# Kinematic limits in 3-body decay

I am considering a three-body decay $p \to p_1 + p_2 + p_3$ in which two of the final particles are massless ($p_1^2 = p_3^2 = 0$). I am referring to two textbooks: Byckling, Kajantie "Particle Kinematics" (BK); Hagedorn "Relativistic Kinematics" (H). Let us put ourselves in the initial momentum rest frame, in which $\vec{\mathbf{p}}=0$.

From Eq. (V.2.3) of BK and (7-50) of H, the kinematical limits of the invariant mass $s_1 = (p_2 + p_3)^2$ are $$(m_2 + m_3)^2 \le s_1 \le (m- m_1)^2,$$ that in our specific case become ($m_1 = m_3 = 0$) $$\tag{1} \label{1} m_2 ^2 \le m_2^2 + 2 p_2 \cdot p_3 \le m^2 \implies 0 \le p_2 \cdot p_3 \le \frac{m^2 - m_2^2}{2}.$$

BK considers this case some paragraphs ahead, and in Fig. V.2.2 we can see the Dalitz plot corresponding to our case (the central figure, for $m_2 \neq 0$). The plot is consistent with \eqref{1}.

Now, if we focus on the first inequality of \eqref{1} we have $$p_2 \cdot p_3 = E_2 E_3 - \sqrt{E_2^2 - m_2^2} E_3 \cos \theta_{23} \ge 0,$$ where $\theta_{23}$ is the angle formed by $\mathbf{\vec{p}_2}$ and $\mathbf{\vec{p}_3}$ in the rest frame. For the $p_2\cdot p_3 =0$ limit we have $$\cos \theta_{23} = \frac{E_2}{\sqrt{E_2^2 - m_2^2}} > 1,$$ since $E_2 > 0$, which is an absurd situation.

My question is: is the Dalitz plot not considering the angular limits (and thus one should add the condition $\cos^2 \theta < 1$ to the equations defining the Dalitz plot)? or is there a mistake somewhere due to the fact that I am considering massless particles?

• What do you get in the ultrasoft $E_3\to 0$ limit? Not $p_2\cdot p_3\to 0$? Oct 8, 2016 at 20:39
• Yes, so in the ultrasoft limit the angular structure does not make any difference? I was looking for details in the angular structure at the kinematic limits. If we put $E_3$ slightly greater than zero we obtain that relation for the angle $\theta$, so how can one picture the configurations at the kinematic limits by looking at the angles formed by the particles' momenta? Oct 8, 2016 at 20:50
• The quasi-2-body decay represents that, otherwise unphysical, limit. Oct 8, 2016 at 21:01

When $p_2\cdot p_3=0$ you just have $E_3=0$. Dividing the two sides of the inequality by $E_3$ is inconsistent.
• You can have $p_2\cdot p_3=0$ if $E\neq0$ and the two momenta are perpendicular.