Updates to flow model calculations

Author: mklilley

Flow model transfer rates.md

@@ -2,22 +2,94 @@
 
 During start-up when the power associated with fusion events is low, we expect to only have excitation transfer via "direct" fusion transitions that go from $\rm D_2 \rightarrow ^4He$. As things get going, contributions from "indirect" fusion transitions starts to dominate - this is when $\rm D_2 \rightarrow (3+1) \rightarrow ^4He$. We represent the transfer rates as $\Gamma_{D_2/He}$ for direct and $\Gamma_{D_2/3+1/He}$ for indirect.
 
+## Reminder from the SI
+
+### Single transition
+
+If we assume there exists a single $\rm Pd$ transition that can accommodate the $\Delta M c^2  \approx 23.85 \, \rm MeV$  fusion energy, then the Dicke enhanced excitation transfer rate from $\rm D_2$ to $\rm Pd$ is given by: 
+
+$$
+\Gamma_{transfer} = {2 \over \hbar} |1 - \eta| {\mathcal{U} \mathcal{V} \over \Delta Mc^2}
+$$
+
+where the Dicke enhanced $a\cdot cP$ matrix elements for the  $\rm D_2$ and $\rm Pd$ transitions are
+
+$$
+\mathcal{U} = |\langle {\rm D}_2  | a_z | ^4{\rm He} \rangle| c \sqrt{\langle P_{\rm ^4He}^2 \rangle}\sqrt{N_{\rm D_{2}}}
+$$
+
+$$
+\mathcal{V} = |\langle {\rm Pd}^* | a_z | {\rm Pd} \rangle|
+c \sqrt{\langle P_{\rm Pd}^2 \rangle} \sqrt{N_{\rm Pd}}
+$$
+
+and $|1 - \eta|$ tells us the degree to which we have broken destructive interference (with 1 being the maximum possible). The square root Dicke factors give the number of  $\rm D_2$ and $\rm Pd$ nuclei inside a coherence domain.
+
+### Multiple transitions
+
+When the transfer of the $\Delta M c^2  \approx 23.85 \, \rm MeV$  fusion energy is shared across a number $\rm Pd$ transitions $n_{tr}$, then the Dicke enhanced transfer rate from $\rm D_2$ to $\rm Pd$ is given by: 
+
+$$
+\Gamma_{transfer} = {2 \over \hbar} |1 - \eta| \,\mathcal{U} \prod_{i=1}^{n_{tr}}\left({\mathcal{V}(\epsilon_i) \over \epsilon_i} \right)
+\label{eq:mutiple_trans}
+$$
+
+where the energy of the participating $\rm Pd$ transitions must add up to the fusion energy  $\sum_i \epsilon_i = 23.85 \, \rm MeV$. 
+
+We write the normalised Dicke enhanced $\rm Pd$ matrix elements
+
+$$
+g_i \equiv {\mathcal{V}(\epsilon_i) \over \epsilon_i}
+$$
+
+so that we can more conveniently write the transfer rate as
+
+$$
+\Gamma_{transfer} = {2 \over \hbar} |1 - \eta| \,\mathcal{U} \prod_{i=1}^{n_{tr}}g_i
+$$
+
+Because the $\rm Pd$ transitions are not suppressed by the Coulomb barrier, we can enter the strong coupling regime from the perspective of the $\rm Pd$ transitions, in other words $g \gtrsim 1$. In the strong coupling regime, the perturbative approach that leads to Eq. $\ref{eq:mutiple_trans}$  no longer holds. We can attempt a crude correction for the strong coupling by effectively capping the $g'\rm s$ in the following way
+
+$$
+\Gamma_{transfer} = {2 \over \hbar} |1 - \eta| \,\mathcal{U} \prod_{i=1}^{n_{tr}}F(g_i)
+$$
+
+where
+
+$$
+F(g) \sim \frac{g}{1+2g}
+$$
+
 ## Direct fusion transitions
 
