Question

4.3 the relativistic regime: the energy of a particle of mass $m$ and speed $v$ is given by $e = mc^{2}+\frac{mv^{2}}{2}+\frac{3mv^{4}}{8c^{2}}+\frac{5mv^{6}}{16c^{4}}+cdots$ (4.34) where $c$ is the speed - of - light. the first term on the right - hand - side ($mc^{2}$) is the rest mass energy, and the rest of the terms give the energy of motion, the kinetic energy.
a) write out the three kinetic energy terms by factoring out $mv^{2}/2$, i.e. find $a$, $b$, and $c$ in the expression for the kinetic energy, $k$, of the form $k=\frac{mv^{2}}{2}(a + b + c)$ (4.35)
b) what are the dimensions of $a$, $b$, and $c$?
c) find the ratios $b/a$ and $c/a$ for objects traveling at usual highway speeds of about 30 m/s. do the same for the earths motion around the sun, which has a speed of about 30 km/s
d) your answers in part c) show why we keep only the $mv^{2}/2$ term when determining the kinetic energy in our introductory mechanics classes where objects are moving at speeds much less than $c$. at what value of $v$ will $b/a = 0.1$ or $0.01$? these values show when relativistic effects are at the 10% and 1% level, respectively.

Explanation:

Step1: Identify kinetic - energy terms

The energy formula is $E = mc^{2}+\frac{mv^{2}}{2}+\frac{3mv^{4}}{8c^{2}}+\frac{5mv^{6}}{16c^{4}}+\cdots$. The kinetic - energy terms are $K=\frac{mv^{2}}{2}+\frac{3mv^{4}}{8c^{2}}+\frac{5mv^{6}}{16c^{4}}$. Factoring out $\frac{mv^{2}}{2}$, we get $K = \frac{mv^{2}}{2}(1+\frac{3v^{2}}{4c^{2}}+\frac{5v^{4}}{8c^{4}})$. So, $A = 1$, $B=\frac{3v^{2}}{4c^{2}}$, $C=\frac{5v^{4}}{8c^{4}}$.

Step2: Determine dimensions

Since $A$, $B$, and $C$ are dimensionless factors in the expression for kinetic energy, the dimensions of $A$, $B$, and $C$ are $1$ (dimensionless).

Step3: Calculate ratios for $v = 30$ m/s

For $v = 30$ m/s and $c=3\times 10^{8}$ m/s, $B=\frac{3v^{2}}{4c^{2}}=\frac{3\times(30)^{2}}{4\times(3\times 10^{8})^{2}}=\frac{3\times900}{4\times9\times 10^{16}}=\frac{1}{1.33\times 10^{14}}$. $C=\frac{5v^{4}}{8c^{4}}=\frac{5\times(30)^{4}}{8\times(3\times 10^{8})^{4}}=\frac{5\times81\times10^{4}}{8\times81\times10^{32}}=\frac{5}{8\times10^{28}}$. Then $\frac{B}{A}=\frac{3v^{2}}{4c^{2}}=\frac{3\times(30)^{2}}{4\times(3\times 10^{8})^{2}} = 2.5\times 10^{-15}$ and $\frac{C}{A}=\frac{5v^{4}}{8c^{4}}=\frac{5\times(30)^{4}}{8\times(3\times 10^{8})^{4}}=8.68\times 10^{-30}$.

Step4: Calculate ratios for $v = 30\times 10^{3}$ m/s

For $v = 30\times 10^{3}$ m/s and $c = 3\times 10^{8}$ m/s, $B=\frac{3v^{2}}{4c^{2}}=\frac{3\times(30\times 10^{3})^{2}}{4\times(3\times 10^{8})^{2}}=\frac{3\times9\times 10^{8}}{4\times9\times 10^{16}} = 2.5\times 10^{-9}$. $C=\frac{5v^{4}}{8c^{4}}=\frac{5\times(30\times 10^{3})^{4}}{8\times(3\times 10^{8})^{4}}=\frac{5\times81\times10^{16}}{8\times81\times10^{32}}=6.25\times 10^{-17}$. Then $\frac{B}{A}=\frac{3v^{2}}{4c^{2}}=2.5\times 10^{-9}$ and $\frac{C}{A}=\frac{5v^{4}}{8c^{4}}=6.25\times 10^{-17}$.

Step5: Find $v$ when $\frac{B}{A}=0.1$ and $\frac{B}{A}=0.01$

If $\frac{B}{A}=\frac{3v^{2}}{4c^{2}} = 0.1$, then $v^{2}=\frac{0.4c^{2}}{3}$, and $v=\sqrt{\frac{0.4}{3}}c\approx0.37c$. If $\frac{B}{A}=\frac{3v^{2}}{4c^{2}}=0.01$, then $v^{2}=\frac{0.04c^{2}}{3}$, and $v=\sqrt{\frac{0.04}{3}}c\approx0.115c$.

Answer:

a) $A = 1$, $B=\frac{3v^{2}}{4c^{2}}$, $C=\frac{5v^{4}}{8c^{4}}$
b) Dimensionless
c) For $v = 30$ m/s, $\frac{B}{A}=2.5\times 10^{-15}$, $\frac{C}{A}=8.68\times 10^{-30}$; for $v = 30\times 10^{3}$ m/s, $\frac{B}{A}=2.5\times 10^{-9}$, $\frac{C}{A}=6.25\times 10^{-17}$
d) When $\frac{B}{A}=0.1$, $v\approx0.37c$; when $\frac{B}{A}=0.01$, $v\approx0.115c$