the same distance from center of same mass (point - black hole, sphere - neutron star) the gravity will be the same.

    That would put the surface of the neutron star below the surface of event horizon of equivalent black hole. Yet I never heard about even horizon of neutron stars.

    Either I made a mistake in my calculations (and if I did, could you point it out?) or... well, why?

    There is an error: where did you get the *5.9* in the equation for the black hole and the 5.2 in the radius of the black hole? You must use 3.2. In this way you get 1.7x10^15 g/cm^3 as density and 9.6km as radius

    Why has this got so many upvotes. It contains a trivial error in the Schwarzschild radius. R_s is 2.96 km per solar mass.

  • AstroFloyd

    AstroFloyd Correct answer

    9 years ago

    As Francesco Montesano points out, using the wrong mass leads to the wrong answer. Also, using the density here seems a complicated way to get to the answer; you could compute the Schwarzschild radius for the NS, and see whether it's smaller than its actual radius.

    Since the density scales as ρ ~ M/R^3 and the Schwarzschild radius as Rs~M, the density of BHs scales as ρ~1/R^2; more massive BHs are less dense and simply testing whether a NS is denser than a BH alone is not sufficient - they must be of the same mass, which means that you are in fact comparing radii.

    +1, though there's another reason why this density is bad: volume is completely frame-dependent. Wiki's density figures use Euclidean volume where the geometry is strongly non-Euclidean. With the metric in the Tolman-Oppenheimer-Volkoff ansantz, a spherically symmetric simple neutron star would have volume $$V_\text{TOV} = \int_0^R \frac{4\pi r^2\,\mathrm{d}r}{\sqrt{1-\frac{2GM(r)}{rc^2}}}\text{,}$$ which is never Euclidean. In another frame, it'd be something else still. We could still use the Euclidean "overall density" to compare neutron stars, but the figure itself doesn't mean much.

    "more massive BHs are less dense" And of course an interesting consequence of this is that assuming a flat and non-expanding space, takings a volume of any positive density and scaling up its size in three dimensions while keeping density within it constant will eventually result in a black hole.

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Content dated before 7/24/2021 11:53 AM