# Another Schwarzschild question

1. Aug 1, 2008

### snoopies622

Recently I read that from the perspective of a distant observer
$$(r \gg r_s =\frac{2GM}{c^2})$$
the speed of a beam of light moving directly towards the center of a spherical (non-rotating, non-charged) object decreases because if we set $$ds^2=d\theta ^2 =d\phi ^2 = 0$$ then
$$\frac {dr}{dt}=c(1-\frac {r_s}{r})$$.

I was wondering why this interpretation is favored over the following one: in the Schwarzschild metric, the speed of light stays the same everywhere, but the distance between successive r coordinates increases asympotically as $$r \rightarrow r_s$$. This would also be consistent with
$$\frac {dr}{dt} \rightarrow 0$$ for light as $$r \rightarrow r_s$$.

Does this interpretation fail somehow?

2. Aug 1, 2008

### Mentz114

If both positions explain the observations equally well, we can't say which one is 'actually correct'. And to me they do just that.

3. Aug 1, 2008

### George Jones

Staff Emeritus
This is a coordinate speed. Any observer, freely falling or acccelerated, at any r that the light whizzes by measures the speed of light to be c.

I don't know what "distance" means. I do know what "coordinate speed" and "measured speed" mean.

4. Aug 1, 2008

### yuiop

Hi Snoopies,
As you know there is usually more than one physical interpretation of a given situation in relativity. The speed of light is always c as measured by a local observer and according to an observer at infinity the speed of light is:

$$c' = \frac {dr}{dt}=c(1-\frac {r_s}{r})$$

The two observations can be combined into one more general equation:

$$c' = \frac {dr}{dt}=\frac{c(1- r_s/r)}{(1 - r_s/R)}$$

where R is the location of the observer and r is the location of the measurement. It is easy to see when R=r the local speed of light c' = c.

At the event horizon the speed of light is zero according to an observer at infinity and c according to a stationary observer located at the event horizon. It is a mute point whether you can actually have a statinary observer at the event horizon. In fact if you look at the general equation for the speed of light that I gave above, the local speed of light at the event horizon is not zero, but 0/0 which means undetermined. Anyway, the event horizon is mathematically tricky. Further down into the black hole below the event horizon things get more interesting. While a local observer (if one could actually exist there) always sees the local speed of a falling photon as positive and always heading from the event horizon to the assumed singularity at the centre, while the observer at infinity see the photon as heading in the opposite direction away from the singularity. Although this by itself does not prove that everything asymptotically "gravitates" towards the event horizon, close inspection of the equations for the acceleration and velocity of particles with rest mass support this view and further suggest there is no mass at the singularity (IMHO).

5. Aug 2, 2008

### snoopies622

Thanks all. I have a follow-up if I may.

Suppose I have a series of ropes of lengths $$x, x+\Delta x, x+2\Delta x, x+3\Delta x..$$..etc. and I bend each one into a circle and place these circles concentrically around our spherical mass. Since in Schwarzschild coordinates $$r= C/2 \pi$$ by definition of r, each of these ropes may be regarded as a coordinate marker for r. If the space were flat, the radial distance between each rope would be $$\Delta x / 2 \pi$$. Am I correct in my suspicion that in the non-flat case with the massive sphere at their center, the radial distance between successive ropes will increase as we approach (but not reach) the event horizon? If it is necessary to define “distance”, I hastily offer, ‘what one measures by lining meter sticks end-to-end and then counting them’.

Last edited: Aug 2, 2008
6. Aug 2, 2008

### yuiop

Call the value $$\Delta x / 2 \pi$$ the coordinate distance.The vertical "distance" as measured using physical meter sticks will be greater than the coordinate distance. The distance as measured by halving the time it takes for a radar signal to travel down to a lower concentric circle and back up again and multiplying by the standard speed of light will be greater than the ruler distance. The distance as measured by halving the time it takes for a radar signal to travel up to a higher concentric circle and back down again and multiplying by the standard speed of light will be less than the ruler distance but still greater than the coordinate distance. The ruler distance might seem a simple and unarguable distance but it gets complicated near an event horizon because it takes an infinite number of meter sticks to extend from the lowest concentric circle to the event horixon. Just for clarity it should be added that the rulers do not length contract horizontally and both the distant and local observers agree on the length of the circumferance of each horizontal circle whether measured by rulers or orbital periods.

