# Change of the minkowski metric

1. Apr 14, 2014

### ChrisVer

If I am not mistaken, the change of the minkowski metric to:
$n_{\mu\nu} \rightarrow g_{\mu\nu}(x)$
will violate the Poincare invariance of (example) the Electromagnetism Action.
However it allows us to define a wider set of arbitrary transformations (coordinate transformations).

The last line confuses me. If I can choose a general coordinate system without a problem (since everything will remain invariant) what restricts my choice not to be of the form:
$x'^{a}(x^{b})= Λ^{a}_{b}x^{b}+p^{a}$?

2. Apr 14, 2014

### Matterwave

Nothing will prevent you from choosing that form of a transformation, if your Lambda matrix is arbitrary. What's special about the Poincare transformations is that the Lambda matrix has special properties (i.e. that it preserves the Minkowski metric). General coordinate transformations will not. Try a change from Cartesian coordinates to Polar coordinates and see what your lambda matrix has to be, and then notice that it is not in the group of Lorentz transformations.

3. Apr 14, 2014

### Geometry_dude

The Poincare group consists of Lorentz transformations ((hyperbolic) rotations in spacetime
by $\Lambda$) and spacetime translations by $a \in \mathbb{R}^4$. Changing you coordinate system from one cartesian one to another one corresponds to letting the Poincare group act on your coordinate system via
$$x'^i=\Lambda^i{}_j x^j + a^i \, .$$
Since
$$\eta = \eta_{ij} \, d x^i \, dx^j$$
you will find that this transformation leaves the metric invariant. Note that I omitted the tensor product, because it is symmetric. Changing to another non-cartesian coordinate system does not (!) change your metric, however, it does change how it looks like in your coordinate system. The EM-field tensor is
$$F= F_{ij} \, d x^i \wedge d x^j$$
and things work in an entirely analogous manner.

4. Apr 14, 2014

### bcrowell

Staff Emeritus
You seem to be assuming that Minkowski coordinates still exist in GR. They don't (except in special cases). Therefore there is no way to define a Lorentz transformation.

5. Apr 14, 2014

### Geometry_dude

Yes, if this is really a GR question, then you're thinking about this in an entirely wrong manner.
Let's get this straight: Special relativity is indeed a special case of general relativity. Spacetime in special relativity is $(\mathbb R ^4 , \eta)$, where $\mathbb R ^4$ is your spacetime manifold and $\eta$ tells you implicitly how far points are from each other, angles and which directions are spacelike, timelike, lightlike/null, etc. The poincare group is the group of active (or passive if interpreted as coordinate transformations) transformations on your manifold that leave the metric $\eta$ the same.
In general relativity, we replace $\mathbb R ^4$ by a general 4-dimensional manifold $Q$ and the Minkowski metric $\eta$ by a general non-constant metric $g$, which is a solution to the Einstein equation. Then the group of transformations that leaves $g$ invariant will not be the Poincare group. In fact, this group might not even exist. If it does, it is called an isometry group, because it doesn't change distances. Hence the Poincare group is the "largest" group of isometries of Minkowski spacetime.

Please read up on basic manifold theory, especially (pseudo-)Riemannian manifolds, before diving into GR.