5 ms·

This is the same physics I was taught in school as for two cars crashing into each other at 70mph I presume? Neither has a 140mph impact, although a 140mph impa

This is the same physics I was taught in school as for two cars crashing into each other at 70mph I presume? Neither has a 140mph impact, although a 140mph impact would involve more than double the energy anyway AIUI.

archgoon 8y ago
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.
maxander 8y ago
No, I don't think so- two cars impacting head-on at 70mph would have a "140mph impact," insofar as that it would be essentially equivalent to one car traveling at 140mph hitting a stationary car.
- kbenson 8y ago
  Hmm, wouldn't a car of the same mass going the same speed joust counteract all the force exactly? If one of the cars had half as much mass, wouldn't that lessen the impact of the larger car and increase it for the smaller car, as the combined center of gravity afterwards would be moving inthe direction the larger car was traveling? It's been far too long since I had a physics class...
  - maxander 8y ago
    Well, maybe we're using different terms- as the article notes, there's technically no such thing as a "such-and-such speed impact." I was speaking in terms of how much kinetic energy is liable to go towards damaging the cars involved (or into the fireball of a particle collision); in that case, if you have X energy in two cars, or 2X energy in one and 0 in the other, the sum is the same. If we consider debris or etc flying away after the collision, then you're right that they're different- the combined velocity of the two cars is 0mph in the both-moving case and 140mph (in some direction) in the one-moving case, and the velocity of all the debris will sum to that velocity. (And the sum of velocities would be weighted appropriately if one of the cars was more massive than the other.)
  - auslander 8y ago
    On a side note, if a massive 4wd heads-on collides with Corolla, the 4wd guy will have scratches and bruises, Corolla people would not suffer, instant death. Choosing a car, choose tons, not stars :)
    - saagarjha 8y ago
      It's a sad reality that just makes the road less safe for everyone and hurts the environment :(
saagarjha 8y ago
Depends on your reference frame. In the reference frame of the approaching car you have a 140 mph impact.
Waterluvian 8y ago
That concept still feels intuitively wrong to me.

archgoon 8y ago

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.

maxander 8y ago

No, I don't think so- two cars impacting head-on at 70mph would have a "140mph impact," insofar as that it would be essentially equivalent to one car traveling at 140mph hitting a stationary car.

kbenson 8y ago
Hmm, wouldn't a car of the same mass going the same speed joust counteract all the force exactly? If one of the cars had half as much mass, wouldn't that lessen the impact of the larger car and increase it for the smaller car, as the combined center of gravity afterwards would be moving inthe direction the larger car was traveling? It's been far too long since I had a physics class...
- maxander 8y ago
  Well, maybe we're using different terms- as the article notes, there's technically no such thing as a "such-and-such speed impact." I was speaking in terms of how much kinetic energy is liable to go towards damaging the cars involved (or into the fireball of a particle collision); in that case, if you have X energy in two cars, or 2X energy in one and 0 in the other, the sum is the same. If we consider debris or etc flying away after the collision, then you're right that they're different- the combined velocity of the two cars is 0mph in the both-moving case and 140mph (in some direction) in the one-moving case, and the velocity of all the debris will sum to that velocity. (And the sum of velocities would be weighted appropriately if one of the cars was more massive than the other.)
- auslander 8y ago
  On a side note, if a massive 4wd heads-on collides with Corolla, the 4wd guy will have scratches and bruises, Corolla people would not suffer, instant death. Choosing a car, choose tons, not stars :)
  - saagarjha 8y ago
    It's a sad reality that just makes the road less safe for everyone and hurts the environment :(

saagarjha 8y ago

Depends on your reference frame. In the reference frame of the approaching car you have a 140 mph impact.

Waterluvian 8y ago

That concept still feels intuitively wrong to me.