# Schwarzschild Metric Question

1. Feb 7, 2004

### franznietzsche

i'm toying around with the schwarzschild metric using it to find distances from a star in sphereical coordinates.

the metric is:

g** =

1/(1-2M/r) 0 0 0
0 r^2 0 0
0 0 r^2sin(theta)^2 0
0 0 0 -(1-2M/r)

Now my question what is the significance of r in this tensor? is it the radisu of the star or the distance from the center of the star? If it is the distance how does it factor into the equation:

$$d^2s = \frac{1}{1-\frac{2M}{r}}d^2r + r^2d^2\theta + r^2sin(\theta)^2d^2\phi + -(1-\frac{2M}{r})d^2t$$

how would i use this to calculate spacetime distances in the presence of a star?

Last edited: Feb 7, 2004
2. Feb 7, 2004

### lethe

it is not distance from the star's center. it is just a radial coordinate.

3. Feb 7, 2004

### HallsofIvy

Staff Emeritus
In my copy of Eddington's "The Mathematical Theory of Relativity", he makes the point that, in fact, there are several "classical" quantities that would correspond to the "r" in the Schwarschild metric and so we are free to choose the simplest. I don't have it with me now but as I recall, his choice for r is not the classical distance. Al Lethe said, it is just a "radial coordinate". You are free to set up your coordinate system as you prefer (and the whole point of the Schwartschild metric is that the coordinates system is not the standard spherical coordinates).

4. Feb 7, 2004

### franznietzsche

but how do i use it the to calculate distance across the manifold?

5. Feb 7, 2004

### Stingray

r is just a coordinate like r in the usual spherical coordinates. You could take r to mean the radius of the star if it is a star (and not a black hole), but its an arbitrary choice.

The only reasonable (coordinate independent) meaning of distance in GR is s in your equation. You just integrate a path to find these distances. For example, if you want to find the distance between two points described by coordinates (t,r1,theta,phi) and (t,r2,theta,phi), then s=int[sqrt(grr),{r,r1,r2}], Sorry I don't know how to make the math look nice.

