7

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.

$$\vec p=\frac{m\vec v}{\sqrt{1-|v|^2/c^2}}\ \rightarrow\ \frac{m\vec v}{\sqrt{1-0}}=m\vec v.$$

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?

Qmechanic
  • 220,844
Nikolaj-K
  • 8,873

3 Answers3

5

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

4

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).

Diego Mazón
  • 7,127
3

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.

Ron Maimon
  • 1