# Coordinate change and gravity

• I

## Main Question or Discussion Point

What I understood is that The Schwazschild metric is obtained by setting spherical symmetry in the metric and solves the field equation in vacuum.

But is the flat metric a solution too, or does it mean that changing the coordinates induces gravity ?

## Answers and Replies

Related Special and General Relativity News on Phys.org
martinbn
The Schwarzschild solution is not a single solution. It is in fact a one parameter family of solutions. If the parameter M is zero you get the Minkowski solution.

Dale
Mentor
But is the flat metric a solution too, or does it mean that changing the coordinates induces gravity ?
If you set M to 0 then you get a flat metric. All of the curvature goes to zero. So there is no gravity induced, it is just spherical coordinates in flat spacetime.

Nugatory
Mentor
But is the flat metric a solution too, or does it mean that changing the coordinates induces gravity ?
The Einstein field equations are differential equations, so they have multiple solutions depending on the boundary conditions that we choose. The solution will be a metric tensor that solves the field equations consistent with the particular boundary conditions we've chosen; it determines a particular spacetime geometry.

Flat spacetime is the geometry we get when we choose empty space everywhere as the boundary condition. We can write this solution in many different coordinate systems; for example Minkowski ##ds^2=-dt^2+dx^2+dy^2+dz^2## and polar ##ds^2=-dt^2+dr^2+r^2(d\theta^2+\sin^2\theta{d}\phi^2)## are two different ways of writing the metric tensor for flat spacetime. No matter what coordinates we choose, the curvature tensors will come out zero, there will be no gravity.

Schwarzschild spacetime is another geometry, the one we get when we choose a spherically symmetric mass distribution as the solution. Again, we can write the metric tensor for that spacetime in many different coordinate systems: Schwarzschild coordinates (of course, and it is unfortunate that we use his name both for the spacetime he discovered and for the coordinates he used to describe it - they're different things), or Kruskal, or Painleve, or .... but no matter which we choose, the gravitational effects calculated from the curvature tensors will be the same.

Other choices for the distribution of stress energy yield yet other spacetimes. Again, the gravitational effects in different spacetimes will be different, but they won't change with our choice of coordinates.

jbriggs444
Homework Helper
2019 Award
Flat spacetime is the geometry we get when we choose empty space everywhere as the boundary condition.
Am I missing something? I would interpret "empty space everywhere" to mean that you are looking for vacuum solutions. But flat space-time is not the only vacuum solution.

Nugatory
Mentor
Am I missing something? I would interpret "empty space everywhere" to mean that you are looking for vacuum solutions. But flat space-time is not the only vacuum solution.
Not missing anything, I'm just being a bit careless about the wording around a non-critical point.

jbriggs444