Question

How fast must you be traveling relative to a blue light \((480 \mathrm{nm})\) for it to appear red \((660 \mathrm{nm}) ?\)

Short answer

Answer: The approximate relative velocity for this phenomenon is 53% of the speed of light, or \(v \approx 0.530c\).

Step-by-step solution

Write down the given information

The given information is: - Wavelength of blue light, \(\lambda_b = 480\,\text{nm}\) - Wavelength of red light, \(\lambda_r = 660\,\text{nm}\) - Relative velocity, \(v\) (we need to find this)

Use the relativistic Doppler shift formula

The relativistic Doppler shift formula is given by: $$ \lambda_r = \lambda_b \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}} $$ where \(\lambda_r\) is the observed (red) wavelength, \(\lambda_b\) is the source (blue) wavelength, \(v\) is the relative velocity, and \(c\) is the speed of light. We will solve for \(v\) using the given \(\lambda_r\) and \(\lambda_b\) values.

Rearrange the formula to solve for \(v\)

We need to isolate \(v\) in the equation. First, we divide both sides by \(\lambda_b\): $$ \frac{\lambda_r}{\lambda_b} = \sqrt{\frac{1 + \frac{v}{c}}{1 - \frac{v}{c}}} $$ Then, we square both sides: $$ \left(\frac{\lambda_r}{\lambda_b}\right)^2 = \frac{1 + \frac{v}{c}}{1 - \frac{v}{c}} $$ Now, let's multiply both sides by the denominator \((1 - \frac{v}{c})\): $$ \left(\frac{\lambda_r}{\lambda_b}\right)^2 (1 - \frac{v}{c}) = 1 + \frac{v}{c} $$ Next, we expand the left side and simplify: $$ \left(\frac{\lambda_r}{\lambda_b}\right)^2 - \frac{v}{c}\left(\frac{\lambda_r}{\lambda_b}\right)^2 = 1 + \frac{v}{c} $$ Now, let's move all terms involving \(v\) to one side of the equation: $$ \frac{v}{c} + \frac{v}{c}\left(\frac{\lambda_r}{\lambda_b}\right)^2 = \left(\frac{\lambda_r}{\lambda_b}\right)^2 - 1 $$ Now we can factor out \(\frac{v}{c}\): $$ \frac{v}{c}(1 + (\frac{\lambda_r}{\lambda_b})^2) = \left(\frac{\lambda_r}{\lambda_b}\right)^2 - 1 $$ Finally, we can solve for \(v\): $$ v = c\frac{ \left(\frac{\lambda_r}{\lambda_b}\right)^2 - 1 }{ 1 + \left(\frac{\lambda_r}{\lambda_b}\right)^2 } $$

Plug in the values and compute \(v\)

Now we can plug in the values for \(\lambda_r\), \(\lambda_b\), and \(c\): $$ v = (3\times10^8\,\text{m/s})\frac{ \left(\frac{660\,\text{nm}}{480\,\text{nm}}\right)^2 - 1 }{ 1 + \left(\frac{660\,\text{nm}}{480\,\text{nm}}\right)^2 } $$ $$ v \approx 0.530c $$ The required relative velocity is approximately 53% of the speed of light, or \(v \approx 0.530c\).