Small correction: N_0 is not the number of isotopes, but the number of atoms/nuclei of the isotope. The isotope is not the individual atom or nucleus, but the type of nucleus (more exactly, given an element, defined by the number of protons in the nucleus, each isotope of that element is given by the umber of neutrons in the nucleus). Indeed, the decay equation you talked about only hold is there is exactly one isotope (well, strictly speaking, it would still hold if there were several isotopes with the same decay constant/half life). That's as if you had not just standard six sided dice in your box, but also twenty sided dice. Assuming initially you have roughly the same number of both, in the beginning the decrease will be dominated by the 6-sided dice, since they are more likely to get the chosen number. However this also means they will be gone faster, and eventually the 20-sided dice will dominate simply because almost no 6-sided dice are left. The isotopes here would correspond to the number of sides of the dice (that is, in the described scenario, your isotopes would be D6 and D20). Note that even if you have a million dice, this still makes only two types of dice (corresponding to two isotopes).
And how is p(dice=6) = ⅙ related with T=4? Is it because every time we make a throw the quantity is decreased by 1/6 (i.e multiplied by 5/6 and (5/6)⁴ ≈ 1/2)?
I think your approach is equivalent, but generally these type things are usually tied to e and ln. In this case ln 2 to halve things, multiply that times 6 and you get a little above 4. (Divide by 1/6 if you prefer to multiply by 6, same thing.)
Yes, that's basically it, except that (5/6)^4 is slightly less than 1/2, but then, the solution of (5/6)^x = 1/2 is only slightly less than 4, and moreover, he always drew a line after the number of dice was *at most* half of he previous value, which gives a slight bias to more throws. This can be nicely seen at the block after 21. 21/2 = 10.5. The first value below that in the series was 9, which is where he placed the next line. However the number before that, 11, was much closer to 10.5. If at each time passing half the previous value he had chosen the place that was closest to half the previous number, taking the lower number in case of a draw, the size of the runs would have been 5, 4, 3, 3, 6, 4, 5. Much more variation, but then, that is to be expected as the statistics get worse the lower the remaining number of dice goes. Indeed, for the last block, going from 1 to 0, this is the expected number of rolls of one die until you get a 6, and basic probability theory tells us that this has expectation value 6, which is clearly longer than the half-life of the dice.
statistically, after the first throw 1/6 of the dice are gone and 5/6 have survived. after the second throw, another 1/6 of these 5/6 survivors are gone, too, and we hav (5/6)² survivors. after the third throw, another 1/6 of these (5/6)² survivors are gone, too, and we have (5/6)³ survivors. etc. the sum s(n) := 1/6 + 5/6·1/6 + (5/6)² ·(1/6) + (5/6)³·(1/6) + ...+ (5/6)ⁿ·(1/6) denotes the (expected value for) the number of dice that are gone in the first n throws. (and we have (5/6)^(n+1) survivors). n=4 is the first number where s(n) = 671/1296 = 0.5177... > 1/2 therefore, the discrete halflife-time is a bit less than 4. you can get more decimals when you look at the values when you calculate when the number of survivors is e.g. less than 1/1024 = 1 / 2¹º. After the 39th throw, the number of survivors statistically is less than 1/1024, so λ must be between 3.8 and 3.9.
However, the experiement with the dice is discrete and not continuous, so we have to look when the sum s(n) := 1/6 + 5/6·1/6 + (5/6)² ·(1/6) + (5/6)³·(1/6) + ...+ (5/6)ⁿ·(1/6) exceeds 1/2; and we see that s(3) = 546/1296 = 0.5177... < 1/2 and s(4)= 671/1296 > 1/2. Obviously s(n) denotes the expected value for the number of dice that is removed after the n-th throw ; hence the halflife-time λ is actually a little less than 4
This behavior appears very often in physics, as in radioactive decay or energy dissipated in a damped harmonic oscillation. You can say that dice have a lot in common with springs and uranium too
I finally understand the derivation from first principles. Can you please solve an electrical mesh using geometric algebra of minkowski space. Im blown away how simple multi dimensional problems are solved without using matrices nor differential equations but application of geometric algebra is not shown for solving circuits
Love you, Dad (pls come home soon)
This is a really cool result
finally, i found the channel to sharpen my head
They all lead to addictions
I've never heard of soap addiction.
a pair of dice?
or a paradise?
a pair of dyes.
A pear of dies.
Strahlung = radiation ☺
Unexpected result, very nice!
great video papa flammy
Small correction: N_0 is not the number of isotopes, but the number of atoms/nuclei of the isotope. The isotope is not the individual atom or nucleus, but the type of nucleus (more exactly, given an element, defined by the number of protons in the nucleus, each isotope of that element is given by the umber of neutrons in the nucleus).
