Lorentz contraction is easy to understand once you realise that it is not a contraction at all. Instead it is a rotation and the length of the object, or more precisely its proper length, doesn't change at all.

To see this take the usual example of a rod of length $2a$ aligned along the $x$ axis. We'll draw the rod at time $t=0$ in its rest frame $\mathbf S$:

So the ends of the rod are at the positions $(t=0, x=-a)$ and $(t=0, x=a)$.

Now consider a frame $\mathbf S'$ moving at a velocity $v$ with respect to $\mathbf S$ and as usual we'll take the origins of the frames to coincide at $t=0$. To find the positions of the ends of the rod in $\mathbf s'$ we use the Lorentz transformations:

$$\begin{align}
t' &= \gamma \left( t - \frac{vx}{c^2} \right ) \\
x' &= \gamma \left( x - vt \right)
\end{align}$$

and with some minor algebra this gives the positions of the ends in $\mathbf S'$ as:

$$ (0,-a) \rightarrow \left(\gamma a \frac{v}{c^2}, -\gamma a \right) $$

$$ (0,a) \rightarrow \left(-\gamma a \frac{v}{c^2}, \gamma a \right) $$

So in $\mathbf S'$ at $t=0$ the rod looks like:

So in the $\mathbf S'$ frame the rod has been rotated. However it is a rotation in spacetime, not just in space, so as well as moving in $x'$ one end of the rod has rotated forward in the $t'$ coordinate while the other has rotated backwards in $t'$.

The proper length of the rod $\Delta s$ is given by:

$$ \Delta s^2 = \Delta x^2 - c^2 \Delta t^2 $$

So:

$$ \Delta x = \gamma a - -\gamma a = 2\gamma a $$

$$ \Delta t = -\gamma a \frac{v}{c^2} - \gamma a \frac{v}{c^2} = -2\gamma a \frac{v}{c^2} $$

And substituting for $\Delta x$ and $\Delta t$ in our expression for the proper length $\Delta s$ gives:

$$\begin{align}
\Delta s^2 &= 4\gamma^2a^2 - c^2\,4\gamma^2a^2 \frac{v^2}{c^4} \\
&= 4\gamma^2a^2 \left(1 - \frac{v^2}{c^2}\right) \\
&= 4\gamma^2a^2 \frac{1}{\gamma^2} \\
&= 4a^2
\end{align}$$

And we find that the proper length of the rod is $\Delta s = 2a$, so the proper length of the rod hasn't changed at all. In fact let me emphasise this:

In $\mathbf S'$ the proper length of the rod hasn't changed at all

So why then do we talk about a Lorentz contraction? It's because if you are the observer in the frame $\mathbf S'$ you are not seeing the two ends of the rod at $t'=-\gamma a v/c^2$ and $t'=\gamma a v/c^2$, you are seeing them both at $t'=0$.

Consider the far end of the rod at $(t'=\gamma a v/c^2, x'=-\gamma a)$. To get the position at $t'=0$ we have to subtract off the distance moved in the time $\gamma a v/c^2$, that is:

$$\begin{align}
x'(t=0) &= -\gamma a + v\,\gamma a v/c^2 \\
&= -\gamma a \left(1 - \frac{v^2}{c^2} \right) \\
&= -\frac{a}{\gamma}
\end{align}$$

And likewise for the other end, though I won't go through the details we get $x'(0)=a/\gamma$, so if we view the ends of the rod at $t'=0$ we find the length is:

$$ \ell = \frac{2a}{\gamma} $$

And this is less than the proper length $2a$, and that's why we say the length of the rod has been decreased by Lorentz contraction. It hasn't really been contracted, it's just that due to the rotation in spacetime we are viewing the two ends at different times.

I misunderstood your question at first. I thought you were asking about the detector downstream of the double slit, where the interference pattern is visible; every practical double-slit experiment includes such a detector. But instead you are asking about a hypothetical detector which could "tag" a particle as having gone through one slit or the other. Most interference experiments do not have such a detector.

The idea of "tagging" a particle as having gone through one slit or the other, and the realization that such tagging would destroy the double-slit interference pattern, was hashed out in a long series of debates between Bohr and Einstein. Most introductory quantum mechanics textbooks will have at least some summary of the history of these discussions, which include many possible "detectors" with varying degrees of fancifulness.

A practical way to tag photons as having gone through one slit or another is to cover both slits with polarizing films. If the light polarizations are parallel, it's not possible to use this technique to tell which slit a given photon came through, and the interference pattern survives. If the light polarizations are perpendicular, the it would be possible in principle to detect whether a given photon went through one slit or the other; in this case the interference pattern is also absent. If the polarizers are at some other angle, it's a good homework problem to predict the intensity of the interference pattern.

## Best Answer

Your worry is not necessary.

In the usual experiment the detector measures the distribution of the time between the muon stopping in the detector and the time of it's decay. Then an exponential curve is fit to the data and the lifetime taken from the fit parameters

Muon decay is a processes analogous to radioactive decay, and (like all exponential processes) it has the special property that the lifetime from the fit

does not depend on the absolute start time.An important thing here is what the "lifetime" means. And what it doesn't mean.

It

mean that muons tend to decay $2.2 \,\mathrm{\mu s}$ after they are created (which would imply that you are looking for a peak and that the absolute start time matters).does notIt

mean that Given a collection of muons, if you wait $2.2 \,\mathrm{\mu s}$ about 67% of them will have decayed. And if you wait another lifetime about 67% of the one that were left will have decayed. And so on. This means thatdoesanytimeyou fit an exponential to the decay time distribution you will always measure the same lifetime.A kind of old school way to see this is to note that $y = A \exp \left( -t/\tau \right)$ implies that (writing $A = \exp \alpha$) $$ \ln y = \alpha -\frac{1}{\tau} t \,,$$ which is a straight line if you plot $\ln y$ as a function of $t$. The slope of the line is $-1/\tau$ and

does not depend on the range of $t$ over which you measure it.