# Why is Schwarzschild metric taught at all?

It seems like the Schwarzschild metric only has historical relevance. Since other coordinates like the (ingoing) Eddington-Finkelstein coordinates are far superior.

It seems like the condition on the coordinates that all lightcones must point in the same coordinate axis leads to the problems of the metric at the horizon.

Allowing the lightcones to tilt in the coordinate system like in EF coordinates solves this in a realistic way and seems like the best coordinates for a black hole formed from a collapsed star.

Further the vierbein in EF coordinates can be put in a very simple form:

$$e^I_\mu(r,\theta,\phi,t) = \begin{bmatrix} 1-\frac{m}{r} & 0 & 0 & \frac{m}{r} \\ 0 & r & 0 & 0 \\ 0 & 0 & r \sin(\theta) & 0 \\ -\frac{m}{r} & 0 & 0 & 1 + \frac{m}{r} \end{bmatrix}$$

This makes me wonder why Schwarzschild coordinates are ever used? Is it just because they seem simpler as they have no diagonal entries? (Where the fact it has no diagonal entries leads to its problems).

• It's easier to work out and still gives correct results for anything outside the horizon, which is what most people use it for – Slereah Mar 25 '20 at 13:34
• Gullstrand-Painleve coordinates en.wikipedia.org/wiki/… work for outgoing as well as ingoing. Why bother teaching the inferior Eddington-Finkelstein coordinates, which need two metrics to cover both cases? – m4r35n357 Mar 25 '20 at 13:36
• Please abstractly justify the form of the metric which is used to then set up the Einstein field equations that possess the metric in Eddington-Finkelstein coordinates as their solution, and then directly solve these EFE to give the EF coordinates, without ever passing through Schwarzschild - and as a bonus please establish Birkhoff's theorem from first principles as can be done when deriving the Schwarzschild solution. – bolbteppa Mar 25 '20 at 13:50
• @bolbteppa Without the (very slightly) sarcastic tone that might make the basis for a reasonable answer. – StephenG Mar 25 '20 at 20:06
• "Eddington-Finkelstein coordinates are far superior" - These coordinate are physically meaningless, as they do not correspond to a frame of any physical observer. "The coordinate transformations such as those given by Eddington and Kruskal possess the singularity as well, and, consequently, cannot be used as the coordinate systems for the metrics [...] The Eddington and Kruskal coordinate transformations are non-differentiable, and are not valid." - newton.ac.uk/files/preprints/ni14098.pdf – safesphere Mar 26 '20 at 7:10

# Schwarzschild coordinates work

Most astrophysical calculations involve things that are outside the event horizon. In these cases Schwarzschild coordinates work just fine. Observed phenomena like gravitational lensing, orbital changes relative to Newtonian gravity, and gravitational redshift can all be understood using Schwarzschild coordinates.

# Schwarzschild coordinates look like spherical coordinates

From a pedagogical standpoint its helpful that Schwarzschild coordinates look like spherical coordinates. Most students learning general relativity have seen Newtonian gravity and electromagnetism described in spherical coordinates. We can draw on this experience to reinforce new results by comparing them to results students are already familiar with. In particular Schwarzschild coordinates are great for exploring weak-field limits of GR.

Schwarzschild coordinates are not really spherical coordinates, so there is some danger of misinterpretations. I would argue the benefits outweighs the risk.

# Schwarzschild coordinates describe what distant observers see

The time coordinate $$t$$ of Schwarzschild coordinates is defined as the proper time for an observer at $$r\rightarrow\infty$$. This makes Schwarzschild coordinates the natural coordinate system for us when we look at astrophysical events occurring near distant black holes.

• The last point is somewhat in accurate. If you want to describe what a distant observer sees, you really want to use outgoing Eddington-Finkelstein coordinates. Also the requirement that $t$ goes to the proper-time of a stationary distant observer is also true for Gullstrand-Painlevé coordinates. – mmeent Mar 25 '20 at 15:14
• It's not simply that $t\rightarrow\tau$ as $r\rightarrow\infty$. The coordinate time of Schwarzschild, $t$, for particles near the black hole is still my proper time. I can naturally parameterize motions near the black hole in terms of my own clock using Schwarzschild. – Paul T. Mar 25 '20 at 16:58
• No, it isn't, and no, you can't. – mmeent Mar 25 '20 at 17:29
• "The coordinate time of Schwarzschild, $t$, for particles near the black hole is still my proper time [as a Schwarzschild observer]. I can naturally parameterize motions near the black hole in terms of my own clock using Schwarzschild [$t$]." - This is correct simply by definition. – safesphere Mar 26 '20 at 7:26
• @safesphere It also equally correct for Gullstrand-Painleve time. And is some sense for retard (Eddington-Finkelstein) time for an observer at future null infinity. – mmeent Mar 26 '20 at 14:13

Schwarzschild (Droste) coordinates have a number of advantages over other coordinates systems for the Schwarzschild metric such as Eddington-Finkelstein (EF) or Gullstand-Painlevé (GP) coordinates.

## 1. Symmetry

One is that in Schwarzschild coordinates the time translation and axial symmetries are explicitly manifest. In these coordinates the vector fields $$\tfrac{\partial}{\partial t}$$ and $$\tfrac{\partial}{\partial \phi}$$ are Killing vector fields. (This applies to EF and GP as well.)

In addition it has manifest time reversal symmetry $$t \mapsto -t$$. This sets Schwarzschild coordinates apart from EF and GP coordinates.

## 2. Simplicity

The Schwarzschild metric takes on its simplest form in Schwarzschild coordinates. It has the least number of non-zero components. This makes it particularly easy to work with. This also makes it a good set of coordinates to start with in a GR class.

This simplicity is very beneficial in proving things like the Birkhoff's theorem.

## 3. Sufficiency

Schwarschild coordinates smoothly cover the entire region outside the event horizon. In many (astro)physical scenarios this is the only region that we care about. These coordinates are therefore sufficient for many practical applications.

Of course, for any calculation we should use coordinates suited to the purpose. For example, if we are interested in understanding something falling into a black hole, ingoing Eddington-Finkelstein coordinates are a good choice. If we are interested in the causal structure of the maximally extended solution, we typically want Kruskal coordinates, etc.

• Well I disagree that time symmetry is an advantage since black holes form from a collapsed star which doesn't seem like a time symmetric thing to happen. Though theoretically I guess a black hole could evaportate and a star appear out of it! – zooby Mar 26 '20 at 13:56
• @zooby It is a symmetry of the Schwarzschild metric. It is therefore useful to have coordinates in which this symmetry is explicit. – mmeent Mar 26 '20 at 14:07