3 ms·

No, not unless you were making relativistic corrections for the vehicles, which would be silly[1]. From the perspective of either vehicle, you do see an oncomi

No, not unless you were making relativistic corrections for the vehicles, which would be silly[1]. From the perspective of either vehicle, you do see an oncoming vehicle traveling at (approximately) 140 miles per hour. When dealing with things traveling close to the speed of light, this doesn't work, because the relativistic correction term becomes significant. The equation they show on the blog (u + v)/(1+uv) represents the velocities in fractions of speed of light. When u and v are much smaller than the speed of light, the denominator can be ignored, as it's close to 1, and we simply get (u+v). [1] But still pedagogically sound when teaching relativity :)

xxxvvv 8y ago
For the people who (like myself) are confused by the missing c^2 in the blog, (u + v)/(1+(uv/(c^2)) is the original. Much easier to understand that the "simplified" version, which is usually the case.
- zinclozenge 8y ago
  In relativity, c=1 is a common convention. It doesn't really add much to the equation because it's just a conversion factor.
  - archgoon 8y ago
    Omitting conversion factors is a bit like omitting type information. If you already know what everything is, it doesn't add information, and can make the code look simpler. If you don't, not omitting them can help make sense of things when you're learning and lack context.
  - wzcard 8y ago
    To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that contains c^2. Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at: c * (a * b) / (1 + a * b) Then you convert the car speeds to fractions, only to multiply back by c=299792458. Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.

xxxvvv 8y ago

For the people who (like myself) are confused by the missing c^2 in the blog, (u + v)/(1+(uv/(c^2)) is the original. Much easier to understand that the "simplified" version, which is usually the case.

zinclozenge 8y ago
In relativity, c=1 is a common convention. It doesn't really add much to the equation because it's just a conversion factor.
- archgoon 8y ago
  Omitting conversion factors is a bit like omitting type information. If you already know what everything is, it doesn't add information, and can make the code look simpler. If you don't, not omitting them can help make sense of things when you're learning and lack context.
- wzcard 8y ago
  To get back a meaningful result for the car example, c=1 does not really help. It is much easier to plug in u and v directly into the simpler equation that contains c^2. Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at: c * (a * b) / (1 + a * b) Then you convert the car speeds to fractions, only to multiply back by c=299792458. Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.