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From cosmological models that involve expansion of the universe, can we not say that there are ever increasing tidal forces felt by solid bodies?

If so, the material in solid bodies like metal blocks, glass rods, skeletal systems will tend to separate causing a strain on the body which counteracts the expansion

If the tidal forces increase with time, will there not come a point in time, when the stress limit is reached and our bodies begin to shatter under intense tidal forces (similar to 'spaghettification')?

If the above is true, using present cosmological data, has anyone calculated how long it will take before our bones shatter?

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As discussed in more detail in this answer, for two test particles released at a distance $\mathbf{r}$ from one another in an FRW spacetime, their relative acceleration is given by $(\ddot{a}/a)\mathbf{r}$. This is observed as an anomalous tidal force. Some people will insist that there is no such effect, but this is simply wrong. A good discussion is given in Cooperstock 1998. What is incorrect is to imagine that bound systems expand in proportion to the FRW scale factor $a$.

The factor $\ddot{a}/a$ is on the order of the inverse square of the age of the universe, i.e., $\sim H^2$, the square of the Hubble constant. So let's say we want to estimate the strain in your thigh bone due to cosmological expansion. The length of the bone is $L$, so the anomalous acceleration of one end of the bone relative to the other is $\sim LH^2$. The corresponding tension is $\sim mLH^2$, where $m$ is your body mass. The resulting strain is

$$ \epsilon \sim \frac{mLH^2}{AE} \qquad , $$

where $E$ is the Young's modulus of bone (about $10^{10}$ Pa) and $A$ is the bone's cross-sectional area. Putting in numbers, the result for the strain is about $10^{-40}$, which is much too small to be measurable by any imaginable technique --- but is not zero! I believe the sign of $\ddot{a}$ is currently positive, so this strain is tensile, not compressive. In the earlier, matter-dominated era of the universe, it would have been compressive. There is no "secular trend," i.e., your leg bone is not expanding over time. It's in equilibrium, and is simply elongated imperceptibly compared to the length if would have had without the effect of cosmological expanson.

If the tidal forces increase with time, will there not come a point in time, when the stress limit is reached and our bodies begin to shatter under intense tidal forces (similar to 'spaghettification')?

These tidal forces are not expected to increase significantly over time. If dark energy is really described by the equation of state of a cosmological constant, then in the vacuum-dominated epoch of the universe, which we are now entering, the tidal forces approach a constant (because $\ddot{a}/a$ approaches a constant). This would not be true, however, in a Big Rip scenario.

Cooperstock, Faraoni, and Vollick, "The influence of the cosmological expansion on local systems," http://arxiv.org/abs/astro-ph/9803097v1

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    $\begingroup$ Clearly, for all practical purposes the expansion of "metal blocks, glass rods, skeletal systems" is completely negligible. So I don't think you are right to say " Some people will insist that there is no such effect, but this is simply wrong." I think it would be better to say " Some people will insist that there is no such effect, and this is essentially correct as the effect is completely negligible, although strictly speaking not exactly equal to zero". $\endgroup$
    – Virgo
    Sep 10, 2013 at 1:38
  • $\begingroup$ No, @Ben, as physicsphile says, the forces are negligible for all bound systems. But for atoms, molecules, solids, glass rods, and skeletal systems, the claimed stretching is exactly zero. For example, there is nothing such as a hydrogen atom with radius equal to 1.00001 times what we are used to. The atomic radius as a proper distance is totally determined by local physics and nothing in cosmology can change anything about it. What cosmology could do in general is to discontinuously ionize or rip atoms or their conglomerates, but this can't be measured by "stress" and is very unlikely. $\endgroup$
    – Luboš Motl
    Sep 10, 2013 at 4:51
  • $\begingroup$ I must also explicitly add that the 1998 paper cited in the completely wrong answer above (that some people still seem to enthusiastically upvote) doesn't apply to atoms, molecules, solids, glass rods, and skeletons which are non-gravitationally bound and therefore their defining proper lengths don't change by the effects of cosmological expansion, not even by an epsilon. They only discuss gravitationally bound local systems (orbiting celestial bodies) for which the expansion is negligible but in principle nonzero. $\endgroup$
    – Luboš Motl
    Sep 10, 2013 at 4:57
  • $\begingroup$ @LubošMotl: Since you've made similar comments on both my answer and your own answer, I'll continue the discussion under your own answer. $\endgroup$
    – user4552
    Sep 10, 2013 at 4:59
  • $\begingroup$ @physicsphile: I agree with all the objective statements in your comment. We disagree only on style and emphasis. $\endgroup$
    – user4552
    Sep 10, 2013 at 5:26
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This is really just a footnote to Luboš's answer, but for completeness (and because it's fun :-) we should note that the equation of state of dark energy has not been determined and it remains possible that the ratio between the dark energy pressure and its energy density is less than or equal to -1. If so, this is known as phantom energy, and it causes an ever increasing acceleration leading eventually to a Big Rip and the destruction of, well, everything!

In this scenario there is indeed a growing stress on anything with a finite size, and our bones will indeed shatter. Fortunately this seems a remote possibility.