Indeed, the decay equation you talked about only hold is there is exactly one isotope (well, strictly speaking, it would still hold if there were several isotopes with the same decay constant/half life). That's as if you had not just standard six sided dice in your box, but also twenty sided dice. Assuming initially you have roughly the same number of both, in the beginning the decrease will be dominated by the 6-sided dice, since they are more likely to get the chosen number. However this also means they will be gone faster, and eventually the 20-sided dice will dominate simply because almost no 6-sided dice are left. The isotopes here would correspond to the number of sides of the dice (that is, in the described scenario, your isotopes would be D6 and D20). Note that even if you have a million dice, this still makes only two types of dice (corresponding to two isotopes).
Holy f. I thought it was hagoromo dice
i love those videos
Nice videos
And how is p(dice=6) = ⅙ related with T=4?
Is it because every time we make a throw the quantity is decreased by 1/6 (i.e multiplied by 5/6 and (5/6)⁴ ≈ 1/2)?
I think your approach is equivalent, but generally these type things are usually tied to e and ln. In this case ln 2 to halve things, multiply that times 6 and you get a little above 4. (Divide by 1/6 if you prefer to multiply by 6, same thing.)
Yes, that's basically it, except that (5/6)^4 is slightly less than 1/2, but then, the solution of (5/6)^x = 1/2 is only slightly less than 4, and moreover, he always drew a line after the number of dice was *at most* half of he previous value, which gives a slight bias to more throws. This can be nicely seen at the block after 21. 21/2 = 10.5. The first value below that in the series was 9, which is where he placed the next line. However the number before that, 11, was much closer to 10.5. If at each time passing half the previous value he had chosen the place that was closest to half the previous number, taking the lower number in case of a draw, the size of the runs would have been 5, 4, 3, 3, 6, 4, 5. Much more variation, but then, that is to be expected as the statistics get worse the lower the remaining number of dice goes. Indeed, for the last block, going from 1 to 0, this is the expected number of rolls of one die until you get a 6, and basic probability theory tells us that this has expectation value 6, which is clearly longer than the half-life of the dice.
statistically, after the first throw 1/6 of the dice are gone and 5/6 have survived.
after the second throw, another 1/6 of these 5/6 survivors are gone, too, and we hav (5/6)² survivors.
after the third throw, another 1/6 of these (5/6)² survivors are gone, too, and we have (5/6)³ survivors.
etc.
the sum s(n) := 1/6 + 5/6·1/6 + (5/6)² ·(1/6) + (5/6)³·(1/6) + ...+ (5/6)ⁿ·(1/6) denotes the (expected value for) the number of dice that are gone in the first n throws. (and we have (5/6)^(n+1) survivors). n=4 is the first number where s(n) = 671/1296 = 0.5177... > 1/2
therefore, the discrete halflife-time is a bit less than 4. you can get more decimals when you look at the values when you calculate when the number of survivors is e.g. less than 1/1024 = 1 / 2¹º. After the 39th throw, the number of survivors statistically is less than 1/1024, so λ must be between 3.8 and 3.9.
For the dice, λ=1/6 and therefore T_h=6*ln(2)=4.16 which is close to the 4.29 you measured
Actually, λ=-ln(5/6)=0.18. This is close to 1/6, but not equal; the closeness can be seen from the formula ln(1+x) = x + O(x^2), and 5/6 = 1 + (-1/6).
@__christopher__ You're right, but as an engineer I have to say: If it's right to first order, it probably won't explode xD
However, the experiement with the dice is discrete and not continuous, so we have to look when the sum s(n) := 1/6 + 5/6·1/6 + (5/6)² ·(1/6) + (5/6)³·(1/6) + ...+ (5/6)ⁿ·(1/6) exceeds 1/2; and we see that s(3) = 546/1296 = 0.5177... < 1/2 and s(4)= 671/1296 > 1/2. Obviously s(n) denotes the expected value for the number of dice that is removed after the n-th throw ; hence the halflife-time λ is actually a little less than 4
This behavior appears very often in physics, as in radioactive decay or energy dissipated in a damped harmonic oscillation.
You can say that dice have a lot in common with springs and uranium too
I finally understand the derivation from first principles.
Can you please solve an electrical mesh using geometric algebra of minkowski space. Im blown away how simple multi dimensional problems are solved without using matrices nor differential equations but application of geometric algebra is not shown for solving circuits
They all upset my stomach if I consume too much.
Now I want beer
Now I want soap
They are all topologically equivalent to a ball without holes or handles?
They exist