Three guys claim that any heavy chunk of matter emits Hawking radiation, even if it’s not a black hole:
• Michael F. Wondrak, Walter D. van Suijlekom and Heino Falcke, Gravitational pair production and black hole evaporation, Phys. Rev. Lett. 130 (2023), 221502.
Now they’re getting more publicity by claiming this means that the universe will fizzle out sooner than we expected. They’re claiming, for example, that a dead, cold star will emit Hawking radiation, and thus slowly lose mass and eventually disappear!
They admit that this would violate baryon conservation: after all, the protons and neutrons in the star would have to go away somehow! They admit they don’t know how this would work. They just say that the gravitational field of the star will create particle-antiparticle pairs that will slowly radiate away, forcing the dead star to lose mass somehow to conserve energy.
If experts thought this had even a chance of being true, it would be the biggest thing since sliced bread—at least in the field of quantum gravity. Everyone would be writing papers about it, because if true it would be revolutionary. It would overturn calculations by experts which say that a stationary chunk of matter doesn’t emit Hawking radiation. It would also mean that quantum field theory in curved spacetime can only be consistent if baryon number fails to be conserved! This would be utterly shocking.
But in fact, these new papers have had almost zero effect on physics. There’s a short rebuttal, here:
• Antonio Ferreiro José Navarro-Salas and Silvia Pla, Comment on “Gravitational pair production and black hole evaporation”, Phys. Rev. Lett. 133 (2024), 229001.
It explains that these guys used a crude approximation that gives wrong results even in a simpler problem. Similar points are made here:
• E. T. Akhmedov, D. V. Diakonov and C. Schubert, Complex effective actions and gravitational pair creation, Phys. Rev. D. 110, 105011.
Unfortunately, it seems the real experts on quantum field theory in curved spacetime have not come out and mentioned the correct way to think about this issue, which has been known at least since 1975. To them—or maybe I should dare to say “us”—it’s just well known that the gravitational field of a static mass does not cause the creation of particle-antiparticle pairs.
Of course, the referees should have rejected Wondrak, van Suijlekom and Falcke’s papers. But apparently none of those referees were experts on the subject at hand. So you can’t trust a paper just because it appears in a supposedly reputable physics journal. You have to actually understand the subject and assess the paper yourself, or talk to some experts you trust.
If I were a science journalist writing an article about a supposedly shocking development like this, I would email some experts and check to see if it’s for real. But plenty of science journalists don’t bother with that anymore: they just believe the press releases. So now we’re being bombarded with lazy articles like these:
• Universe will die “much sooner than expected,” new research says, CBS News, May 13, 2025.
• Sharmila Kuthunur, Scientists calculate when the universe will end—it’s sooner than expected, Space.com, 15 May 2025.
• Jamie Carter, The universe will end sooner than thought, scientists say, Forbes, 16 May 2025.
The list goes on; these are just three. There’s no way what I say can have much effect against such a flood of misinformation. As Mark Twain said, “A lie can travel around the world and back again while the truth is lacing up its boots.” Actually he probably didn’t say that—but everyone keeps saying he did, illustrating the point perfectly.
Still, there might be a few people who both care and don’t already know this stuff. Instead of trying to give a mini-course here, let me simply point to an explanation of how things really work:
• Abhay Ashtekar and Anne Magnon, Quantum fields in curved space-times, Proceedings of the Royal Society, 346 (1975), 375–394.
It’s technical, so it’s not easy reading if you haven’t studied quantum field theory and general relativity, but that’s unavoidable. It shows that in a static spacetime there is a well-defined concept of ‘vacuum’, and the vacuum is stable. Jorge Pullin pointed out the key sentence for present purposes:
Thus, if the underlying space-time admits a everywhere time-like Killing field, the vacuum state is indeed stable and phenomena such as the spontaneous creation of particles do not occur.
