Chapter 33, Problem 29P

To determine

To show:The Doppler shift in wavelength is $\frac{\Delta \lambda }{{\lambda }_0}\approx \frac{v}{c}$ for $v<<c$ .

Explanation of Solution

Given:

  ${\lambda }_0$ is the emitted wavelength as seen in a reference frame at rest with respect to source, $\lambda$ is the wavelength measured in a reference frame moving with velocity v away from source along the line of sight, and c is the speed of the light.

Formula used:

The expression of binomial theorem is,

  ${\left(1-x\right)}^{-1/2}=\left(1+\frac{1}{2}x+\frac{3}{8}{x}^2+\cdots \right)$

Calculation:

The expression for Doppler effect is,

  $\begin{array}{l}\lambda ={\lambda }_0\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}\\ \lambda ={\lambda }_0\frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}\end{array}$

Multiply the numerator and denominator by $\sqrt{1+\frac{v}{c}}$ .

  $\begin{array}{l}\lambda ={\lambda }_0\frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1-\frac{v}{c}}}\times \frac{\sqrt{1+\frac{v}{c}}}{\sqrt{1+\frac{v}{c}}}\\ \lambda ={\lambda }_0\frac{1+\frac{v}{c}}{\sqrt{1-\frac{v^2}{c^2}}}\\ \lambda ={\lambda }_0\left(1+\frac{v}{c}\right){\left(1-\frac{v^2}{c^2}\right)}^{-1/2}\end{array}$

Apply binomial theorem.

  $\lambda ={\lambda }_0\left(1+\frac{v}{c}\right)\left(1+\frac{v^2}{2c^2}+\cdots \right)$

For $v<<c$ , neglect higher terms.

  $\begin{array}{l}\lambda ={\lambda }_0\left(1+\frac{v}{c}\right)\left(1+\frac{v^2}{2c^2}\right)\\ \lambda ={\lambda }_0\left(1+\frac{v^2}{2c^2}+\frac{v}{c}+\frac{v^3}{2c^3}\right)\\ \lambda ={\lambda }_0\left(1+\frac{v}{c}+\frac{v^2}{2c^2}+\frac{v^3}{2c^3}\right)\end{array}$

Neglect higher term of the above expression.

  $\begin{array}{l}\lambda \approx {\lambda }_0\left(1+\frac{v}{c}\right)\\ \frac{\lambda }{{\lambda }_0}\approx \left(1+\frac{v}{c}\right)\\ \frac{\lambda }{{\lambda }_0}-1\approx \frac{v}{c}\\ \frac{\lambda -{\lambda }_0}{{\lambda }_0}\approx \frac{v}{c}\\ \frac{\Delta \lambda }{{\lambda }_0}\approx \frac{v}{c}\end{array}$

Conclusion:

Thus, the Doppler shift in wavelength is $\frac{\Delta \lambda }{{\lambda }_0}\approx \frac{v}{c}$ for $v<<c$ .