In general relativity,we will define a unique connection on a manifold with a metric g\mu\nv by introducing some additional properties.There are torsion-free and metric compatibility as assumption.
But we know metric compatibility means that the covariant derivative of the metric with respect to that connection is everywhere zero.This assumption really constrains the freedom of the metric g\mu\nv ,which would contribute to the constraints of 20 freedom of Riemann tensor.But metric compatibility is only a assumption(or not,it maybe have some intrinsic mechanism).So it cannot make that contribution while metric compatibility was a choice of metric,just like using surface with different shape to employ Guass Law.
Following statements are my thought of this problem.If metric compatibility has physics meaning,it might relate to the topology of the manifold with definite metric.Such as in flat spacetime,we have Lorentz symmetry so that we can make some confinement of the energy and momentum.As we all know ,there would be conserved quantities such as invariant mass.So is there some symmetry of manifold contribute to the metric compatibility?
I don't think we have any constrains on metric. For EACH given metric we have restricted connection. Try think this way.
No, since it doesn't refer to quantities that are invariant under general coordinate transformations. It's just a convenience, since it simplifies calculations.
Metric compatibility has measurable consequences if we are using that metric to compare physical dimensions of objects, which is just what we do. Suppose we measure the dimensions of two identical cubes, then transport them about in a spacetime in which the metric is not compatible. Then when we bring the cubes back together we will find they have distorted relative to one another as measured by that metric. We do not experience such distortions using our usual measuring tools.
Another important point about metric compatibility is that if we treat the metric and the connection as independent variables in general relativity, varying the action with respect to each separately, the increased number of equations precisely imply metric compatibility.
Dear James,
I think your explanation really good. It's really related to the dimensions of the system.And if there were some corresponding information about the measurable consequences of Metric compatibility , I want to know more about that.It's interesting.
Palatini formalism is one of the complete explanations which describe that metric compatibility is just constrain of "gauge freedom" .That mean we can choose different connection with the same metric and have the same physical consequence.Explicitly,We can make a transform to connection :
\Gamma'^{\rho}_{\mu\nu}=\Gamma^{\rho}_{\mu\nu}+A_{\mu}\delta^{\rho}_{\nu}
Where A_{\mu} is a constant four vector , \delta^{\rho}_{\nu} get 1 when \rho=\nu,else cases get 0. Above transformation is just like the case of choosing a connection that is not torsion-free,choosing a connection that is not metric-compatible.And through this transformation, we don't influence the physical result and explicitly the Riemann tensor.
Did the metric compatibility in General Relativity really have some physics meaning? - ResearchGate. Available from: https://www.researchgate.net/post/Did_the_metric_compatibility_in_General_Relativity_really_have_some_physics_meaning#583479d348954c0f6f278a31 [accessed Nov 23, 2016].
The statements abut the cubes are wrong, because they don't involve quantities that are invariant under general coordinate transformations. Similarly, choosing a torsion-free connection in general relativity doesn't imply any assumption about physics. It's a choice that simplifies calculations. It is possible to choose a connection with torsion and do calculations and find that quantities that are invariant under general coordinate transformations do not depend on this choice.
There do exist cases, where torsion does play a role, namely in extensions of general relativity, where the spacetime geometry is determined, not only, by the metric, but by additional fields. However, there also, metric compatibility amounts to a choice of coordinates.
It's a theorem of Riemannian geometry that there exists a unique connection that's torsion-free and metric compatible-which may be taken to imply that these conditions fix the associated gauge freedom completely.
Metric compatibility is actually a construct. When you states that \Nabla_{\alpha}(g_{\mu\nu})=0, this allows you to construct the Levi-Civita connection \Nabla for the metric g. As is well known, for every metric g we can do a conformal transformation and obtain a new metric \tilde{g} related to g by \tilde{g}=F(x)g for some positive definite function of the coordinates F(x). It is easy to see that \Nabla_{\alpha}(\tilde{g}_{\mu\nu})=g_{\mu\nu}\Nabla_{\alpha}F(x), thus the new metric \tilde{g} is not compatible with the connection \Nabla. However, \tilde{g} is compatible with a Levi-Civita connection \tilde{\Nabla} constructed from the "compatibility condition" \tilde{\Nabla}_{\alpha}(\tilde{g}_{\mu\nu})=0.
This is in fact an inherent issue of the so called Einstein frame versus Jordan frame.
Some more few words: The physical meaning of this is still a subject discussed in the scientific literature. The question is: which frame is the physically significant? The Einstein frame or the Jordan frame? If you start in the Jordan frame, do a conformal transformation and enter the Einstein frame, the results obtained here have any physical meaning? Many researchers advocate that the Einstein frame is where we can do the physical interpertatio of our equations, while many others advocate on the contrary.
A word of caution: I named the greek letter \Nabla as a "connection" in the same sense as the mathematicians do.
Einstein's happiest but wrong thought!
“Einstein's equivalence principle is wrong because the gravitational force experienced locally is caused by negative energy, gravitons energy and the force experienced by an observer in a non-inertial (accelerated) frame of reference is caused by positive energy”
Adrian Ferent
Because Einstein's equivalence principle is wrong, Einstein’s gravitation theory is wrong.
“Einstein’s gravitation theory is wrong, because is limited to the speed of light” Adrian Ferent
Einstein described his discovery of Equivalence Principle as the "happiest thought of my life".
Looks like the "happiest thought of Einstein’s life" was a wrong thought.
What is worst in 100 years all the scientists (because they did not comprehend Gravitation) did not understand that this thought is a wrong thought. Your professors taught you a wrong theory.
Hilbert was a mathematician, like Einstein he did not understand Gravitation, but they discovered the General theory of Relativity, in my view a wrong theory.
What is the difference between time dilation in Special theory of Relativity and Gravitational time dilatation?
Time dilation in Special theory of relativity is caused by positive energy.
“Gravitational time dilatation is caused by negative energy”
Adrian Ferent
https://www.researchgate.net/publication/310589580_Einstein%27s_equivalence_principle_is_wrong
Article Einstein's equivalence principle is wrong
Yes, it does. It has a deep conceptual meaning.
If the covariant derivative acting on the metric tensor vanishes, it means that during the parallel transport of any two vectors u, v (that is, vectors living on the tangent space at a point of the manifold) along any curve, the inner product between them is covariantly constant (which in general does not mean numerically contant), i.e. =constant.
This also means that under parallel transport of vector v, its own length is also covariantly preserved (set u=v above).
Now, since the generalization of straight lines are geodesic curves (which are locally perceived as straight lines to a travelling observer), this condition mean that parallel transporting the tangent vector to a curve γ means that its length is preserved. Now, a tangent vector to a curve in real space, means that the vector is the velocity vector. So the above is the "curved generalization" of the statement that when the sum of external forces acting on a body is zero, then its velocity is constant (locally it is indeed constant -not only in lenght- when viewed by a local observer since he/she views his path along a geodesic as following a straight line. Although again, it is not numerically constant).
EDIT: Metric compatibility is something we impose on physical grounds, since General Relativity is formed using the Levi-Civita connection. And this is imposed by the principle of equivalence since you can start from the geodesic equation of a locally Minkowiski observer (so d^2(y^μ)/dτ^2=0 is the geodesic equation, such as Newton's law when no forces act on the particle) and then change coordinates from y^μ to χ^μ and you get the full geodesic equation of GR, with the gamma symbols being symmetric (i.e. the Chrystoffel symbols), implying that the principle of equivalence imposes the Levi-Civita connection, which is unique (torsion-free and metric-compatible).
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