This condition of having an “everywhere time-like Killing field” says that a spacetime has time translation symmetry. Ashtekar and Magnon also assume that spacetime is globally hyperbolic and that the wave equation for a massive spin-zero particle has a smooth solution given smooth initial data. All this lets us define a concept of energy for solutions of this equation. It also lets us split solutions into positive-frequency solutions, which correspond to particles, and negative-frequency ones, which correspond to antiparticles. We can thus set up quantum field theory in way we’re used to on Minkowski spacetime, where there’s a well-defined vacuum which does not decay into particle-antiparticle pairs.
The Schwarzschild solution, which describes a static black hole, also has a Killing field. But this ceases to be timelike at the event horizon, so this result does not apply to that!
I could go into more detail if required, but you can find a more pedagogical treatment in this standard textbook:
• Robert Wald, Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics, University of Chicago Press, Chicago, 1994.
In particular, go to Section 4.3, which is on quantum field theory in stationary spacetimes.
I also can’t resist citing this thesis by a student of mine:
• Valeria Michelle Carrión Álvarez, Loop Quantization versus Fock Quantization of p-Form Electromagnetism on Static Spacetimes, Ph.D. thesis, U. C. Riverside, 2004.
This thesis covers the case of electromagnetism, while Ashtekar and Magnon, and also Wald, focus on a massive scalar field for simplicity.
So: it’s been rigorously shown that the gravitational field of a static object does not create particle-antiparticle pairs. This has been known for decades. Now some people have done a crude approximate calculation that seems to show otherwise. Some flaws in the approximation have been pointed out. Of course the authors of the calculation don’t believe their approximation is flawed. We could argue about that for a long time. But it’s scarcely worth thinking about, because no approximations were required to settle this issue. It was settled over 50 years ago, and the new work is not shedding new light on the issue: it’s much more hand-wavy than the old work.
, consider the following question:
be a finite subset of 
to the tiling equation 
, must there exist a periodic solution 
to the same equation 
? 
(trivially, since all sets are periodic in this case), one-dimensional groups 
with 
, but it can fail for 
for certain finite 
for sufficiently large 
; see 
tilings 
for a given natural number 
for some periodic function 
. 
and 
where 
are given integer-valued functions (with 
finitely supported).
. Here the results are very satisfactory:
to 
that is not identically zero. 
to 
that is not identically zero. 
for some non-trivial character 
of finite order.
, let 
.
. 
.
), but it also remains open whether Theorem 
that are one-periodic (periodic in a single direction). This already explains why tiling is easy to understand in one dimension, and why the two-dimensional case is more tractable than the case of general dimension. This structure theorem can be obtained by averaging a dilation lemma, which is a somewhat surprising symmetry of tiling equations that basically arises from finite characteristic arguments (viewing the tiling equation modulo 
for various large primes 
: if 
also solves the same equation 
. This freedom to take finite differences one to selectively eliminate certain one-periodic components 
to 
to convert a vanishing Fourier coefficient 
into an integer solution to 
) by polynomials of one variable. Applying a retraction homomorphism, one can make the coefficients of these polynomials rational, which makes the polynomials periodic. This turns out to reduce the original tiling equation 
at integers, so a certain amount of technical manipulation (in particular, passing at one point to a weak limit of the original tiling) is needed to avoid ever having to encounter this discontinuity.
are positive reals such that 
and 
, prove that 
.
and 
imply 
:
(implemented as 
and positive reals 
such that 
and 
, and the task is to conclude that 
:
that depends on one or more asymptotic parameters (such as 
) – for instance, whether the quantity 
equipped with a non-principal 
that describes the subsets of parameter space that are “sufficiently large” (e.g., the 
, or the cocompact filter on 
). We will use 
of the filter 
that are defined for sufficiently large 
, one can then define the following notions:
, 
, or 
(and sometimes 
) if there exists a constant 
such that 
for all sufficiently large 
, 
, or 
(and sometimes 
) if, for every 
, one has 
for all sufficiently large 
(and sometimes 
or 
) if 
, or equivalently if there exist constants 
such that 
for all sufficiently large 
instead of (iii). However, we will use the Hardy-Landau notation exclusively in this blog post.
on the real line. For instance, the symbols 
behave very much like 
, 
, 
, with properties such as the following:
, then 
. 
. 
is true.
