# Gravitational radiation from hydrogen atom [duplicate]

Background: It is known that orbital motion is accelerated motion under the influence of a central force; and that in a hydrogen atom the electron orbits the proton and is therefore accelerated; and that accelerated charges emit radiation; and that therefore the electron emits radiation, loses energy and falls into the proton; and that therefore hydrogen cannot exist. The paradox is resolved by saying that classical electrodynamics alone is not an adequate description of phenomena within the atom, so a quantum description is required. Or, to put it another way, hydrogen exists although it oughtn't, "because quantum physics". All this is uncontroversial.

The proton and electron form a mutually gravitating system, with the gravitational attraction between them being some $10^{-43}$ times the electrostatic attraction. Like any mutually orbiting masses, they must therefore emit gravitational radiation. Blindly following the formula in Wikipedia, the power radiated appears to be $7.4\times{10^{-188}}$ watts, or $4.6\times{10^{-169}}$ eV per second.

I don't have time to wait several times $10^{169}$ seconds to see how hydrogen atoms age, so I thought it would be quicker to ask here.

1. Do we genuinely predict that hydrogen will emit gravitational radiation at approximately this rate, or does the "because quantum physics" override apply to gravitational radiation just as it applies to electromagnetic radiation?

2. What would it even mean, in the context of a hydrogen atom, where electrostatic forces predominate, for the atom to lose energy by gravitational radiation? How would an old atom look different from a young one? (The absurdity of this question suggests that gravitational radiation may indeed be forbidden at the atomic level).

## marked as duplicate by ACuriousMind♦, Gert, CuriousOne, heather, honeste_vivereAug 6 '16 at 15:08

• The solution is exactly the same as in case of electromagnetism: in order for an excited state to decay into a lower energy state or the ground state, the atom has to interact with an external field that can absorb the angular momentum change. In case of electromagnetic fields that's a change by $\hbar$, for gravitons it would be $2\hbar$, if gravity was quantized in the naive way. Since the coupling to gravitational fields is much smaller than to electromagnetic fields, this probably happens at such a low rate that it would be unobservable, as the classical approximation shows. – CuriousOne Aug 2 '16 at 10:57
• You would simply get a different selection rule than for optical transitions, but it would be just that, except, of course, that there would be no observable photon. I think the more likely experiment would be to have a high energy graviton excite an atom, which then decays by emitting two photons of different energy. Having said that... you won't find enough matter in the universe for a $10^{160}$ atom detector... :-) That's 80 orders of magnitude short... not to mention that we don't have a 10eV (or so) graviton source! – CuriousOne Aug 2 '16 at 11:05
• BTW, apparently a duplicate of physics.stackexchange.com/q/92173/80649 or closely related. – LLlAMnYP Aug 2 '16 at 14:44
• The electron doesn't orbit the proton. There's no net motion between the proton and the electron. You seem to have things backward - quantum description is not a "fix" for a "paradox" - the paradox was only there because the model of the atom was wrong. When we found the correct model, the paradox no longer existed. Saying "hydrogen atoms shouldn't exist" is like saying "bumblebees shouldn't be able to fly". When the model is broken, you need a new model - you don't change reality. – Luaan Aug 2 '16 at 16:38
• People voting to close as a duplicate of physics.stackexchange.com/q/92173 : is there any chance we could close the other question instead? This question is of much better quality. – Nathaniel Aug 3 '16 at 6:48

As John Rennie says the hydrogen atom is a case where the electron is in the s shell, which means it has no angular momentum. We can think of the electron if it were measured as being at a point above the proton, where the electron as a wave would then spread into a spherical shape around the proton. One might imagine a sort of Zeno machine that keeps the electron wave function reduced so it remains at a point, or a set of points that hop around. If the Zeno effect is a set of measurements occurring at time intervals much shorter than the spreading time of the electron wave function then this hopping of the point can be minimized.

Atomic physics is a bit complicated. The wave function has a radial and angular part to it. The angular part defines the shells, s, p, d, f etc. These are also a series in multipole moments, s = spherical, p = dipolar, d = quadrupolar and so forth. So we need to consider the quadrupolar terms. This occurs with atoms in the transition metals, with scandium being the first. This would have a single electron in the outer d shell.

How would one then consider gravitational radiation produced by the quadrupole motion of a particle? This is a sketch of how to look at weak field gravitational radiation. A full treatment is a bit longer. We start with the metric $g_{\mu\nu}~=~\eta_{\mu\nu}~+~h_{\mu\nu}$ where $\eta_{\mu\nu}$ is the flat Minkowski background metric and $h_{\mu\nu}$ is the perturbation on that. We further look at the traceless components of this metric perturbation which we label as $\bar h_{\mu\nu}$. These traceless metric components have two non-zero terms $h_{ii}~=~A_+(t,r)$ and $h_{ij}~=~A_\times(t,r),~i~\ne~j$ for the components indicies $i,~j$ running over two spatial dimensions. These traceless metric components obey the inhomegenous wave equation $$\square\bar h_{\mu\nu}~=~-\frac{16\pi G}{c^4}T_{\mu\nu}$$ Now set $G/c^4~=~1$ for simplicity. These metric coefficients are then written according to a Green's function with $$\bar h_{\mu\nu}~=~-16\pi\int G_{\mu\nu}^{\alpha\beta}(t,{\bf r},t{\bf r}')T_{\alpha\beta}(t,{\bf r}')$$ We now expand the Green's function according to spherical harmonics and consider the quadrupolar terms the traceless metric terms are then approximately $$\bar h_{\mu\nu}^{\alpha\beta}~=~-4\int d^3r \frac{Q_{\mu\nu}(t-|{\bf r}-{\bf r}'|,{\bf r}')}{|{\bf r}-{\bf r}'|}.$$

