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I'm taking a GR course, in which the instructor discussed the 'Hamiltonian constraint' of spherical Friedmann cosmology action. I'm not quite clear about the definition of 'Hamiltonian constraint' here. I searched internet but cannot find a satisfying answer. Could anybody help me a bit?

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The Hamiltonian constraint comes from but one component of Einstein's equations.

$$\underline G(a) = 8 \pi \underline T(a)$$

The Einstein tensor $\underline G(a)$ deals with the metric and its derivatives--terms involving curvature. The stress-energy tensor $\underline T(a)$ describes the matter and energy in a point of space. This equation relating the two contains the great physical content of Einstein's theory--it is the relationship between matter/energy and spacetime curvature.

The Hamiltonian constraint comes from looking at the time-time component of this equation. That is

$$\text{Hamiltonian constraint: } \underline G(e_t) \cdot e^t = 8 \pi \underline T(e_t) \cdot e^t$$

This is referred to as a constraint because it involves no time derivatives in the equation--only spatial derivatives are buried or hidden within $\underline G(e_t)$. It is called Hamiltonian because it is closely related to energy and the overall Hamiltonian formulation of GR.

The Hamiltonian constraint is important because any description of a system at some initial time must obey the constraint to be physically possible, and it should do so at all times. Hence, ensuring that some initial data obeys the constraint is a key consideration when solving Einstein's equations using, for example, numerical simulations.

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When you decompose Einstein's equation into a Hamiltonian formalism using the ADM approach, you get four equations to solve: Two equations of motion in the three-metric of your initial spacelike section, and its extrinsic curvature in the enveloping four-dimensional spacetime.

Additionally, you get two constraint equations, the Hamiltonian constraint:

$16\pi\rho = {}^{3}R - K^{ij}K_{ij} + K^{2}$,

and the momentum constraint:

$8\pi j^{i} = \bar \nabla_{j}\left(K^{ij} - {}^{3}g^{ij}K\right)$

Any more detail than this takes considerable effort. It is a chapter in Wald, and it took up a whole chapter in my dissertation, which I can point you to on the arxiv if that's what you feel you need.

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A nice answer, but I think the question is more about the physical meaning of the Hamiltonian constraint than its mathematical definition. One perhaps could add: the name 'Hamiltonian constraint' comes about from the fact that one can apply the Hamiltonian equations on this constraint (function,) and derive the equation of motion of its dynamic variables. I hope you agree. –  JKL Feb 11 '13 at 22:02
    
@John: or alternately, that the Hamiltonian constraint is literally that the Hamiltonian is zero. –  Jerry Schirmer Feb 11 '13 at 22:18
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Thanks for your helpful answer! but I think what Muphrid offered is more like what I want to know. –  Simon Feb 21 '13 at 0:55
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