My question involves broadly why calculating different (Einstein) relative velocities give different answers:

Say we have car 1 traveling in front of car 2. Car 1 goes at velocity $3c/4$, while car 2 goes at velocity $c/2$, both relative to the ground. Now, car 2 launches a projectile at car 1, with velocity $c/3$ relative to car 2. The question is, whether this projectile will reach car 1. This is a relative velocity with special relativity problem, though I am very confused why different approaches give different conclusions:

I. Compute $v_{p1}$, or the relative velocity of the projectile relative to car 1. Using the formula for relative velocities with Einstein correction, we can write $$v_{p1}=\frac{v_{p2}+v_{21}}{1+v_{p2}\cdot v_{21}/c^2}.$$ Plug in the numbers (classical relative velocities), we have $v_{p2}=c/3$, $v_{21}=-c/4$, so numerator is positive, (denominator clearly also positive), so $v_{p1}>0$, in other words the projectile hits car 1.

II. Compute $v_{pg}$, or the velocity of the projectile relative to the ground. Following the above formalism, $$v_{pg}=\frac{v_{p1}+v_{1g}}{1+v_{p1}\cdot v_{1g}/c^2}.$$ Here, we have $v_{p1}=c/3$ and $v_{1g}=c/2$ as givens in the problem. Evaluating the whole fraction, we have $v_{pg}=5c/7$, which is less than the velocity of car 1 relative to the ground $3c/4$, so projectile never hits car 1.

I just can't see what's wrong with either approach, though clearly the results are contradictory... Some guidance would be much appreciated!

  • $\begingroup$ I should add that there are definitely no arithmethic errors made during this comparison, though likely some misuse of the relativistic relative velocities formula was done in one of the two approaches. $\endgroup$ Apr 3, 2021 at 22:57
  • $\begingroup$ After revisiting your question, I updated my answer. $\endgroup$
    – robphy
    Apr 4, 2021 at 8:01

2 Answers 2



I had a chance to look at your problem more carefully.

  • As @BillN said, there are some errors in your velocities.
    Part of the problem is that velocities are not additive in special relativity.
  • I think your second equation should have labels $v_{p2}$ (not $v_{p1}$) and $v_{2g}$ (not $v_{1g}$).

I'll use "A" instead of "1", and "B" instead of "2".

I've drawn a spacetime diagram [to scale] centered at the projectile launch event. I translated $A$'s worldline back to make the velocity comparisons easier.
Graphically, we see that worldline-of-$P$ has a lab-frame velocity less than that of $A$, and so worldline-$P$ will never meet worldline-$A$.
Let's do the calculation.

I have shaded in sectors that mark the given "rapidities" (the signed Minkowski angle, whose hyperbolic-tangent gives the relative velocity between the worldlines).
Unlike velocities, rapidities are additive.
Note that (for example) $$ v_{PA}=\tanh{\theta_{PA}}\equiv\tanh(\theta_{PB}+\theta_{BA})\equiv \frac{\tanh\theta_{PB}+\tanh\theta_{BA}} {1+\tanh\theta_{PB}\tanh\theta_{BA}} =\frac{ v_{PB}+v_{BA} }{1+ v_{PB}v_{BA} } $$


  • To find $v_{PA}=\tanh\theta_{PA}$, use an expression involving $\theta_{PA}$.
    \begin{align}\theta_{PA} &=\stackrel{\checkmark}{\theta_{PB}}+\stackrel{?}{\theta_{BA}}\\ &={\theta_{PB}}+(\theta_{BG}- \theta_{AG}) =\mbox{arctanh}\displaystyle\frac{1}{3} +\left(\mbox{arctanh}\displaystyle\frac{1}{2} -\mbox{arctanh}\displaystyle\frac{3}{4}\right)\approx -0.0770753 \end{align} So, $v_{PA}=\tanh\theta_{PA}=-\displaystyle\frac{1}{13} \approx -0.076923 \quad <\quad 0$. Worldline-$P$ won't meet worldline-$A$.

  • To find $v_{PG}=\tanh\theta_{PG}$, use an expression involving $\theta_{PG}$.
    \begin{align}\theta_{PG} &=\theta_{PB}+\theta_{BG}\\ &=\mbox{arctanh}\displaystyle\frac{1}{3}+\mbox{arctanh}\displaystyle\frac{1}{2}=0.8958 \qquad\mbox{note: }\mbox{arctanh}\displaystyle\frac{3}{4}\approx 0.972955 \end{align} So, $v_{PG}=\tanh\theta_{PG}= \displaystyle\frac{5}{7} \approx 0.714 \quad < \quad 0.75 $. Worldline-$P$ won't meet worldline-$A$.

One could work solely with the velocity-composition formula.
However, making reference to rapidity allows one to use aspects of Euclidean geometry that carry over into Minkwoskian geometry.

[original version of my answer]

This notation suggests "velocity addition" or "velocity composition" $$v_{p1}=\frac{v_{p2}+v_{21}}{1+v_{p2}\cdot v_{21}/c^2}.$$ Think about walking forward on a train that is also moving forward.
What is your velocity with respect to the ground?

Here's an analogue with angles,
$$(\theta_p-\theta_1)= (\theta_p-\theta_2)+(\theta_2-\theta_1).$$

"Relative velocity" is expressed as $$v_{p1}=\frac{v_{p2}-v_{12}}{1-v_{p2}\cdot v_{12}/c^2},$$ Think differences from a common frame.
Think about both you and train moving forward with respect to the ground. What is the relative velocity of you with respect to the train?

Here's an analogue with angles,
$$(\theta_p-\theta_1)= (\theta_p-\theta_2)-(\theta_1-\theta_2).$$

  • $\begingroup$ Hi robphy, I appreciate your response - I am wondering how the two formulas of $v_{p1}$ you've written down are different? since $v_{12}$ is just $-v_{21}$? $\endgroup$ Apr 4, 2021 at 0:31
  • 1
    $\begingroup$ @Houndbobsaw, they are equivalent, geometrically. But you get relative sign errors if you don’t handle the accounting correctly. $\endgroup$
    – robphy
    Apr 4, 2021 at 0:35
  • $\begingroup$ Hi @robyphy, thanks so much for your detailed solution!! Accepting, and it makes so much more sense now. $\endgroup$ Apr 4, 2021 at 22:35

Your relative velocity formula is incorrect. Also, you must transform (via Lorentz) the velocity of the projectile from ref. frame 1 to ref. frame 2. It is not merely subtraction. In your second approach, the velocities $v_{p1}$ and $v_{1g}$ are incorrect.

You need to be extremely careful with negative signs and relative velocity symbols. Those are the sources of most mistakes in Lorentz transformations.

  • $\begingroup$ Thanks @BillN for your answer - I think you bring up a good point, though this approach is an approach given in Griffiths Intro. to particle physics solution manual, which is why i'm confused! I originally took the first approach in solving this problem and got a contradiction with the solution. $\endgroup$ Apr 4, 2021 at 0:30
  • $\begingroup$ Now. upon second thought, it seems approach 2 is very natural and consistent with velocity addition formula. The formula was most originally derived for a partile moving with speed u' in frame S' that moves at v with respect to relatively stationary frame S, and the LHS is just this speed u as seen in S. Essentially, approach 2 takes the ground frame as S, car 2 frame as S', and calculates bullet speed (the u') as seen in ground frame (the u). Then this is compared with car 1 speed in ground frame. On the other hand, I don't see how approach 1 is problematic! :p $\endgroup$ Apr 4, 2021 at 0:34

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