, making asymptotic notation a kind of “
(or explicit decay rates) from the asymptotic notation. However, there are some cases in which the tradeoff is worthwhile. For instance, symbolic computations tend to be easier to perform in algebraic settings than in orthodox analytic settings, so formal computations of orders of infinity (such as the ones discussed in the 
, exactly one of the two statements “
to be the space of all non-negative functions 
defined for sufficiently large 
defined previously. That is to say, a nonstandard order of infinity is an equivalence class 
defined on elements 
itself is an order of infinity, as is 
, 
, 
, 
, and so forth. But we exclude 
; it will be important for us that the order of infinity is strictly positive for all sufficiently large 
. It is easy to see that this is well-defined, by verifying that if 
and 
, then 
. However, because of our positivity requirement, we do not define subtraction on 
and 
; we do not need to worry about division by zero, thanks to the positivity requirement. Again it is easy to verify this is well-defined. 
is a real number, we define 
. Again, this is easily checked to be well-defined. 
if 
if 
and 
is a totally ordered set. 
is a 
playing the role of the zero vector, multiplication 
playing the role of vector addition, and scalar exponentiation 
playing the role of scalar multiplication. (Of course, division would then play the role of vector subtraction.) To avoid confusion, one might refer to 
is a 
for all 
. 
for all 
. (It would therefore be natural to call ![\displaystyle N^{O(1)} := \bigcup_{k \geq 0} [\Theta(N)^{-k}, \Theta(N)^k]](https://s0.wp.com/latex.php?latex=%5Cdisplaystyle+N%5E%7BO%281%29%7D+%3A%3D+%5Cbigcup_%7Bk+%5Cgeq+0%7D+%5B%5CTheta%28N%29%5E%7B-k%7D%2C+%5CTheta%28N%29%5Ek%5D&bg=ffffff&fg=000000&s=0&c=20201002)
which are of polynomial type in the sense that 
for some 
that assigns to each order of infinity 
such that 
, that is to say for all 
for all sufficiently large 
of this surjection is then the log-subspace of quasilogarithmic orders of infinity – 
for all 
of non-empty closed intervals has a non-empty intersection, which is a property closely tied to the more familiar definition of completeness as the assertion that Cauchy sequences are always convergent. This claim of course fails for open intervals: for instance, 
for 
is a nested sequence of non-empty open intervals whose intersection is empty. However, in the nonstandard orders of infinity 
be a nested sequence of non-empty intervals in 
is non-empty. 
, although much the same argument would also work for closed or half-open intervals (and then by the pigeonhole principle one can then handle nested sequences of arbitrary intervals); we leave this extension to the interested reader.
of each 
, then we have 
whenever 
. In particular, one can find a set 
in the ultrafilter such that 
and 
. If we now define 
to equal 
for 
(and leave 
undefined outside of 
), one can check that 
for all 
, thus 
lies in the intersection of all the 
space); but for this discussion I will focus on the simpler situation of asymptotic estimates involving a finite number of positive real numbers, combined using arithmetic operations such as addition, multiplication, division, exponentiation, and minimum and maximum (but no subtraction). A typical inequality here might be the weak arithmetic mean-geometric mean inequality 
are arbitrary positive real numbers, and the 
here indicates that we are willing to lose an unspecified (multiplicative) constant in the estimates.
is comparable to 
up to constants, so it suffices to determine if 
; 
; and 
. Suppose for instance that 
, 
. The task of determining such a linear combination is a standard linear programming task, for which many computer software packages exist.
obeying the constraints
, 
, and 
are the maximum, median, and minimum of 
respectively, and similarly for 
, 
, and 
, and 
. This particular bound could be dispatched in three or four lines from some simpler inequalities; but it took some time to come up with those inequalities, and I had to do a dozen further inequalities of this type. This is a task that seems extremely ripe for automation, particularly with modern technology.
that involve complex expressions such as 
, as opposed to hypotheses that only involve atomic variables such as 
, 
, but I can at least handle such complex expressions in the left and right-hand sides of the estimate I am trying to verify.
