Schwarzschild Metric

1. Feb 5, 2008

keithh

Hi,
I am new to this forum so I apologise if a similar thread already exists.

I am trying to resolve the implications on space and time as you approach a black hole event horizon with respect a distant observer and an onboard observer. My issue relates to combining the effects of both special and general relativity, i.e. velocity and gravitational effects.

If a free-falling observer acquires velocity on approaching the event horizon, then special relativity suggests that a distant (stationary) observer will perceive time dilation and space contraction in the direction of motion in the free-falling frame of reference. However, while the Schwarzschild metric appears to support the concept of time dilation when approaching the event horizon, it suggests that the space expands, not contracts, due to the increasing curvature of space near the horizon. Therefore, my question is whether both these spatial effects have to be taken in consideration?

Would appreciate any insights to this issue.
Thanks

2. Feb 5, 2008

A.T.

You mean length contraction of the moving object?
Yes the distances between space coordinates are greater. But this is not countering the length contraction of the object, but rather further increasing it (for the distant observer).

3. Feb 5, 2008

Mentz114

Hi Keithh,
welcome to the PF. I suspect you're looking at the metric and have observed how g00 and grr are changing with r. This is not curvature. The curvature is measured by the Riemann tensor, which is a multivalued object, which depends on the second derivatives of the metric. I happen to have the relevant equations set up in latex -

Christoffel symbols ( comma ',a' indicates partial differentiation)
$$\Gamma^{m}_{ab} = \frac{1}{2}g^{mk}(g_{ak,b}+g_{bk,a}-g_{ab,k})$$
Riemann tensor
$$R^{r}_{mqs} = \Gamma ^{r}_{mq,s}-\Gamma ^{r}_{ms,q}+\Gamma ^{r}_{ns}\Gamma ^{n}_{mq}-\Gamma ^{r}_{nq}\Gamma ^{n}_{ms}$$

which may be found in any textbook.

There is a host of movies, papers and popular articles exacty answering your question, try Googling 'falling into a black hole'. The first dozen hits are all good.

4. Feb 6, 2008

keithh

Thanks to both A.T. and Mentz114 for their input.

AT: An interesting point about space versus length contraction. Certainly, within special relativity the emphasis is on length contraction, but the implication is still that the forward length is contracted thereby maintaining the invariance of light speed [c]. While this length contraction is only observed by the distant observer, it suggests that the distance between two points in space, traversed by the moving observer is actually contracted, otherwise time dilation could give the perception of a faster than light velocity, i.e. v = x/t = x'/t'.

Mentz114: You give me too much credit. While I have started to come to grips with some of the maths associated with general relatvity, I have a long way to go. I was actually just looking at the Schwarzshild metric in spherical coordinate form, as per Edwin Taylor book Exploring Black Holes. In this respect, I was trying translate each component of the metric, i.e. dt, dr & d(angle), back into an energy equation that could be compared to a classical orbit. The implication of the metric is that there is an additional component to gravity, i.e. GMm/r -> GMm/r(1-Rs/r). However, coming back to your reply, the 2nd term of the metric (1-Rs/r)^-1*dr suggests that the proper distance travelled by an onboard observer increases, irrespective of velocity, while special relativity suggests that v->c will decrease the distance perceived by the onboard observer, otherwise velocity would not be invariant.

Maybe I could express my issue in a slightly different way:
If the onboard observer was free-falling under gravity [g] towards a black hole, he could measure his proper (onboard) time and could then determine his velocity from the force of acceleration F=ma=mg. Although I am not sure whether this acceleration would be perceived at constant given time dilation? From this he could determine distance travelled, i.e. x'=vt'. So would the spacetime interval simply increase as per dr/(1-Rs/r) or
would his velocity have an effect?

I think you are both saying no, but I am not sure why. Sorry.

I am attempting to read other sources, but sometimes it is the basic assumptions that we need to really understand. Thanks again.

