Could someone briefly explain Hayward black holes to me? I tried looking them up on Wikipedia and there’s not much there.

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    $\begingroup$ Did you look at the five external references on the Wikipedia pages? At least some are freely accessible. $\endgroup$
    – hft
    Jul 5, 2023 at 18:27

2 Answers 2


The Hayward metric is a non-singular black hole metric in the sense that a central singularity is absent. It satisfies the condition $g_{tt}=1/g_{rr}$ (hence it is sourced by some particular theory that satisfies $G_{t}^{~t} = G_{r}^{~r} \to T_{t}^{~t} = T_{r}^{~r}\to -\rho = p_{r}$ where $\rho$ and $p_r$ are the energy density and the radial pressure of the corresponding energy momentum tensor that sources this spacetime) and is given by the line element $$ds^2 = -f(r)dt^2 + \frac{dr^2}{f(r)} +r^2d\Omega^2$$ with $$f(r) = 1-\frac{2GMr^2}{r^3+2GMl^2}$$ where $M$ is the conserved black hole mass, $G$ is the Newton constant and $l$ is a length scale responsible for the regularity of spacetime.

The regularity can be seen by calculating the components of the Riemann tensor, to see that all of them are finite for $r \ge 0$. This might also be checked by evaluating the Kretschmann scalar which is the contraction of the Riemann tensor with itself (the reasoning behind this can be found on page 5 in the paper Regular black holes sourced by nonlinear electrodynamics):

$$K = R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta} = \frac{48 G^2 M^2 \left(72 G^2 l^4 M^2 r^6-16 G^3 l^6 M^3 r^3+32 G^4 l^8 M^4-8 G l^2 M r^9+r^{12}\right)}{\left(2 G l^2 M+r^3\right)^6}~,$$ which is finite everywhere and near the origin behaves as $$K(r\to0)\sim \frac{24}{l^4}-\frac{84 r^3}{G l^6 M}+O\left(r^6\right)~.$$ This metric has a de-Sitter core (the expansion of the metric near the origin resembles a de Sitter metric) which is the reason of finiteness near the origin.

This metric possesses two black hole horizons a Cauchy and an event horizon and consequently suffers from the mass inflation instability Inner-horizon instability and mass inflation in black holes .

Recently, it was found that the Hayward metric can be derived using a particular non-linear electromagnetism Lagrangian, Remarks on nonsingular models of Hayward and magnetized black hole with rational nonlinear electrodynamics, which one can argue, that from a field theory point of view it may be problematic, since this Lagrangian contains the integration constants of the solution.

The corresponding Lagrangian may be found, according to the paper cited above, as $$\mathcal{L} = -\frac{6GM^2l^2(2\mathcal{F})^{3/2}}{(q_m^{~3/2}+2^{7/4}GMl^2\mathcal{F}^{3/4})^2}$$

where $q_m$ is the magnetic charge and $\mathcal{F} = F^{\mu\nu}F_{\mu\nu}$.

As one can see in this Lagrangian, the mass and magnetic charge appear. A Lagrangian can include fields and constants. Fields are allowed to vary (and hence we can obtain equations of motion for them) while constants are keeped fixed. The fact that this Lagrangian contains the black hole mass raises questions about the thermodynamic nature of this black hole (if the mass is fixed from the theory, then this object cannot lose mass and emit Hawking radiation and hence only eternal cold black holes with zero temperaure are allowed). Anyway, the author managed to rexpress the Lagrangian in such a way that the theory will give black holes with fixed mass to magnetic charge ratio, see equation (13), so the mass is allowed to change alongside the magnetic charge so their ration $\alpha$ will always be kept constant.

