Question

Derivation of the Doppler formula for light. A source and receiver of EM waves move relative to one another at velocity \(v .\) Let \(v\) be positive if the receiver and source are moving apart from one another. The source emits an EM wave at frequency \(f_{\mathrm{s}}\) (in the source frame). The time between wavefronts as measured by the source is \(T_{\mathrm{s}}=1 / f_{\mathrm{s}}\) (a) In the receiver's frame, how much time elapses between the emission of wavefronts by the source? Call this \(T_{\mathrm{T}}^{\prime} .\) (b) \(T_{\mathrm{T}}^{\prime}\) is not the time that the receiver measures between the arrival of successive wavefronts because the wavefronts travel different distances. Say that, according to the receiver, one wavefront is emitted at \(t=0\) and the next at \(t=T_{\mathrm{o}}^{\prime} .\) When the first wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}\) When the second wavefront is emitted, the distance between source and receiver is \(d_{\mathrm{r}}+v T_{\mathrm{r}}^{\prime} .\) Each wavefront travels at speed \(c .\) Calculate the time \(T_{\mathrm{r}}\) between the arrival of these two wavefronts as measured by the receiver. (c) The frequency detected by the receiver is \(f_{\mathrm{r}}=1 / T_{\mathrm{r}} .\) Show that \(f_{\mathrm{r}}\) is given by $$f_{\mathrm{r}}=f_{\mathrm{s}} \sqrt{\frac{1-v / \mathrm{c}}{1+v / \mathrm{c}}}$$

Short answer

Answer: The Doppler formula for light is given by: $$f_\text{r} = f_\text{s} \sqrt{\frac{1 - v/c}{1 + v/c}}$$ where \(f_\text{r}\) is the frequency detected by the receiver, \(f_\text{s}\) is the source frequency, \(v\) is the relative velocity between the source and receiver, and \(c\) is the speed of light.

Step-by-step solution

Find time interval between wavefront emission in the receiver's frame

Considering that the source is moving with respect to the receiver at a velocity \(v\), we can use the time dilation formula relating the source frame and the receiver frame: $$T_{\text{T}}^{\prime} = \frac{T_s}{\sqrt{1 - \frac{v^2}{c^2}}}$$ Now substitute \(T_s = \frac{1}{f_s}\) and simplify the expression for \(T_\text{T}^{\prime}\).

Calculate the time between the arrival of two wavefronts at the receiver \(T_\text{r}\)

Let's denote the time taken by each wavefront to reach the receiver as \(t_1\) and \(t_2\). According to the receiver, the first wavefront is emitted at \(t = 0\) and the second at \(t = T_\text{o}^{\prime} = T_{\text{T}}^{\prime}\). Therefore, we can write equations for the distances traveled by each wavefront: 1. \(d_\text{r} = c \times t_1\) 2. \(d_\text{r} + v T_\text{r}^{\prime} = c \times t_2\) Subtract equation 1 from equation 2 to find the time difference between the arrival of two wavefronts: $$T_\text{r} = t_2 - t_1$$

Determine the frequency detected by the receiver, \(f_\text{r}\)

The frequency detected by the receiver is the reciprocal of the time between the arrival of the two wavefronts: $$f_\text{r} = \frac{1}{T_\text{r}}$$ Substitute the expression for \(T_\text{r}\) in terms of \(T_{\text{T}}^{\prime}\), and then find the ratio of the received frequency to the source frequency: $$\frac{f_\text{r}}{f_\text{s}} = \frac{1}{T_\text{r}} \cdot \frac{T_\text{s}}{1}$$ Finally, express \(f_\text{r}\) in terms of \(f_\text{s}\), \(v\), and \(c\) using the results from steps 1 and 2, and obtain the Doppler formula for light: $$f_\text{r} = f_\text{s} \sqrt{\frac{1 - v/c}{1 + v/c}}$$