Hello LinearAq,
Does not even make sense in this context. If one can obtain different results using 1/4-life, 1/2-life, and 1/20483-life, who's to say which results are "right" or "wrong"? So long as they're all based upon observed decay rates, they're all valid abstractions of those decay rates.
And there you have it....CTD claims that you obtain different results using the radioactive decay formula.
I will admit that he didn't claim that the formula resulted in a linear graph, but the method he used to show 1/4-life to be different than 1/2-life REQUIRED the decay to be linear in order to be valid.
Yes, he keeps compounding his initial mistake - or keeps repeating his original deception/misunderstanding.
I forgot to add that CTD's math comments were correct in that two 1/4 lives would leave more material than one 1/2 life. However in
post 39563 on page 2 of the same thread:
Yes, the curve is exponential, and it's not hard to play tricks with the math. One could calculate 1/4 lives, 1/3 lives, or 1/277 lives for decaying elements. All will result in different exponential curves.
This is the statement that he is trying to support with his tricky math in an attempt to discredit the calculations involved in radioactive decay.
Yes this is the initial lie. The fact is that he never ended up with different exponential decay curves, just false results by NOT using the decay curve. All he demonstrates is that he is playing tricks on people that can't see his mistake because they don't fully understand the math.
You can find his post in the "How old was that again?" thread on page 3. It is the third post down on that page. It's
post number 39635. I have quoted the relevant portion below, but you can look at the whole thing and determine for yourself is I am quoting out of context.
This is pretty simple math. After the first 1/4 life, 3/4 remain. After the next 1/4 life, 9/16 remain. 9/16 ain't 1/2.
1/4 x 1 = 1/4 First ql
1/4 x 3/4 = 3/16 Second ql
1/4 + 3/16 = 7/16 Total
-7/16 + 1 = 9/16 Result
So 2 quarterlives do not give the same result as one halflife. Other fractions will do this too. Care to see four 1/8 lives?
Note that CTD is saying that two 1/4-lives SHOULD result in the same amount of radioactive decay as one 1/2-life in order for the radioactive decay dating methods to be valid. However a simple look at the formula by any mathematician will assure you that this is not true.
Yes, and any mathematician\scientist\engineer\etc. would also tell you that if you get different results from the original input data, the data that was used to
derive the formula, that the math is wrong, not the data. But let's look at where he defines his methodology:
In post 41757 on the "Evolution: The Big Joke" thread You & Linear's claim is just like claiming interest compounded annually = interest compounded weekly merely because you can take the annual rate and backtrack it to any given week and produce a result. The results will not be the same, and anyone who really cares can find out for themselves. This isn't hard at all.
This is where he first claims that the exponential decay curve is "just like" compound interest.
Then after I showed that converting compound interest from one period to another could produce results similar to the decay curve, he clarified
which compound interest formula he meant. In
post 41901 he said:
Because he references the "Effective Interest" article on wikipedia (the second link)
and links to a site that calculates compound interest from a nominal interest rate, there is no question that the equation he means is "just like" the exponential decay curve is:
A = P(1 + (r/n))^(nt)
Where
* P = principal amount (initial investment)
* r = annual nominal interest rate (as a decimal)
* n = number of times the interest is compounded per year
* t = number of years
* A = amount after time t Let's check this against his original post to confirm it:
This is pretty simple math. After the first 1/4 life, 3/4 remain. After the next 1/4 life, 9/16 remain. 9/16 ain't 1/2.
1/4 x 1 = 1/4 First ql
1/4 x 3/4 = 3/16 Second ql
1/4 + 3/16 = 7/16 Total
-7/16 + 1 = 9/16 Result
So 2 quarterlives do not give the same result as one halflife. Other fractions will do this too. Care to see four 1/8 lives?
We'll start with the half life to set the constant values in the formula with the original input data:
* P = inital amount = 100%
* r = annual nominal growth/decay rate (as a decimal)
* n = number of times the interest is compounded per half-life = 1
* t = number of half-lives = 1
* A = amount after 1 half-life = 50% (by definition)
Solve for r:
50 = 100(1+(r/1))^(1*1) = 100(1+r)
(1+r) = 50/100 = 0.50
r = 0.50-1.00 = -0.50 (negative 50%, right on cue, matches the original input dats)
Now we solve for his value (not the decay curve value) at half of a half-life, where n = 2:
A = 100(1+(-0.50/2))^(2*1) = 100(1-0.25)^2 = 100(0.75)^2 = 56.25% == 9/16
Yep, it matches the CTD calculation above. This does not, however, match the original input data used to develop the rate, so it is wrong, but where does the error come from?
