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Saturday, February 25, 2012

Minkowski metric

why is the Minkowski metric not the solution to the Einstein equations for empty space? (or equivalently, why is the empty space of special relativity not the empty space of general relativity?)

Very nice question, I try to give some qualitative argument:

In special relativity spacetime is rigid and described by the Minkowski metric $\eta_{\mu\nu}$. Thus gravity is still given by the non-covariant Newtonian force.

In general relativity the metric becomes a dynamical field $\g_{\mu\nu}$ and spacetime a four-dimensional manifold $\mathcal{M}$. Thus $g_{\mu\nu}$ generally depends on coordinates that can be locally defined on $\mathcal{M}$.

Empty spacetime in special relativity thus describes a subset of the rigid Minkowski spacetime in which no material, but eventually a gravitational force field coming from a mass outside that subset is present

Empty space in general relativity should be a subset $\mathcal{D}$ of the manifold $\mathcal{M}$ defined by a vanishing energy-momentum tensor ($T_{\mu\nu}(x)=0$ for all $x \in \mathcal{D}$). According to the field equations this means $R_{\mu\nu}=0$ for all points within that subset. Since at least in the Levi-Civita connection the Ricci tensor is a function of second- and first-order partial derivatives of $g_{\mu\nu}$, integration allows integration constants that are chosen such that the Newtonian limit can be reproduced for weak fields (see derivation of the Schwarzschild solution). Because of this, $R_{\mu\nu}=0$ has non-trivial vacuum solutions.

This is of course a very heuristic argument. It will be nice to know, if general relativity provides this degree of freedom already by its structure.

the two answers are quite to the point, as well as the post on facebook. Minkowski-space is rigid, masses do not introduce any distorsion, and therefore there can't be any gravitational force acting on particles, because geodesics are always straight lines and the Christoffel symbols trivially zero. if one insists to work out a homogeneous and isotropic FLWR-metric as a solution of an empty universe, the solution is a critically curved cosmology with $\Omega_K=1$, which of course is not static, but shows a constant expansion velocity (and no acceleration).

The Minkowski metric is a solution to the Einstein equations for empty space, though not uniquely. It's the same solution as the FLRW solution for an empty universe, but in different coordinates. Exercise: Solve the Friedmann equation for an empty universe, put the solution into the RW metric, and find the coordinate transformation that makes this metric into the Minkowski metric.

This is not a unique solution. For example, the metric for gravitational waves and the Schwarzschild metric are also vacuum solutions, though this is for a vacuum around a source i.e. Tmunu is not zero everywhere.

Very nice question, I try to give some qualitative argument:

ReplyDeleteIn special relativity spacetime is rigid and described by the Minkowski metric $\eta_{\mu\nu}$. Thus gravity is still given by the non-covariant Newtonian force.

In general relativity the metric becomes a dynamical field $\g_{\mu\nu}$ and spacetime a four-dimensional manifold $\mathcal{M}$. Thus $g_{\mu\nu}$ generally depends on coordinates that can be locally defined on $\mathcal{M}$.

Empty spacetime in special relativity thus describes a subset of the rigid Minkowski spacetime in which no material, but eventually a gravitational force field coming from a mass outside that subset is present

Empty space in general relativity should be a subset $\mathcal{D}$ of the manifold $\mathcal{M}$ defined by a vanishing energy-momentum tensor ($T_{\mu\nu}(x)=0$ for all $x \in \mathcal{D}$). According to the field equations this means $R_{\mu\nu}=0$ for all points within that subset. Since at least in the Levi-Civita connection the Ricci tensor is a function of second- and first-order partial derivatives of $g_{\mu\nu}$, integration allows integration constants that are chosen such that the Newtonian limit can be reproduced for weak fields (see derivation of the Schwarzschild solution). Because of this, $R_{\mu\nu}=0$ has non-trivial vacuum solutions.

This is of course a very heuristic argument. It will be nice to know, if general relativity provides this degree of freedom already by its structure.

I'd say ultimately because of the Equivalence Principle.

ReplyDeleteDirectly from the Equivalence Principle you can derive observable effects like gravitational redshift which are forbidden in flat space!

the two answers are quite to the point, as well as the post on facebook. Minkowski-space is rigid, masses do not introduce any distorsion, and therefore there can't be any gravitational force acting on particles, because geodesics are always straight lines and the Christoffel symbols trivially zero. if one insists to work out a homogeneous and isotropic FLWR-metric as a solution of an empty universe, the solution is a critically curved cosmology with $\Omega_K=1$, which of course is not static, but shows a constant expansion velocity (and no acceleration).

ReplyDeleteThe Minkowski metric is a solution to the Einstein equations for empty space, though not uniquely. It's the same solution as the FLRW solution for an empty universe, but in different coordinates. Exercise: Solve the Friedmann equation for an empty universe, put the solution into the RW metric, and find the coordinate transformation that makes this metric into the Minkowski metric.

ReplyDeleteThis is not a unique solution. For example, the metric for gravitational waves and the Schwarzschild metric are also vacuum solutions, though this is for a vacuum around a source i.e. Tmunu is not zero everywhere.