However, the themrodynamic interpetation of regular black holes seems to be under debate up to this date, so i refer you to the following papers for relevant discussions:

Corrected form of the first law of thermodynamics for regular black holes

Regular black holes: A short topic review

Regular Compact Objects with Scalar Hair

  • $\begingroup$ And at a early undergraduate level. $\endgroup$ Jul 7, 2023 at 0:36
  • $\begingroup$ What do you mean? $\endgroup$
    – Noone
    Jul 7, 2023 at 3:56
  • $\begingroup$ Could you tone it down to a first/second year undergrad level? $\endgroup$ Jul 7, 2023 at 5:24
  • $\begingroup$ The basic idea is that at the center of the Schwarzchild metric (see en.wikipedia.org/wiki/Schwarzschild_metric) lies a singularity. Singularities are points, or regions in spacetime where a physical theory ceases to hold. Since the Schwarzchild black hole is a solution of the Einstein field equations in vaccum this, the singularity implies that Einstein's theory of gravity fails to describe physical phenomena at $r=0$. One has to resort to other gravity theories that can cure the problem of singularity. $\endgroup$
    – Noone
    Jul 7, 2023 at 7:51
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    $\begingroup$ The Hawyard metric is a line element that is not singular at the point $r=0$ neither at any other point, and is the minimal deformation one can perform to the Schwarzchild metric to make the metric regular at $r=0$, so it cures the problem of singularity. I think that in order to go further i have to refer to upper undergrad level mathematics. $\endgroup$
    – Noone
    Jul 7, 2023 at 7:53

The original paper is Hayward, Sean, "Formation and evaporation of non-singular black holes" (arXiv, DOI).

As Noone's answer said, he starts with the family of metrics $$ds^2 = -F(r)\,dt^2 + \frac{dr^2}{F(r)} + r^2 d\Omega^2, \tag 1$$ which includes Schwarzschild black holes, and he observes that if $F(0)=1$ and $F'(0)=0$ then the metric is well-behaved at $r=0$. "For definiteness", he picks a specific $F$ with that property that also approaches $1-2m/r$ as $r\to\infty$, namely $$F(r) = 1 - \frac{2mr^2}{r^3+2l^2m} \tag 5$$ where $l$ is some constant.

I want to make a couple of points that Noone's answer didn't.

First, (5) appears to be just an example, and not a defining property of the solutions that he wants to consider—i.e., it's not really the case that the "Hayward metrics" are given by (1) and (5). All that really matters is that the metric be well-behaved near $r=0$, and black-hole-like for large $r$.

Second, his choices seem strange, and not physically well-motivated. I see no reason why the coefficients of $dt^2$ and $dr^2$ should be given by the same function $F$ at all (much less the one he wrote down). A number of vacuum solutions of GR fit this pattern, but the Schwarzschild interior metric doesn't. He says that in solutions of the form $(1)$ that are well-behaved at $r=0$, "there is an effective cosmological constant at small distances". I think this is probably an artifact of his vacuum-like Ansatz. There has to be mass there, and a cosmological constant is the only way to have mass in a vacuum, but it's not the only way to have mass.

The other important claim of the paper, as suggested by the title, is that you can get a classical model of black hole evaporation out of this. He does it by switching to an Eddington-Finkelstein time coordinate $v$, and then making $m$ a function of $v$. The result is an ingoing positive energy flux when $m$ increases with time, and an ingoing negative energy flux (violating energy conditions) when $m$ decreases. This also lacks physical motivation. It's similar to Alcubierre's warp drive or Kip Thorne's traversible wormhole, where you just write down the spacetime shape you want, with no cause-and-effect mechanism relating its shape at different times. These are not solutions to GR except in the trivial sense that every metric is a solution if you don't constrain the stress-energy tensor.

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    $\begingroup$ +1. I think that this is a minimal model in the sense that it has the minimal assumptions and parameters (only lambda) in order to support regularity. I agree with you that this metric is more like a discovery of the desired line element rather than a metric that has some interesting physical interpretations beyond this. Maybe one could find a corresponding Lagrangian with a better physical motivation, than the magnetically one I cite in my answer, but that fact that g_{tt}=1/g_{rr} restricts even more the theories one could consider. $\endgroup$
    – Noone
    Jul 6, 2023 at 18:37

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