physicists: reducing Newtonian Mechanics to Special Relativity

[gusl]

Special Relativity (SR) is more general than Newtonian Mechanics (NM), right?

We say that the laws of SR reduce to the laws of NM when (v/c)^2 goes to 0, correct?

But (v/c)^2 = 0 implies that v = 0 (which makes sense, since this is required for the theories to agree exactly).

So how can I formalize the fact that NM is an approximation to SR at small speeds? Is the above good enough just because (v/c)^2 decreases faster than v as v -> 0?

Comments

[smandal.livejournal.com]

It's not that v->0, but c->infinity

[spoonless.livejournal.com]

That is correct. In the limit of (v/c) approaches zero, both of the theories agree. This means that for small values of (v/c), or v/c << 1, the two theories agree very well. The closer to zero it is, the better they agree... where at zero the agreement is exact, although as you point out physics is not very interesting when v/c equals zero so just knowing that doesn't help you much.

So how can I formalize the fact that NM is an approximation to SR at small speeds? Is the above good enough just because (v/c)^2 decreases faster than v as v -> 0?

You show that two theories agree in a limit like this by keep the first non-zero terms in the Taylor expansion for all equations involved. If there is a first-order term which is non-zero and they both agree, then you use that. If the first-order term is zero, then you have to use the second-order term. The first non-zero term is also called the "dominant" term. It controls the behavior for small values of the expansion parameter (in this case, v/c). Does that help?

[pbrane.livejournal.com]

As spoonless says, you can just Taylor expand SR about v=0.

For example, the famous Einstein equation E = mc^2 is really

E = (1-v^2/c^2)^-1/2mc^2 = mc^2 + (1/2)(v/c)^2*mc^2 + O[(v/c)^4]

The first (nonconstant) term simplifies to the familiar (1/2)mv^2 kinetic energy of newtonian mechanics (NM), and so the rigorous statement is "SR is equivalent to NM up to terms of order (v/c)^4". If you want precision in addition to rigor, you can do the "post-newtonian approximation" which entails using the coefficients of the further terms in the taylor expansion.