Lab energy available in threshold (endothermic) reactions

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SUMMARY

The discussion centers on the energy available in the laboratory frame for threshold endothermic reactions, specifically examining the reaction dynamics of nuclei and particles. It establishes that while the kinetic energy available in the center-of-mass frame is zero for endothermic reactions (where Q < 0), the laboratory frame reveals a non-zero energy availability due to relativistic effects, expressed as Eavail = E3 + E4 = (γ - 1)mf. The example of the reaction 10B(n,d)9Be demonstrates an available energy of 441 keV, highlighting significant implications in nuclear physics. The discussion also addresses the inconsistencies of non-relativistic kinematics when Q ≠ 0.

PREREQUISITES
  • Understanding of relativistic physics, specifically Lorentz transformations.
  • Familiarity with nuclear reaction dynamics and conservation laws.
  • Knowledge of center-of-mass (CM) frame calculations.
  • Basic concepts of kinetic energy and mass-energy equivalence.
NEXT STEPS
  • Explore relativistic kinematics and its implications on particle reactions.
  • Study the conservation of momentum in both laboratory and center-of-mass frames.
  • Investigate the significance of the Q-value in nuclear reactions.
  • Learn about the applications of energy calculations in nuclear physics experiments.
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Physicists, nuclear engineers, and students studying particle physics or nuclear reactions who seek to understand the implications of relativistic effects on energy availability in laboratory settings.

sirapwm
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TL;DR
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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