7. Aug 2, 2008

### George Jones

Staff Emeritus
For what it's worth, this time is something measured on *one* clock, and is calculated in

8. Aug 2, 2008

### snoopies622

Wow, there's more to this than I thought. Thank you both for such wonderfully detailed answers!

9. Aug 2, 2008

### MeJennifer

How do you reason that as it is clear that the clock rates of local and distant observers are not identical?

10. Aug 2, 2008

### yuiop

Derivation of the coordinate horizontal velocity $c_H$ of a photon.

Starting with the Schwarzschild metric:

$$c^2(dtau)^{2}=(1-{r_s}/{r}) c^2(dt)^{2}- (1-{r_s}/{r})^{-1}(dr)^{2}-r^{2} (d\theta)^{2}-r^{2}\sin^{2}(\theta)(d\phi)^{2}$$

The proper time rate for a photon is zero so dtau=0.
For horizontal velocity dr=0.
By setting $d\phi$ to zero we are left with:

$$0=(1-{r_s}/{r}) c^2(dt)^{2}-r^{2} (d\theta)^{2}$$

which rearranges to:

$$r (d\theta)/dt = c\sqrt{1-r_s/r} = c_H$$

which is the coordinate tangential velocity of light according to an observer at infinity.

At the photon orbit the speed of light according to a local observer is c and this corresponds to circumference = orbital period*c = t*c

According to the observer at infinity the circumference is proportional to the coordinate orbital period multiplied by the coordinate tangential velocity of light which is

$$\frac{t}{\sqrt{1-r_s/r}}*c\sqrt{1-r_s/r} = t*c$$

which is the same as the photon orbit circumference as measured by the local observer at r.

A similar result can be deduced for other radii by assuming an elaborate system of mirrors to guide the photon path. The circumference is always measured to be the same by the observers at infinity and locally using a light signal.

Although I won't prove it here, the same is true for circumferences measured from the orbital period and velocities of satellites with rest mass. Clocks slow down low in the gravity well but so do orbital velocities by the same amount and the end result is that the circumference is unchanged. Orbital circumference can be thought of as a sort of gravitational invariant.

Perhaps you thought I implied orbital periods where the same no matter who measures them? I can see that what I said might be ambiguous. I meant orbital circumference is the same if calculated from orbital period multiplied by the orbital velocity, no matter whether the measurement is done by a local observer or one at infinity and is the same as the circumference of a Snoopy Ring at the same location measured by local rulers, laid end to end horizontally. Sorry for any confusion.

11. Aug 3, 2008

### MeJennifer

In fact they can work as a kind of clock counting the number of full orbits!

The catch is, I think, let me know if you disagree, that in certain spacetimes not all observers might agree on when there is one full circumference.

12. Aug 3, 2008

### snoopies622

I wonder if this invariance of circumference is simply a direct consequence of $$g_{\theta \theta}$$ being the same in the Schwarzschild metric as it is in flat space.

13. Aug 3, 2008

### yuiop

Hi Jennifer,

I only have a few minutes, so I will have to come back to this later when I have more time

There are many kinds of spacetimes (or choices of coordinate systems) so I am sure what you are saying is almost cetainly true. Did you have a particular spacetime in mind when you made that statement?

One that comes to mind is the point of view of an observer orbiting the stationary body. Where would they think the start and finish of a orbit is located relative to a test particle at the same radius orbiting in the opposite direction?

There seems to be a difficulty in defining a location in space to return to. How do we put a "marker" there? Do we define a point in space as the location of an object that no acceleration acting on it? That does not seem to work as a body in orbit and a body falling are clearly not stationary from other points of view. Maybe a non inertial observer has a better point of view of a stationary location. For example am observer on a snoopy ring will observer the greatest proper acceleration downwards on an accelerometer, when he has no orbital motion forward or backwards.

One glimmer of light in the turbulent sea of different spacetimes is that it is not true that "Everything is relative". Rotational motion has an absolute nature. A satellite in geosynchronous orbit with the Earth is NOT the same as a stationary body above a stationary Earth although it might appear to be superficially if the observer in geosynchronous orbit declares himself to be stationary.

14. Aug 5, 2008

### snoopies622

And yet someone falling into a black hole reaches the event horizon in finite proper time. How is this possible?