. Another task would be to automatically verify the ability to estimate some multilinear expression of various functions, in terms of norms of such functions in standard spaces such as Sobolev spaces; this is a task that is particularly prevalent in PDE and harmonic analysis (and can frankly get somewhat tedious to do by hand). As speculated in that MO post, one could eventually hope to also utilize AI to assist in the verification process, for instance by suggesting possible splittings of the various sums or integrals involved, but that would be a long-term objective.
) do for our civilization that’s worth defending.
gates—
︎
due to a projection map 
, and some map 
from a third space 
into the base space 
of 
such that 
. In many applications we would like to have 
. So it is natural to impose the requirement that 
in the fiber of each point 
, and one can lift any 
. Conversely, to build a choice function for a surjective map 
to 
from 
, but the identity map 
cannot be lifted continuously up to 
be the same space 
through the “projection map” 
(that maps each point to itself) is if 
and complement 
are both closed, hence compact, subsets of 
is another CH space, which has an obvious surjective continuous projection map 
to 
. One easily checks that 
ot the second component; hence 
, and so 
and 
. Any continuous lift of the identity map 
, when projected onto 
. Let us call this surjection 
of 
of 
in 
must contain at least one fiber 
of 
, and by shrinking this neighborhood one can ensure the base point is arbitrarily close to 
. Thus, every open neighborhood of 
, contradicting the Hausdorff property.
is a homeomorphism, and one can lift the identity map to the inverse map 
. 
in 
for all 
and then 
are disjoint open sets that both have 
is not extremally disconnected, even though it is 
be a 
of 
) is an extremally disconnected CH space. 
of 
give a basis of the topology given by the clopen sets 
. Because the Boolean algebra is complete, we see that the closure of the open set 
for any family 
of sets is simply the clopen set 
, which obviously open, giving extremal disconnectedness. 
of 
equipped with the discrete topology, or equivalently the Stone dual of the power set 
(i.e., the ultrafilters on 
. The construction is the same, but instead of working with the entire power set 
be a bounded linear functional. Then there is a (unique) 
on 
-algebra
) such 
for all 
. 
is a continuous surjection between CH spaces. If the Riesz representation theorem is true for 
is an isometry, hence every bounded linear functional on 
, and hence by the Hahn–Banach theorem it extends to a bounded linear functional on 
is then a Radon measure on 
induces a finitely additive measure 
. This is in fact a 
matrix. So I’d rather learn about the many simplified models of nuclei people have created, which offer partial understanding… and lots of beautiful math.
is 1-dimensional. A spin-2 particle has 5, since the spin-2 representation is 5-dimensional.
symmetry! And part of why I got interested in this stuff was that it would be fun to see a rather large group like 
centered at the origin is 6-dimensional, because these ellipsoids are described by equations like
are small. If our nucleus were classical, we’d want equations that describe how these numbers change with time as our little droplet oscillates.
, and introduce corresponding momentum operators 
, obeying the usual ‘canonical commutation relations’:![[q_j, q_k] = [p_j, p_k] = 0, \quad [p_j, q_k] = - i \hbar \delta_{j k}](https://s0.wp.com/latex.php?latex=%5Bq_j%2C+q_k%5D+%3D+%5Bp_j%2C+p_k%5D+%3D+0%2C+%5Cquad+%5Bp_j%2C+q_k%5D+%3D+-+i+%5Chbar+%5Cdelta_%7Bj+k%7D+&bg=ffffff&fg=333333&s=0&c=20201002)
or the ‘Fock space’ on 
: the Hilbert space completion of the symmetric algebra of 
that we’re using to describe ellipsoids.
splits into two summands:
to our equation
and 
to get annihilation and creation operators for s-bosons and d-bosons. If we want, we can think of these bosons as nucleon pairs. But we don’t need that microscopic interpretation if we don’t want it: we can just say we’re studying the quantum behavior of an oscillating ellipsoid!
of ellipsoid shapes, and then form its cotangent bundle 
, and then quantize that and start studying ‘quantum ellipsoids’.
or 
become negative things go haywire: our ellipsoid can turn into a hyperboloid! The approach I’ve described is probably fine ‘perturbatively’, i.e. when 
