Time to prove that you do not have to assume the
following in radiometric dating.
1) There was no daughter isotope present at the starts.
2) There has been no gain from or loss to outside the system.
We will look at two cases. Case 1 will be where #2 above
is true and case 2 is were #2 above is false.
Case 1: Completely closed system: no loss or gain of
the isotopes involved.
P will stand for parent isotope; D will stand for the daughter.
P decays into D if it was not already obvious. Orig will
be short for original. k will be the decay constant. Usually
the Greek letter lambda is used for it, but using k avoids
browser compatibility problems and is easier to type.
e is the base of the natural logarithms and is about 2.718281828.
t is the amount of time that has passed.
P[sub]now[/sub] = P[sub]orig[/sub] e[sup]-kt[/sup]
This is the basic equation of the behavior of radioactive
decay. Anyone who has taken calculus should be familiar
with it. Elementary textbooks in many subjects cover it
as well.
Now remember in this case we are assuming no loss or gain of
atoms to or from the outside enviroment (which will be dealt
with in case 2). The following should be obvious since
it is a direct consequence of the assumption:
P[sub]orig[/sub] = P[sub]now[/sub] + D[sub]formed[/sub]
where the last term is amount of the daughter formed due
to radioactive decay. (The actual amount of daughter will
either be equal to or more likely greater than this amount.)
The second equation can be substituted into the first equation.
P[sub]now[/sub] = ( P[sub]now[/sub] + D[sub]formed[/sub] ) e[sup]-kt[/sup]
P[sub]now[/sub] e[sup]kt[/sup] = P[sub]now[/sub] + D[sub]formed[/sub]
D[sub]formed[/sub] = P[sub]now[/sub] e[sup]kt[/sup] - P[sub]now[/sub]
D[sub]now[/sub] - D[sub]orig[/sub] = P[sub]now[/sub] ( e[sup]kt[/sup] - 1 )
D[sub]now[/sub] = D[sub]orig[/sub] + P[sub]now[/sub] ( e[sup]kt[/sup] - 1 )
Now for any given element there are more than one isotopes. We will call
D[sub]i[/sub] the amount of non-radioactive isotope of the D's element.
This isotope does not radioactively decay, nor is is the product of
another radioactive decay. Since in case 1 there is not gain or loss
of isotopes and since it is not affected by radioactive processes, this
is a constant.
Divide both sides by it with a few rearrangment of terms:
D[sub]now[/sub]/D[sub]i[/sub] =(e[sup]kt[/sup]-1)(P[sub]now[/sub]/D[sub]i[/sub]) + (D/D[sub]i[/sub])[sub]orig[/sub]
This above equation is in the form of
y = mx + b
which if you recall from junior high is the equation of a line. The
date of the rock can be determined by the slope (m) of the line and
the amount of daughter present when the rock first solidified from
lava can be determined from the y-intercept (b). Basically
the isotope ratios will have to be measured for multiple locations
of the rock.
This proof does depend on one assumption that I have not mentioned
so far. It assumes that every part of the rock when formed the radio
of the radioactive daughter to the non-radioactive isotope of the same
element as the daughter is the same. This is quite reasonable.
The different isotopes of an element are chemically identical and there
physical properties are almost identical especially for heavy elements
(which we are dealing with) where the difference in mass between the
isotopes is trivial. That this assumption is true is seen everyday
in the real world and is true in lava formed today. This is extremely
well understood stuff. Besides if it was not true, the line would
not form. We don't have to assume the parent to daughter
ratio will initally vary though it would be a very safe assumption do to
basic physics and chemistry. The reason for this is if they did not
the plot would just be a single point, if we see a line formed that must
have varied.)
Solving for the age of the rock is math is covered in high school and will
be left to the reader.
Thus if the rock does not gain or loose atoms to or from the outside
enviroment we have shown that we can date the rock without knowledge
of the initial composition of the rock.
Case 2: The rock is an open system: atoms lost or gained
This one is really quite simple. If the system was not closed the
above proof is invalid and there is no reason to expect that a line
will form. Simple playing with graphs should be convincing that
any gain or loss will destroy the line. It would take a huge
coincidence for the line to be preserved -- highly improbably.
Thus when the data is plotted the if a line is formed, the slope is
the key to the age. And if a line is not formed, one knows that the
the system has be disturbed and the test will not give a result.
A graphical approach without the mathematical formalism to this
can be found at the Isochron Dating
FAQ. This article also deals with creationists objections and potential
problems.
I have not exhausted what geologists can do do date rocks either.