 ### Multiple transitions, single receiver
 
-We will consider the excitation transfer from $\rm D_2$ to a single receiver state made up of four $\rm Pd$ transitions. For the sake of this calculation, we assume that there exists four palladium transitions whose combined energy matches the $\approx 24 \, \rm MeV$ fusion energy. The transfer rate has been calculated by Peter to be:
+We will consider the excitation transfer from $\rm D_2$ to a single receiver state made up of four $\rm Pd$ transitions. The transfer rate has been calculated by Peter to be:
+
 $$
 \Gamma_{D_2/He}^{(0)} \sim 1.3 \times 10^2 \,
-\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right|  \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3/4} g^4
-\sqrt{ \frac{P_D}{1\,\mathrm{W}}  } \sqrt{\frac{N_{D_2}}{N}} \sqrt{N_{^4 He}}
-\,\sec^{-1} 
+\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right|  \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3/4}
+\sqrt{ \frac{P_{diss}}{1\,\mathrm{W}}  } \sqrt{\frac{N_{D_2}}{N}} \sqrt{N_{^4 He}} F(g)^4
 \label{eq:gamma_direct}
 $$
 
 > Note that we're not sure on the helium Dicke factor yet.
 
-Here $g$ is the normalised coupling constant for the $\rm Pd$ transitions:
+The asymmetry factor $\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right|$ arrises due to fusion loss and for this "direct" fusion transition Pete calculated this to be:
+
+$$
+\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right| \approx \frac{1}{80}
+$$
+Substituting back into Eq. $\ref{eq:gamma_direct}$ rate we get:
+$$
+\Gamma_{D_2/He}^{(0)} \sim 1.6 \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3/4}
+\sqrt{ \frac{P_{diss}}{1\,\mathrm{W}}  } \sqrt{\frac{N_{D_2}}{N}} \sqrt{N_{^4 He}} F(g)^4
+\label{eq:gamma_direct2}
+$$
+
+Technically we should have  $F(g_1)F(g_2)F(g_3)F(g_4)$ instead of $F(g)^4$ to allow for the possibility of four different $\rm Pd$ transitions. We will however treat them in an averaged sense for now so that $\epsilon  \approx 24 \, \rm MeV / 4$ ; Peter uses $\epsilon = \ 6.75 \, \rm MeV$.
+
+To get an expression for $g$ , we need to express the matrix element in terms of acoustic phonon mode energy $E_A = P_{diss}\tau_A$
 $$
 \begin{equation}
 \begin{aligned}
@@ -26,26 +98,23 @@ g
 \sqrt{\langle P_{\mathrm{Pd}}^2 \rangle} }{ \epsilon}
 \sqrt{N_{\mathrm{Pd}}} \\[6pt]
 &= \frac{ |\langle \mathrm{Pd}^* | a_z | \mathrm{Pd} \rangle| }{ \epsilon}
-\sqrt{\frac{M_{Pd}c^2 P_D \tau_A}{N}}
+\sqrt{\frac{M_{Pd}c^2 P_{diss} \tau_A}{N}}
 \sqrt{N_{\mathrm{Pd}}} \\[6pt]
 &= \frac{ |\langle \mathrm{Pd}^* | a_z | \mathrm{Pd} \rangle| \,
-\sqrt{M_{Pd}c^2 P_D \tau_A} }{ \epsilon}
+\sqrt{M_{Pd}c^2 P_{diss} \tau_A} }{ \epsilon}
 \sqrt{\frac{N_{\mathrm{Pd}}}{N}} .
 \end{aligned}
 \end{equation}
 $$
-Technically we should have  $g_1g_2g_3g_4$ instead of $g^4$ to allow for the possibility of four different $\rm Pd$ transitions. We will however treat them in an averaged sense for now so that $\epsilon = 6.75 \, \rm MeV \approx 24 \, \rm MeV / 4$.
 
