# Do the momentum conservation equations have a unique solution?

In high energy physics, one often encounters conservation of momentum and energy equations of the following form:

$$\begin{array}{rcr} \sum_i (-1)^{\alpha_i}\sqrt{k_i^2 + m_i^2} = 0 \\ \sum_i (-1)^{\alpha_i}k_i\cos{\theta_i} = 0 \\ \sum_i (-1)^{\alpha_i}k_i\sin{\theta_i}\sin{\phi_i} = 0 \\ \sum_i (-1)^{\alpha_i}k_i\sin{\theta_i}\cos{\phi_i} = 0 \end{array}$$

Assume moreover that: $$k_i \ge 0,\ m_i \ge 0, \ \sqrt{k_i^2 + m_i^2} >0$$ And that the signs $\alpha_i\in\{0,1\}$ are known.

Assume all but 4 variables are known, and that there is at least one solution, is this solution unique?

By unique I mean distinct physical solutions, so that degeneracy due to spherical coordinates should not be considered as lack of uniqueness. From a physics perspective, it seems like it should be unique, but for some configurations it seems like two answers are some times possible. So can one prove / disprove the uniqueness?

I have looked into using the inverse function theorem, but I can't seem to find an expression for the lower bound of the radius of the ball in which an inverse (and therefore a unique solution) exist.

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I don't see how the first equation would have any real solutions unless all the particles are massless? May be I misinterpreted something? Anyway, if there is a solution, where one of the particles is left motionless, then surely the values of the variables of the direction of motion of that particle are left undecided. Here, if $k_i=0$ for some $i$, then $\theta_i$ and $\phi_i$ are left undetermined. Admittedly that is an artifact of using spherical coordinates. But this does show that you need to define the problem more carefully. – Jyrki Lahtonen Nov 25 '12 at 8:44
@JyrkiLahtonen - fixed the question to address your comments. – nbubis Nov 25 '12 at 8:57
Some mathematician friends noted that of course this has infinite solutions, since $\theta_i$ can take on any multiple of $2\pi$. I hope it's understood I meant only distinct physical solutions, so that degeneracy due to spherical coordinates should not be considered as lack of uniqueness. – nbubis Nov 25 '12 at 15:23
Any particular reason you have written these out in spherical coordinates instead of keeping them in the simpler (in my opinion) vectorial form $\sum_i(-1)^{\alpha_i}\sqrt{\lVert \mathbf k_i\rVert^2+m_i^2}=0$ and $\sum_i(-1)^{\alpha_i}\mathbf k_i=0$? – Rahul Feb 9 '13 at 8:12
@ℝⁿ. - yes, because that is the general form often used in experiments. If you can show the two are equivalent, and that your form has a unique solution, I'll accept that to. – nbubis Feb 9 '13 at 8:28

No, it need not be unique in some circumstances.

Assuming not all $m_i = 0$ and not all $k_i = 0$, following is a rough analysis of most of the cases.
(most but not all because if some $m_i = 0$, it might not be possible to find a C.M. frame to carry out the analysis as given below).

Case 1: All $(-1)^{\alpha_i}$ have the same sign.

The set of equations clearly have no solution.

Case 2 A pair of $(-1)^{\alpha_i}$ is positive and the other pair negative.

Let's say $(-1)^{\alpha_1} = (-1)^{\alpha_2} = -1, (-1)^{\alpha_3} = (-1)^{\alpha_4} = 1$, we have: $$\sqrt{k_1^2 + m_1^2} + \sqrt{k_2^2 + m_2^2} = E = \sqrt{k_3^2 + m_3^2} + \sqrt{k_4^2 + m_4^2}$$ If one look at the process in C.M. frame, the net momentum of both incoming and outgoing pair vanish. This means: \begin{align} \vec{k_1} + \vec{k_2} &= \vec{0} \implies k_1 = k_2\\ \vec{k_3} + \vec{k_4} &= \vec{0} \implies k_3 = k_4 \end{align} and hence:

$$\sqrt{k_1^2 + m_1^2} + \sqrt{k_1^2 + m_2^2} = E = \sqrt{k_3^2 + m_3^2} + \sqrt{k_3^2 + m_4^2}$$

Since L.H.S are strictly monotonic increasing functions for $k_1$, R.H.S are strictly monotonic increasing function for $k_3$, both $k_1$ and $k_3$ are uniquely determined by a single number $E$, the energy in the C.M. frame. So in this case, the configuration is physically unique.

Case 3: One of $(-1)^{\alpha_i}$ has different sign from others.

Let's say $(-1)^{\alpha_1} = (-1)^{\alpha_2} = (-1)^{\alpha_3} = -1$ but $(-1)^{\alpha_4} = 1$.