6. Feb 7, 2004

### franznietzsche

i'm assuming you mean:
$$s = \int_(r^1)^(r^2) \sqrt{g^{\ir ir}} dr [\tex] however i'm not sure what you mean by $$g^{ir ir} [\tex] 7. Feb 8, 2004 ### Stingray By grr I mean the rr component in your matrix: 1/(1-2M/r). 8. Feb 8, 2004 ### franznietzsche ah, ok. 9. Nov 28, 2006 ### Chris Hillman Multiple competing operationally significant notions of disance Way back in February 2003, franznietsche asked: The radial coordinate here keeps track of "distance" from the object which is the source of the field in this spacetime model, and it is partially but not fully analogous to the usual radial coordinate familiar from euclidean geometry. One geometric interpretation of the Schwarzschild radial coordianate can be deduced by comparing the line element you wrote down: $ds^2 = -(1-2m/r) \, dt^2 + \frac{1}{1-2m/r}dr^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right),$ $-\infty < t < \infty, \; 2m < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$ with the line element for Minkowski spacetime in polar spherical coordinates: $ds^2 = -dt^2 + dr^2 + r^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right), \; -\infty < t < \infty, \; 0 < r < \infty, \; 0 < \theta < \pi, \; -\pi < \phi < \pi$ In both cases, fixing [tex]t=t_0, \, r=r_0$$ gives the metric of a round sphere of surface area $$A = 4 \pi \, r_0^2$$, namely $d\sigma^2 = r_0^2 \, \left( d\theta^2 + \sin(\theta)^2 \, d\phi^2 \right), 0 < \theta < \pi, \; -\pi < \phi < \pi$ A possibly more important interpretation is that rests upon a more sophisticated concept, the "optical expansion scalar". For those who have seen this term, the obvious outgoing radial null congruence has (in both cases) optical expansion scalar $$1/r$$. Next, consider two observers who hover at $$r = r_1, \, r_2$$ where $$2m < r_1 < r_2 < \infty$$. Indeed lets put $$r_1 = R, \, r_2 = R+h$$ and consider h to be some small quantity. Then we might ask: what is the length of a radial curve in a hyperslice $$t=t_0$$ which connects the two observers? This quantity will will be the distance measured by a slowly moving walker who walks along a radial wire suspended between the two observers. Let's compute this "pedometer distance" in the case where we have Putting $$dt = d\theta = d\phi = 0$$, we obtain $$ds = \frac{dr}{\sqrt{1-2m/r}}$$, which we integrate: $\int_{r_1}^{r_2} \frac{1}{\sqrt{1-2m/r}} \, dr$ The answer is rather complicated! To understand it, we expand the multivariable Taylor expansion in the quantities $$m, h, 1/R$$, obtaining: $$\rm{distance} = h + \frac{m \, h}{R} - \frac{m \, h^2}{2 R^2} + \frac{3 m^2 \, h}{2 R^2}$$ which is both simpler and more perspicuous than the exact result. But there is another, even more nature notion of distance which is very often employed: light travel time distance. This is defined very simply: one observer sends a radar pip which strikes the second observer and returns to the first, who divides by twice the elapsed time as measured by an ideal clock which he holds. Let's compute the light travel time measured by the closer observer. The radial null geodesics satisfy $$ds^2 = 0, \, d\theta = 0, \, d\phi = 0$$, so we can write: $\pm dt = \frac{dr}{1-2m/r}$$ With some care, you can see that we have [itex] \frac{\Delta t}{2} = \int_R^{R+h} \frac{1}{1-2m/r}$
This is half the elapsed coordinate time, so we need to multiply by the factor which converts an elapsed coordinate time to elapsed proper time (measured by a static observer), namely $$\sqrt{1-2m/r}$$. We find
$\frac{\Delta s}{2} = \sqrt{1-2m/R} \cdot \; 2 m \, \log \frac{R-2m+h}{R-2m} \approx h + \frac{m \, h}{R} - \frac{m \, h^2}{R^2} + \frac{3 m^2 \, h}{2 R^2}$

Let's compare this with the light travel time measured by the farther observer. Proceeding similarly we find a result larger by a factor of
$\frac{\sqrt{1-2m/(R+h}}{\sqrt{1-2m/R}}$
which yields the multivariable Taylor expansion
$\frac{\Delta s}{2} \approx h + \frac{m \, h}{R} + \frac{3 m^2 \, h}{2 R^2}$
We can summarize our results as follows: to second order in m,h, 1/R, we have
$\rm{distance}_{\rm{LTT, near to far}} = \rm{distance}_{\rm{pedometer}} + \frac{m \, h^2}{2 R}$
$\rm{distance}_{\rm{LTT, far to near}} = \rm{distance}_{\rm{pedometer}} - \frac{m \, h^2}{2 R}$
$\rm{distance}_{\rm{pedometer}} = h + \frac{m \, h}{R} - \frac{m h^2}{2 R^2} + \frac{3 m^2 h}{2 R^2}$

With more thought and more effort, we could add further notions to this list, such as an "optical disk distance".

There are several points of interest here. First, the light travel time is not symmetric: in general, the distance from A to B is not the same as the distance from B to A. Second, there are multiple distinct but operationally signficant notions of distance which we can adopt in this spacetime model. These all give very similar results over small distances, but can differ appreciably over larger distances. In this case, all three distances which we computed are to first order $$\rm{distance} \approx h + h \, \frac{m}{R^2} \, R$$, where I wrote the result in a manner which suggests extracting the Newtonian gravitational acceleration $$m/R^2$$ and comparing with similar compuations in the well-known Rindler metric for Minkowski spacetime. I'll leave that as an exercise, but we find the same conclusion: even in flat spacetime, there are multiple reasonable but distinct operationally significant notions of distance which accelerating observers can employ, and these are in general not even symmetric.

Failure to recognize this key point is a source of many common misconceptions, even by some physicists.

