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Example:

When one studies the spin statistics theorem, one of the phrases that's been repeated a lot was that "the spin statistics theorem was derived from relativistic physics... there's no way to prove it in non-relativistic physics."

However, if it could not be derived from non-relativistic physics, why does one assume that it is true for the non-relativistic case as well?

Part of me tried to argue that, with a switch of reference frame, non-relativistic physics becomes relativistic.

However, another case - from particle physics - is that electro-weak only works in a certain energy interval. And the fact that classical mechanics works just fine with first order configuration space! It doesn't seem to be a requirement that a result of relativistic physics must hold for non-relativistic physics.

Must conclusions from relativistic physics hold for non-relativistic physics?

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    $\begingroup$ I've removed a number of comments that were attempting to answer the question and/or responses to them. Commenters, please keep in mind that comments should be used for suggesting improvements and requesting clarification on the question, not for answering. $\endgroup$
    – David Z
    May 6, 2020 at 17:30
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    $\begingroup$ I don't quite understand the meaning of "must" and "requirement" you're asking about. Nonrelativistic and relativistic models are mathematically contradictory, so obviously there's no logical reason that conclusions of one should hold in another. We know that nonrelativistic models are not completely accurate and we can accept that, so there's no "moral" requirement either. But you probably know this already? $\endgroup$
    – JiK
    May 7, 2020 at 10:28

6 Answers 6

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Physics is not just a branch of math: it is a method for modeling phenomena in the real world. If a fact is proven experimentally, but a theory fails to account for it, it is a problem with the theory, rather than with the reality.

E.g., spin arises naturally in relativistic theory, but there is no reason why it should exist in non-relativistic quantum mechanics. Yet, we do include the Zeeman term in the Schrödinger equation, since otherwise we wouldn't be able to describe the spin-related phenomena. Same is true for symmmetrizing/antisymmetrizing the wave functions.

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    $\begingroup$ "it is a problem with the theory, rather than with the reality" - or the observation, of course. $\endgroup$
    – corsiKa
    May 7, 2020 at 3:28
  • $\begingroup$ @corsiKa You mean problems in a particular experiment? Or observation as a part of a theory, as in relativity and QM? $\endgroup$
    – Roger V.
    May 7, 2020 at 7:38
  • $\begingroup$ You know, I hadn't thought of that aspect. I was considering something along the lines of faulty equipment, unsynchronized clocks, and the like. $\endgroup$
    – corsiKa
    May 8, 2020 at 1:24
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It boils down to two things:

  1. You're right that in a general nonrealativistic theory, spin-statistics theorem does not necessarily hold.
  2. But we assume that our actual nonrelativistic physics is really only an approximation to a more fundamental and relativistic nature. Thus, the nonrelativistic theories we use to describe our world carry an imprint of the underlying relativistic theory.
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The relativistic case is a more general description of how the universe operates, and the "non-relativistic" case is a simplified approximation that only works in certain cases.

It's important to note that, contrary to your phrasing of "switch of reference frame", exotic effects such as time dilation or unification of forces aren't something you simply switch on or off—instead, they're phenomena that vary in degree based on (velocity, temperature, gravity, other parameter). We have investigated these phenomena in a variety of conditions, and we have models that appear to reliably describe the behavior of reality.

As an analogy, consider the function $f(x) = \sin x$. We know that this function is a periodic curve, but in the very special case where $x$ is close to zero, the function looks pretty close to linear, and sometimes we can use the approximation $f(x) = \sin x \approx x$. (Undergraduate classical mechanics would be impossible without the magic phrase "for small angle $\theta$"!) Any model that claims to represent the behavior of a sine-based function at large or all angles must, in the special case of $x \approx 0$, produce results very close to the approximation, but that's not because we "switch" our model, it's because the approximated model is intentionally trading a small amount of accuracy for tractability.

So yes, the conclusions from relativistic physics must hold in non-relativistic situations specifically because we are saying that "this is a more complex model that describes the way things work in a wider variety of situations than the simpler model". If the more complicated model does not (after performing a lot more calculations!) produce answers that are so close to the simple everyday model that we can't tell the difference, then its claim to describe the universe fails. (In fact, this is how we know that neither GR nor QM as we currently have them is "correct"—they don't extend into each other's domains.)