Transforming to the C.M. frame, we have $k_4 = 0$ and:

\begin{align} m_4 &= \sqrt{k_1^2 + m_1^2} + \sqrt{k_2^2 + m_2^2} + \sqrt{k_3^2 + m_3^2}\tag{1}\\ \vec{0} &= \vec{k_1} + \vec{k_2} + \vec{k_3} \end{align} The equation $(1)$ clearly has no answer unless $m_4 > m_1 + m_2 + m_3$. In order for the 3 vectors $\vec{k_1}, \vec{k_2}, \vec{k_3}$ sum to $\vec{0}$, they must lie on the same plane and their magnitude satisfies the triangle inequalities:

$$k_1 + k_2 \ge k_3,\, k_2 + k_3 \ge k_1 \text{ and } k_3 + k_1 \ge k_2$$

Conversely, if you have given 3 numbers $k_1, k_2, k_3$ which satisfies the triangle inequalities, a unique triangle can be constructed. This in turn allow one to construct the 3 vectors $\vec{k_1}, \vec{k_2}, \vec{k_3}$, unique up to rotations in $\mathbb{R}^3$, which sum to $\vec{0}$.

Since $k_1, k_2, k_3$ need to satisfy $(1)$, the problem essentially have two degrees of freedom.

Case 3a if two momentum are given, say $k_1$ and $k_2$, one can obtain a unique $k_3$ by solving $(1)$. The physics is unique in this case.

Case 3b if two angles are given, say $\measuredangle(\vec{k_1},\vec{k_2})$ and $\measuredangle(\vec{k_2},\vec{k_3})$, the third angle and the ratios of the sides are known: \begin{align}\measuredangle(\vec{k_3},\vec{k_1}) &= \pi - \measuredangle(\vec{k_1},\vec{k_2}) + \measuredangle(\vec{k_1},\vec{k_3})\\ k_1 : k_2 : k_3 &= \sin(\measuredangle(\vec{k_2},\vec{k_3})) : \sin(\measuredangle(\vec{k_3},\vec{k_1})) : \sin(\measuredangle(\vec{k_1},\vec{k_2})) \end{align} Under scaling $k_i \mapsto \lambda k_i, i = 1,2,3$, the R.H.S of $(1)$ is a strictly monotonic increasing function in $\lambda$. This means given the ratios of the sides, a unique scaling factor $\lambda$ can be chosen to satisfy $(1)$ and we can determine $k_1, k_2, k_3$. The physics is again unique in this case.

Case 3c if only one momentum and an angle of that momentum with another momentum is given, say $k_1$ and $\theta = \measuredangle(\vec{k_1},\vec{k_2})$, the physics no longer need to be unique. First, $$\vec{k_3} = - ( \vec{k_1} + \vec{k_2} ) \implies k_3^2 = k_1^2 + k_2^2 + 2 k_1 k_2 \cos(\theta)$$ Plug this into $(1)$, we find $k_2$ will be a solution for $x$ in the equation: $$\sqrt{x^2 + m_2^2} + \sqrt{ k_1^2 +x^2 + 2 k_1 x \cos(\theta) + m_3^2} = m_4 - \sqrt{k_1^2 + m_1^2}\tag{2}$$ If you throw in some random parameters for $k_1, \theta, m_2, m_3$ and plot the L.H.S of $(2)$ as a function of $x$, you will find it no longer need to be monotonic. For some parameters, there are more than one solution for a fixed value of R.H.S. Whenever this happens, the physics is no longer unique.

Case 3d if only one momentum and the angle between other two momenta is given, say $k_1$ and $\theta = \measuredangle(\vec{k_2},\vec{k_3})$, solutions also need not be unique. It is not hard to visualize this when $m_2$ and $m_3$ are close to massless.

Start with a triangle $ABC$ with $\overline{AB} = k_1$ and let $\overline{BC} = k2, \overline{CA} = k_3$. If one impose the condition onto $\measuredangle{BCA}$ equal to the constant $\pi - \theta$, $C$ will be lying on a circular arc passing through $A$ and $B$. In the massless limit of $m_2, m_3$, Equation $(1)$ reduces to: $$k_2 + k_3 \sim \sqrt{k_2^2+m_2^2} + \sqrt{k_3^2+m_3^2} = m_4 - \sqrt{k_1^2+m_1^2}$$ This is the equation for an ellipse having $A$ and $B$ as foci. In general, the circular arc will intersect the ellipse at either zero or two points. This means in this particular limit, if the equation has an solution, the solution is unlikely to be unique.

Conclusion: the configuration is unique if all momenta are given. If you have mixed information about momenta and angles, the solutions need not be physically unique.

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