Question

Question: Apply the binomial theorem (Appendix E) to the last part of Eq. 37-52 for the kinetic energy of a particle. (a) Retain the first two terms of the expansion to show the kinetic energy in the form

$\mathit{K}\mathbf{=}\left(firstterm\right)\mathbf{+}\left(secondterm\right)$

The first term is the classical expression for kinetic energy. The second term is the first-order correction to the classical expression. Assume the particle is an electron. If its speed vis c/20, what is the value of (b) the classical expression and (c) the first-order correction? If the electron’s speed is 0.80s, what is the value of (d) the classical expression and (e) the first-order correction? (f) At what speed parameter $\mathit{\beta }$does the first-order correction become 10%or greater of the classical expression?

Short answer

Answer

a. The first two terms of the kinetic energy are

$K=\frac{1}{2}{E}_o{\frac{v}{c^2}}^2+\frac{3}{8}{E}_o\frac{v^4}{c^4}$

b. For c/20;$K_1=0.639KeV$

c. For c/20;$K_2=1.2eV$

d. For 0.80c;$K_1=0.164MeV$

e. For 0.80c;$K_2=1.3\times {10}^{-14}\text{J}$

f.$\beta =0.365$

Step-by-step solution

Step 1: Identification of the given data

The kinetic energy is, $K=\left(firstterm\right)+\left(secondterm\right)$.

The speed of electron is, 0.80c .

Significance of kinetic energy

A particle or an item that is in motion has a sort of energy called kinetic energy. An item accumulates kinetic energy when work, which involves the transfer of energy, is done on it by exerting a net force.

Step 3(a): Expand the kinetic energy relation

The relativistic kinetic energy of a particle is given by

$$\begin{gathered} K=\left(\gamma -1\right)m{c}^2 \\ =\left(\frac{1}{\sqrt{1-{\beta }^2}}-1\right)m{c}^2 \\ =\left[{\left(1-{\beta }^2\right)}^{\frac{-1}{2}}-1\right]m{c}^2 \end{gathered}$$

The binomial expression for ${\left(1+x\right)}^n$is

${\left(1+x\right)}^n=1+\frac{nx}{1!}+\frac{n\left(n-1\right)}{2!}{x}^2+...$

Using this expression to solve kinetic energy relation further,

$$\begin{gathered} K=\left[\left(1+\left(-\frac{1}{2}\right)\left(-{\beta }^2\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}-1\right)}{2}{\left(-{\beta }^2\right)}^2+...\right)-1\right]m{c}^2 \\ =\left[\left(1+\frac{{\beta }^2}{2}+\frac{3{\beta }^4}{8}+...\right)-1\right]m{c}^2 \\ =\left(\frac{{\beta }^2}{2}+\frac{3{\beta }^4}{8}+...\right)m{c}^2 \end{gathered}$$

Considering only first two terms of the expansion,$K=\left(\frac{{\beta }^{2}m{c}^2}{2}+\frac{3{\beta }^{4}m{c}^2}{8}\right)$

But,$\beta =\frac{v}{c}$ and$E_0=m{c}^2$

$$\begin{gathered} K=\left[\frac{{\left(v/c\right)}^2{E}_0}{2}+\frac{{\left(v/c\right)}^4{E}_0}{8}\right] \\ =\frac{1}{2}{E}_o{\frac{v}{c^2}}^2+\frac{3}{8}{E}_o{\frac{v}{c^4}}^4 \end{gathered}$$

Where E₀ is the rest mass energy which has the value of $0.511MeV$for an electron.

The first term here is for classical limit and the second term is the second order correction which is added to the first term for relativistic velocities.

Step 4(b and c): Determine first term and second term for the electron with the speed c/20 = 0.005c.

The first term i.e., the classical expression for an electron with speed is

$K_1=\frac{1}{2}{E}_0\frac{v^2}{c^2}$

Substitute all the value in the above equation.

$$\begin{gathered} K_1=\frac{1}{2}\left(0.511MeV\right)\left(\frac{{\left(0.05c\right)}^2}{c^2}\right) \\ =0.639KeV \end{gathered}$$

Hence the classical expression for an electron with speed is, 0.05c and 0.639 KeV.

The second order correction is

$K_2=\frac{3}{8}{E}_0\frac{v^4}{c^4}$

Substitute all the value in the above equation.

$$\begin{gathered} K_2=\frac{3}{8}\left(0.511MeV\right)\left(\frac{{\left(0.05c\right)}^4}{c^4}\right) \\ =1.2eV \end{gathered}$$

Hence the second order correction is, 1.2 eV .

The second order correction is miniscule as the velocity is much less than the speed of light. Now if the speed of the electron is 0.8 c, the second order is substantial which is shown in the next step.

Step 5(d and e): Determine first term and second term for the electron with the speed 0.80 c .

The first term i.e., the classical expression for an electron with speed is

$K_1=\frac{1}{2}{E}_0\frac{v^2}{c^2}$

Substitute all the value in the above equation.

$$\begin{gathered} K_1=\frac{1}{2}\left(0.511MeV\right)\left(\frac{{\left(0.80c\right)}^2}{c^2}\right) \\ =0.164MeV \end{gathered}$$

Hence the classical expression for an electron with speed is,0.80c ,164MeV .

The second order correction is

$K_2=\frac{3}{8}{E}_0\frac{v^4}{c^4}$

Substitute all the value in the above equation.

$$\begin{gathered} K_2=\frac{3}{8}\left(0.511MeV\right)\left(\frac{{\left(0.80c\right)}^4}{c^4}\right) \\ =1.3\times {10}^{-14}\text{J} \end{gathered}$$

Hence the second order correction is, $1.3\times {10}^{-14}\text{J}$.

Step 6(f): Determine the speed parameter for part(f)

The value of speed parameter is to be determined at which the second order expression is equal or greater than 10 5of classical expression. It can be mathematically expressed as

$$\begin{gathered} K_2⩾0.1K_1 \\ \frac{3}{8}{\beta }^4{E}_o=0.1\left(\frac{{\beta }^2{E}_o}{2}\right) \\ {\beta }^2=\frac{0.4}{3} \\ \beta =0.365 \end{gathered}$$

Hence, at $\beta =0.365$the second order expression is equal or greater than 10 % of classical expression