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Non-relativistic phenomena are perfectly described by a relativistic theory. One can use either the full relativistic theory or instead study the small $v/c$ limit and both have to agree (up to neglected $O(\tfrac{v}{c})$ terms).

The spin-statistics theorem holds automatically for fields whose S-matrix is Lorentz-invariant. We expect fundamental particles to be described by such a theory and thus in principle it could be used to describe anything, including for example a "non-relativistic" condensed matter system. However, in practice this is far too difficult and a different effective theory needs to be used. This effective theory does not necessarily have to have a Lorentz-invariant S-matrix. And thus the spin-statistics theorem does not necessarily have to hold.

On the other hand the spin-statistic theorem also follows when you require that the Hamiltonian is bounded from below. So for a stable system the spin-statistic theorem will still hold.

Recommended reading: For example chapter 12 of Quantum Field Theory and the Standard Model By M.D. Schwartz.

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Well the my answer would be that relativistic conclusions should also hold in nonrelativistic physics(which always should be included in the relativistic theory by an low energy limit) but if the effect depends on the energy of the system, it's not necessary that the effect is part of the nonrelativistic theory. For example does relativity of time and space make no sense in newtonian mechanics, because it follows from the infinite speed of light limit of special relativity, where the lorentz transformations reduce to galilei transformations, which never mix space and time. But this doesn't mean that the relativity isn't there. Its effect is just much much smaller than at high velocities. The nonrelativistic case therefore shouldn't explain all phenoma that happen because it comes from an limit where you "exclude" information about nature.

Otherwise nonrelativistic quantum mechanics in form of the pauli equation follows from the dirac equation, where spin remains a relevant thing. I would say it's safe to say that the old fashioned schroedinger equation without spin just lacks this information about spin from the more general dirac theory and therefore shouldn't be able to describe all phenomena as the spin-statistics connection. But spin survives the low energy limit to the pauli equation and therefore the spin-statistics connection should also remain true in nonrelativistic physics.

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A simple counterexample: There is no frame in GR where gravitation travels with infinite velocity.

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    $\begingroup$ This is a bad counterexample. It is not just the issue of frame but also of degrees of freedom. In a certain gauge for a given frame, a specific degree of freedom of gravitational field (corresponding to Newtonian gravitational potential) will propagate with infinite speed. So, "-1" for the answer. $\endgroup$
    – A.V.S.
    May 6, 2020 at 19:35
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    $\begingroup$ Newtonian gravitational potential propagate with infinite speed, only in the nonrelativistic aproxximation. No matter which you consider the Newtonian gravitational field component of the gravity tensor $\endgroup$
    – user263319
    May 7, 2020 at 3:49
  • $\begingroup$ Let's use electrodynamics for an example, since it is easier to find equations for it elsewhere and there are no nonlinearities. In Coulomb gauge the scalar EM potential propagates with infinite speed: the equation for it $\Delta \Phi = -4\pi \rho$ has no time derivatives. In full EM theory it is a constraint equation. Likewise, in GR the generalization of Coulomb gauge would have a similar Poisson equation for Newtonian gravitational potential. Incidentally, when Laplace concluded in 1805 that “the speed of gravity” exceeds $\sim 10^7\,c$ he in fact measured such constraint propagation. $\endgroup$
    – A.V.S.
    May 7, 2020 at 5:29
  • $\begingroup$ There is propagation, in electrostatic, and in Newton Universal Gravitation. But the effect is that the fied is static because the field is changed in this point with other with the same value. If you move for an external force, the electrostatic emission of a charge change the factor $ \frac {1} {r^2} $ for a delayed potential . You can see here: en.wikipedia.org/wiki/Retarded_potential. Doing my degree they told us that really it was a consequence for interchange of virtual particles, but we never were descript this method. Anyway, it is a consequence of classical Maxwell Eq $\endgroup$
    – user263319
    May 7, 2020 at 7:33

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