15. Aug 6, 2008

### DrGreg

Schwarzschild's parabola

Kev has got this wrong. It's true that, for radial motion,

$$\frac {ds}{dr} = \frac {1} {\sqrt{1 - r_s/r}} \rightarrow \infty$$ (1)​

as $r \rightarrow r_s$. However the ruler distance to the event horizon is the integral

$$s = \int_{r_s}^r \frac {ds}{dr} dr$$ (2)​

which is finite.

It turns out that, for a constant t coordinate, the curvature of space outside the event horizon can be visualised as extrinsic curvature in a higher dimensional Euclidean space.

Take the equation for the radial metric (when $dt = d\theta = d\phi = 0$ and r>rs),

$$ds^2 = \frac {dr^2} {1 - r_s/r}$$ (3)​

and substitute

$$Z = 2 \sqrt{r_s} \sqrt{r - r_s}$$ (4)​

It is not hard to show that

$$ds^2 = dr^2 + dZ^2$$ (5)​

Note that (4) is the equation of a parabola in (r, Z) coordinates, and (5) is saying that the ruler distance s is just the Euclidean arc length along the parabola. This arc length is finite, despite dZ/dr becoming infinite at r=rs.

Z is to be thought of as a "phantom" coordinate in a fourth space dimension, with Schwarzschild space (suppressing time) as a three dimensional surface of revolution of the parabola around the Z axis.

There's a diagram of this to be found at the end of http://www.bun.kyoto-u.ac.jp/~suchii/schwarzschild.html [Broken].

Note that as the time dimension has been suppressed, this visualisation does not tell the whole story; nor does it work inside the event horizon.

Last edited by a moderator: May 3, 2017
16. Aug 7, 2008

### snoopies622

Thanks, DrGreg. I was wondering about that integral

$$\int_{r_s}^r \frac {dr} {\sqrt{1 - r_s/r}}$$.

I entered it into this anti-derivative finder

http://integrals.wolfram.com/index.jsp

but the result was something that was both more complex than I expected and which diverged at $$r=r_s$$ so I assumed that the distance was in fact infinite. Your parabola substitution -- however -- looks very interesting to me.

17. Aug 7, 2008

### yuiop

Hi jennifer, Nice observation Orbiting satellites do seem to provide a big easy-to-observe natural clock for comparisons. Only trouble is, there are no natural stable orbits below the photon horizon , maybe slightly higher?

Hi Snoopies,

Infinity is a tricky subject which is paradoxical to many people including me :tongue2: One example of the uncomfortable coexistence of infinite and finite is the logarithmic spiral that has the peculiar property that "Starting at a point P and moving inwards along the spiral, one has to circle the origin infinitely often before reaching it; yet, the total distance covered on this path is finite. " See http://www.nationmaster.com/encyclopedia/Logarithmic-spiral
I am sure there are many other examples. I do not pretend to understand the exact meaning and significance of infinity and I am still coming to terms with it. I may be wrong about there being an infinite number of physical rulers extending down to the event horizon and DrGreg seams to have raised a valid objection (See below). On the other hand a finite "distance" to the event horizon does not necessarily mean it can be reached in a finite time. I put "distance" in quotes because it has many meanings especially in GR.

Hi DrGreg,

I think you have raised a good point here and I might learn something from it :) There are however a number of issues that are puzzling me with this interpretation.

First, I wish to understand the physical meaning of s or ds in this context. From the Schwarzschild metric it would appear to be ds = c*dtau where tau is the proper time, but the proper time according to whom? If we are measuring the distance from coordinate radius r1 to r2 is it the proper time according to:

a) A stationary observer located at r1?
b) A stationary observer located at r2?
c) An observer free falling from r2 to r1?
d) The sum of infinitesimal proper time intervals made by a large number of stationary observers spread out from r1 to r2?

I think case (c) can safely ruled out because we are assuming the coordinate time interval between measurements to be zero and that would require the free falling observer to fall with infinite coordinate velocity. I ask, because the proper time varies with location. It may be that the variation of proper time with radius is already taken into account by the integration, in line with case (d) but I just wanted to be sure.

Next, there seems to be a problem with imaginary values popping up when the "ruler distance" is derived from the Schwarzschild metric.

Starting with the simplified Schwarzschild metric (no rotation):

$$c^2 {dt}^{2} = \left(1 - \frac{r_s}{r} \right) c^2 dT^2 - \left(1-\frac{r_s}{r}\right)^{-1}{dr^2}$$

where dt is a proper time interval and dT is coordinate time interval.

