Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

This site http://what-if.xkcd.com/14/ states that during a helium flash, "the reaction rate is proportional to the 40th power of the temperature".

Taking for granted that this is true, how can such a large exponent arise in a physic equation? It just makes no sense to me that some physical process could give rise to this phenomenon.

An explanation of how it can arise for a helium flash would be satisfactory, but the question is really more general. How can a physical process accumulate enough factors to get up to x^40? Thermodynamic processes tend to be in the x^2 or x^3's and geometry can only add a few more x^3's.

Note that we are not talking about the scale of things (obviously the 40th power is not a big deal in the length or time scales), we are talking about a physics /equation/.

share|improve this question
1  
That's a good question. High powers may occur but I still think that the explanation here is different. The dependence isn't an exact 40th power in any sense. In fact, it is exponential – the fusion depends on quantum tunneling that probes the exponentially decreasing "tails" of the distribution and at higher temperatures, one may get much further in the tail, thus speeding up the rate exponentially - and the (approximately) 40th power is arguably just some best interpolation of the intrinsically exponential function for some interval of parameters. –  Luboš Motl Oct 4 '12 at 15:05
1  
I would assume it's roughly the value of the power derivative $\frac{d ln(f(r))}{d ln(r)}$, which gives the exponent of a the local approximate power law. It's probably an exponential law like @LubošMotl said. I remember once I did a calculation about stellar power production that yielded a power derivative of about 22, so why not 40? –  Benji Remez Oct 4 '12 at 18:32
1  
What I meant is that the quantity in question is my be an exponential function of the form $y(x) = e^x$. Both quantum tunneling and Boltzmann factors are exponential, and they both come into play in stellar fusion. Try looking up the Gamow Peak. Now, if you have some power law %f(r) = r^n%, notice that its logarithmic derivative %\frac{d(log(f(r))}{d(log(r))}$ is equal to %n%, so this yields the exponent of the law. In principle, it may be applied to any general function, and its value would give the appropriate exponent should you wish to locally approximate that function as a power law. –  Benji Remez Oct 5 '12 at 6:12
1  
When you have exponential behavior which goes as exp(ax - b x^2), where a is positive and b is positive, you can approximate the thing in the exponent as a logarithm, by scaling by a constant, and then the power can be whatever you like. –  Ron Maimon Oct 5 '12 at 7:39
1  
I don't have time to furnish a detailed answer now but I refer you to a free textbook, Collins's Fundamentals of Stellar Astrophysics. Section 3.3 discusses nuclear reactions rates. Table 3.4 provides approximate power-laws for various reaction rates. Note that different power-laws approximate different temperature ranges, and 40 is relevant at the temperature of core at the start of the flash, when the reaction rate is rising most steeply. –  Warrick Oct 5 '12 at 11:33
show 4 more comments

1 Answer

Expanding the comments above:

The quantity in question probably doesn't obey a power law in the traditional sense. It is most likely an exponential law $y(x) = e^x$, since it involves thermodynamic Boltzmann factors and quantum tunneling probabilities, and both are exponential.

Having said that, it can still be locally approximated as a power law. Consider a function $f(r) = r^n$. If you take its logarithmic derivative $\dfrac{d(ln\ f(r))}{d(ln\ r)}$, you recover the exponent $n$. You may also apply this derivative to any general function. The value you obtain (as a function of $r$) is then the appropriate exponent should you wish to use a power-law local approximation.

Check out this document; it both shows the exponential calculations and the use of this derivative.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.