-
-Peter averages over all the possibe $\rm Pd$ transitions to get 
+Now we can bring in Peter's calculation for $\Upsilon$ averaged over all the $\rm Pd$ transitions
 
 $$
 \bar \Upsilon = |\langle \mathrm{Pd}^* | a_z | \mathrm{Pd} \rangle|^2 \frac{M_{Pd}c^2}{\epsilon} \approx 8.7 \times 10^{-8}
 $$
 
-This $\bar\Upsilon$ needs to be divided by four - the number of transitions.
+This $\bar\Upsilon$ needs to be divided by four - the number of transitions. We can connect $\bar \Upsilon$ to $g$ through:
 
-We can connect $\bar \Upsilon$ to $g$ through:
 $$
 \frac{|\langle \mathrm{Pd}^* | a_z | \mathrm{Pd} \rangle| \sqrt{M_{Pd}c^2}}{\epsilon} = \sqrt{\frac{\bar \Upsilon}{4 \epsilon}} \approx \sqrt{\frac{8.7 \times 10^{-8}}{4\times6.75 \times 10^6 \times 1.6\times 10^{-19}\, \rm J}} \approx 142
 $$
@@ -54,7 +123,7 @@ $$
 This gives
 
 $$
-g \approx \ 142 \sqrt{\frac{P_D \tau_A}{1 \, \rm J}} \sqrt{\frac{N_{\mathrm{Pd}}}{N}}
+g \approx \ 142 \sqrt{\frac{P_{diss} \tau_A}{1 \, \rm J}} \sqrt{\frac{N_{\mathrm{Pd}}}{N}}
 $$
 
 The phonon lifetime is given by
@@ -63,47 +132,59 @@ $$
 \tau_A = 10^{-12} \left( \frac{10^{-7} f_A}{1 \, \text{MHz}} \right)^{-3/2} \, \text{sec}
 $$
 
-For the transfer rate we need $g^4$:
+#### Small g
+
+In the weak coupling regime ($g \ll 1$) we have  $F(g) \approx g$ and we can get a nice expression for the transfer rate. For the transfer rate we need $g^4$
 
 $$
-g^4 \approx 4\times 10^8 \left(\frac{P_D }{1\, \rm W}\right)^2 10^{-24}\left(\frac{1 \, \rm  MHz}{10^{-7}f_A}\right)^3 \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2 
-=  4\times 10^5 \left(\frac{P_D}{1\, \rm W}\right)^2 \left(\frac{1 \, \rm  MHz}{f_A}\right)^3 \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2
+g^4 \approx 4\times 10^8 \left(\frac{P_{diss} }{1\, \rm W}\right)^2 10^{-24}\left(\frac{1 \, \rm  MHz}{10^{-7}f_A}\right)^3 \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2 
+=  4\times 10^5 \left(\frac{P_{diss}}{1\, \rm W}\right)^2 \left(\frac{1 \, \rm  MHz}{f_A}\right)^3 \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2
 $$
 
-Substituting back into Eq. $\ref{eq:gamma_direct}$ gives:
+Substituting back into Eq. $\ref{eq:gamma_direct2}$ gives:
 
 $$
-\Gamma_{D_2/He}^{(0)} \sim 5.2 \times 10^7 \,
-\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right|  \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3.75}
+\Gamma_{D_2/He}^{(0)} \sim 6.4 \times 10^5  \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3.75}
 \left(\frac{P_D}{1\,\mathrm{W}}  \right)^{2.5} \sqrt{\frac{N_{D_2}}{N}} \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2\sqrt{N_{^4 He}}
-\,\sec^{-1}
-\label{eq:gamma_direct2}
+\label{eq:gamma_direct3}
 $$
 
