[chris (OP)]

If I had a lump of radioactive material that was ejecting particles such as alpha or beta particles that I could measure, and I accelerated the material to a very high speed, to an observer travelling with the decaying material the decay rate (count rate) would be the same.

But for me, the observer, time is running faster, so what count rate would I detect, if I collected and measured the decay products?

[jeffreyH]

That sounds a lot like observations of muon lifetimes by Rossi and Hall back in 1940. This was the topic of a post in another forum recently. The lifetimes are longer than expected. Would this relate to your question I wonder?

[chiralSPO]

My understanding is that the rate would be "normal" for the observer in the same reference frame as the sample, but that observers in different frames of reference would find different rates.

Let's imagine that there is a radioactive sample inside a shielded chamber in a spaceship, which is traveling at a very high velocity relative to a space station. An experimenter opens the shielded chamber for some period of time, and then closes it again. Both the experimenter and the an observer on the space station measure the released radiation. Once we account for the distance between the observer on the station and the sample, both observers will count the same number of particles being released from the deshielded sample--only they will disagree about how long the chamber was open--thereby calculating different rates. (NOTE: this is not my field of expertise; if I am in error, please correct me publicly!)

[Colin2B]

if I am in error, please correct me publicly!

It sounds right to me. All processes are seen to run slow and that includes radioactive emission rate and particle decay. As Jeffery says, good example is the muons, it has also been confirmed in particle accelerators.

[Space Flow]

Totally agree with all the answers given.
Every measurable timed process of any reference frame observed to be in relative motion is seen to be dilated by a factor called Gamma "γ" which is given by;
 [ Invalid Attachment ]

[chris (OP)]

So the half life I would see as the external observer would appear to be different to the half life measured by the person travelling with the radiation source?

[Colin2B]

So the half life I would see as the external observer would appear to be different to the half life measured by the person travelling with the radiation source?

Yes. It also changes depending where you are observing from in a gravity field, decay is faster in zero g as seen by an earth observer.
Interestingly the GR effect is not symmetrical - as with SR - so the observer in zero g sees decay on earth's surface to be slower.

[evan_au]

so what count rate would I detect, if I collected and measured the decay products?

With all this counting and timing, you need to take into account the delay for the light that reaches you.
If the radioactive source is moving away from you, the light reaching you at the end of the measurement period must travel further than light that left at the start of the measurement period.

After you subtract this correction, there is still a residual time correction due to the Lorentz factor in gamma (γ), above.

[jeffreyH]

So the half life I would see as the external observer would appear to be different to the half life measured by the person travelling with the radiation source?

Yes. It also changes depending where you are observing from in a gravity field, decay is faster in zero g as seen by an earth observer.
Interestingly the GR effect is not symmetrical - as with SR - so the observer in zero g sees decay on earth's surface to be slower.

Of course the asymmetry is due to the non-linear nature of the gravitational field. Which is exactly why we have time dilation or length contraction. A gravitational monopole field has no counterbalance like the opposite poles of both the electric and magnetic fields.

This also implies a time dependence on the rate of the increase of entropy in a varying strength gravitational field. All these factors will play a part.