Its often said QM is for the very small and GR for the very large. This brings to mind that there should be some limit at which one starts to apply and the other stops. Now I know there are more substantial differences, GR being a continuous description and QM using Planck length, and Planck time, and so on...

But keeping only to the distance scale difference, is there a hard limit where either theory suddenly stops working? Is there an overlap where both theories have similar predictive accuracy? Or is there a bad gap, a distance scale where neither theory is very useful?

Secondly, I imagine QM calculations can quickly become cumbersome when the number of particles increase. Is this image correct, is QM merely computationally impractical for considering planets and satellites etc. Are there other conceptual/practical problems, such as entanglement with something in another galaxy?

Edit: The suggested duplicate question asks whether a theory is quantum. This question asks whether an application of a theory (should) be QM or GR. I can now say from answers that the question is really more about gravity than the dimension of distance.

However it is also clear that although I tried to restrict it to a length scale, there are inextricable considerations. For instance, collision dynamics of very large molecules, large group behavior of particles, gravitational effects on quantum scales inside dense matter structures, etc.

The second part can be clarified: If all relevant particles could be accurately tracked, and all relevant interaction calculations performed, would QM give similar, or better, results over GR? (regarding the orbit of the moon aroundEarth)

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    $\begingroup$ What level of study are you at? $\endgroup$ Feb 10, 2019 at 18:13
  • $\begingroup$ Well, the Planck length is a good place where GR and QM clash: The BH horizon of a system recedes behind its Compton radius, so the black hole has "hair", a no-no. This is a scale inaccessible to both theories. $\endgroup$ Feb 10, 2019 at 23:38
  • $\begingroup$ There is no line. It is not about the scale, but about the nature of things. For example, in GR, light has energy and thus must produce gravity. However, in QFT, photons cannot decay by emitting gravitons, so light does not produce gravity regardless of the scale. $\endgroup$
    – safesphere
    Feb 11, 2019 at 3:38
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    $\begingroup$ Possible duplicate of What makes a theory "Quantum"? $\endgroup$
    – user191954
    Feb 11, 2019 at 6:34
  • $\begingroup$ @Chair maybe as a subset, but this is about the limit region where application of QM or GR transfer. $\endgroup$
    – christo183
    Feb 11, 2019 at 8:48

3 Answers 3


At any size, you must apply QM to have an understanding of the objects. Stars are huge, but the chemical reactions, radiation distribution, etc, are described exclusively by QM (And can be approximated by the rules of Chemistry, but GR is entirely irrelevant). When it comes to analyzing how objects such as stars will behave under the gravitational force, you can approximate GR by simply considering QM under some pressure exerted on the particles, and make predictions such as the radius of brown dwarfs based on that. (Such calculations rely heavily on QM such as Pauli Exclusion, which remains the leading repulsive force that fights gravity for all objects that don't have fusion - Even you would be much tinier without Pauli Exclusion! Fusion is still described by QM though. It's all QM)

GR only really applies when calculating the trajectory of large objects, or gravitational lensing. It doesn't have to do with size. A small black hole changes the trajectory of tiny photons, even if a larger neutron star requires QM to accurately describe. Saying "small QM, big GR", is missing a lot of what goes into which theory to use where.

An interesting consideration is the Plank Epoch, or more accurately any temperature above $10^{34}$ Kelvin, since then GR would actively affect QM highly significantly, in a way that current physics cannot predict. But that's irrelevant, since we can't answer questions even simpler than that. Our knowledge of QM doesn't work at the much more forgiving $10^{27}$ Kelvin, even if GR wouldn't be relevant at that level, since we do not know how electroweak and the strong force might merge.

  • $\begingroup$ I do not think I have ever considered stellar nucleosynthesis as chemistry. $\endgroup$
    – Kyle Kanos
    Feb 12, 2019 at 10:59
  • $\begingroup$ @Kyle Kanos I was taught it in chemistry, so idk. Same with absorption lines, and other properties of matter which are dependent on QM (And are properties of large mass "GR" objects). Of course it's only skimming the surface, which is why chem got a footnote and QM was mentioned a dozen times. $\endgroup$ Feb 19, 2019 at 4:22

Our current understanding of GR (that is, how gravity works) has been tested by observation from length scales typical of galaxy superclusters down to distances of a few centimeters. If there is a small-distance cutoff for GR (as probed for by experiment) we haven't found it yet.

Our present understanding of QM was derived by experiment and observation is that it correctly describes atoms and nuclei and its predictions blend into those of newtonian dynamics for length scales of order ~bigger than an atom. So in this case, there is an effective crossover from one description of nature to the other.

QM is not only computationally impractical for describing the behavior of planets and people, but it is also unnecessary in the sense that the bigger the object becomes, the smaller the effects of QM become, until you are left with newtonian picture in the large-object limit.


I would say that quantum is for small numbers of particles (which usually implies small length scales, but there are a few exceptions), while GR becomes important when one considers a large curvature of spacetime.

It is reasonable to expect one theory to smoothly merge into another other at some relevant scale. In my opinion your question is similar to asking 'when does thermodynamics begin?' (e.g. how many molecules do we need in a system before we can start speaking of thermodynamic quantities like pressure, temperature, etc?)

Generally you would not want to treat problems like planetary orbits quantum mechanically since a planet is made of many particles, and calculating $10^{\text{many}}$ Schrodinger equations would be incredibly unpleasant. Perturbing a handful of atoms in the Earth won't really affect the Earth's orbit around the Sun in any observable way. The classical physics description of orbits is 'good enough' for any observable that I can think of.

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    $\begingroup$ "I would say that quantum is for small numbers of particles" how is that so? $\endgroup$ Feb 10, 2019 at 18:23
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    $\begingroup$ super conductors cover kilometers and are described by quantum mechanical equations $\endgroup$
    – anna v
    Feb 10, 2019 at 18:50
  • $\begingroup$ QM is not only for small numbers of particles - even ignoring exotic things like superfluid helium ordinary metals require a QM description precisely because there are many electrons! $\endgroup$
    – jacob1729
    Feb 10, 2019 at 19:19
  • $\begingroup$ I agree that QM is 'not exclusively for small numbers of particles' (I did say there are exceptions to this....such as in a BEC, and the double slit experiment has been performed on some very large molecules). However, imo it is useful to begin by saying that QM should deal with small numbers of particles. Considering a single particle allows the quantum behaviour (and necessity of QM) to be very evident. Macroscopic objects are made of many particles that all follow the rules of QM, and there will be cumulative effects on macroscopic objects that are quantum mechaincal in nature. $\endgroup$
    – Bob
    Feb 10, 2019 at 20:29

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