What is the topology of phase space of $n$ free relativistic particles in center of mass frame?

Question

Consider an ensemble of $n$ relativistic particles of fixed masses $m_i \geq 0$, $i=1,\ldots,n$ with four momenta $p_i$ such that $p_i^2=m_i^2$. In center of mass frame they sum up to $$P=p_1+\cdots+p_n=(E,0,0,0),$$ where $E>m_1+...+m_n$ is fixed number (last inequality is to exlude trivial case in which all particles are at rest).

The set of all possible configurations $(p_1,...,p_n)$ satisfying these two constraints, i.e. $n$-tuples of vectors such that each is on mass shell and they sum up to our fixed vector is a compact subspace of $\mathbb R^{4n}$. If all masses are strictly greater than zero then the implicit function theorem implies that it is actually a smooth submanifold of dimension $3n-4$, but if some masses are nonzero I'm not sure if one can exclude existence of cusps.

For example for $n=2$ three spatial components of equation $P=p_1+p_2$ and mass-shell condition imply that they are of the form $p_1=(\sqrt{m_1^2+\vec p ^2}, \vec p)$, $p_2=(\sqrt{m_2^2+\vec p^2}, - \vec p)$. Energy equation $E=\sqrt{m_1^2+\vec p ^2} + \sqrt{m_2^2+\vec p ^2}$ fix length of $\vec p$ uniquely. However angles of $\vec p$ remain unspecified. Therefore set of all solutions is a sphere.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Blazej

Comments

Should one assume that there exist a relationship between E and M (such as E=Mc^2)? Why is the dimension of the submanifold 3n-4 and not 4n-4?

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user flippiefanus

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Each four momentum satisfies Einstein's relation $p_i=E_i^2-\vec p_i ^2=m_i^2$. This is one constraint for each particle. $p_1+....+p_n=P$ is another four constraints. All these constraints are easily seen to be independent except so we end up with $4n-n-4=3n-4$ dimensions, except for special cases such as $E=m_1+...+m_n$ when all particles are at rest and therefore space of solutions is a point.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Blazej

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How do you figure that the set you're interested in is compact? For $n=2$, it includes the subset $$\left\{\left(\left(p_{1x},0,0,\sqrt{m_1^2+p_{1x}^2}\right),\left(-p_{1x},0,0, \sqrt{ m_1^2+p_{1x}^2}\right)\right) : p_{1x}\in\mathbb R\right\},$$ which is unbounded, so the set is not compact. This then makes your claim that the $n=2$ set is a sphere dubious - can you clarify what you mean there, and why you think that?

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Emilio Pisanty

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I'd also take issue with your constraints count. If each particle mass is predetermined then sure, that's $n$ restrictions, but the setting to $P$ is only three constraints unless you pre-specify the total energy, which I don't think is the case. The total dimension should be $3n-3$ by my count.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Emilio Pisanty

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@Blazej. If the masses m_i were fixed then I would agree that one can use the Einstein relation as a constraint that removes a degree of freedom. However, if I understood your explanation correctly then the masses are not fixed. The only requirement is that they are not zero. So then the Einstein relation contains these masses as an uncertainty, which leaves the energy and the three momentum components as separate degrees of freedom. Hence, 4 degrees of freedom per particle. Do I perhaps understand this incorrectly?

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user flippiefanus

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@EmilioPisanty I editted my question. Perhaps now it's clear what I meant.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Blazej

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So you're fixing the energy? That's definitely not the topology for '$n$ free relativistic particles'. It's probably a more interesting question, but it does need further trimming and clarifying. (If you don't fix the energy, the topology is simple - it's homeomorphic to $\mathbb R^{3n-3}$, since you can always specify the first $n-1$ momenta and that uniquely determines the last one.)

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Emilio Pisanty

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May move on to n =3, using standard 3-body decay discussion.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Cosmas Zachos

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Superficially, the 3-body decay looks like a 5-sphere, topologically: You might, of course, recast it in Jacobi coordinates, but you can intuit it directly. The three decay products lie on a plane (which may rotate by two angles). On that plane, the magnitude of the 3rd decay product is bounded by the sum and difference of magnitudes of the other two, with the angle between those ranging from 0 to π. These two magnitudes are bounded above (and below) by your energy constraint. Two bounded magnitudes and three angles in all.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Cosmas Zachos

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I don't really understand the argument that this space is $S^5$. Locally it looks like a product of a disk with $SO(3)$ because we can use $E_1$ and $E_2$ (two cordinates in compact region of plane) and three Euler angles to parametrize it. However this parametrization is singular in some regions of phase space. In particular if you fix values of energies which correspond to situation when one of the particles is at rest in center of momentum frame, remaining two momenta lie in a line rather than a plane. For these fixed energies you have only a sphere of solutions.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Blazej

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This subtlety doesn't make any difference in usual calculation where one just inegrates over this phase space. The region where usual coordinates are singular is of measure zero and can be neglected. However for the question of global topology of this space it is very important to carefully check how is this space "glued together" in places of overlap of two different coordinate charts.

This post imported from StackExchange Physics at 2016-10-30 15:22 (UTC), posted by SE-user Blazej

Answer 1

In the relativistic case, a classical multiparticle picture is inconsistent. This is a well-known no-go theorem by Currie, Jordan and Sudarshan, Reviews of Modern Physics 35.2 (1963): 350. Hence there is no sensible associated phase space. 

Feynman (before he worked on QED), created a classical relativistic but nonlocal many-particle theory, With the right (''absorbing'') boundary conditions you get more or less classical electrodynamics. But (like any classical multiparticle picture I have seen) its interpretation defies any rational sense of physics. In the present case, the dynamics of any particle depends on the past and future paths of all other particles!
Eric Gourgoulhon who discusses the theory in his book ''Special Relativity in General Frames'', mentions on p. 375 the disadvantage that it leads to integro-differential equations that have no well-defined Cauchy problem. This leads to interpretational difficulties.

In the quantum domain, the situation is a little bit better, since fields are not absolutely unavoidable - however, things get exceedingly messy. See the topic ''Is there a multiparticle relativistic quantum mechanics?'' from Chapter B1 of my theoretical physics FAQ. You can also find a discussion (together with a proposed solution that I don't find convincing) in the arXiv book http://arxiv.org/abs/physics/0504062.