Is an atomic nucleus dense enough to cause significant bending of the spacetime?

Question

We know that an atomic nucleus is very dense, but is it dense enough to create some interesting relativistic effects?

Obviously, in the classic limit, where we just assume that nucleons are little balls.

Answer 1

The density of nuclear matter is on the order of $2 \times 10^{17}\, {\rm kg/m^3}$, while the Planck density--the density at which quantum gravity is "interesting"--is:

$$\rho_P = \frac{m_P}{l^3_P}= \frac{c^5}{\hbar G^2} \approx 5 \times 10^{96}{\rm kg/m^3}$$

70 orders of magnitude larger. So it may be safe to assume gravity is no big deal.

One could also calculate the Schwarzschild radius of a proton and find that it is 35 orders of magnitude smaller than the proton radius.

Answer 2

Gravity isn't dependent solely on density, but on both density and size. A very dense object that is very, very small - like an atomic nucleus - may not have a lot of gravity.

To see this, note that before you reach for General Relativity, you should first reach for Newtonian gravity. And in Newtonian gravity, we have the formula for the magnitude of gravitational field, $g$,

$$g(r) = \frac{GM}{r^2}$$

for a spherically symmetric mass $M$ and distance $r$ from the center and exterior. We can rephrase this with density by noting $M = \frac{4}{3} \pi r^3 \rho$, where $\rho$ is the density:

$$g(r) = \frac{4\pi G \rho r}{3}$$

so you see that the gravity decreases linearly with radius, even if the density is still the same. Thus while a nucleus and, say, a neutron star, may have similar density, since a nucleus is on the order of $10^{18}$ times smaller in radius, its gravitation is on the order of $10^{18}$ times weaker - in fact, you can calcultae that for a uranium nucleus, the "surface gravity" should be only about 1.05 µm/s^2, that's microns per second per second, or equivalently 1.05 m/s/Ms - that is, to gain just the speed of a slow shuffle, you'd have to be under this gravitational influence for a full #megasecond - 11 days & 49.6 kiloseconds.

If the Newtonian gravitation is that weak, then given that general relativity very closely approximates it in this case, you can pretty much say the "bending of space-time" is practically nil, so the answer to your question is no, it doesn't.

ADD!: The commenter below gave a very good point about the fact that relative to the size of the nucleus, this is actually a substantial gravitational pull. Which made me decide to investigate the question using General Relativity after all. And we actually don't need much - since a nucleus is typically fairly spherical, we can approximate its gravitation via the Schwarzschild metric for a point "just outside" of it. And if we do that, a good measure of "curvature of spacetime" in a single number - since more generally we have a 4x4x4x4 Riemann curvature tensor, which gives more detailed directional information thereabout - is the Kretschmann scalar

$$K = R_{abcd} R^{abcd} = \frac{48 G^2 M^2}{c^4 r^6}$$

and it turns out that - surprisingly - if you use the mass and radius of a uranium nucleus again (395 yg, 5 fm) you get that $K \approx 2.65 \times 10^{-16} \mathrm{m^{-4}}$, but for the surface of the Earth, with mass 5972 Yg and radius 6.37 Mm, we get $K$ is approximately $1.41 \times 10^{-44}\ \mathrm{m^{-4}}$, or almost $2 \times 10^{28}$ times smaller! In fact, the $K$ we obtained from the $^{238}_{92}\mathrm{U}$ nucleus is almost exactly the same as the scalar at the surface of a neutron star of radius 10 km and mass 1.4 $M_\odot$ ($2.8 \times 10^9\ \mathrm{Yg}$)!

Moreover, note that if you do the algebra as in the previous section substituting appropriately, you find that in general $K$ depends only on the average density of the material in the enclosed shell of [Schwarzschild coordinate] radius $r$ [heuristically: note $M^2$ gives you $r^6$ on the numerator]. That is to say, "gravity" in accelerative sense doesn't solely depend on density, but spacetime curvature does.

