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I am following "A Shape Dynamics Tutorial, Flavio Mercati" (https://arxiv.org/abs/1409.0105), and have problems understanding the hamiltonian formulation of $N$-particle dynamics as sketched on page 32 onward ("6.3 Application to our systems, Differential almost-Hamiltonian formulation").

No problem do I have with the lagrangian formulation as sketched on page 26 onward ("The Newtonian $N$-body problem"). Here I recapitulate the hamiltonian formalism, as I tried to follow it. We start with the lagrangian \begin{align} S = &\int d\mathcal{L} = 2 \int \sqrt{E - V} \sqrt{\frac{1}{2}\sum_i m_i \| \mathfrak{D}q^i \|^2 },\tag{55} \cr \mathfrak{D}q^i =& dq^i + d\omega \times q^i + d\theta.\tag{56} \end{align} In Mercatis book (or preprint as linked above) it is not stated clearly, but I assume that we treat the variable $q^i$ as those with which we do the canonical analysis (deriving canonical momenta, doing Legendre transformations, Poisson brackets ... etc), whereas we leave $d\omega$ and $d\theta$ as spectators (or simple Lagrange multipliers) as far as the canonical analysis is concerned. The canonical momenta are \begin{align} p_i = \frac{\partial\ d\mathcal L}{\partial q^i} = \frac{m_a \mathfrak{D}q^a}{d\chi}. \end{align} This gives us immediately a primary constraint \begin{align} H - E = 0, \qquad\qquad H := \sum_a \frac{p_a^2}{2m_a} + V(q). \end{align} So far so good, but know I am getting confused. If I do a legendre transformation with respect to $dq$, I get as the canonical hamiltonian \begin{align} \sum_i p_i\ dq^i\ - d\mathcal{L} = -d\theta \cdot P - d\omega \cdot L \end{align} with $P = \sum_i p_i$ and $L = \sum_i q^i \times p_i$. Incoorporating the primary constraint I get the total hamiltonian \begin{align*} -d\mathcal{A} &= -d\lambda\ (H - E) - d\theta \cdot P - d\omega \cdot L. \end{align*} This, it seems to me, is (apart from a sign) the differential almost hamiltonian (as cited on page 32). Also I have changed notation: I write $d\lambda$ instead of $d\chi$. That is because I think at this stage of the dirac-analysis, $d\lambda$ is either arbitrary (gauge parameter) or will be constrained as a result of the diract consisteny algorithm. I has not yet the meaning of the "differential of the instant" (page 27) \begin{align} d\chi := \frac{\sqrt{\frac{1}{2}\sum_i m_i \| \mathfrak{D}q^i \|^2 }}{\sqrt{E - V}}. \tag{58} \end{align} As defined above, $d\chi$ depends upon the "velocities" $dq^i$, and those shall not appear explicitly in the hamiltonian. But somehow, the above expression for $d\chi$ should reappear in the hamiltonian anaylsis ?!

I hoped that it will appear through the dirac constraint analysis (constraints shall propagate), but sofar I only got \begin{align*} 0 = \lbrace H, -d\mathcal{A} \rbrace &= d\omega \cdot \lbrace H, L \rbrace + d\theta \cdot \lbrace H, P \rbrace \\ &= d\omega \cdot \sum_a q^a \times \frac{\partial V}{\partial q^a} + d\theta \cdot \sum_a \frac{\partial V}{\partial q^a} \end{align*} \begin{align*} 0 = \lbrace \alpha P, - d\mathcal{A} \rbrace &= -d\lambda \lbrace \alpha P, H \rbrace + d\omega \cdot \lbrace \alpha P, L \rbrace \\ &= d\lambda\ \alpha \cdot \sum_a \frac{\partial V}{\partial q^a} + \alpha \cdot d\omega \times P \end{align*} \begin{align*} 0 = \lbrace \alpha L, -d\mathcal{A} \rbrace &= -d\lambda \lbrace \alpha L, H \rbrace + d\theta \cdot \lbrace \alpha L, P \rbrace \\ &= d\lambda\ \alpha \cdot \sum_a q^a \times \frac{\partial V}{\partial q^a} - d\theta \cdot \alpha \times P \end{align*} where $\alpha$ is simply a (3d)vector-constant, that avoids me to write vector-indizes.

And another thing that I don't understand: The above equation for the propagation of constraints seem to imply other constraints \begin{align} \sum_a q^a \times \frac{\partial V}{\partial q^a} = 0 \\ \sum_a \frac{\partial V}{\partial q^a} = 0. \end{align} But I do not remember them from doing the Lagrangian analysis?

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The (hamiltonian) equation of motion give \begin{align} dq^i = \lbrace q^i, d\mathcal{A} \rbrace = d\lambda\ \frac{p_i}{m_i} - d\omega \times q^i - d\theta \end{align} Solving for $p_i$, putting that into the primary constraint \begin{align} \sum_i \frac{p_i^2}{2 m_i} + V - E = 0 \end{align} and solving for $d\lambda$ gives \begin{align} d\lambda = \frac{\sqrt{\frac{1}{2}\sum_i m_i \|\mathfrak{D}q^i\|^2}}{\sqrt{E - V}} \end{align} as expected from the lagrangian formalism.

As to the secondary constraints, $\sum_i q^i \times \frac{\partial V}{\partial q^i} = 0$ and $\sum_i \frac{\partial V}{\partial q^i} = 0$: They are nothing else than the statements that total momentum and angular momentum are conserved, using the equation of motion from the lagrangian: \begin{align*} \frac{\mathfrak{D}p^a}{d\chi} &= - \frac{\partial V}{\partial q^a} \end{align*}

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