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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?

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  • $\begingroup$ A radioactive isotope doesn't care at what relative speed you are traveling. It doesn't even know that you exist, hence it won't change its decay pathways depending on your relative velocity. Time dilation and length contraction will, of course, apply, and they are being used in accelerator facilities all the time. $\endgroup$ – CuriousOne Jun 10 '16 at 11:58
  • $\begingroup$ So in the brothers paradox where one gets on a spaceship travelling near the speed of light and the other brother stays home, even though they will age differently, the brother on the spaceship measuring emissions from a local radioactive particle will count the same number as his brother pointing a geiger counter at the orbiting spaceship? $\endgroup$ – Justin S. Jun 10 '16 at 12:36
  • $\begingroup$ Will decays look the same except for the Doppler effect? Yes. Scalars don't transform under Lorentz transformation, only vectors and tensors do, so whenever you have a simple number (like in this case the number of decay products), it's always the same in all observer systems. $\endgroup$ – CuriousOne Jun 10 '16 at 12:39
  • $\begingroup$ So the brother on the spaceship will perceive the particle as releasing far more radioactivity in a short amountil of time due to his time dilation from travelling so fast. Wouldn't this follow from the earth brother and the spaceship brother seeing the same number of decay events during the spaceship brother's journey? $\endgroup$ – Justin S. Jun 10 '16 at 12:45
  • $\begingroup$ They will both be seeing "the same amount" of radioactivity. Whether it looks more blue or red shifted depends on the direction of travel. You are mistaking time dilation for what one would "see", but they are not the same. The observational aspects of relativistic boosts are given by the Doppler effect, the Lorentz transformation is merely a coordinate system transformation. $\endgroup$ – CuriousOne Jun 10 '16 at 12:50
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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:

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