That said (ADD 2, thanks @Ghoster!), note that this Kretschmann scalar can be thought of as a sort of "volumetric curvature" - note that the units are $\mathrm{m}^{-4}$, or per space-time volume, so curvature over one quartic meter of space-time volume. In particular, while the curvature in a unit volume is the same as a neutron star, a nucleus has far less volume - on the order of $10^{-45}\ \mathrm{m}^3$ spatially and around $10^{-18}\ \mathrm{s}$ (the timescale of strong interactions) temporally, for a total of $10^{-71}\ \mathrm{m}^4$. Much like how that the Earth has a radius of curvature of about 6370 km, but a human is only 1.69 m in size, and hence at scales around a human the flat Earth model works well, so too, this is completely negligible in terms of total curvature.

Hence, the conclusion is a bit more complicated than the prior two I gave: the result from Newton's gravity and GR thus agree that it does not cause significant total "bending", but each unit of 'stuff' is "just as powerful" a bending "agent" for a nucleus as for a neutron star.

ADD 3!: Aand as @KP99 suggests, yet another wrinkle is put into place regarding the fact that the nucleus has charge. They link to a paper with a complicated(!) expression for $K$, which you can follow on for the link. For the uranium case, we fortunately(!) have spin zero, so $a = 0$, but the charge $Q = +92\ \mathrm{e} \approx 14.7\ \mathrm{aC}$. The difference is not large - but the more accurate value, taking into account the charge, is about 2% smaller than the one quoted above.

(Note that this paper apparently uses normalized units [$G = c = 1$] but I am not sure of the normalization used for charge; the calculation above assumes the Planck normalization, i.e. $\frac{1}{4\pi \epsilon_0} = 1$, but the author doesn't say which they use; I assume this due to the absence of factors of $4\pi$ hanging around. It doesn't drastically change the result, but don't trust that 2% figure exactly.)

Answer 3

The Einstein field equations for general relativity has on the right hand side an object, $T_{\mu\nu}$ called the stress-energy tensor. The $T_{00}$ component of this object is the energy (density). The LHS of these equations has what is called the Einstein tensor $G_{\mu\nu}$ and represents spacetime curvature$^1$. One may conclude that since a nucleus has energy (mass), then the spacetime around it should curve.

The problem with this assertion is that we are not sure that general relativity works at such small scales. It is possible that the laws of gravitation breakdown at the quantum level. That is to say, we have no consistent quantum theory of gravitation at all energies and scales.

But as pointed out in comments, we do have effective field theories, or "approximations" to QG, that are well-behaved and produce consistent results only at "normal" energies. In the realm of very high energy physics, the same is not true.

$^1$ This tensor has the form $${\bf G_{\mu\nu}}={\bf R_{\mu\nu}}-\frac 12 {\bf g_{\mu\nu}}R$$ where $\bf R_{\mu\nu}$ is the Ricci curvature tensor, $R$ is the Ricci scalar curvature, and $\bf g_{\mu\nu}$ is called the metric tensor.

EDIT: @ohwilleke has correctly pointed out I have misinterpreted the intent of the OP, and that this question is about whether a nucleus is dense enough to cause "interesting relativistic effects", and to summarize, rather than delete this answer, we know that quantum gravitational effects become discernable at the Planck scale. So, we can use the Planck mass, and Planck length to get a measure of the density at which such effects are noticeable. Since your question also assumes that the nuclei are spheres (which is not accurate from quantum mechanical perspective), we can compute $$\text{Planck density}=\frac{m_p}{\frac 43 \pi(\frac{l_p}{2})^3}$$ which gives a number $\text{~} 10^{94}\ kg \ m^{-3}$ We know that the density of a nucleus is of the order $\text{~}10^{16}\ kg \ m^{-3}$, we can safely conclude that there will be no "interesting" relativistic effects.

Answer 4

The density of a mass $m$ distributed over space with scale parameter $r$ goes like

$$ \rho \sim \frac{m}{r^3} $$

(For example, a spherical distribution replaces the squiggle with a factor of $4\pi/3$.) Meanwhile, the Schwarzchild radius goes like

$$ R_\text{grav} \sim \frac{GM}{c^2} $$

These relationships mean that density alone isn't enough to ask whether gravitational effects are significant: the total mass, or equivalently the size of the dense distribution, is important as well.

