Muon decay in muons frame of reference, time muon expects us to record

I have already asked a question regarding this concept but it was flagged as a duplicate. I know this misconception is very common for special relativity but I haven't found a question that talks about the misconception I'm having. Or atleast I haven't made the link between their question and mine. My question is why does it look like in this case the factor by which the clocks are slower is $$γ^2$$?

On the earth lets say a muon has to travel a distance $$d$$ at a speed $$v$$. We expect the time it takes for the muon to reach us as time $$t$$, where $$t=d/v$$, but factoring in time dilation of running clocks, the muon has only experienced time $$t/γ$$ in our frame of reference, hence we can observe it on Earth.

But in the muons frame of reference, the distance it has to travel is shorter by a factor of gamma, so the time it experiences to travel to earth will be $$t_2$$, where $$t_2=v/(d/γ)$$. This is the same time as we record in our frame of reference. But according to the muon, we too have our clocks running hence we should record in our labs, a dilated time, $$t_2/γ$$. This is our original expected time $$t$$, but reduced by a factor $$γ^2$$.

Why is this so? I am sure it has something to do with the simultaneity of events but I don't know where to start.

• Don't you mean $t=d/v$? – PM 2Ring Feb 18 at 22:36

Indeed, the muon will think that our clock is not only running slow, but running ahead, by the usual relativistic $$dv/c^2$$ delay. The final reading we take down, according to the muon, is $$\frac{t}{\gamma} + \frac{dv}{c^2} = \frac{t}{\gamma} + \frac{v^2}{c^2} \frac{t}{\gamma} = \frac{t}{\gamma} \gamma^2 = \gamma t$$ which is perfectly consistent with what we calculate in our frame.