Emissions of radioactive particles travelling near the speed of light vs stationary particles Sorry in advance if this is a noob question. I just heard another explanation of special relativity and was wondering about it.
If a radioactive particle was passing by near the speed of light, would it emit the same amount of particles as a function of a stationary observer's time than a stationary particle of the same material would?
Also, could an experimentalist at a particle accelerator lab use this property to calculate the speed of particles in the accelerator if he knew what their rate of decay should be?
 A: Let's be a little more clear about what we mean here.
We can go down to a radioactive beam facility (these exist), or to a muon storage ring like Muon g-2 (now at Fermilab), or even climb a steep mountain and use cosmic muons (Halfdome is a scenic location, let's imagine doing it there).1 In any case, call the lifetime of the beam particles $\tau$. (For muons this will be $\tau_\mu \approx 2.2\,\mathrm{\mu s}$.)
In any case, we'll put up two flags. A green one near the source of the radioactive particle and a red one farther away.
Then we count the number of decays that happen in the beam between the green and red flags. Every observer will agree on this number (always assuming the detection equipment is up to the task), and if we have enough statistics they will all agree that it is the number that they expected. 
However, they may have different expectations for the reasons they expected that number of decays between the flags.
A person standing in the lab (or beside Halfdone) will reason thus:

The distance between the flags is $\ell$, and the beam is moving at $v \approx c$ such that the Lorentz factor is $\gamma$. The time it takes for the beam to get from the green flag to the red is $t = \ell/v$, but the time measured by a clack co-moving with the beam would be $t' = t/\gamma = \ell/(\gamma v)$, so the fraction of the beam that decays should be 
  $$ f = 1 - \exp \left[ - \frac{\ell}{\gamma v \tau}\right] \,. $$

A hypothetical particle-physics fairy riding a beam particle would reason thus: 

The lab is moving past us at speed $v \approx c$ such that the Lorentz factor is $\gamma$, so the flags are actually separated by $\ell' = \ell/\gamma$. That means that the time between the green flag passing and the red flag passing is $t' = \ell'/v = \ell/(\gamma v)$. As a result the fraction of the beam that exist as the green flag passes that has decayed by the time the red flag passes is
  $$ f' = 1 - \exp \left[ - \frac{\ell}{\gamma v \tau} \right] \,.$$

In short the predictions are the same, but the reason for the predictions differ. In the lab, there is a long length to be covered, but the clocks of the moving particle run slowly; from the beams perspective their clocks run normally, but he distance to be covered is much shorter.

1 I include this case because it is a common textbook example (though using a generic mountain rather than Halfdome in particular) and was the earliest test case for time-dilation. This example also shows up in a few other places on Physics:


*

*Muon demonstration of time dilation

*“Reality” of length contraction in SR
