# Triple-alpha reaction rate

I'm trying to find a general equation for the fusion rate of the triple alpha process. I found this equation:

The rate of energy generated by this process as a power law in $T$, centered around $10^8\,\rm K$ is $$\epsilon_{3\alpha}\approx\epsilon_{0,3\alpha}'\rho^2X_{He}^3\left(\frac{T}{10^8\,\rm K}\right)^{41}$$

But I couldn't find any value for the $\epsilon_{0,3\alpha}$ constant. I also found this equation:

$$\epsilon_{3\alpha}\approx3\times10^{11}X_4^3\rho^2T_8^{-3}e^{-43.5T^{-1}_8}\qquad[\rm erg\,s^{-1}\,g^{-1}]$$

But I cannot understand the negative -43.5 exponent, given that the rate of fusion is supposed to exponentially increase as temperature increases from what I have read. Can anyone help me understand what is going on?

EDIT: Source for first link (something from Duke): http://webhome.phy.duke.edu/~mkruse/PHY105_S11/Stellar_Reactions_2.pdf

Source for second (from Princeton - it was a .ps file, so I downloaded, converted to pdf, and uploaded to Dropbox.): https://www.dropbox.com/s/1vxuepijyloiu68/rates.three%20%281%29.pdf?dl=0

I should also mention that $\rho$ is density and X is the concentration of Helium.

• I don't really know your topic but I can say that the function for which $T \mapsto \frac{1}{T}$ is decreasing. So $T \mapsto \frac{-43.5}{T}$ is increasing. Yet we know that $x \mapsto e^{x}$ is increasing. Therefore $T \mapsto e^{\frac{-43.5}{T}}$ is an increasing function as it it the composition of two increasing functions. Assuming that $T_8$ is the temperature. It doesn't solve the problem of the expronential growth but at least it is not the fault of the exposent. Aug 21 '18 at 11:03
• Can you give references for the two formulas you wrote down? Aug 21 '18 at 12:39
• @JohnDonne Just added references. Aug 21 '18 at 22:43

The first equation is an approximation of the second, that is valid when $T \simeq 10^{8}$ K. The exponential factor in the second equation is $\exp(-43.5/T_8)$, where $T_8$ is the temperature in units of $10^{8}$K, so the energy generation does indeed increase rapidly as the temperature goes up.
The trick then is to try and recast this as a power law function that will be roughly correct in some narrow temperature range - i.e. $$\exp(-43.5/T_8) \sim A T_8^{\gamma}$$
If we differentiate both sides with respect to $T_8$ $$\frac{43.5}{T_8^2} \exp(-43.5/T_8) \sim \gamma A T_8^{\gamma-1}$$ $$\frac{43.5}{T_8^2} AT_8^{\gamma} \sim \gamma AT_8^{\gamma -1}$$ and so $$\gamma \sim 43.5 T_{8}^{-1}$$ and at $T_8=1$, then $\gamma = 43.5$. Then, since the second equation also has a leading factor of $T_{8}^{-3}$, the overall temperature dependence of the energy generation rate would be $\propto T_8^{40.5}$. I believe this is the origin of the approximate temperature dependence of $T^{41}$ in the first equation.
The first and second equations should give equivalent energy generation rates at $10^8$ K. So if you want to find out what the constant is in the first equation just set $T_8=1$ in both equations, equate them to each other, and obtain: $$\epsilon'_{0,3\alpha} = 3\times10^{11} \exp(-43.5) =4\times10^{-8}$$ to give an energy generation rate in ergs s$^{-1}$ g$^{-1}$, and where I have only used one significant figure as per your first equation.