1
$\begingroup$

First I must let you know that I don't have much understanding of neither GR nor quantum mechanics, and therefore this question.

I've mentally pictured Newtonian physics, GR and quantum mechanics all somewhat competing (in terms of use case) physics models. For one situation Newtonian physics model may be sufficient, but for another (e.g. interplanetary calculations) general relativity is needed. And then there's quantum mechanics.

Is it possible that the exact same exercise can be solved in Newtonian physics, GR and quantum mechanics? Under "solved" I mean that it can be calculated according to the model (Newtonian, GR or QM) but the answer would come different and hence illustrating the need of why and for what GR was required and the same for QM.

Basically what I mean is that is there any problem that could be solved with different physical models? For example if I would have a problem A, then by applying a solution based on Newtonian physics (NP), GR or QM, I would get different results, e.g:

NP(A) = x
GR(A) = y
QM(A) = z

If it's possible, can someone please give an concrete example?

$\endgroup$
1

2 Answers 2

2
$\begingroup$

The Kepler problem: the motion of a probe mass around a massive (spherical non-moving) body.

Newtonial mechanics gives Keplerian orbits (ellipses, parabolas and hyperbolas).

General Relativity modifies these orbits (some acquire perihelion precession, some change the period, and some become infalling spirals).

And Quantum Mechanics states that a position of the probe mass is not a point at all, but is rather a wave-packet, which travels about the same mean path, but expands, and falls apart into hydrogen-like orbitals, overlapping, being sumed up, and oscillating somehow.

These answers are such that NM answer is an approximation to the GR answer, and it is an approximation to the QM answer. GR and QM answers are irreducible to each other. There could be some 4th answer, for which both GR and QM answers are approximations, given by some quantized gravity theory. But such theory is not built yet (in some sense).

$\endgroup$
0
$\begingroup$

I'm not very keen on GR, but there are lots of problems that can be solved both in QM and NP; there are, for example, the free particle, the harmonic oscillator, the box potential and the infinite potential well. Solutions are very different; in fact, they are described by totally different mathematical tools (points on a phase space in Classical mechanics, functions in an hilbert space in QM). This sometimes gives rise to Quantum effects (interference, tunnel effect etc); these effects, if you look at systems of appropriate scale, are observable! And that's why you need QM to study the world at a certain scale. You could also study a 1 meter long pendulum with QM, and you'll find that it has discrete energy levels; however they are so close that any macroscopical observer would agree that the energy spectrum of a pendulum is a continuum.

$\endgroup$
4
  • $\begingroup$ Could you please give some concrete example? $\endgroup$
    – user315648
    Commented May 10, 2013 at 14:57
  • $\begingroup$ Sure! Any specific example of the ones I listed? $\endgroup$
    – Alex A
    Commented May 10, 2013 at 15:04
  • $\begingroup$ If you solve the harmonic oscillator with quantum mechanics, you'll find that the energy levels are discrete and equispaced, with the spacing being $\hbar \omega$; now, a pendulum is, if you impose the small angle approximation, an harmonic oscillator with $\omega = \sqrt{\frac{g}{L}}$. If you put $L = 1 m$, the spacing between energy levels is of the order of $10^{-34} J$ (a bit smaller, actually). I challenge you to measure it! :-D $\endgroup$
    – Alex A
    Commented May 10, 2013 at 15:17
  • $\begingroup$ If you solve the box potential in NM, you find out that a particle that doesn't have the energy to overcome the potential barrier has zero probability of crossing it; if instead you look at QM solutions, they give rise to the tunnel effect. It should be noticed, however, that it's higly suppressed when you insert in the transmission coefficient physical constants of macroscopic scale, hence you do not observe tunneling in CM, that describes the macroscopic world. $\endgroup$
    – Alex A
    Commented May 10, 2013 at 15:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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