16
$\begingroup$

On earth and in our solar system we do not notice any effects of a non-zero cosmological constant. The accelerating expansion of the universe was only detected by observing the most distant supernovae.

There is no accelerated expansion of our solar system and our galaxy.

Is this, because the cosmological constant is so tiny and therefore gravity simply "wins the fight" against cosmological constant? Or is the cosmological constant locally, say in our solar system, zero and only in between galaxies non-zero?

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ Somewhere I've seen that the effective repulsive force goes as $\sim \Lambda r$ and therefore, only relevant at large cosmological scales. $\endgroup$
    – SRS
    Commented Jun 6, 2017 at 8:08
  • $\begingroup$ Related: physics.stackexchange.com/q/2110/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Commented Jun 6, 2017 at 11:50
  • $\begingroup$ I can't find the paper, but I recall reading you'd need an increasing cosmological constant (i.e., a scale factor with a non-zero third time derivative) to break apart gravitationally bound systems. $\endgroup$
    – Robbie Rosati
    Commented Jun 6, 2017 at 16:06
  • 1
    $\begingroup$ @RobbieRosati: For a circular orbit, the secular trend in the orbital radius is given by $\dot{r}/r=\omega^{-2}(d/dt)(\ddot{a}/a)$. See physics.stackexchange.com/questions/70047/… . So it's not exactly a third derivative, but close to that. The actual derivative expression $(d/dt)(\ddot{a}/a)$ vanishes exactly for a cosmology containing only dark energy. $\endgroup$
    – user4552
    Commented Jun 6, 2017 at 23:17
  • $\begingroup$ @BenCrowell Thanks for the reference, this seems to be what I was thinking of. $\endgroup$
    – Robbie Rosati
    Commented Jun 6, 2017 at 23:32

3 Answers 3

Reset to default
30
$\begingroup$

No. The cosmological constant is indeed a constant. It is what it is - both locally and globally. The fact that we don't notice the effects of a non-zero cosmological constant at small scales is certainly because the cosmological constant is very small. In principle, the non-zero cosmological constant does contribute to the expansion of the solar system as well. But certainly, the effect is ridiculously negligible at the scale of the solar system.

Improve this answer
$\endgroup$
5
  • 2
    $\begingroup$ It's not even really noticeable at the galactic level, or even the intergalactic level (galaxies in your relative neighborhood). It doesn't really become relevant until you're looking at separated galactic clusters separated by a few million light years. $\endgroup$
    – Shufflepants
    Commented Jun 6, 2017 at 21:03
  • 7
    $\begingroup$ The fact that we don't notice the effects of a non-zero cosmological constant at small scales is certainly because the cosmological constant is very small. This is not correct. No matter how big you make the cosmological constant, its effects become undetectable at small scales, because the fictitious force of expansion goes like $r\ddot{a}/a$, where $a(t)$ is the cosmological scale factor and $r$ is the distance between two particles whose relative acceleration we're measuring. It also doesn't make sense to say that the cosmological constant is small unless you say what you're comparing to. $\endgroup$
    – user4552
    Commented Jun 6, 2017 at 23:23
  • $\begingroup$ @Shufflepants We are separated from our nearest galaxy by a few million light years (two and a half), which I am sure would qualify as "relative neighbouthood". Did you mean a few hundred million light years or a few billion light years? $\endgroup$
    – Arthur
    Commented Jun 7, 2017 at 12:45
  • $\begingroup$ @BenCrowell The current best measurement of its value is 1.5*10^(-25). I'd call that small. I wouldn't call it big until we could see its effects in our own solar system. Very big when you could notice it on the surface of the planet :) $\endgroup$
    – Shufflepants
    Commented Jun 7, 2017 at 15:00
  • $\begingroup$ @Arthur Yeah, I should probably have said "a few hundred million". $\endgroup$
    – Shufflepants
    Commented Jun 7, 2017 at 15:01
11
$\begingroup$

As I have stated in my answer to a similar question here the cosmological model is not about bound states, and even galaxies are bound states by the gravitational attraction. Certainly bound states are presumed to overcome any distortions of space time; if it were not so all our cosmological observations of velocities which depend on identifying the Doppler shifted spectra of known elements come into doubt. Also the standard Big Bang model becomes nonsense: if space between bound particles expanded, with the cosmological constant , we would not be able to measure the expansion.

At the local level in order of strength, a proton is bound by the strong force, the hydrogen atom by the electromagnetic, the earth to the sun by gravitation. A look at the relative contribution to the space curvature by Λ (equation 1 here ) shows that Λ is order of 10^20 smaller than the contribution of the gravitational constant.

At the quantum level of the interactions binding atoms and molecules, Λ would appear as an extra dispersing potential, modifying the quantum levels measured . The smallness of the number locally ( look at the answer by Ben Crowell here) assures that within the widths of the bound states the effect is infinitesimal. It is similar to not considering the gravitational attraction between electron and proton when calculating the bound states of hydrogen. For the reason that the effect of changes in gravitational attraction are not seen , see this answer by Ben Crowel, .

So it is only at cosmological levels that the expansion driven by Λ can have a measurable effect, not at a local one.

Improve this answer
$\endgroup$
2
  • 1
    $\begingroup$ This is incorrect. See physics.stackexchange.com/questions/70047/… $\endgroup$
    – user4552
    Commented Jun 6, 2017 at 23:24
  • $\begingroup$ @BenCrowell I expanded and referenced your answers that give numbers $\endgroup$
    – anna v
    Commented Jun 7, 2017 at 3:42
0
$\begingroup$

The cosmological constant or dark energy is present everywhere and at all scales according to theory. (Experimentally we don't know that it is, though it certainly doesn't clump very much if at all.) But it doesn't, even in theory, cause a time variation of orbital parameters.

The easiest way to understand its effect is to think of it as matter with a uniform density of roughly $ρ_Λ \approx -10^{-26}\text{ kg}/\text{m}^3$ (or around −10 yoctograms per m3 if that's easier to remember).* Its effect on orbits is therefore (by the shell theorem) roughly equivalent to a decrease of the central mass by $\frac43 π r^3 ρ_Λ$ where $r$ is the radius of the orbit. Because this decrease doesn't vary with time, it doesn't have a time-varying effect on the orbit. It merely makes it a little smaller or a little slower than it would have been otherwise.

If $r = 1\text{ AU}$, then the decrease is about $10^8\text{ kg}$. To put this in perspective, the sun loses around $5\times10^9\text{ kg}/\text{s}$ in the form of light, neutrinos, and solar wind. That continuous loss actually should in principle cause a time variation in orbital parameters. The cosmological constant doesn't increase the rate of the change; it merely shifts it in time by a fraction of a second.

The radius at which the sun's mass is entirely canceled, meaning there are in theory no solar orbits beyond that radius, works out to about $100\text{ pc}$. In practice this is meaningless because it assumes that there are no other gravitating bodies within that distance of the sun.

* The dark energy density is actually positive. The reason this "effective density" is negative is that the acceleration in GR is proportional to $ρ+3p$, with $p=-ρ$ for dark energy. In Newtonian gravity only $ρ$ gravitates, but you can simulate the GR acceleration by multiplying $ρ$ by $-2$.

Improve this answer
$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.