That is the classical theory. We want to quantize this. We then have a wave function(al) of the form $\Psi[h]$. To make this simple we then consider this wave function(al) as expanded according to a radial and angular part. The $\frac{1}{|{\bf r}-{\bf r}'|}$ part of the metric means we will have a radial part similar to the Laguerre polynomials in atomic physics, and we then consider the quadrupole term as giving the Legendre polynomial term $Y_\ell^m(\theta,\phi)$ for $\ell~=~2$. We may then proceed with an atomic physics calculation, which below I will only gives a few pointers on.

The stress-energy term $T^{00}~=~\rho$ the energy density, is then expressed as $T^{00}~=~\hbar\omega/volume$. The coupling term for gravitation is $G/c^4$ and so there is the factor $G\hbar/c^4$ associated with the perturbation of the d shell due to gravitation. An atomic transition due to the emission of a graviton would be the emission of a spin-$2$ particle and the transition in $\ell~=~2$ to $\ell~=~0$, so the entire quadupole term is carried off by the graviton. This would have a coupling term $\sim~G\hbar/c^4$, which is $8.7\times 10^{-79}m-s$. This is very small.

Since this is computed for a quadrupole moment or the d shell, one would either have to work with excited hydrogen atoms that remain in that state long enough to perform measurements, or one has to work with a transition metal such as scandium. In the first case this would be tough to measure the perturbation of the d shell by gravitation in a time within the transition time for the atom to relax to the s-shell. If one works with a transition metal that problem is replaced by the fact the underlying electronic configuration will have a lot of complexity that needs to be computed the very high order in perturbation theory, such as Hartree-Fock method. Either way this is a tough call, but not absolutely impossible. I think working with higher Rydberg atomic states of a hydrogen atom would be most likely bear fruit.

The physics that is most likely relevant will then be the perturbation of the d shell by gravitational physics. This will be very small. There could of course be a transition that produces a soft graviton as presented by Weinberg, but the coupling is very small and the probability of such a transition in any reasonable period of time extremely small.

• Wouldn't transition metals not work either, simply because the underlying orbitals are all occupied? – LLlAMnYP Aug 2 '16 at 14:54
• The underlying orbitals of scandium are all s and p orbitals. Since these all have $\ell~=~0,~1$ there would not be gravity content. These would though form an underlying electronic complexity in a form factor that would have to known to an exceedingly high level of accuracy. The single d electron would then be a qudrupole orbit. – Lawrence B. Crowell Aug 2 '16 at 15:32
• You misunderstood: you say that an atomic transition emitting a graviton would cause an electronic transition from l=2 to l=0. But the d-electron in scandium cannot do that, because all lower lying states with l=0 are occupied by other electrons and we don't care about how much they complicate the physics, since they preclude any possible graviton-emitting transition. – LLlAMnYP Aug 2 '16 at 15:38
• @LLlAMnYP Does an excited state of a d-electron also have a qudrupole orbit? – OrangeDog Aug 2 '16 at 16:00
• @OrangeDog not if its $3d \rightarrow 4p$ – LLlAMnYP Aug 2 '16 at 16:18

It is known that orbital motion is accelerated motion under the influence of a central force; and that in a hydrogen atom the electron orbits the proton and is therefore accelerated

Your initial assumption is wrong. Electrons do not orbit atoms like a tiny planetary system. The angular momentum of the electron in the ground state of hydrogen is zero. So the electron is not in accelerated motion - in fact it is not in motion at all.

And this neatly explains why hydrogen atoms emit neither EM nor gravitational radiation. To emit gravitational radiation requires the mass distribution to have an oscillating quadrupole moment, and the electron density in hydrogen atom has a quadrupole moment that is not only not oscillating but is zero (the dipole moment is also zero, which is why it doesn't emit EM radiation either).

• I believe, the OP acknowledged that. However, we do observe electromagnetic radiation by excited hydrogen atoms relaxing into the ground state. So there is a dipole moment present at some point. Along the lines of comments by @CuriousOne, shouldn't we allow for a possibility of a quadrupole moment in an excited state? – LLlAMnYP Aug 2 '16 at 14:42
• @LLlAMnYP: my interpretation of the question is that it is about a possible decay of the ground state. – John Rennie Aug 2 '16 at 14:48
• Hmm, you seem to be right; the question is posed as such. – LLlAMnYP Aug 2 '16 at 14:54
• ...but this seems like a worthwhile extension! – Martin Kochanski Aug 2 '16 at 15:49

There is a good review of the possible transitions, and (to a minimal degree) their practicability in experiment, given in https://arxiv.org/ftp/cond-mat/papers/0208/0208276.pdf.