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I'm trying to work on an exercise in Wolfgang Rindler's book "Introduction to Special Relativity" and I'm stuck on the following exercise:

Two particles move along the x-axis of S at velocities 0.8c and 0.9c, respectively, the faster one momentarily 1m behind the slower one. How many seconds elapse before collision?

My confusion comes from the discrepancy between two different approaches I took:

Version A: Don't switch frames: The first object moves with $x_1(t) = 0.8c\cdot t$, the second one with $x_2(t) = -1\ \mathrm{m} + 0.9c\cdot t$. The event where $x_1(t) = x_2(t)$ is $x = 7.9992\ \mathrm{m}$ and $t=3.3333\times 10^{-8}\ \mathrm{s}$. For this I don't use any special relativity properties which makes me slightly suspicious of this approach.

Version B: Choose a comoving frame $S'$ with the slower particle in its origin: We set $S$ and $S'$ such that $x=x'$, $y=y'$ and $z=z'$ for $t=0$. The relative velocity of $S'$ to $S$ is the velocity of the slower particle, i.e. $v=0.8c$ in direction of $x$. We drop $y$ and $z$ coordinates now because we won't need them.

We call the slower object's coordinate $x_1$ in $S$ and $x_1'$ in $S'$ and the other one $x_2$ and $x_2'$, respectively. Then $x_1'=0$ for all times as we choose object 1 as the origin of $S'$.

By velocity transformation, $$u_2' = \frac{u_2-v}{1-u_2v/c^2} = 0.35714c,$$ which is the relative velocity of the faster object to the slower object.

The initial position of the second object is by Lorentz transformation given by $$x_2'(0) = \gamma \cdot x_2(0) = -1.6667.$$

Now we need to find the time $t'$ in $S'$ when $x_2'$ hits the origin (and thus the other particle): $$x_2'(t') = -1.6667+0.35714c\cdot t' = 0,$$ hence $$t' = 1.5556\cdot 10^{-8}\ \mathrm{s}.$$

The collision event in $S'$ is thus given by $x'=0$ and $t'= 1.5556\cdot 10^{-8}\ \mathrm{s}$. Transforming back to $S$, we get $$x = \gamma\cdot(x'+vt') = 6.2223\ \mathrm{m}$$ and $$t = \gamma\cdot(t'+0) = 2.5926\cdot 10^{-8}\ \mathrm{s}.$$ This is not the same as Version A. Where did I go wrong?

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  • $\begingroup$ I'm not sure offhand where your mistake is, but version A is definitely a correct procedure. $\endgroup$ – David Z Nov 23 '15 at 10:14
  • $\begingroup$ In which frame is the 1 metre difference defined? $\endgroup$ – drglove Nov 23 '15 at 10:42
  • $\begingroup$ The 1m is probably understood in the stationary frame $S$, but I don't have any more information than what it is stated in the question. $\endgroup$ – Mercury Bench Nov 23 '15 at 10:52
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I think the problem is in your $x'_2(t')$ equation in Version B.

Let the equation for the faster object be ($\beta_2 = 0.9$) $$ x_2(t) = x_2^0 + \beta_2 ct $$ in S. Then direct transformation of $(x_2(t), ct)$ to S' coordinates $(x'_2(t'), ct')$ gives ($\beta = 0.8$) $$ x'_2 = \gamma (x_2 -\beta ct) = \gamma x_2^0 + \gamma(\beta_2 - \beta) ct\\ ct' = \gamma(ct -\beta x_2) = \gamma(1-\beta\beta_2)ct - \beta\gamma x_2^0 $$ From the latter get $$ ct = \frac{ct' + \beta\gamma x_2^0}{\gamma(1-\beta\beta_2)} = \frac{1}{\gamma(\beta_2 - \beta)}u'_2(ct' + \beta\gamma x_2^0) $$ and eliminate $ct$ in the first: $$ x'_2(t') = (1+\beta u'_2)\gamma x_2^0 + u'_2 ct' $$ Now the first term amounts to $(1+\beta u'_2)\gamma x_2^0 = - (1 + 0.8\times 0.35714)\times 1.6667 = -2.1429$ and the collision takes place at $ct' \approx 6$. Take this back into the S frame and get $x = 1.6667 \times 0.8 \times 6 \approx 8$, $ct = 1.6667\times 6 \approx 10$, etc.

Ok now, so how come?

Basically the difference in the first term comes from the linearity of the Lorentz transform. When transforming $x_2(t)$ to S' you basically used $$ (x_2(t), ct) = (x_2^0, ct=0) + (\beta_2 ct, ct) \rightarrow \mathcal{L} (x_2^0, ct=0) + \mathcal{L}(\beta_2 ct, ct) $$ where $\mathcal{L}$ denotes the Lorentz transform to S'. The correct form reads instead, by virtue of the linearity of $\mathcal{L}$, $$ (x_2(t), ct) = (x_2^0, ct) + (\beta_2 ct, ct) \rightarrow \mathcal{L} (x_2^0, ct) + \mathcal{L}(\beta_2 ct, ct) $$ and is not so pretty to calculate, after all, since $ct$ now appears in both terms.

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