Chris Hillman

Last edited: Nov 28, 2006
10. Nov 30, 2006

### Jim Graham

Hi Chris Hillman. This is an interesting concept, but I am concerned about how “Pedometer” distance is expressed here. The value of “h” is defined only at one end point. Shouldn’t you use an expression defined at each point on the line? If you did, would you get the same results as the Light Travel Time measurement for a round-trip pedomenter measurement? (I really don’t know – I’m pretty naïve about this stuff)…Jim

11. Nov 30, 2006

### Chris Hillman

Problem with parameter?

Hi, Jim,

I don't understand why you say that: $$h$$ is just a parameter chosen so that for small values all three quantities which I discussed reduce to $$h$$.

It might help to recognize that I was discussing two operationally defined methods of measuring distance over substantial regions in the exterior (static) region of the Schwarzschild vacuum, and indeed in other spacetime models. One turns out to be nonsymmetric (but is of course approximately symmetric in small regions, as Riemann would lead us to expect), hence I computed three quantities, not two.

Chris Hillman

12. Dec 2, 2006

### Jim Graham

I was thinking that h was defined as the distance between r1 and r2, as observed at r1. I guess that does not make sense, since you’ve calculated this distance as the Light Travel Time (LTT) distance as seen from r1. So who observes the distance h? Not even the traveler with the pedometer in your example sees this distance.

I would expect GTR to say that any frame of reference is equally valid (this is just an assumption on my part). I take that to mean that the distance is the distance observed at r1 using the LTT method AND it is also the distance observed at r2 using the same method AND it is also the distance seen by a variably accelerated observer like the guy with the pedometer. It sounds to me like you’re saying that one of these observers sees “reality” while the others see “illusions”, but I don’t know which observer it is. Even the guy with the pedometer is not physical – he’s moving in zero time (or maybe I’ve misunderstood the condition t=0, maybe t is not purely time?)

Now I'm just making myself more confused. Sorry if I'm being obtuse - it's not intentional, I just need a few years to study the math and physics.

Jim

13. Dec 2, 2006

### Jim Graham

By the way – I did think about the round-trip travel distance for the guy with the pedometer, and convinced myself that he would see a different round-trip distance than the stationary observer using the LTT method. The symmetry of the intermediate measurements of the three observers lead me to this conclusion.

Your assertion that the measurements at r1 and r2 are not symmetric should probably be amended to say that they are not “equal”. I’m pretty sure that each observer measures the same distance from A to B as from B to A. The pedestrian in your example would see a continuously variable distance A to B as he traveled his path. He would also see a continuously variable distance B to A as he traveled the path. Always, he would see the distance A to B equal to the distance B to A. (A pedestrian walking from B to A is a different observer from one going the other way, but he would have the same kind of observations along the way, even though he would report a different pedometer reading at the end).

My gut feeling is that the distance measurement has everything to do with the observer – I would go so far as to question the existence of “distance” in the absence of an observer. The distance is not so much a single distance that is “operationally measurable” as it is a different entity for each observer. (At least that’s what my intuition is telling me).

Jim

14. Dec 4, 2006

### Chris Hillman

What is h in my comparision of various "distances"?

Hi, Jim,

To answer the question in the title: in my post I described two distinct operationally significant notions of "distance" valid for static (or quasistatic) observers, one of which is nonsymmetric. I computed these three quantities for a pair of static observers hovering outside the event horizon of a nonrotating static black hole (modeled by the Schwarschild vacuum solution in gtr). I then compared these three values and found they are not distinct but even satisfy two inequalities.

In my computations, h is just a parameter, corresponding to differences in Schwarschild radial coordinate for two radially aligned static (or quasistatic) observers, not something one would measure directly. However, as I pointed out, to first order in h, all three quantities do agree. This is just a more precise way of stating that, as we would expect, while there are multiple operationally significant notions of distance which can be used by accelerating observers (even in flat spacetime, as studying the Rindler model reveals), these all agree in very small regions.

A picture would be worth a thousand words here! I am still new to the forum, so I haven't yet learned how to upload a sketch (I confess that simple laziness has also played a role here).

