3 ms·
Doesn't that assume the distribution at time 1 is also the distribution at time 2, which isn't entirely true since the source is already sightly less radioactiv
by klipt 5y ago
Doesn't that assume the distribution at time 1 is also the distribution at time 2, which isn't entirely true since the source is already sightly less radioactive then?
- Animats 5y agoThat's a good point. The effect is small but nonzero.
- teraflop 5y agoStill works, with a slight variation: instead of taking two measurements, you can take four successive measurements [A,B,C,D] over equal periods of time, and then do the aforementioned procedure with the values A+D and B+C as input. If the activity is decreasing at a linear rate, then A+D and B+C should have exactly the same distribution. In reality, they will be ever so slightly different because the decay is exponential rather than linear, but over short periods of time (compared to the half-life) the linear approximation is extremely good.
- OscarCunningham 5y agohttps://en.wikipedia.org/wiki/Thue%E2%80%93Morse_sequence#Equitable_sequencing https://en.wikipedia.org/wiki/Thue%E2%80%93Morse_sequence#Eq...
- tedunangst 5y agoCoin flips aren't fair either, because gravity will change slightly between successive flips.
- r00tanon 5y agoOK, but if you toss the coin higher, relativistic time effects come into play.
- dotancohen 5y agoAnd if you're flipping near the Earth's surface, there will be aerodynamic effects as well. Which is more streamlined: The Queen's side or the Platypus' side?