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To show that energy in special relativity reduces to $E=m+mv^2/2$ for low velocities, if we make a Taylor expansion of $m\gamma$ around $v=0$ we get $$E=m+mv^2/2+3mv^4/8+\cdots$$ But why can we cutoff after the second term? How does one know all the other terms don't add up to something significant?

We get $E = \text{rest energy}+\text{kinetic energy}+\frac{3mv^4}{8} + \cdots$ What kind of energy is the $3mv^4/8$ term?

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You've dropped some powers of c from the higher terms. They are important. (Or alternately do it in c=1 units where $v = \beta$, and consider the size of $\beta$.) – dmckee Jun 23 '12 at 3:46
Yea, I set c=1. Btw I often encounter this argument, where they ignore higher order terms, is there a general justification for this? – HarryPotter Jun 23 '12 at 3:51
You are ready to discover the general justification. How big is $v$ now that you've picked (1) human velocity scales and (2) $c=1$ units? How big does that make the $v^4$ term by comparison to the $v^2$ term? What about the higher degree terms? – dmckee Jun 23 '12 at 3:55
Ok cool. But the two first terms correspond to two different forms of energy, what is the third? – HarryPotter Jun 23 '12 at 3:57
It's just the first non-Newtonian term in the kinetic energy. As far as I know it does not have a name. – dmckee Jun 23 '12 at 3:59
up vote 4 down vote accepted

Because the radius of convergence of the expansion is 1, so you know the Taylor series is good until a sizable fraction of v=1 (units where c=1). The heuristic for Taylor series is that for an analytic function (or one with poles and cuts), your expansion is good in a circle in the complex plane which extends until the first singularity. Further, inside the circle, away from the edge, the convergence of the Taylor series is like the convergence of a geometric series, the error is about the same size as the first neglected term. The more precise statement is that the size of the consecutive terms are approximately reduced by the ratio of the distance from the center to the distance to the first singularity, this is the comparable geometric series.

So in this case, for $1\over \sqrt{1-v^2}$, you have a branch cut starting at $v=1$, so starting at v=0, the radius of convergence is 1, and the error for small v in taking the first two terms is just well estimated by the third term. You are far from the cut.

But in this case, you can also show it explicitly without complex analysis ideas, just by estimating the difference between the function and the first two terms in the expansion using Taylor's formula (derived by repeated integration by parts):

$$ |{1\over \sqrt{1-v^2}} - 1 - {v^2\over 2} | = | \int_0^v {d^3\over du^3} ({1\over\sqrt{1-u^2}}) {u^2\over 2} du |$$

and then note that the third derivative of the function in question is bounded in the given interval by, say, 10, so that the error is at most 10 times the neglected term.

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We get E=rest energy+kinetic energy+3mv48+⋯ What kind of energy is the 3mv4/8 term?

Not quite. That should be:

E = invariant energy + (relativistic) kinetic energy

In other words, what you've labeled "kinetic energy" is actually just "Newtonian kinetic energy".

Once you see this, your question evaporates.

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