1
$\begingroup$

Nakahara in his book on the Geometry and Topology introduces the pseudo-Riemannian metric as a type of (0,2) tensor which contains some properties which I interpreted them as a kind of multiplication factor. On the other hand, as we know, the metric is a solution of Einstein field equations, hence it describes the gravitational field (maybe it is better to say that it is an auxiliary field since the connection could have a better description of this field on the equivalence principle.). I want to know, how can be connect these two definitions on metric? Why the factor that change the upper indices to lower one and vice versa, is known as a gravitational field?

$\endgroup$
2
$\begingroup$

Don’t think of the metric tensor as just a “multiplication factor”, or as something to raise and lower tensor indices. Think of it as what determines distance in spacetime. It’s the coefficients of the terms in the Pythagorean Theorem!

For example, in flat spacetime (and taking $c=1$) the four-dimensional distance between nearby points is

$$ds^2=dx^2+dy^2+dz^2-dt^2$$

but at a particular point in a curved spacetime it could be something like

$$ds^2=1.12\,dx^2+0.05\,dx dy+0.97\,dy^2+1.27\,dz^2-0.85\,dt^2.$$

In General Relativity, particles free of non-gravitational forces move on geodesics through curved spacetime. Geodesics are paths of minimum/maximum/stationary distance, so knowing the metric tensor means you can find out how things move under gravity.

Spacetime is curved by the density and flow of energy and momentum. The Einstein field equations specify how the energy-momentum tensor determines the metric tensor.

So the “big picture” is: energy and momentum cause distance in spacetime to be more complicated than $ds^2=dx^2+dy^2+dz^2-dt^2$. In other words, they cause a non-Minkowskan metric tensor. This curved geometry then causes the “straightest possible lines” in it to be non-trivial gravitational trajectories (including Earth’s elliptical orbit!).

$\endgroup$
0
$\begingroup$

Any manifold has a tangent and cotangent space, they are dual to each other, hence am inner product is defined between them resulting in scalars. In physics we can only observe scalars, hence vectors have to have their image in the tangent and cotangent spaces to form measurable quantities.This is the role of the metric tensor.

$\endgroup$
0
$\begingroup$

I did not understand what you exactly by "multiplication factor" mean.

But, a (pseudo-Riemannian) metric do not necessarily solve the Einstein field equation (EFE)! A geometer does not necessary need EFE and still he can do a lot in geometry. The EFE relates geometry to physics! That is, geometrical solutions of EFEs are special solutions and configurations in geometry, but not all possible ones! In other words, the EFE restricts geometry to "physical" geometry!

A pseudo-Riemannian metric is a metric which is nondegenerate, whereas a Riemannian metric is a nondegenerate positive definite metric. For instance, a Lorentzian metric is a pseudo-Riemannian metric. A pseudo-Riemannian metric does not satisfy EFE. So the metric is not a solution of EFE.

And the metric is sometimes called "gravitational field" or "gravitational potential"? It has, I think, two reasons. First, because of technical reasons, i.e., the relations that the metric satisfy. Second, the General Relativity is a theory of spacetime, so it is plausible to regard the most important variable in this theory as "potential".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.