Even if many interesting similarities between the classical and the quantum mechanical framework have been worked out, e.g. in the subject of deformation quantization, in general, there are some mathematical problems. And in the conventional formulation, you don't want to make things like $\hbar\rightarrow 0$ for the expression $P=-\text i\hbar\tfrac{\partial}{\partial x}$.

In special relativity there are many formulas where one optains the non-relativistic formula by taking the naive limit $c\rightarrow \infty$, e.g.

$$\vec p=\frac{m\vec v}{\sqrt{1-|v|/c}}\ \rightarrow\ \frac{m\vec v}{\sqrt{1-0}}=m\vec v.$$

I wonder if it is know that you can always do that. Is there a formulation of special relativity (maybe it's the standard one already), where the starting assumptions/axioms/representations of objects of discourse involve the constant $c$, and as you take them with you to do all the standard derivations, you always end up with results which reduce to the Newtonian mechanics if you take that limit?

  • $\begingroup$ Do you have an example where this fails in the usual framework? As far as I can tell you can always take the non-relativistic limit, as long as its appropriate. i.e. We can talk about velocity addition at speeds much lower than c and it's a reasonable thing to talk about, but there isn't any reasonable way to talk about a photon non-relativistically. $\endgroup$ – DJBunk Jun 29 '12 at 18:52
  • $\begingroup$ @DJBunk: I don't know any counterexmaples where there is a relativistic theory and classical considerations and something bad happens. But I could imagine such cases. E.g. when $c$ stands in $\tfrac{v}{c}$ and you don't really want to kill the velocity or if there is a term multiplied by $\tfrac{1}{c^2}$ in the Maxwell equations and when you just take the limit then the term vanishes together with some function whose existence you need if you model it non-relativistically. Electric forces are used in classical considerationas after all. Maybe these are always be saved by $c\tfrac{1}{c}=1$ etc. $\endgroup$ – Nikolaj-K Jun 29 '12 at 19:00
  • $\begingroup$ @NickKidman: Wait, what's wrong with your last example? THe going away of derivative terms is just the same as hbar to zero. I posted it just now as an answer, but it seems you considered this and rejected it. $\endgroup$ – Ron Maimon Jun 30 '12 at 2:35
  • $\begingroup$ @RonMaimon: Yes, I guess I "rejected" it by saing if you have that time derivative of the E-field term, then you're not in electrostatics (your Laplace equation), but in a theory with propagating waves and this is not a model of classical non-relativistic physics. I basically accepted is as a phenomenon which isn't even in the Newtonian physics and so it doesn't have to be taken into accound. The $\hbar$ problem is more serious in the sense that while you don't need waves, you need momentum in all classical models. $\endgroup$ – Nikolaj-K Jun 30 '12 at 15:48

Suppose I have a scalar field. The equation for the field evolution is

$${1\over c^2} \partial_t^2 \phi -\nabla^2\phi =0 $$

So the issue with taking the limit $c\rightarrow\infty$ is exactly the same as taking $\hbar$ to zero in quantum mechanics, a derivative term is going away.

The reason you think $\hbar\rightarrow 0$ is somehow more difficult is because of the abstractness of the quantum formalism. If you rewrite $p=\hbar {\partial\over\partial x}$ as $p={h\over \lambda}$ (which is the same thing for plane waves), the small $\hbar$ limit becomes more obvious--- the wavelength goes to zero holding p fixed, so that the diffraction effects go away.

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  • $\begingroup$ I think this is actually an example where $c\rightarrow\infty$ gives the right answer. The wave equation given is for a massless scalar field (not just any scalar field). In the nonrelativistic limit, we don't expect to be able to see wave disturbances of massless fields, so only static fields should exist. This was developed explicitly, for electromagnetism, in the paper by le Bellac referenced in my answer. $\endgroup$ – user4552 Aug 20 '13 at 18:24
  • $\begingroup$ @BenCrowell: I agree, I was using it to argue that hbar to zero is similarly not problematic. $\endgroup$ – Ron Maimon Aug 20 '13 at 19:39
  • $\begingroup$ OK, that's fine. I can't remove my downvote unless you make some edit to your answer. If you want to make an edit to clarify what your interpretation is, that would work. $\endgroup$ – user4552 Aug 20 '13 at 19:45

The problem arises when one naively takes the limit of an expression as a constant, such as $c$ or $\hbar$, goes to a value (or infinity). What these limits physically mean is that a dimensionless ratio between a characteristic magnitude and that constant goes to certain value (or infinity).

