Conserved and invariant quantities in a nuclear reaction

Question

For a reaction $$a+X\rightarrow Y+b$$ in the book, Introductory Nuclear Physics by Krane, he has written that the total energy and the total momentum both in RHS and LHS remains conserved in Lab frame, i.e. $$E_a+E_X=E_b+E_Y$$ where $E$ represents the total energy i.e. kinetic energy and the mass energy. Also, he claims that $$p_a+p_X=p_b+p_Y$$ Now if we transform the equations to a new inertial frame, the equations might not hold true and actually the quantity $$E^2-p^2(=m_0^2)$$ should be conserved both sides in the reaction. But the conservation of $E^2-p^2$ comes from the Lorentz in variance of the rest mass.

What if the rest mass of the participating species isn't conserved in the nuclear reaction, then which quantity remains conserved in the reaction?

Answer 1

You are being misled by Krane's wording. Energy and momentum are conserved in the Lab frame - but they are also conserved in any other frame. He just mentions the Lab frame as that's where it's easiest to do the measurements.

Given $E_a+E_X=E_b+E_Y\qquad \vec p_a+\vec p_X=\vec p_b+\vec p_Y$

then $E_a'+E_X'=E_b'+E_Y'\qquad \vec p_a'+\vec p_X'=\vec p_b'+\vec p_Y'$

for any set of primed quantities obtained from the unprimed quantities by a Lorentz transformation. If this is not obvious, just write out the Lorentz-transform expressions for the primed quantities and watch everything cancel.

It is also true that particle masses are invariant: $E_a^2-\vec p_a^2 c^2=m_a^2c^4=E_a'^2-\vec p_a'^2c^2$ and likewise for $b,X,Y$

Also the total energy and momentum form an invariant which can be useful as this is both conserved and the same in any frame $(E_a+E_X)^2-(\vec p_a+\vec p_X)^2c^2= M^2c^4=(E_b+E_Y)^2-(\vec p_b+\vec p_Y)^2c^2=(E_a'+E_X')^2-(\vec p_a'+\vec p_X')^2c^2=(E_b'+E_Y')^2-(\vec p_b'+\vec p_Y')^2c^2$

$M^2c^4$ is often written $s$

Answer 2

Definition: The invariant mass of a system is defined through the four vector

$$ P_\mu = \sum_i \left( E_i , \mathbf{p}_i\right) = (E_{tot}, \mathbf{p}_{tot}). $$

where the sum $i$ can be taken over all of the initial particle's momenta (or final). The invariant mass $M^2$ is then given by

$$ M^2 = P_\mu P^\mu. $$

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If rest mass isn't conserved, that means it's not conserved in its own rest frame. This implies that $P_\mu P^\mu$ is not Lorentz invariant and hence Lorentz invariance is violated.

When you violate Lorentz invariance, you better have a good reason; special relativity breaks, everything breaks, and you should question the assumptions that led you there.