I Lab energy available in threshold (endothermic) reactions

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The discussion focuses on the energy available in the laboratory frame during threshold endothermic reactions, highlighting a surprising relativistic effect. In such reactions, while the center-of-mass frame shows zero energy available to products, the lab frame reveals a non-zero energy due to relativistic effects, calculated as E_avail = E_3 + E_4. The conversation emphasizes that non-relativistic kinematics fail to account for energy conservation when Q is not zero, leading to inconsistencies in momentum conservation. Participants reflect on the implications of these findings, particularly regarding Galilean boosts and their inadequacy in this context. The discussion concludes that understanding energy conversion in reactions is complex and challenges traditional non-relativistic interpretations.
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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.
 
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Why is this surprising?

You have the reaction A (beam) + B (target) → X. The momentum of the initial state is p (whatever it had) so the momentum of the final state is p as well. Since it has momentum, it has kinetic energy.
 
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You’re right about lab frame momentum conservation making it unsurprising.

But I claim it’s zero non-relativistically, which is still a bit surprising.

But maybe it shouldn’t be. For ##Q\ne 0##, the non-relativistic kinematics aren't consistent. We can see this if we try to show that the total momentum is conserved in the center of mass.

In any case, if the statement that the available energy in the lab is zero non-relativistically can be refuted, that would be helpful.
 
sirapwm said:
You’re right about lab frame momentum conservation making it unsurprising.

But I claim it’s zero non-relativistically, which is still a bit surprising.

But maybe it shouldn’t be. For ##Q\ne 0##, the non-relativistic kinematics aren't consistent. We can see this if we try to show that the total momentum is conserved in the center of mass.

In any case, if the statement that the available energy in the lab is zero non-relativistically can be refuted, that would be helpful.
Never mind. That’s stupid. I see the refutation. The problem I’m having is with Galilean boosts. And they’re jus wrong when ##Q\ne 0##.
 
What does a reaction converting energy into mass even mean non-relativistically.
 
Vanadium 50 said:
What does a reaction converting energy into mass even mean non-relativistically.
Right. That's what this statement indicated: "For ##Q\ne 0##, the non-relativistic kinematics aren't consistent. We can see this if we try to show that the total momentum is conserved in the center of mass."
 
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