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$D$ is traveling at .995c with respect to $C$ who is traveling at .995c with respect to $B$ all in the same direction. We want to compute $D$'s velocity as observed from $B$. Note that the Lorentz contraction/time dilation factor for .995c is $\sqrt{1-0.995^2}$ which I will approximate as 1/10. Now:

  • $C$ announces "my clock reads 10 seconds, and $D$ is .995*10 or 9.95 light seconds ahead of me".

  • $B$ converts this reading to his own frame of reference:

    • 10 seconds of $C$ time is 100 seconds of my time.

    • 9.95 light seconds of $C$ distance is 0.995 light seconds of my distance.

    • Thus, the distance between $C$ and $D$ is 0.995 light seconds when my clock reads 100 seconds.

    • Since I know $C$ is 99.5 light seconds away at 100 seconds, and I know (from above) that $D$ is .995 light seconds further out.

    • Thus $D$ is 99.5+.995 = 100.495 light seconds away from me at 100 seconds.

    • However, this would mean $D$ is traveling faster than light in my reference frame (100.495 light seconds in 100 seconds = 1.00495c), which is impossible.

What have I done wrong?

I know the correct formula for relativistic velocity addition (which would give $D$ a speed of 0.999987c), but just want to know why I didn't get that answer.

I also realize there are several similar questions, but I don't think there's one that exactly addresses this (mis)-derivation.

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closed as off-topic by Kyle Kanos, CuriousOne, Bill N, ACuriousMind, John Rennie Mar 13 '16 at 15:16

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You forgot the relativity of simultaneity.

If C thinks that D is at a certain time and place $a$ at the same time that C is at a certain time and place $b$, then others will not agree with that.

They will say they are talking about an event that happened earlier or later. They will say that $a$ and $b$ happened at different times.

Time dilation, length contraction, relativity of simultaneity. Forget any one and you'll get completely wrong answers.

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  • $\begingroup$ OK, but isn't this similar to the way the true relativity velocity addition formula is derived? I'm only transforming between two reference frames: $B$ and $C$. Even though $D$ has its own reference frame, I never use it. I just consider $D$ to be an object with a certain position/velocity in $C$'s frame. $\endgroup$ – barrycarter Mar 12 '16 at 21:25
  • $\begingroup$ @barrycarter There are two frames. The frame B in which you specify the velocity of C, and the frame C in which you specify the velocity of D. And note that you desire to find the velocity of object D in frame B. You can't assume that simultaneity in B is the same as simultaneity in C, they are different. $\endgroup$ – Timaeus Mar 12 '16 at 21:41
  • $\begingroup$ OK, but isn't the velocity addition formula derived this way? I mean, if they can do it, why can't I? $\endgroup$ – barrycarter Mar 13 '16 at 12:54
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    $\begingroup$ @barrycarter Your method wouldn't even show that $v$ plus $-v$ gives zero. And I told you that you forgot about the relativity of simultaneity and your response is to a) continue to not use it and not even try to use it (thanks) and b) insinuate that no one else uses it either. Seriously? $\endgroup$ – Timaeus Mar 13 '16 at 15:34
  • $\begingroup$ OK, I will check out that v and -v thing. I've seen the derivation for speed addition, and it doesn't mention simultaneity. Also, how exactly would I use it? You're saying that there is no such thing as a fixed time (which is true), but not how I would convert between the times? Could you give me a quantitative understanding (or pointer) to simultaneity? $\endgroup$ – barrycarter Mar 13 '16 at 15:39

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