For nuclear matter, at our one-squiggle factor-of-a-few precision, an atomic nucleus has the same density as a neutron star. Let's replace one of our two variables $M,R$ with the constant density $\rho$:

$$ \rho\sim \frac{M}{R^3} \sim \frac{R_\text{grav}}{R^3} = R^{-2} \qquad\text{or}\qquad R_\text{grav} \sim \rho^{-1/2} $$

You frequently find this statement turned around. If we pretend like a black hole is a constant-density sphere whose surface is its event horizon (holding our nose to discourage relativists from coming along to explain why this idea is flawed), we find that large black holes are less dense than small black holes. Pop-science descriptions of black holes will get into "death by spaghettification," and point out that a small black hole will spaghettify passersby outside the event horizon, while a large black hole won't exert significant tidal forces until after the event horizon has been crossed.

A careful answer to your question is that objects with the same density as a nucleus can produce significant spacetime-curvature effects. But that doesn't mean that a nucleus produces significant spacetime curvature: you need a solar mass worth of nuclear matter, compacted into the size of a city.

Answer 5

There is plenty of good information in existing answers, but I think we can add something directly relevant to the question. The question asks about "significant bending" so we have to quantify how much curvature would be "significant". I would like to translate this into the question "if we use Euclidean geometry to relate distances, areas, and volumes near an atomic nucleus, how accurate will our calculations be?"

Here is a precise way to answer this. Well, precise if you will allow me to model a nucleus as a sphere of uniform density.

For a sphere of uniform density one can derive from the Einstein field equations the following useful result: $$ r - \sqrt{A/4\pi} = \frac{G M}{3 c^2} $$ where $M$ is the mass of a uniform sphere of radius $r$ and surface area $A$. To be precise, $r$ here is the distance from the centre to the surface of the sphere, as indicated by a set of standard rulers laid end-to-end, and $A$ is the surface area as indicated by area measurements using standard rulers distributed over the surface of the sphere. Euclidean geometry would give $A = 4 \pi r^2$, general relativity says $A \ne 4 \pi r^2$. The above formula gives the leading order correction to the Euclidean result. It is called a radius excess, which is to say, how much more radius the sphere has compared with what you might have guessed from its surface area. Alternatively, one can say that the surface area is a little smaller than one might have guessed from the radius.

Note that $$ G / 3 c^2 \simeq 2.5 \times 10^{-28} \mbox{metres per kilogram} $$

I will put numbers in for three examples: planet Earth, a neutron star, and an atomic nucleus. $$ \begin{array}{lll} \mbox{object} & \mbox{mass} & \mbox{radius excess} \\ \mbox{Earth} & 5.97 \times 10^{24}\; {\rm kg} & 1.48\; \mbox{millimetres} \\ \mbox{neutron star} & 3 \times 10^{30}\; {\rm kg} & 743\; \mbox{metres} \\ \mbox{neutron} & 1.67 \times 10^{-27}\; {\rm kg} & 4 \times 10^{-55}\; \mbox{metres} \end{array} $$ You can compare these results with the radius of the object in question: about 6400 km for Earth, about 10 km for a neutron star, about $0.8$ fm for a neutron.

In terms of missing area (i.e. $4 \pi r^2 - A$) we get $$ \begin{array}{ll} \mbox{Earth} & 0.24\;{\rm km}^2 \\ \mbox{neutron star} & 194\;{\rm km}^2 \\ \mbox{neutron} & 8 \times 10^{-69}\;{\rm m^2} \end{array} $$ Thus the missing area for a neutron star is about 15% of the surface area. The missing area for planet Earth is about $10^{-9}$ of the area, but small as it is this is enough to make 33 football pitches. The missing area for a neutron is about $10^{-39}$ of the area, so this is negligible.

Since the radius excess is proportional to the mass (for a uniform sphere) you can easily calculate results for a nucleus of your choice. A heavy nucleus such as uranium-238 has a mass of 238 atomic units. One finds that for such a nucleus the missing surface area is negligible compared to the surface area. Therefore, on this measure, the atomic nucleus does not cause significant spacetime bending.

