# Why are the ADM momentum and angular momentum such defined?

In Sec. 4.3 of the book "A relativist's toolkit" by Eric Poisson, it is explained how to extract ADM angular momentum from the ADM Hamiltonian. He says that it suffices to replace the lapse $N$ and the shift $N^a$ with the components of the vector field associated to the quantity you are looking for. For example, the angular momentum is recovered by replacing $N=0$ and $$N^a=\phi^a$$ where $\phi^a$ is the generator of rotations at asymptotic infinity. (Similarly, many references say that the ADM momentum $P_b$ is recovered for $N=0$ and $$N^a=\delta^a_b$$. )

I can see that this "trick" gives you back the correct momenta, but I don't understand what is the rationale behind it.

I understand the procedure for the ADM energy, in which you replace $N=1$ and $N^a=0$, because of course the Hamiltonian is associated with the energy. But I don't understand the connection of the same Hamiltonian with the other momenta.

• Because energy is the generator of time evolution, momentum is the generator of space translation, and angular momentum is the generator of space rotation. – Eric Yang Jun 10 '17 at 6:23
• No. I already know this. What I don't understand is why the same Hamiltonian functional $H[N]+\vec{H}[\vec{N}]$, generating time evolutions for appropriate lapse and shift, generates also translations and rotations for other lapse and shift. In the usual Hamiltonian formalism the Hamiltonian is just the generator of time translations. Maybe it has to do with general covariance? I think the answer must come from someone expert in general relativity.. – Mike Robert Jun 11 '17 at 10:38
• You must now the Hamiltonian in ADM is the generator of $t^{\alpha}$. Whether this evolution is time, translation or rotation is dependent on how you choose $N$ and $N^a$. I post an answer to your question. – Eric Yang Jun 11 '17 at 12:31

In ADM form of General Relativity, we must apply 3+1 decomposition to the entire space-time and how to evolute time-slice is arbitrary. This arbitrariness is reflected on the arbitrariness of the vector $(N,N^a)$.
Suppose that the "space coordinate" of an particle is fixed and denote the tangent vector of its world line as $t^{\alpha}$, we have (equation 4.36 of your textbook) $$t^{\alpha} = N n^{\alpha} + N^a e_a^{\alpha}$$ And the Hamiltonian is the generator of the evolution in the direction $t^{\alpha}$.
In asymptotic flat space-time, we have asymptotic Minkowski coordinates $(\bar{t},\bar{x},\bar{y},\bar{z})$. Roughly speaking, for a rest observer at infinity, his watch is $\bar{t}$ and his ruler is $\bar{x},\bar{y},\bar{z}$. From now on, we will denote this coordinate as $x^{\alpha}$.
Now, let us choose the initial time slice as $\Sigma_{\bar{t}}$. In this time slice, we choose $y^a$ the same as $\bar{x},\bar{y},\bar{z}$. So at infinity, which is flat, we have $$n^{\alpha} = \delta^{\alpha}_{0}, e_a^{\alpha} = \delta^{\alpha}_a.$$ The evolution vector now become $$t^{\alpha} = N\delta^{\alpha}_0 + N^a\delta^{\alpha}_a.$$ If $N = 1$ and $N^a = 0$, we have $$t^\alpha = (1,0,0,0)$$ So, the Hamiltonian is the generator of the evolution in direction $\bar{t}$. It is just the energy of the entire space-time, at least for the observer at infinity.
If $N = 0$ and $N^a = \delta^a_b$, we have $$t^\alpha = \delta^{\alpha}_b$$ So, the Hamiltonian is the generator of the evolution in direction $x^b$. It is just the momentum in direction $x^b$ of the entire space-time, at least for the observer at infinity.
If $N = 0$ and $N^a = \phi^a = \frac{\partial y^{a}}{\partial \phi}$, we have $$t^\alpha = \frac{\partial x^{\alpha}}{\partial \phi}$$ For example, for rotation in $\bar{z}$ direction, we have $$\bar{x} = r\sin\theta\cos\phi, \; \bar{y} = r\sin\theta\sin\phi, \; \bar{z} = r\cos\theta$$ we then have $$t^{\alpha} = r\sin\theta(0,-\sin\phi, \cos\phi,0)$$ So, the Hamiltonian is the generator of the evolution rotation around $\bar{z}$. It is just the angular momentum in direction $\bar{z}$ of the entire space-time, at least for the observer at infinity.