Don't make the mistake of assuming that the notion of "reference frame" from str carries over to gtr. On very small scales, of course it does--- that is the idea behind frame fields--- but on larger scales, curvilinear coordinates and various curvature effects are unavoidable in curved spacetimes.

No No No!

Forget the guy with the pedometer, especially since I didn't try to explain that in detail, just provided a hint sufficient (or so I believe) for a sophisticated reader to figure out what I must have meant. Let's just concentrate on understanding why the "radar distance" (aka "light travel time distance") is operationally significant but nonsymmetric.

I guess someone, probably me, really does need to provide a sketch and upload it.

Don't worry, Jim, I think the problem is that I've been too lazy to provide the neccessary figure, which I think would clarify the confusion.

Er.... I confess that I am hoping that some kind member will step in and save me the trouble of drawing and uploading my own figure!

Chris Hillman

Last edited: Dec 4, 2006
15. Dec 4, 2006

### Daverz

I don't think this is a particularly good drawing, but this is what I found by googling [URL [Broken] parabaloid"[/url].

Last edited by a moderator: May 2, 2017
16. Dec 4, 2006

### Chris Hillman

Flamm paraboloid

Thanks, Daverz!

That's not at all what I had in mind (I meant that someone, probably me, needs to draw a figure showing the world lines of a round-trip radar pip from A up to B, and from B down to A, with some tiny light cones drawn to scale so that it becomes visually apparent that radar distance is not symmetric in this situation, or in most other situations).

Fortunately, looking at the Flamm paraboloid should help clear up the confusion about the meaning of the parameter h!

First, for those who don't know, this is what you get when you embedd a hyperslice of constant Schwarzschild time coordinate (it misses the interior region entirely but has a grazing intersection with the event horizon) into a flat embedding space.

In the figure which daverz found (Wikipedia is unstable by design, so one should cite a specific version http://commons.wikimedia.org/w/index.php?title=Image:Flamm's_paraboloid.jpg&oldid=2593207 by hitting the "permanent link" button in the sidebar!), horizontal circles correspond to spheres of constant Schwarzschild radial coordinate (with one dimensional supressed), and the smallest circle, at the bottom, corresponds to the horizon.

OK, next, in the figure of the embedded paraboloid, draw a short radial arc on the surface and also its projection to the horizonal plane in the euclidean embedding space (which is of course not part of our original spacetime model!). In this figure, the Schwarzschild radial coordinate of a point E on the paraboloid is $$r= r_0$$, where $$C = 2 \pi \, r_o$$ is the circumference of the circle $$t=t_0, \; r = r_0$$ drawn through E.
Then, the parameter h, which is just the difference in the Schwarzschild radial coordinate for two radially seperated observers, corresponds to "radial distance" in the horizonal plane in the unobservable and artificial embedding space, while the "pedometer distance" agrees with the length of the arc as measured out in the paraboloid itself.

Recall that in my post, I showed that the radar distance from the near observer up to the far observer is larger than the pedometer distance, but the distance from the far observer down to the near observer is smaller than the pedometer distance, by the same amount. Both radar distance and pedometer distance have operational significance in our spacetime model. In contrast, h is just the artificial parameter I used to compare these three quantities. (Three, not four, because pedometer distance is symmetric in this circumstance, unlike radar distance.)

I must stress that the intrinsic geometry of the paraboloid is what we want to get intuition for. There are always infinitely many ways of embedding things in higher dimensional spaces; all such embeddings are purely artificial. They may have some value in helping us get intuition for the intrinsic geometry, but it is important to clearly understand that the intrinsic geometry but not the embedding has operational significance in our thought experiment.

In addition, I should stress that there are infinitely many ways of choosing hyperslices; constant Schwarzschild time coordinate is just one of these. If we choose constant Painleve time slices, we obtain hyperslices which are flat, right down to the curvature singularity! There are some beautiful pictures in MTW illustrating the point that we get quite different looking Monge patch embeddings if we use different hyperslices. (Monge embedding includes what you would naturally do to embedd a radially symmetric surface in three dimensional Euclidean space. Not all two manifolds can be embedded this way, however.)

Chris Hillman

Last edited by a moderator: May 2, 2017