Answer 6

While the effects may be weak (see other answers for dimensional analysis arguments), the gravitational structure of nucleons is a field of active research. As you know, each nucleon is composed of three quarks, either two ups and a down for the proton or two downs and an up for the neutron. In either case, the bare mass of the constituent quarks accounts for only about 10% of the total nucleon mass. The fundamental theory of the strong interactions, quantum chromodynamics (QCD), becomes strongly coupled at low energies, and so it is exceptionally difficult to compute observable quantities of the nucleons from their constituent quarks. For example, only in recent years has it become computationally feasible to compute the nucleon axial coupling constant $g_A$ (a crucial quantity for predicting the lifetime of a free neutron) to within 1% accuracy using lattice QCD [1].

The quantity that is relevant to gravity is the energy-momentum tensor $T^{\mu\nu}$. When probed at low energies—relative to the Planck energy—there is no conceptual issue in defining this quantity for hadronic systems (one of the other answers is wrong on this point). In general, $T^{\mu\nu}$ receives contributions from gluons, valence quarks, and sea quarks within the target nucleon.

Knowing that parity and time-reversal are symmetries of QCD and that the proton and neutron are each spin-$\frac{1}{2}$ particles, we can expand matrix elements of the energy-momentum tensor in terms of three form factors corresponding, respectively, to momentum $A(t)$, angular momentum $J(t)$, and pressure and shear forces $D(t)$ [2]. Each form factor depends only on the magnitude of the four-momentum-transfer $t = (p'-p)^2$. At $t=0$, the momentum form factor reduces to $A(0)=1$, corresponding to the invariant mass, and the angular momentum form factor reduces to the total spin $J(0)=\frac{1}{2}$. Only the value of the shear and pressure term $D$ is generically unconstrained at $t=0$. In contrast to $A(0)$ and $J(0)$, $D(0)$ is not related to any external properties of the nucleon. Constraining these form factors (at $t=0$ and at nonzero $t$) provides crucial information about the internal structure of nucleons. For example, one interesting observable is the mean-square-radius of the energy density \begin{equation} \begin{split} \langle r^2\rangle_E &\equiv \frac{\int d^3r\;r^2\;T_{00}(r)}{\int d^3r\;T_{00}(r)}\\ &=6A'(0)-\frac{3}{2m_N^2}D(0) \end{split} \end{equation} Analogous to the proton charge radius, this quantity tells us how matter/energy is distributed within a nucleon.

To relate this back to the original question, if one knows the energy-momentum tensor $T^{\mu\nu}$ of a nucleon then one can calculate the resulting spacetime curvature using Einstein's field equations. As others have pointed out, these gravitational effects will be extremely weak relative to the electromagnetic, strong and weak nuclear forces. The later three forces all make major contributions to the physics of nucleons. Although the gravitational effects are weak, the energy-momentum tensor of a nucleon contains valuable information about its internal structure.

Answer 7

Note: please see article below on the Aharonov-Bohm effect revealing strong experimental support for quantum gravity.

I think the easiest way to approach this is to first acknowledge that the density of a bare nucleus must be relatively close to the density of a neutron star. This is true because the neutron star maintains its size as the result of neutron degeneracy pressure. A neutron star is likely to be slightly more dense than a free neutron because of gravity.

A quick search shows that neutron stars can cause gravitational lensing, but I think in the case a single nucleon it would be very difficult for anyone to be able to observe similar gravitational effects because the scattering of one or more photons on a single nucleus can be very complicated.

To the point about quantum gravity, there is no sure way using current approaches to ensure you captured all the potential interaction diagrams to perform a calculation with sufficient precision to test for the effects of gravity.

So in short, while the density of the nucleus is high enough to consider bending of light near the nucleus, the "near the nucleus" part is very muddy and there is no clear way to account for all the quantum effects that would tend to dominate the effects of gravity (I suppose I should add, "at least at low energies")

Neutron Degeneracy Pressure https://www.nustar.caltech.edu/page/neutron_stars

Effects of gravitational potential, in this case they were able to observe this effect using atoms in a gravitatonal potential, it is a slightly different set up but it is one of the strongest peices of evidence supporting quantum gravity to date - Observation of a gravitational Aharonov-Bohm effect - https://www.science.org/doi/10.1126/science.abl7152

https://bigthink.com/starts-with-a-bang/quantum-gravity/