Isotopic 15N compisition of soil
Beni Stocker
4/2/2018
Case 1
Problem: Find the steady-state isotopic signature of (bulk) soil \(\delta_S\).
Nitrogen is added to (bulk) soil with an isotopic signature of \(\delta_I\). \[ \delta = \Bigg( \frac{R}{ R_{\text{Std}} } - 1 \Bigg) \cdot 1000 \] \[ R = \Big( \frac{^{15}N}{^{14}N} \Big) \] \(R_{\text{Std}}\) is the molar ratio of a given standard (here, atmospheric N\(_2\)).
N is lost through a single process (1st-order decay) with an isotopic discrimination factor of \(\varepsilon\). \[ \varepsilon = \Bigg( \frac{^{14}k}{^{14}k} -1 \Bigg) \cdot 1000 \] \(^ik\) is the rate constant of the decay process for isotopes \(^iN\), \(i=14, 15\).
Steady-state of a first-order decay process: \[ \frac{d^iN}{dt} = ^iN_I - ^ikN_S = 0 \]
Amount of \(^{15}N\) entering soil: \[ ^{15}N_I = (\delta_I/1000 + 1)\; R_{\text{Std}}\; ^{14}N_I \]
Steady-state amount of soil \(^15N\): \[ \begin{align} ^{15}N_S &= \frac{1}{^{15}k}^{15}N_I \\ \\ &= \frac{1}{ \frac{^{14}k}{\varepsilon/1000 + 1} } (\delta_I/1000 + 1)\; R_{\text{Std}}\; ^{14}N_I \\ \\ &= \frac{(\varepsilon/1000 + 1)\;(\delta_I/1000 + 1)\;R_{\text{Std}}\; ^{14}N_I}{^{14}k} \end{align} \] This can be expressed as a \(\delta\): \[ \delta_S / 1000 + 1 = (\varepsilon/1000 + 1)\;(\delta_I/1000 + 1) \] Now, we can use the approximation \(\log(1 + \varepsilon) = \varepsilon\), valid for small \(\varepsilon\), take the natural logarithm on both sides of the equation and write \[ \delta_S = \varepsilon + \delta_I \]
Case 2
Problem: Find the steady-state isotopic signature of (bulk) soil \(\delta_S\).
All as above, but now N is lost through two different processes, gaseous and leaching pathways, which fractionate with \(\varepsilon_G\) and \(\varepsilon_L\), respectively. The fraction of N lost through gaseous pathways is \(f\), the fraction through leaching pathways is \((1-f)\). \(k\) is the decay rate of combined losses of \(^{14}N\).
- Steady-state (analogous to case 1): \[ \frac{d^{15}N_S}{dt} = \; ^{15}N_I - \Big( \frac{f}{\varepsilon_G/1000+1} + \frac{(1-f)}{\varepsilon_L/1000+1} \Big) \;k\; ^{15}N_S = 0 \] Solving for \(^{15}N_S\) and expressing \(^{15}N_I\) in terms of \(^{14}N_I\): \[ ^{15}N_S = \frac{(\delta_I/1000+1)\;R_{\text{Std}}\;^{14}N_I}{k\;\Big(\frac{f}{\varepsilon_G/1000+1} + \frac{1-f}{\varepsilon_L/1000+1} \Big)} \] And expressing this as a \(\delta\): \[ \delta_S / 1000 + 1 = \frac{\delta_I/1000+1}{\frac{f}{\varepsilon_G/1000+1} + \frac{1-f}{\varepsilon_L/1000+1} } \]
The question is: Why is above equation identical to, or an approximation, to the following equation: \[ \delta_S = \delta_I + f\varepsilon_G + (1-f)\varepsilon_L \] ???
I don’t understand the steps in between. I guess I have to use again \(\log(1 + \varepsilon) = \varepsilon\), valid for small \(\varepsilon\), but what else?
I can see how this is true by solving the steady-state numerically, and analytically using the exact and approximative solutions. Here, using \(\delta_I=0\), \(\varepsilon_L=0\), \(\varepsilon_G=10\), and \(f=0.6\)
source("n15soil.R")## [1] "delta of input: 0"
## [1] "Soil delta: 5.97609561752988"