# I Non-coordinate bases

1. Nov 16, 2017

### Silviu

Ah, I see. Makes sense now. Thank you! One more question. The way you play with the coordinates is independent from a point to another? I.e. can you at a point keep the coordinates still while at a nearby point on the manifold you rotate the coordinate system by 90 degrees? Or the transformation should be smooth along the manifold? So if the basis vectors at a point transform using a matrix M, is this matrix a smooth function M(p), or it can take any value at any point on the manifold?

2. Nov 16, 2017

### Orodruin

Staff Emeritus
Take two sets of coordinates $x^a$ and $y^i$. Based on the chain rule for derivatives
$$\newcommand{\dd}[2]{\frac{\partial #1}{\partial #2}} \dd{}{y^i} = \dd{x^a}{y^i} \dd{}{x^a}$$
so the matrix you are looking for just consists of partial derivatives of the smooth coordinate functions.

3. Nov 16, 2017

### Silviu

But this is only for coordinate basis (where the transformation matrix is the Jacobian). However you said that non-coordinate basis need not be partial derivatives. So in that case is the transformation matrix still smooth?

4. Nov 16, 2017

### Orodruin

Staff Emeritus
If the basis is a basis of smooth vector fields.

5. Nov 16, 2017

### Silviu

Thank you! One more question (sorry for taking so long...). So in the book it says that we pick the non-coordinate basis in such a way that $g(\hat{e}_\alpha,\hat{e}_\beta) = e_\alpha^\mu e_\beta^\nu g_{\mu \nu}=\delta_{\alpha \beta}$. So, based on what I understood this is defined for just a point on a manifold, i.e. you can't find a smooth transformation such that the metric at that point is the flat metric. By this I mean that you can apply a transformation such that $Mg=\delta$, but this can't hold over the whole manifold. Is this correct? So the equation above is local.

6. Nov 16, 2017

### pervect

Staff Emeritus
Sorry to come in late, but I thought I'd mention that one of the important properties of vectors is that they can be added together, and that the addition is commutative. So if A and B are vectors, A+B = B+A.

On a general manifold, displacements do not add like vectors, so for example if you go 100 miles north and 200 miles east (in that order) on the curved manifold representing the Earth's surface, you do not wind up in exactly the same spot as if you went 200 miles east first, then went 100 miles north.

This means the operation (which I haven't formally defined) of going "100 miles east", the operator I am calling a displacement cannot be a vector, because it doesn't add with other displacements. One of the things we study in GR is the failure of displacements to commute.

In this specific example of polar coordinates, it turns out that the displacements do commute because the plane is flat (the curvature tensor vanishes), but relying on this specific feature of the specific problem will cause problems if you assume it's always true.

I'm not sure if this will help, there are a lot of semantic problems with the meaning of words. I hope that this informal note helps, but I'm not sure it will.

7. Nov 17, 2017

### PeroK

If you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ globally then what you started with was flat spacetime in some or other coordinate system.

In general, for each point, you can find a coordinate transformation that transforms the metric to $\eta_{\alpha \beta}$ at that one point.

In fact, you can also ensure that in these coordinates $\frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} = 0$. Again only at that one point.

This means that in these coordinates the Christoffel symbols vanish at that point and the equations of motion reduce to those of flat spacetime (at that point). Hence the term "Local Inertial Frame".

8. Nov 17, 2017

### Orodruin

Staff Emeritus
Just to be clear, this refers to a holonomic basis. It is often possible to find a non-holonomic basis where the metric is $\eta_{\alpha\beta}$ in an extended region.