# Antisymmetric Metric: Exists in GR?

• micomaco86572
In summary, the metric in the classic GR is always symmetric. However, in special relativity, the Minkowski metric is dss=-1dtt+dxx, which is symmetric. The gravity field in GR is always symmetric, but it has an interpretation as a metric only if it is symmetric.
micomaco86572
asymmetric metric

The metric in the classic GR is always symmetric. What if it is not metric? It has to be symmetric? Does there exist some space with an asymmetric metric? Thx.

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In special relativity, the Minkowski metric is dss=-1dtt+dxx, which is symmetric. We would like GR to approximate SR in small regions of space or when spacetime is almost flat. So the GR gravity field is symmetric. Also the gravity field only has an interpretation as a metric if it is symmetric, a metric is just your dot product which measures the squared length of a vector.

One way of interpreting the metric is that if you're given a certain (timelike) world-line, the metric tells you the amount of time on a clock that follows that world-line. If the metric was asymmetric, it seems as though either the asymmetry wouldn't matter (because the line element would only depend on gij+gji) or would somehow make the clock reading double-valued or something.

There is something vaguely like what you're thinking of in Weyl's gauge theory, in which he attempted to unify gravity and E&M. There is a good nontechnical description available here: http://books.google.com/books?id=uU1WAAAAMAAJ&pg=PA167#v=onepage&q&f=false The quantity $\kappa$ is sort of like an asymmetric metric in the sense that it doesn't stay the same under a complete spacetime inversion $t\rightarrow -t$, $x\rightarrow -x$, etc.

[EDIT] Another thing along these lines is that in any theory with a preferred reference frame, there is some local property of spacetime that is not invariant under spacetime inversion. An example is Jacobson's Einstein-Aether theory, http://arxiv.org/abs/0801.1547

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See for instance http://en.wikipedia.org/wiki/Symplectic_manifold"

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There seems to be a way to arrive at the metric in a completely antisymmetric way so that it may not be necessary to take the inner product of vectors to obtain displacements. As antisymmetric operators are certainly asymmetric these would fit the bill to arrive at the square displacement, in one sort of manner, though I don't know quit what you are asking.

In my way of thinking, objects that multiply themselves are suspect as physical entities; measures of spacetime should be developed upon 'torsional' objects only. Of particular focus in this concern is the metric itself, as it is a symmetrical object that cannot be ignored. Either the connectivity of spacetime would need adaptation or the metric should be explained by antisymmetric operators alone.

In the following, without advancing any theory that would redefinition locality of events within spacetime, but keep the metric intact as a meaningful obect, consider the following-

A is an infinitessimal displacement. AgA defines the local square interval in Minkowski space. But in pseudoRiemann coordinates, applicable to general relativity, equations in p-differential-form tensor densities are one better--valid even for non-local objects without invoking the Christoffel connection.

Let A’ be the dual vector density of the displacement vector A. (*) is the Hodge star operator. (/\) is the wedge product.

Does *(A’/\*A’) = AgA?

A’ is a one-form. *A’ is a three-form. This, wedged with A’, is a 4-form. The Hodge star of this is a scalar. It may not work out in spacetime where the metric trace is (-,+,+,+). I'm looking into it.

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## 1. What is an antisymmetric metric in general relativity?

An antisymmetric metric is a mathematical concept used to describe the geometry of spacetime in the theory of general relativity. It is a mathematical object that expresses the curvature of spacetime in terms of a set of four coordinates.

## 2. How does an antisymmetric metric differ from a symmetric metric in general relativity?

In general relativity, an antisymmetric metric differs from a symmetric metric in that it has a skew-symmetric or anti-symmetric form. This means that the elements of the metric tensor are related by a specific mathematical relationship, as opposed to being equal to each other as in a symmetric metric.

## 3. What is the physical significance of an antisymmetric metric in general relativity?

The antisymmetric metric is used to describe the curvature of spacetime in general relativity, which is the fundamental theory of gravity. It is a key component in Einstein's field equations, which relate the curvature of spacetime to the distribution of matter and energy.

## 4. How is an antisymmetric metric calculated in general relativity?

An antisymmetric metric is calculated through the use of tensor calculus, a branch of mathematics that deals with the manipulation of tensors. In general relativity, the metric tensor is calculated by solving the Einstein field equations, which relate the curvature of spacetime to the distribution of matter and energy.

## 5. Can an antisymmetric metric exist in other theories of gravity?

Yes, an antisymmetric metric can also exist in other theories of gravity, such as modified theories of gravity or alternative theories of gravity. However, the specific form and significance of the antisymmetric metric may differ depending on the theory being studied.

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