Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 0 down vote favorite

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 classical case if you take that limit?

share|improve this question
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. – DJBunk Jun 29 '12 at 18:52
@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. – Nick Kidman Jun 29 '12 at 19:00
@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. – Ron Maimon Jun 30 '12 at 2:35
@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. – Nick Kidman Jun 30 '12 at 15:48

1 Answer

up vote 0 down vote accepted

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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