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sirapwm
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- TL;DR Summary
- Interesting, strictly relativistic effect at threshold
I was surprised this morning when I got off on a tangent regarding the amount of energy available in the laboratory frame just at threshold. It reveals an interesting relativistic effect.
Consider a reaction (I'm thinking in terms of nuclei and/or particles) ##1 + 2 \to 3 + 4 + \cdots## with masses ##m_1, m_2## in the initial state and any number of final state particles with masses ##m_3, m_4, \ldots##. Let $$m_i = m_1 + m_2, ~ m_f = m_3 + m_4 + \cdots.$$ The value ##Q= m_i - m_f## is positive, negative or zero for the cases exothermic (or exoergic -- I'll consider these interchangeable), endothermic, and elastic, respectively.
In the center-of-mass (CM) frame, the kinetic energy available due to the reaction is the sum of the energies of the products. We call this ##E'_{avail} = E_3' + E_4'##, where these are the kinetic energies of the products and the primes indicated the CM frame. Conservation of energy in CM frame gives $$m_i + E_1' + E_2' = m_f + E_3' + E_4'.$$ This gives $$E'_{avail} = Q + E_i',$$ where ##E'_i = E_1' + E_2'##.
The endothermic case ##Q<0##, at the threshold for the reaction ##E_i' = -Q## has zero energy available to the products: $$E'_{avail}(E_i'=-Q) = 0.$$ The question is then: what is the energy available in the lab? The surprising (maybe just to me) answer is non-zero: $$E_{avail} = E_3 + E_4 = (\gamma - 1)m_f \approx \tfrac{1}{2} \beta^2 m_f,$$ where ##\gamma = (1-\beta^2)^{-1/2}, \beta = v/c##. Note that this is a strictly non-relativistic effect! And this effect can be relatively large, at least on the scale of nuclear physics. Take ##^{10}B(n,d)^{9}Be##. The energy available in the lab frame is 441 keV, which isn't exactly chicken feed.
Consider a reaction (I'm thinking in terms of nuclei and/or particles) ##1 + 2 \to 3 + 4 + \cdots## with masses ##m_1, m_2## in the initial state and any number of final state particles with masses ##m_3, m_4, \ldots##. Let $$m_i = m_1 + m_2, ~ m_f = m_3 + m_4 + \cdots.$$ The value ##Q= m_i - m_f## is positive, negative or zero for the cases exothermic (or exoergic -- I'll consider these interchangeable), endothermic, and elastic, respectively.
In the center-of-mass (CM) frame, the kinetic energy available due to the reaction is the sum of the energies of the products. We call this ##E'_{avail} = E_3' + E_4'##, where these are the kinetic energies of the products and the primes indicated the CM frame. Conservation of energy in CM frame gives $$m_i + E_1' + E_2' = m_f + E_3' + E_4'.$$ This gives $$E'_{avail} = Q + E_i',$$ where ##E'_i = E_1' + E_2'##.
The endothermic case ##Q<0##, at the threshold for the reaction ##E_i' = -Q## has zero energy available to the products: $$E'_{avail}(E_i'=-Q) = 0.$$ The question is then: what is the energy available in the lab? The surprising (maybe just to me) answer is non-zero: $$E_{avail} = E_3 + E_4 = (\gamma - 1)m_f \approx \tfrac{1}{2} \beta^2 m_f,$$ where ##\gamma = (1-\beta^2)^{-1/2}, \beta = v/c##. Note that this is a strictly non-relativistic effect! And this effect can be relatively large, at least on the scale of nuclear physics. Take ##^{10}B(n,d)^{9}Be##. The energy available in the lab frame is 441 keV, which isn't exactly chicken feed.
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