Special relativity

The so-called non-relativistic limit (the name is horrible because Galilean physics is as relativistic as special relativity) of special relativity consists of taking the limit as the ratio $v/c$ or $p/(mc)$ goes to zero, with $c$ fixed. As $c$ only appears through these ratios very often, it is formally equivalent to the limit $c$ going to infinity (in these cases where $c$ only appears through these quotients). However, there are cases, involving for instance field theories, where one has to work out a little the expression to find these ratios. (By the way, Carroll kinematics is also characterized by the limit $v/c \rightarrow 0$, but in this case $c$ itself additionally goes to zero (as $\sqrt v$), something with little physical meaning as far as I know).

Quantum mechanics

The classical limit is frequently trickier because it is more difficult to identify characteristic magnitudes with dimensions of $\hbar$. In some cases, however, is pretty clear. For example in the path integral formulation of quantum mechanics the dimensionless exponent $S/\hbar$ determine the classical limit. When this quotient goes to infinity (or formally when $\hbar$ goes to zero) all path contributions cancel each other except the one that minimizes $S$, the classical path. Something similar happens in Sommerfeld quantization for large quantum numbers. Although the issue is the same one as in special relativity, there are fewer expressions that involve simple dimensionless ratios and this makes more difficult the classical limit than the non-relativistic limit. And in fact, one can read in tens of good books and papers wrong things like the identification of the classical limit with the tree level contribution (zeroth order in perturbation theory), which is not true in all cases (even though it is right in most cases).

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  • $\begingroup$ it is formally equivalent to the limit c going to infinity If it's formally equivalent, what's the point of worrying about which way we do it? $\endgroup$ – user4552 Aug 20 '13 at 22:59
  • $\begingroup$ One has to be careful when $c$ doesn't appear directly through those dimensionless ratios. For example, in the Larmor formula or in the massive Klein-Gordon equation. @BenCrowell $\endgroup$ – Diego Mazón Aug 20 '13 at 23:03
  • $\begingroup$ Can you give an example of how your reasoning applies to the Larmor formula? Sorry, but I'm still not getting your point. $\endgroup$ – user4552 Aug 20 '13 at 23:08
  • $\begingroup$ Larmor formula is already non-relativistic. Likewise, if you take the naive limit $c\rightarrow \infty$ in the massive Klein-Gordon equation, you don't get the Schr. equation. Does this help? $\endgroup$ – Diego Mazón Aug 20 '13 at 23:11
  • $\begingroup$ Interesting points -- maybe you could integrate them into your answer? I'm still not following your logic over all. $\endgroup$ – user4552 Aug 20 '13 at 23:18

Nice question!

There is a funky limit called Carroll kinematics, introduced in Levy-Leblond 1965. Baccetti 2011 is a freely available paper that describes it.

There are two different Galilean limits of electromagnetism (Le Bellac 1973). See de Montigny 2005 for a description.

So since the Galilean limit is often nonunique, it should be pretty clear that we can't obtain it in all cases just by taking $c\rightarrow\infty$.

Another way to see that this approach runs into problems is that in the $c\rightarrow\infty$ limit, the metric becomes degenerate. All of the standard machinery of relativity, e.g., the ability to raise and lower indices, is predicated on the assumption that the metric is not degenerate.

Baccetti, Tate, and Visser, 2011, "Inertial frames without the relativity principle," http://arxiv.org/abs/1112.1466

Le Bellac M and Levy-Leblond J M 1973, "Galilean electromagnetism," Nuov. Cim. B 14 217-233

Levy-Leblond, "Une nouvelle limite non-relativiste du group de Poincaré," Ann. Inst. Henri poincaré 3 (1965)

Marc De Montigny, Germain Rousseaux, 2005, "On the electrodynamics of moving bodies at low velocities," http://arxiv.org/abs/physics/0512200

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