Schwarzschild spacetime in Kruskal coordinates

[PeterDonis]

the ingoing and outgoing Painleve charts overlap in region I.

Yes. And also, if you read the Insights article you referenced, you will see that I say that there should also be another pair of Painleve charts that overlap in region III (the left exterior region). Then the two "ingoing" charts will overlap in the black hole (region II) and the two "outgoing" charts will overlap in the white hole (region IV).

 

[Orodruin]

So, the ingoing and outgoing Painleve charts overlap in region I. It makes sense since generally in an atlas charts may overlap.

Not only are charts allowed to overlap, it is essential that there are chart overlaps. Otherwise you would simply not describe how different parts of the manifold are stitched together.

 

[cianfa72]

I'm still confused about geodesic affine parameterization. For a timelike geodesic any affine parameter $\lambda$ is related to the proper time $\tau$ along the curve via $\lambda =a\tau + b$. So I believe the Schwarzschild coordinate time $t$ (since is not related via an affine map to the proper time of free-falling observers) cannot be used as affine parameter for timelike geodesics in Schwarzschild spacetime.

 

[PeterDonis]

I believe the Schwarzschild coordinate time (since is not related via an affine map to the proper time of free-falling observers) cannot be used as affine parameter for a timelike geodesic in Schwarzschild spacetime.

You are correct. One easy way to see it is to note that Schwarzschild coordinate time along an ingoing timelike geodesic increases without bound as the horizon is approached, where of course proper time along the geodesic is finite.

 

[cianfa72]

The same should apply for geodesic spacelike curves. In the sense that an affine parameterization has to be related by an affine map to the proper lenght along the spacelike geodesic ?

 

[PeterDonis]

The same should apply for geodesic spacelike curves. In the sense that an affine parameterization has to be related by an affine map to the proper lenght along the spacelike geodesic ?

Yes.

 

[cianfa72]

So when we write down the geodesic equation as $$ \frac {D} {d\lambda} \frac {dx^{\mu}} {d\lambda} =0$$
$\lambda$ is implicitly an affine parameter.

 

[Orodruin]

So when we write down the geodesic equation as $$ \frac {D} {d\lambda} \frac {dx^{\mu}} {d\lambda} =0$$
$\lambda$ is implicitly an affine parameter.

Yes

Edit: For a non-affine parameter, the RHS would be proportional to $dx^\mu/d\lambda$

Edit 2: … and in that case you can use the chain rule to obtain an ODE expressing the relationship between $\lambda$ and proper time.

 

[cianfa72]

So a solution of the equation $$\frac {D} {d\lambda} \frac {dx^{\mu}} {d\lambda} = K \frac {dx^{\mu}} {d{\lambda}}, K \neq 0$$ gives a geodesic $x^{\mu} ({\lambda})$ implicitly parametrized by a non-affine parameter $\lambda$.

 

[cianfa72]

Btw $\lambda$ can be viewed as a real-valued smooth function defined along the curve on the manifold. Therefore we can evaluate the vector field of components $\frac {dx^{\mu}} {d\lambda}$ in $\frac {\partial} {\partial x^{\mu}}$ coordinate basis on it (in another thread I asked for a clarification on the fact that such evaluation is actually well-defined for functions defined only along a curve and not on open neighborhoods of each point along the curve itself). By definition such vector field is the pushforward of $\frac {d} {d\lambda}$ via the map $x^{\mu} (\lambda)$ hence the evaluation should always return $\frac {d\lambda} {d\lambda} = 1$.

Does the above result hold for any kind of parametrization (affine or not) ?

ps. Happy new year 2024 !