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I suspect a flaw in the reasoning below, but am unable to pinpoint it: Is there something inconsistent in terms of the application of conservation of momentum and energy? Thanks for any hints in this regard.

Let's consider the 2-body decay:

$$Y \to B^++B^-$$

where $m_Y=10.58\,\text{GeV}/c^2$ and $m_{B^{\pm}}=5.279\,\text{GeV}/c^2$. Let's suppose that the momentum of $Y$ (in the laboratory reference system) before the decay is given by:

$$\vec{p}_Y=(-p_Y,0,0), \quad p_Y=5.9\,\text{GeV}/c$$

Then the energy of $Y$ is given by:

$$E_Y=\sqrt{m_Y^2c^4+p_Y^2c^2}=12.1\,\text{GeV}$$

Now we ask ourselves if it is possible that $B^+$ is emitted with a momentum given by:

$$\vec{p}_{B^+}=(p_{B^+},0,0)$$

with $p_{B^+}>0$, namely that $B^+$ is emitted in the positive direction of x axis. Using conservation of total momentum we have:

$$\vec{p}_Y=\vec{p}_{B^+}+\vec{p}_{B^-} \quad \Rightarrow \quad \vec{p}_{B^-}=\vec{p}_Y-\vec{p}_{B^+}=(-p_Y-p_{B^+},0,0)$$

Then we have:

$$|\vec{p}_{B^-}|=p_{B^-}=p_Y+p_{B^+} \quad \Rightarrow \quad p_{B^+}=p_{B^-}-p_Y$$

By now we have only used the conservation of total momentum. Now let's apply the conservation of energy:

$$E_Y=E_{B^+}+E_{B^-} \quad \Rightarrow \quad E_Y=\sqrt{m_B^2c^4+p_{B^+}^2c^2}+\sqrt{m_B^2c^4+p_{B^-}^2c^2}$$

From this equation, with some calculation (I repeated the calculations several times carefully) and using the fact that $p_{B^+}=p_{B^-}-p_Y$, we find that:

$$(4m_Y^2c^6)p_{B^-}^2-(4p_Ym_Y^2c^6)p_{B^-}+(4m_Y^2m_B^2c^8+4p_Y^2m_B^2c^6-m_Y^4c^8)=0$$

This is a second degree equation in $p_{B^-}$, and solving it we find:

$$p_{B^-}=2.56\,\text{GeV}/c, \quad p_{B^-}=3.34\,\text{GeV}/c$$

Then if $p_{B^-}=3.34\,\text{GeV}/c$ we have:

$$p_{B^+}=p_{B^-}-p_Y=-2.56\,\text{GeV}/c$$

while if $p_{B^-}=2.56\,\text{GeV}/c$ we have:

$$p_{B^+}=p_{B^-}-p_Y=-3.34\,\text{GeV}/c$$

This is impossible, since $p_{B^+}$ must be positive. Then we conclude that $B^+$ can't be emitted in the positive direction of x axis, I guess. However the exercise I am doing tells me to suppose that $B^+$ is emitted in the positive direction of x axis.

This contradiction suggests a misunderstanding of something conceptual about the conservation laws. Thanks for any pointers.

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    $\begingroup$ I think that the decision of closing my question is wrong. I wasn't asking to check my computations or doing my homework, I was asking a more conceptual thing: if I obtain negative momentum using momentum and energy conservation, then I must conclude that it is impossible for $B^+$ particle to be emitted in the positive x axis direction. $\endgroup$
    – Leonardo
    Commented Jun 22 at 10:23
  • $\begingroup$ Your question is essentially asking "where did I go wrong", which is off-topic. There is nothing "inconsistent" about the conservation of momentum. As a suggestion, it is easier to do this in the center of mass frame. Then both particles will have equal and opposite momenta. $\endgroup$ Commented Jun 23 at 0:42
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    $\begingroup$ Anyway, I can confirm that your calculations are correct. With the given mass and energy, the momentum will be negative. There is no inconsistency. $\endgroup$ Commented Jun 23 at 1:07
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    $\begingroup$ Since the original momentum is negative, in order for one of the resultant particles to have positive momentum, their mass must decrease so that they have more kinetic energy. In summary the particles do not have sufficient separation velocity for one of them to reverse direction. $\endgroup$ Commented Jun 23 at 1:14
  • $\begingroup$ @VincentThacker Thank you for the answer. Then the text of my exam (this is not just an exercise, but an university exam I did a few days ago) is wrong, not my application of conservation laws. Maybe I should just ask the following question to be on-topic: "if the application of energy and momentum conservation law gives negative momentum, then I must conclude that the starting data are wrong?" $\endgroup$
    – Leonardo
    Commented Jun 23 at 9:20

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