It comes from using the wrong formula to calculate the value at time t = 1/2 half-life.
Rather than show that the decay rate changes, all CTD has done is confirm that the interest formula is the wrong formula to use. This means that he has not shown that the
Nt = No*e^(-?t)
Where
Nt = The amount of material at a particular time
No = The initial amount of material.
t = The time expended
? = the decay rate(where ? differs between materials, but constant for each material)
changes with different "compounding" periods, as was his original, falsified, claim.
Not surprisingly the real exponential decay curve matches the original input data every time. No matter where you start on the curve, when you add 1 half-life you will have half the radioactive isotope than at the start of that half-life. CTD's formula does not do this either.
If you ask a mathematician to explain it he will tell you that in the formula
Nt = No*e^(-?t)Nt is a function of one variable, t, usually expressed as Nt =
f(t)
And he will tell you that in the formula
A = P(1 + (r/n))^(nt)A is a function of two variables, n and t, usually expressed as A =
f(n,t)
When you look at it this way it should be obvious that if you have different results when you change n but keep t the same, that the change in A is not due to the change in t or any of the constants, but to the change in n, a factor absent in the exponential decay curve.
Is there a value for n that results in the correct value? Yes, and you could solve for it, but we know that one value where it works is n = 1 (see above). Like a broken watch that shows the correct time twice a day, having only 1 value of n that gives correct decay amounts for the original input data means that n is not a variable but a constant == 1 in order to compare with the exponential decay curve.
A = P(1+r)^(t)
Where
* P = inital amount = 100%
* r = -0.50 (from above)
* t = number of half-lives
* A = amount after 1 half-life = 50% (by definition)
A = P(1-0.5)^t = P(1/2)^tor if we want to measure t in years, compared to half-lives
A = P(1/2)^(t/Th)
Where
* P = inital amount
* r = -0.50 (from above)
* t = number of years
* Th = half-life in years
* A = amount after time tNot surprisingly this
is a valid formula for calculating decay at time t, and it matches the original input data every time. No matter where you start on the curve, when you add 1 half-life you will have half the radioactive isotope than at the start of that half-life. Can we show this is so?
From his wiki link
http://en.wikipedia.org/wiki/Compound_interest Continuous compounding
Continuous compounding can be thought as making the compounding period infinitely small; therefore achieved by taking the limit of n to infinity. One should consult definitions of the exponential function for the mathematical proof of this limit.
A(t) = Ao*e^rt
A common mnemonic device considers the equation in the form
A = P*e^rt
called 'PERT' where P is the principal amount, e is the base of the natural log, R is the rate per period, and T is the time (in the same units as the rate's period), and A is the final amount.
The effective interest rate per year is
i = e^r ? 1
Using this i the amount function can be written as:
A(t) = Ao(1 + i)^t
or
A = P(1 + i)^t
Of course in our comparison to the exponential decay curve we would no longer use r as the ratio after 1 half-life (-0.5), but i as the ratio after 1 half-life, as it is the "effective interest rate" for that period, so i=-0.5, and we have the same formula as above (shock!).
The other time when the broken watch is correct is when n = infinity.
REALITY CHECKJust for fun, lets assume that CTD by some quirk of fate actually is right, and there is some n in nature that needs to be accounted for in the formula.
Let's also pretend we are CTD and insist that we use r = -0.5 even though we
KNOW the actual rate generated will be different, what will be the difference?
A = P*e^rt
Where
* P = inital amount = 100%
* r = -0.50 (from above)
* t = number of half-lives = 1
* A = amount after 1 half-life
A = 100*e^(-0.5) = 100/e^0.5 = 60.65%
You can run a lot of calculations from n = 1 to n = infinity and you will see that A goes from 0.50 to 0.6065 along another exponential curve, thus the value will always be between 0.50 and 0.6065 with this formula.
What does that mean for radioactive decay rates and dating by radioactive decay? It would mean that the time necessary to reduce the original amount to 50% increases:
* P = inital amount = 100%
* r = -0.50 (from above)
* t = number of half-lives
* A = 50% after time t
50 = 100*e^-0.5t
Solve for t
50/100 = 0.5 = e^-0.5t
ln(0.5) = -0.5t
-0.693 = -0.5t
0.693/0.5 = t = 1.19 half-lives ...
Thus it take 20% longer for the amounts to decay to the levels used to calculate the age of the earth, and the age of the earth is increased by 20%. Thus CTD's manipulation of formulas does not make the earth younger.
Curious, then, that he is so adamant that he is correct eh? Maybe he is an evolutionist mole?
Enjoy.
