Boundary conditions in ##\delta I=0## to derive Einstein's equations

  • #1
Kostik
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TL;DR Summary
In making the variation ##\delta I=0##, where the ##g_{\mu\nu}## are considered the 'coordinates' of the Lagrangian, Dirac assumes the ##g_{\mu\nu}## and their first derivatives are constant on the boundary. How does this correspond to the usual action principle method?
Dirac derives Einstein's field equations from the action principle ##\delta I=0## where $$I=\int R\sqrt{-g} \, d^4x$$ (##R## is the Ricci scalar). Using partial integration, he shows that $$I=\int L\sqrt{-g} \, d^4x$$ where ##L## involves only ##g_{\mu\nu}## and its first derivatives, unlike ##R##. Clearly ##L\sqrt{-g}## is both the action density in four dimensions as well as the Lagrangian density in three dimensions, since $$I=\int L\sqrt{-g} \, d^4x = \int dt \int L\sqrt{-g} \, d^3x.$$ Dirac considers the ##g_{\mu\nu}## to be the 'coordinates', and their time derivatives ##g_{\mu\nu,0}## the 'velocities', so by the action principle he determines the actual 'path' ##g_{\mu\nu}## which turns out to be Einstein's field equations (in the same way that the Euler-Lagrange equations determine the equation of motion).

When performing the variation ##\delta g_{\mu\nu}## he assumes ##g_{\mu\nu}## is constant on the boundary of the four-dimensional volume ##D## in the action integral
$$I=\int_D R\sqrt{-g}d^4x.$$ This parallels the classical variation method where ##I=\int_{t_0}^{t_1} L ( q^i, {\dot{q}}^i ) \, dt##, and one assumes ##q^i(t_0)=q^i(t_1)## for all paths, i.e., all paths are constant at the endpoints (the 'boundary').

However, Dirac also assumes that we keep "the ##g_{\mu\nu}## and their first derivatives constant on the boundary." This is essential in two separate places where the divergence theorem is used, because there are Christoffel symbol terms that contain derivatives of ##g_{\mu\nu}##.

How does one explain the need to assume that both ##g_{\mu\nu}## and their first derivatives are constant on the boundary (other than the fact that it's required for the action principle method to work)? Transitioning from the usual 'principle of stationary action' in 1D to the same principle in 4D, is it clear how the spatial derivatives of the 'coordinates' ##g_{\mu\nu}## come into play?
 
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  • #2
You do not need that requirement. The Euler-Lagrange equations need to be locally satisfied regardless of boundary conditions.
 
  • #3
The boundary condition requirement is used in varying the action. I’m not concerned with the E-L equations per se, which are local. I am asking about the boundary conditions in the 4D case versus the classical 1D case.
 
  • #4
Kostik said:
The boundary condition requirement is used in varying the action. I’m not concerned with the E-L equations per se, which are local. I am asking about the boundary conditions in the 4D case versus the classical 1D case.
They are no more required in the 4D case than they are in the 1D case.
 
  • #5
You are misunderstanding my statement. In varying the action, the endpoints of all paths remain fixed. That is the boundary condition I am referring to. Notice how the 4D is different than the 1D case - the derivates of the dynamical variables are also fixed.
 
  • #6
Kostik said:
In varying the action, the endpoints of all paths remain fixed.
No, they are not in general. There is no need to do this. Not in one dimension and not in four dimensions.
 
  • #7
They remain fixed so that the boundary term disappears in the integration by parts. I think we are speaking about different things. Perhaps another reader can shed some light here.
 
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  • #8
Kostik said:
They remain fixed so that the boundary term disappears in the integration by parts. I think we are speaking about different things. Perhaps another reader can shed some light here.
No, this is a misunderstanding of the actual requirements necessary.
 
  • #9
To put that into context:

Many texts will require some boundary conditions to be satisfied but this does not impact the actual derivation of the EL equations. Yes, you end up with some boundary terms that are not zero for particular variations, but you need the variation of the action to be zero for all variations. Including those for which the boundary terms vanish.

The only reason not to include the boundary terms from the beginning is that you can just hand wave those terms away. In fact, if you do not impose boundary conditions, the boundary terms of the variations vanishing will provide you with boundary conditions in order for the variation to vanish.
 
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