# Free fall time of collapsing non-rotating star in GR according to a distant observer

Suppose a star with a spherically symmetric mass distribution, mass $2\cdot M_{sun}$ and initial radius $R$, undergoes a gravitational collapse after all gas pressure has vanished. The final radius will be about 15 km, typical of a neutron star. A Newtonian derivation of the free fall time gives $$t_{ff}=\tfrac{\pi}{2}\sqrt{\frac{R^3}{2\cdot G\cdot M}}=\sqrt{\frac{3\cdot\pi}{32\cdot G\cdot\rho}}.$$

Since we end up with a compact object, general relativity might lead to a different free fall time according to a distant observer.

Question: what is the expression for the free fall time in GR in this case?

An object in "freefall" spends most of its trajectory at large radii (since acceleration ensures that the velocity increases more than linearly with time), so the collapsing proto-neutron star will spend most of the collapse period close to the size of its original electron-degenerate progenitor. For a hot, degenerate iron core, this is of order $R> 1000$ km.

The dimensionless ratio $GM/Rc^2$ will then be $<0.003$ for most of the collapse, so the influence of GR on the freefall timescale will be rather negligible. Even when the neutron star is close to its final radius (and I think 10 km is too small for the collapse to be considered "free fall", because the separation of the neutrons will be $\sim 10^{-15}$m at this radius, which is comparable to the range of the strong nuclear force, and also because the neutrons will be highly degenerate and neutron degeneracy pressure would not be negligible.), the gravitational time dilation factor for freely falling object in the Schwarzschild metric, $(1 - r_s/r)^{-1}$, would only reach a factor of $\sim 2$.

My intuition would therefore be that a Newtonian approximation for the free fall timescale is quite good here.

An analytic expression would follow an object on the surface of the (non-spinning) collapsing star. Assuming spherical symmetry, then by Birkhoff's theorem, the Schwarzschild metric can be used throughout.

The geodesic for an object freely falling from rest at $t_0, r_0$ is given by (e.g. e.g. see Chapter 25 in "Gravitation" by Misner, Thorner & Wheeler, 2017, Princeton University) $$\frac{c(t-t_0)}{r_s} = \ln \left| \frac{ (r_0/r_s -1)^{1/2} + \tan (\eta/2)}{(r_0/r_s -1)^{1/2} -\tan(\eta/2)}\right| + \left(\frac{r_0}{r_s}-1\right)^{1/2} \left( \eta + \frac{r_0}{2r_s}(\eta + \sin \eta)\right). \tag{1}$$ The "cycloid parameter" $\eta(r)$ is defined by $$r = \frac{r_0}{2}(1 + \cos \eta)$$

If we assume that $r_0 \gg r$ and $r_0 \gg r_s$ and $t_0=0$, then $\eta \simeq \pi$, and the time to fall from $r_0$ to $r$ (your free fall time) is $$\tau \simeq \frac{\sqrt{r_s r_0}}{c} \left( \pi + \frac{\pi r_0}{2r_s}\right) \simeq \frac{\pi r_0^{3/2}}{2 r_s^{1/2} c}$$ Substituting $r_s = 2GM/c^2$, we do recover the Newtonian formula.

e.g. If $r_0 = 1000$ km, $R= 10$ km and $M = 2M_{\odot}$, the Newtonian formula gives a free fall time of 0.0680 s, but the GR formula above gives 0.0685 s - so longer by 0.7%.

• @Jeffries Could you please explain why 10 km is too small for the collapse to be considered "free fall"? – gamma1954 May 8 '18 at 12:50
• @gamma1954 See edit. Degeneracy pressure would not be negligible and the neutrons are essentially "touching" at these densities. This is why you get core bounce in a supernova. – Rob Jeffries May 8 '18 at 13:10
• Expression (1) also gives coordinate time $t$ of an object in free fall to non-rotating black hole starting from $t_0=0$ as measured by an observer at rest at a great distance. Imagine a black hole with $2\cdot M_\text{sun}$. If the object starts falling from $r=2\cdot r_\text{s}$, coordinate time $t$ diverges as $r\to r_\text{s}$, as expected, because the denominator in the logarithm approaches 0. $t$ becomes large only when the object is VERY close to the horizon (ln). E.g. root finding doesn't give $\eta$ for which $t=2$ seconds. Does this reflect great time dilation near the horizon? – gamma1954 May 22 '18 at 19:37
• @gamma1954 I don't understand what you are asking. There is of course a solution for any value of $t$ and it becomes increasingly close to $r_s$. If you want your collapse to halt arbitrarily close to $r_s$ (it can't - see something called the Buchdahl limit) then obviously the answer is that it can take as long a (coordinate) time you like. Your question was about collapse to a neutron star, which is what my answer pertains to. – Rob Jeffries May 22 '18 at 19:48
• I apologize for being unclear. The original question was about collapse to form a neutron star. But equation (1) is also valid for the coordinate time $t$, measured by a distant observer, of an object falling from e.g. $r_0=2 r_\text{s}$ towards a pre-existing black hole with $M_\text{bh}=2 M_\text{sun}$ (forget about NS). The coordinate time goes to infinity as $r\to r_\text{s}$, so for some $r$ slightly larger than $r_\text{s}$ I expect to find e.g. $t=1$ second (or larger). But even for $r_\text{final}=1.0000000001 r_\text{s}$ equation (1) doesn't give such a 'large' $t$. This puzzles me. – gamma1954 May 22 '18 at 20:53