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  • $\begingroup$ @dj_mummy: for Schwarzschild-De-Sitter space, this is governed by the dimensionless parameter $\Lambda M^{2}$, where $M$ is the mass of the central body. So, given a fixed $\Lambda$, you'll be able to find some $M$ such that $\Lambda M^{2}$ still lies in the range where orbits are allowed. $\endgroup$
    – Zo the Relativist
    Sep 9, 2013 at 19:20
  • $\begingroup$ Right. And the effect is nonvanishing even when the equation of state is a cosmological constant rather than the type that would cause a big rip scenario. $\endgroup$
    – user4552
    Sep 9, 2013 at 21:08
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There is no stress whatsoever trying to disrupt solid bodies – or bound states of any sort – caused by the expansion of the Universe.

The individual atoms or pieces of solid bodies are arranged to minimize the total energy and have stable relative positions for that reason. Equivalently, it is possible to parallel transport a body in the timelike direction, along the world line, and that's exactly what these bodies are doing as they evolve in time.

This becomes particularly clear in the de Sitter space – the exponentially accelerating expansion caused by the positive cosmological constant, a phase that we have been entering in recent billions of years. The isometry of $dS_4$ is $SO(5,1)$ which also includes a boost-like generator that plays exactly the same role as the time translations in the flat Minkowski space. So this isometry tells you how bound states evolve in time.

Your assumption reflects a widespread misconception about what is actually expanding. What is expanding is the Universe itself, not the size of the objects. The size of atoms, molecules, and even planets etc. stays the same which really means that the expanding Universe is able to harbor an increasing number of atoms, molecules, and/or planets. It is really expanding. It's not just some vacuous change of units that wouldn't change anything material.

Whether the distance between two objects is increasing as the result of the expansion of the Universe depends on what determines their location. If they're just "attached" to some regions of space, like galaxies, they will expand with the space itself. But bound objects' molecules or components have positions determined by the equilibrium of various forces, especially attractive forces, acting inside them. So they're not attached to "independent regions of space" which is why the distance between them isn't increasing, surely not by the same factor as the factor that stretches the distances between galaxies.

Some intermediate situations, like clusters of galaxies that are "partly/loosely bound", would deserve a special discussion. They may expand a bit and it's calculable how much. However, it's important that the systems dominated by the attractive binding forces, e.g. electromagnetic forces that keep solid matter connected, surely don't suffer from the same rate of expansion as the Universe itself. At the same moment, I have to emphasize that the "cosmological stretching" of atoms, molecules, solids, glass rods, and skeletons is exactly zero because all these proper distances are fully determined by local physics governed by non-gravitational forces. For example, there is nothing such as a hydrogen atom that is 1.00001 times the usual radius and the same observation holds for the other tightly bound states, too. This is elementary quantum mechanics. Some people and some papers may err about this basic point but they will never change the radius of the atom or other tight bound stats.

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  • $\begingroup$ Small addition--it turns out that if you have a cosmological constant present in the universe, then orbits around a central massive body, in addition to having an innermost stable circular orbit, will also have an outermost stable circular orbit, beyond which they will be pulled into the surrounding cosmology. $\endgroup$
    – Zo the Relativist
    Sep 9, 2013 at 15:10
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    $\begingroup$ Nope, this is wrong, for the reasons given in my answer. Some intermediate situations, like clusters of galaxies that are "partly bound"[...] may expand a bit and it's calculable how much. Yes, this is right. However, it's important that the systems dominated by the attractive binding forces, [...] surely don't suffer from the same rate of expansion as the Universe itself. This is also true, but the effect is nonvanishing. This sentence also contradicts the first sentence of the answer, which asserts that such stresses vanish identically. $\endgroup$
    – user4552
    Sep 9, 2013 at 20:56
  • $\begingroup$ Dear @dj_mummy, the interatomic distances in molecules or solids (and electron-nucleus distances in atoms) aren't increasing at all because they're determined by the completely local laws of quantum mechanics. There is nothing such as an atom with the radius 1.0001 of the normal hydrogen atomic radius, for example, so at least for atoms, molecules, and solids which are completely bound, Ben's comments are totally wrong, not just morally wrong. $\endgroup$
    – Luboš Motl
    Sep 10, 2013 at 4:48
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    $\begingroup$ Hi, Lubos -- Perhaps this would be a nice time to sit back and appreciate the sheer silliness of engaging in a spirited debate over whether a certain effect is $10^{-40}$ or exactly 0. Much more fun than debating the number of angels that can dance on the head of a pin. Anyhow, I've pointed out that you've contradicted yourself on a number of points, so you might want to start off by editing your answer and sorting out exactly what your claims are. If I'm understanding your current iteration correctly, you seem to think that the effect vanishes for individual hydrogen atoms, [...] $\endgroup$
    – user4552
    Sep 10, 2013 at 5:06
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    $\begingroup$ It is a pretty straightforward exercise to evaluate the gravitational corrections to various atomic properties. Just write the general Dirac equation in curved spacetime, plug in Schwarzschild-de Sitter for the metric, take nonrelativistic limits etc etc etc. The effect does not vanish identically, but you will never ever be able to measure it. The point is it is computable and nonzero. No arm waving or arguing required. $\endgroup$
    – Michael
    Sep 10, 2013 at 5:13

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