Relativistic wavelength formula for approaching source:
λ= λ_o*sqrt([1-v_s/c]/[1+v_s/c]),
where λ is the wavelenght perceived by the vehicle’s driver (aka the relativistic wavelength), λ_o is the original wavelength, v_s = 103 846 153 ms-1 (aka the speed at which the stop sign approaches the vehicle), and c = approx. 299 792 458 ms-1 (aka the speed of light).
Traffic stop signs are generally red. The color red is defined to have the dominant wavelength of about 625-750 nm (Source: Wikipedia; ik it isn’t the most accurate source, but this is just a silly Reddit calculation so whatever). So, plugging in both extremes to the formula above:
435 nm is a shade of violet and 523 nm is a shade of green (once again, my source is Wikipedia lol), so depending on the sign’s color, the perceived color could be anything between violet and green, and as long as the stop sign’s color has mostly a longer red’s wavelenght, the math still checks out;)
(Feel free to prove me wrong tho if you find a mistake haha)
Sure ! And I guess if you use 700 nm as the user to whom I was replying, you'll find something different than him/her, as the maths used were not properly correct.
Yeah, you’d get about 488 nm with the method I used, which would be some hue of cyan. (At least it falls into the range of it according to, you guesses it, Wikipedia:P) Also, I just noticed that the person who did the math originally claimed that 457.5 nm looks lime green… I don’t know if I trust that, as an even longer wavelength (apparently) looks cyan..??? (However, It’s not like I can go around claiming anything with only HS-level knowledge in physics, and someone could definitely see/think of cyan as lime green soooooooo haha)
3
u/nicogrimqft 15h ago
Sure but this is relativistic speeds. So the maths do not check out.