By setting dT to zero:

$$c^2 {dt}^{2} = - \left(1-\frac{r_s}{r}\right)^{-1}{dr^2}$$

there is a problem when taking the square root of both sides because one side of the equation always ends up as an imaginary number no matter what side the negative sign is placed. Taking the integral of $$\frac{c dt}{dr} = \sqrt{\left(\frac{-1}{(1-{r_s}/{r})}\right)}$$ yields values for c*dt that are also imaginary for radii greater than the Schwarzschild radius. How is this issue resolved?

Could you show how equation (5) is obtained?

Last edited by a moderator: May 3, 2017
18. Aug 8, 2008

### DrGreg

I confess I never worked out the integral. I knew it had to be finite because of the parabola argument.

But, using the website you quoted, the integral is finite. Did you enter it correctly? If you replace r by x and rs by a, the site gives formula that works perfectly well at x=a.

Integrate[1/Sqrt[1 - a/x], x] == Sqrt[1 - a/x]*x + (a*Log[-a + 2*(1 + Sqrt[1 - a/x])*x])/2

19. Aug 8, 2008

### DrGreg

I was a bit lax in my notation, and this is indeed a bit of a notational grey area.

To keep things simple, let's just consider flat 2-d spacetime (i.e. 1 space + 1 time coord, with no gravity). Different authors will tell you that the metric is given by one of the following equations.

$$ds^2 = c^2 dt^2 - dx^2$$ (I)
$$ds^2 = dt^2 - dx^2 / c^2$$ (II)
$$ds^2 = dx^2 - c^2 dt^2$$ (III)
$$ds^2 = dx^2 / c^2 - dt^2$$ (IV) ​

(I) and (II) are referred to as a (+---) metric signature, while (III) and (IV) are referred to as a (-+++) signature. When you get more experienced in the subject, you might even mentally switch from one definition to another without explicitly saying so.

What is s? Well you have to measure it along a curve in spacetime, and it depends what sort of curve it is.

If the curve is a timelike curve, i.e. represents the wordline of a particle, then equation (II) gives you the proper time $s = \tau$ experienced by that particle. (By the way, use "\tau" in tex.) Equivalently equation (I) gives you $s = c\tau$, and rather more uglier, (III) gives you $s = ic\tau$ and (IV) gives you $s = i\tau$.

On the other hand, if the curve is spacelike and lies within the surface of simultaneity of an observer, then equation (III) gives you the distance s = d along that curve, according to that observer (with appropriate variants for the other metric equations I-IV). In GR this has to be modified slightly to say that the distance is measured using local rulers, i.e. at each point along the curve you use a local observer at that point to make the infinitesimal measurement ds. The curve has to lie parallel to each local observer's surface of simultaneity at that point, in other words, each local observer is stationary relative to the line in space being measured.

(If the curve is null or lightlike, s = 0, and the curve is the worldline of a photon.)

In my previous post, I was implicitly assuming the (III) version of the metric, because that version is more convenient for measuring distance as opposed to time. By putting dt = 0 I was using the metric to measure distance along a spacelike curve rather than time.

(It might have been less confusing if I'd used a different letter e.g. d instead of s to denote ruler distance. Then, using the metric signature (I) that you had used in previous posts, d = is.)

Does that make more sense now?

From my original equation (4)

$$dZ = 2 \sqrt{r_s} \cdot \frac{1}{2}(r - r_s)^{-1/2} dr$$
$$dZ^2 = \frac {r_s}{r - r_s} dr^2$$
$$dr^2 + dZ^2 = \frac{(r-r_s) + r_s}{r - r_s} dr^2$$
$$dr^2 + dZ^2 = \frac{dr^2}{1 - r_s/r}$$​

(It is also possible to work backwards to obtain (4) from (5), which is what I originally did.)

Indeed. It depends which "time" you mean. For the Schwarzschild coordinate time t, or the time of any "stationary" observer, it takes an infinite time. But in the traveller's proper time $\tau$, it takes a finite time. The difference is due to gravitational time dilation (which isn't included in the "Schwarzschild parabola" space-only visualisation).

20. Aug 8, 2008

### snoopies622

I guess not. I did not know to use "a" to represent a constant, so I just tried different integers instead and the results were always in agreement with each other. For some reason my way produces a different anti-derivative which has $$\sqrt{\frac{(r-r_s)}{r}}\sqrt{r}$$ for the denominator. I will accept that yours is correct, but for now I don't know why they differ.