I'm solving a problem in which a muon with a half life of $\tau=10^{-6}s$, generated at 30km of altitude, travels at an unknown velocity. I'm asked to calculate its velocity and then the distance in the muon's reference system. What I've done so far is the following: Having the first event be $$ x^{'\mu}(0,\vec{0})=x^\mu(0,\vec{0})=0^\mu,\tag{1} $$ I should be able to get the velocity and distance in the primed system by the transformation $$ x^{'\mu}=\Lambda^\mu_\nu x^\nu.\tag{2} $$ Taking the velocity as $\vec{v}=v\hat{x}$, then the Lorentz tensor should have components $\Lambda_0^0=\Lambda_1^1=\gamma$, $\Lambda_0^1=\Lambda_1^0=-\gamma\beta$, $\Lambda_2^2=\Lambda_3^3=1$ and every other component should be zero. Having $x^2=y,x^3=z$ be always zero, I come to the relations $$ (c t)'=\gamma(c t-\gamma\beta x) \qquad;\qquad x' = -\gamma\beta(c t) + \gamma x. $$ I'm going to skip a bit, because I think I'm not doing anything groundbreaking. Noticing that $t = x/(\beta c)$, $t'=\tau$ and $x$ is known, I can simply NSolve this and get that $\beta=0.9998$, which is correct. Using the same relations, though, I come to the conclusion that $$ x' = -\gamma\beta (x/\beta) + \gamma x = -\gamma x + \gamma x = 0, $$ which, if I interpret this result to be the distance that the muon travels in its own system of reference, makes absolutely no sense. Obviously, then, my interpretation is incorrect. How could I explain this result and how could I, with tensor/4-vector notation, obtain the lenght dilation that special relativity without tensors just shows to be $\gamma^{-1} x$?
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2$\begingroup$ Just for clarity why should the muon move at all in its own frame? $\endgroup$– user196418Dec 5, 2020 at 3:31
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2$\begingroup$ @ggcg I don't understand how I could miss something so obvious. Thank you. $\endgroup$– condoszDec 5, 2020 at 3:33
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2$\begingroup$ Relativity does that to you. I am not sure there are not mistakes but being able to say something is zero in a frame makes calculations fall out. Do the same for photons, they live in the Null hyper space, zero magnitude in all frames. $\endgroup$– user196418Dec 5, 2020 at 3:35
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I'm going to answer my own question because I reckon someone may get something out of this, someday.
As ggcg pointed out, it would indeed make no sense for the position of the muon in its own reference system to be different than zero, so the result I showed before is correct and intuitive.
The correct way to solve this problem is (I realise now this should have been obvious) to take the inverse Lorentz transformation, $$ \Lambda^{-1} x' = x $$ where the only parameters that change (for this problem) are $\Lambda_0^1=\Lambda_1^0=\gamma\beta$. That way, the correct answer is given by the relation $$ \gamma\beta c \tau + \gamma x' = x. $$