Is There a Logical Derivation for Einstein's Field Equations?

[LHS1]

I realized that there is no strict derivation of Einstein's Field Equations. However I found no 'derivation' that make me feel 'comfortable' and 'logical'. Could anyone post a 'derivation' with smooth logical sense ? Thank you.

 

[Mentz114]

Hi,
there are several derivations of the EFE. Start with the 'Einstein-Hilbert action' here -

en.wikipedia.org/wiki/Einstein-Hilbert_action

or is this one that makes you uncomfortable ?

M

 

[George Jones]

I realized that there is no strict derivation of Einstein's Field Equations. However I found no 'derivation' that make me feel 'comfortable' and 'logical'. Could anyone post a 'derivation' with smooth logical sense ? Thank you.

The paper The Meaning of Einstein's Equation by Baez and Bunn,

http://arxiv.org/abs/gr-qc/0103044

answers a different question, but you might still find it interesting.

 

[v.dinesh]

there can be a derivation of EFE but makin it logically complete a good amount of paper work is needed...but is there any such attempt available??

 

[Ben Niehoff]

If there were an absolutely logical, mathematical derivation, then this wouldn't be physics. Ultimately, one has to simply postulate something and treat it as a working hypothesis to be verified by experiment.

First we assume the equivalence principle, which tells us how matter is affected by curved spacetime. This is essentially the same thing as the basic postulate of Riemannian geometry: in the limit as displacements become small, spacetime looks flat and inertial observers travel in straight lines. For motions that are no longer infinitesimal, these "straight lines" extrapolate to geodesics.

Next one needs the other piece of the puzzle: how is spacetime affected by matter? We assume that some sort of matter-energy density must be the source of curvature. We know that matter-energy satisfies a local conservation law (due to the equivalence principle), so we choose the simplest object constructed out of matter-energy that obeys such a conservation law in relativistic mechanics: the stress-energy tensor. This object has two indices, is symmetric, and has zero divergence. To couple it to curvature, we need to form some object out of the curvature tensor that also has two indices, is symmetric, and has zero divergence. The simplest such object is the combination $R_{\mu\nu} - \frac12 R g_{\mu\nu} + \Lambda g_{\mu\nu}$. So, we set this object proportional to the stress-energy tensor, and work out the constant of proportionality by looking at the weak-field limit.

The weak-field limit gives an additional constraint: the theory must reduce to Newtonian gravity for weak fields. Historically, these steps were worked backwards, starting from the weak field limit and then guessing what kinds of metric theories might produce it. There were competing theories besides Einstein's (notably Nordstrom's) that predicted different phenomena, either slight variations or drastic ones (Nordstrom's predicts no bending of light by gravity, for example).

 

[v.dinesh]

yes Ben,what u said is absolutely true,in my previous post i assumed equivalance principle and i was talking about the formal derivation through action principle.

 

[atyy]

yes Ben,what u said is absolutely true,in my previous post i assumed equivalance principle and i was talking about the formal derivation through action principle.

I think for the LHS there is - most general 2 index tensor from metric and derivatives up to second order (Lovelock 1972) p100 http://books.google.com/books?id=YA...ge&q=geometrical relativity ludvigsen&f=false


Also see Carroll's comments just before 4.33 http://nedwww.ipac.caltech.edu/level5/March01/Carroll3/Carroll4.html

 

[v.dinesh]

sorry if i sound stupid...is noethers theorem essential to say that stress energy tensor is divergence free?? :-8

 

[dx]

The Einstein equations can be justified in many ways, but I think there is still much to learn about their basis, since they apparently express analogies with thermodynamical relationships whose significance is still not understood.

 

[Altabeh]

sorry if i sound stupid...is noethers theorem essential to say that stress energy tensor is divergence free?? :-8

No! The "divergence free" property of the energy-stress-momentum tensor is intrinsic in the sense that due to an empirically proven nature of the conservation laws of momentum and energy through the initial theoretical picture that these quantities remain invariant under respectively the translations of spatial coordinates and time coordinate, spatial relativity suggests the formula

$$T^{\mu\nu}_{,\nu}=0,$$

to hold for a general contravariant energy-momentum tensor $$T^{\mu\nu}.$$ This tells us that if the universe is filled with a symmetrically distributed static dust with pressure $$p_i$$ and mass density $$\rho$$ where the index $$i$$ runs over 1,2,3 each showing the pressure along a spatial axis and the gravitational field is so weak, then

$$T^{0\nu}_{,\nu}=\rho_{,0}=0,$$
$$T^{k\nu}_{,\nu}=p^i_{,i}=0,$$ (Do not sum over $$i$$!)

so that $$\rho=\rho_0$$ is probably constant (if not dependent on the position) and the spatial derivatives of the pressure must be zero in order to just get the energy of dust conserved in this model of the universe we are considering here! Therefore the dust would have a constant pressure along all spatial axes but the density might be position-dependent if not constant.

AB

 

[v.dinesh]

well explained for that particular case@Altabeh
But in search for a second order symmetric tensor for the RHS which should contain the material property can we precisely prove that there exists only one such tensor?? is my question too vague??

 

[Mentz114]

well explained for that particular case@Altabeh
But in search for a second order symmetric tensor for the RHS which should contain the material property can we precisely prove that there exists only one such tensor?? is my question too vague??

If you can write the Lagrangian of the source in terms of a field $\phi$ and its derivatives, then there is one canonical EMT for that Lagrangian given by

$$T^{\mu\nu}=\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\partial^\nu \phi - \frac{1}{4}g^{\mu\nu}\mathcal{L}$$

and $T^{\mu\nu}_{,\nu}=0$

( I've butchered the indexes so better check this ). As you can see [tex]g^{\mu\nu}[/itex] is in there so it's not much help if you don't know g.