-The asymmetry factor $\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right|$ arrises due to fusion loss and for this "direct" fusion transition Pete calculated this to be:
+Let's use some numbers from the SI, and setting $N_{^4He}=1$ for now:
+$$
+f_A \sim 5 \, \rm MHz, \quad \frac{N_{D_2}}{N} \sim 0.25 \times \frac{1}{9}, \quad \frac{N_{{Pd}}}{N} \sim 0.25
+$$
+
+The acoustic mode lifetime is $\tau_A\approx \, \rm 3ms$  and so
 
 $$
-\left|1 - \eta_{D_2/^4\mathrm{He}}\eta_{P_d}\right| \approx \frac{1}{80}
+g = 3.8\sqrt{\frac{P_{diss}}{1 \, \rm W}}
+$$
+
+
+For $P_{diss} \lesssim 1 \, \rm mW$ we are in the perturbative regime where Eq. $\ref{eq:gamma_direct3}$ can be used. Let's look at transfer rates for $1\, \rm mW$ and $1\, \rm \mu W$
+
+
+$$
+\Gamma_{D_2/He}^{(0)}\, (1 \, \rm mW) \approx 5\times 10^{-7} \, \rm s^{-1}
 $$
-Substituting back into Eq. $\ref{eq:gamma_direct2}$ rate we get:
 
 $$
-\Gamma_{D_2/He}^{(0)} \sim  6.5\times 10^5  \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3.75}
-\left(\frac{P_D}{1\,\mathrm{W}}  \right)^{2.5} \sqrt{\frac{N_{D_2}}{N}} \left(\frac{N_{\mathrm{Pd}}}{N}\right)^2\sqrt{N_{^4 He}}
-\,\sec^{-1}
-\label{eq:gamma_direct3}
+\Gamma_{D_2/He}^{(0)}\, (1 \, \rm \mu W) \approx 1.6\times 10^{-14} \, \rm s^{-1}
 $$
 
-Let's use numbers from the SI, and setting $N_{^4He}=1$ for now:
+#### Large g
+
+In the strong coupling regime ($g \gtrsim 1$) we have $F(g) \approx \frac{1}{2}$ and we can also get a nice expression for the transfer rate. Substituting into Eq. $\ref{eq:gamma_direct2}$ gives
 
 $$
-f_A \sim 5 \, \rm MHz, \quad P_D \sim 1 \, W, \quad \frac{N_{D_2}}{N} \sim 0.25 \times \frac{1}{9}, \quad \frac{N_{{Pd}}}{N} \sim 0.25
+\Gamma_{D_2/He}^{(0)} \sim \frac{1}{10} \left( \frac{1\,\mathrm{MHz}}{f_A} \right)^{3/4}
+\sqrt{ \frac{P_{diss}}{1\,\mathrm{W}}  } \sqrt{\frac{N_{D_2}}{N}} \sqrt{N_{^4 He}}
 $$
 
-The rate then evaluates to:
+Using $P_{diss}\sim 1 \, \rm W$ from the SI and setting $N_{^4He}=1$ for now, we get a rate of
 
 $$
-\Gamma_{D_2/He}^{(0)} \approx 16 \sec^{-1}
+\Gamma_{D_2/He}^{(0)}\, (1 \, \rm W)\approx 2\times 10^{-3} \, s^{-1}
 $$
 
 ### Multiple transitions, multiple receivers
@@ -137,7 +218,7 @@ $$
 &= \hbar \omega_A
 \left( 2 n_j \bar{Y}_j |e^{(A)}|^2 n_A \right)^{1/3} \\
 &= 8.4 \, |e^{(A)}|^{2/3} \, (n_j)^{1/3}
-\left( \frac{P_D^{(A)}}{1\,\mathrm{W}} \right)^{1/3}
+\left( \frac{P_{diss}^{(A)}}{1\,\mathrm{W}} \right)^{1/3}
 \left( \frac{f_A}{1\,\mathrm{MHz}} \right)^{1/6}
 \;\mathrm{meV}.
 \end{aligned}
@@ -151,7 +232,7 @@ $$
 $$
 
 
-This gives us an transfer rate of:
+This gives us an transfer rate of
 $$
-\Gamma_{D_2/He} \sim \Gamma_{D_2/He}^{(0)}  \times 6.8\times 10^5 \sim 10^7 \, \sec^{-1}
+\Gamma_{D_2/He} \sim \Gamma_{D_2/He}^{(0)}  \times 6.8\times 10^5 \sim 1400 \, s^{-1}
 $$