Talk:Bateman equation

There is both a summation and a product involving index j, which is rather confusing. Would someone please modify this to use a third index variable?

Bateman Equation

The equation seems to have an error in the product term, an error inherited from reference 3. The two methods of noticing the inconsistency in the old formula:

- the decay constant ( $\lambda$ ) has units of 1/time, and so the final equation must have equal numbers of $\lambda$ terms in the numerator and denominator

- considering the N=2 case, the Bateman equation should reduce to:

N_{2}(t)=\lambda _{1}N_{1}(0)\left({\frac {e^{-\lambda _{1}t}}{\lambda _{2}-\lambda _{1}}}+{\frac {e^{-\lambda _{2}t}}{\lambda _{1}-\lambda _{2}}}\right)+N_{2}(0)e^{-\lambda _{2}t}

instead the previous version of the equation would simplify to:

N_{2}(t)=\lambda _{1}N_{1}(0)\left({\frac {e^{-\lambda _{1}t}}{\lambda _{2}-\lambda _{1}}}+{\frac {e^{-\lambda _{2}t}}{\lambda _{1}-\lambda _{2}}}\right)+\lambda _{1}N_{2}(0)e^{-\lambda _{2}t}

where the extra $\lambda _{1}$ in the final term presents both a unit problem and gives a illogical answer at t=0:

N_{2}(t=0)=\lambda _{1}N_{2}(0)

I do not have a source to confirm this math, but it checks out thoroughly in test cases and never will result in a unit problem. Further I will note that that both the new and old are valid solutions to the differential equations, because multiplication by constants will yield the same derivative; so no clarity can be obtained from there.

Since this equation is such a mess, I've added a lot of parentheses and brackets to clarify order of operation and extend of the sums/products. This may violate some style guide and I would welcome improvement in presentation. The rearrangement of the summation over i and product over j is important, as the product should depend on the range of i.

These edits do not explicitly address comment 1 above, but the additional parentheses may clarify the use of the j notation to render the concern accounted for. — Preceding unsigned comment added by 130.132.173.207 (talk) 20:12, 7 August 2015 (UTC)[reply]

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E.g. for the simple case of a chain of three isotopes with only the grand-parent present at time 0, the corresponding Bateman equation reduces to:

$A{\xrightarrow {\lambda _{A}}}B{\xrightarrow {\lambda _{B}}}C$

N_{A}=N_{A_{0}}e^{-\lambda _{A}t}

N_{B}=N_{A_{0}}{\frac {\lambda _{A}}{\lambda _{B}-\lambda _{A}}}(e^{-\lambda _{A}t}-e^{-\lambda _{B}t})

N_{C}=N_{A_{0}}\lambda _{B}\lambda _{A}\left({\frac {e^{-\lambda _{A}t}}{(\lambda _{B}-\lambda _{B})\lambda _{B}}})+{\frac {e^{-\lambda _{B}t}}{(\lambda _{B}-\lambda _{B})\lambda _{B}}})+{\frac {1}{(\lambda _{A}\lambda _{B}}})\right)