Ain't math grand!
following in radiometric dating.
1) There was no daughter isotope present at the starts.
2) There has been no gain from or loss to outside the system.
We will look at two cases. Case 1 will be where #2 above
is true and case 2 is were #2 above is false.
Case 1: Completely closed system: no loss or gain of
the isotopes involved.
P will stand for parent isotope; D will stand for the daughter.
P decays into D if it was not already obvious. Orig will
be short for original. k will be the decay constant. Usually
the Greek letter lambda is used for it, but using k avoids
browser compatibility problems and is easier to type.
e is the base of the natural logarithms and is about 2.718281828.
t is the amount of time that has passed.
P[sub]now[/sub] = P[sub]orig[/sub] e[sup]-kt[/sup]
This is the basic equation of the behavior of radioactive
decay. Anyone who has taken calculus should be familiar
with it. Elementary textbooks in many subjects cover it
as well.
Now remember in this case we are assuming no loss or gain of
atoms to or from the outside enviroment (which will be dealt
with in case 2). The following should be obvious since
it is a direct consequence of the assumption:
P[sub]orig[/sub] = P[sub]now[/sub] + D[sub]formed[/sub]
where the last term is amount of the daughter formed due
to radioactive decay. (The actual amount of daughter will
either be equal to or more likely greater than this amount.)
The second equation can be substituted into the first equation.
P[sub]now[/sub] = ( P[sub]now[/sub] + D[sub]formed[/sub] ) e[sup]-kt[/sup]
P[sub]now[/sub] e[sup]kt[/sup] = P[sub]now[/sub] + D[sub]formed[/sub]
D[sub]formed[/sub] = P[sub]now[/sub] e[sup]kt[/sup] - P[sub]now[/sub]
D[sub]now[/sub] - D[sub]orig[/sub] = P[sub]now[/sub] ( e[sup]kt[/sup] - 1 )
D[sub]now[/sub] = D[sub]orig[/sub] + P[sub]now[/sub] ( e[sup]kt[/sup] - 1 )
Now for any given element there are more than one isotopes. We will call
D[sub]i[/sub] the amount of non-radioactive isotope of the D's element.
This isotope does not radioactively decay, nor is is the product of
another radioactive decay. Since in case 1 there is not gain or loss
of isotopes and since it is not affected by radioactive processes, this
is a constant.
Divide both sides by it with a few rearrangment of terms:
D[sub]now[/sub]/D[sub]i[/sub] =(e[sup]kt[/sup]-1)(P[sub]now[/sub]/D[sub]i[/sub]) + (D/D[sub]i[/sub])[sub]orig[/sub]
This above equation is in the form of
y = mx + b
which if you recall from junior high is the equation of a line. The
date of the rock can be determined by the slope (m) of the line and
the amount of daughter present when the rock first solidified from
lava can be determined from the y-intercept (b). Basically
the isotope ratios will have to be measured for multiple locations
of the rock.
This proof does depend on one assumption that I have not mentioned
so far. It assumes that every part of the rock when formed the radio
of the radioactive daughter to the non-radioactive isotope of the same
element as the daughter is the same. This is quite reasonable.
The different isotopes of an element are chemically identical and there
physical properties are almost identical especially for heavy elements
(which we are dealing with) where the difference in mass between the
isotopes is trivial. That this assumption is true is seen everyday
in the real world and is true in lava formed today. This is extremely
well understood stuff. Besides if it was not true, the line would
not form. We don't have to assume the parent to daughter
ratio will initally vary though it would be a very safe assumption do to
basic physics and chemistry. The reason for this is if they did not
the plot would just be a single point, if we see a line formed that must
have varied.)
Solving for the age of the rock is math is covered in high school and will
be left to the reader.
Thus if the rock does not gain or loose atoms to or from the outside
enviroment we have shown that we can date the rock without knowledge
of the initial composition of the rock.
Case 2: The rock is an open system: atoms lost or gained
This one is really quite simple. If the system was not closed the
above proof is invalid and there is no reason to expect that a line
will form. Simple playing with graphs should be convincing that
any gain or loss will destroy the line. It would take a huge
coincidence for the line to be preserved -- highly improbably.
Thus when the data is plotted the if a line is formed, the slope is
the key to the age. And if a line is not formed, one knows that the
the system has be disturbed and the test will not give a result.
A graphical approach without the mathematical formalism to this
can be found at the Isochron Dating
FAQ. This article also deals with creationists objections and potential
problems.
I have not exhausted what geologists can do do date rocks either.
Ain't math grand!
