How to "derive the leading correction" to an energy expression?

Question

The kinetic energy of motion of a particle is the relativistic total energy minus the rest energy.

(a) A particle has rest mass $M$ and speed $v$. If $v \ll c$, then show that the kinetic energy due to motion is approximated by the well-known non-relativistic expression for the kinetic energy.

(b) Derive the leading correction to the non-relativistic expression.

In the above problem, I've done part (a) by expanding the Lorentz factor as a power series and then truncating the "negligible" terms.

However, I'm stumped for deriving the "leading correcton". I've read the Wikipedia article on leading corrections: https://en.wikipedia.org/wiki/Leading-order_term, but I still don't understand what the problem is asking for exactly?

Answer 1

I'm not sure what expanding the Lorentz factor does. I do this:

$$ T = E-E_0 = \sqrt{(pc)^2+(mc^2)^2} -mc^2 $$

$$ T = mc^2[(1+\frac{(pc)^2}{(mc^2)^2})^{\frac 1 2}-1]$$

$$ T = mc^2[(1+(\frac{p}{mc})^2)^{\frac 1 2}-1]$$

Now for Taylor expanding:

$$ f(x) = (1+x^2)^{\frac 1 2}-1= \frac{x^2} 2 - \frac{x^4}8 +O(x^6)$$

Note that I included the subtraction of "one" outside the radical--this means the leading order term is $O(x^2)$ and the leading correction is $O(x^4)$.

Plugging it in with $x=p/(mc)$:

$$ T = mc^2[\frac{p^2}{2m^2c^2} - \frac{p^4}{8m^4c^4}]$$

$$ T = \frac{p^2}{2m} - \frac{p^4}{8m^3c^2}$$

$$ T = \frac{p^2}{2m} - \frac{p^4}{8m^3c^2} = T_0 + T_1$$

The leading term, $T_0=p^2/2m$, is the non-relativistic term, and the leading correction, $T_1$, depends on $c$, as it should.

The leading correction can be rewritten:

$$ T = T_0(1 - \frac 1 2 \frac{T_0}{E_0}) $$

showing that it depends on the ratio of the kinetic energy to the rest mass. If you are interested in a velocity dependent expression, you directly substitute it in without further consideration:

$$ T(v) = \frac 1 2 m v^2 (1-\frac 1 2 \frac{\frac 1 2 m v^2}{mc^2} )$$

and that is:

$$ T = \frac 1 2 m v^2(1-\frac 1 4 \beta^2)$$