5. Feb 6, 2008

Ich

There is a fundamental difference between the concept of coordinates in GR and "classical" SR. While coordinates have a direct physical meaning in SR, they are just numbers in GR. You have to extract the actual physics by using the metric.
For example, if you define the speed of light c' as dr/dt in Schwarzschild coordinates, you will find that c' changes with r. You will also find that c''=r dtheta/dt is different from c' at the same r. Finally, you will find that the speed dr/dt of an infalling body is actually decreasing, approaching 0 at the event horizon. Similarly, its length will decrease to 0.
You can find some information regarding different coordinates and the applicability of "classical" SR in these coordinates at http://en.wikipedia.org/wiki/Gullstrand-Painlev%C3%A9_coordinates

6. Feb 6, 2008

Mentz114

Keithh,
Ich has hit the nail squarely on the head, so to speak. The r in the Schwarzschild metric is not the coordinate we would measure from a distance. One can't 'interpret' the metric the way you are trying.

I have to pick you up on this
I don't understand what you mean by 'measure his proper ...time'. A freely-falling observer can only measure time by using local clocks and would not be aware of any changes that might be seen by a distant observer. I haven't put that very well, but don't think that 'time dilation' is detectable in the local frame, any more than length contraction is. It's all relative.

7. Feb 6, 2008

keithh

Thanks to both Ich & Mentz114 for taking the trouble to help me with my questions. It is much appreciated.

I realise that an objects falling towards the event horizon can be subject to different interpretations depending on the frame of reference, e.g. distant observer, shell observations at different radii [r] and the aggregated view based on a series of local Schwarzschild solutions for dt, dr and d-theta. While I am presently trying to work through a maze of mathematical QED’s, I am also trying not to lose sight of some form of physical interpretation.

First to clarify to Mentz114 about my use of the term proper time'. I was implying the time measured onboard a ship moving into the gravity field at velocity [v].

Second, a thanks to Ich for the reference about Gullstrand-Painlevé coordinates as it looks very useful and readable.

I guess I don't want to belabour the issue of the net effect of both general and special relativity but, as indicated, I was trying to anchor my understanding on some real physical interpretation. By way of example, I understand that GPS has to account for both general & special relativity. The clocks onboard GPS satellites have to account for its clocks running faster than on Earth due to the general relativity, i.e. they are further from the centre of mass, while the orbital velocity causes the clocks to slow down due to special relativity. This seems to be a real example, where the effects of both forms of relativity have to be aggregated.

However, I accept the under-text of your response. I do need to read further into all these topics. Again, thanks.

8. Feb 7, 2008

Ich

This is a good example. Usually you find
$$\frac{\tau_s}{\tau_h}=1 - \frac{v^2}{2c^2} (SR)+\frac{g h}{c^2}$$
for the clock rates of a satellite with velocity v and orbit height h, splitted into a SR part and a GR part, compared to the clock rate of Houston.
Actually, GR includes SR. You can calculate the above approximation from the Schwarzschild metric
$$ds^2=(1-\frac{2GM}{c^2r}) dt^2 - r^2d\phi^2$$,
where ds means the "real" (i.e. proper) time and length intervals as a function of the coordinate intervals.
You find easily that for a static observer (dphi=0) at radius r the real time
$$d\tau=\sqrt{ds^2}=\sqrt{1-\frac{2GM}{c^2r}} dt$$.
Now, with the help of $$v=\frac{rd\phi}{d\tau}$$, calculate dphi for the satellite as a function of dt. Then calculate the real time of the satellite as a function of dt from the metric:
$$d\tau_s=\sqrt{1-\frac{2GM}{c^2r}}\sqrt{1-v^2/c^2} dt$$
Divide this by the real time of a static observer at some different r'.
After extensive first order expansions, with the use of
$$g=\frac{GM}{rr'}$$
you find the first formula, where you are free to call one term "SR" and the other "GR". But you never had to explicitly build SR into GR, because it's already there.

Last edited: Feb 7, 2008
9. Feb 7, 2008

keithh

Ich, many thanks for some excellent feedback. Basically I agree that the special relativity effect is built into the Schwarzshild metric, which in the GPS example comes from the angular velocity term associated the orbital velocity of the satellite. Of course, while it is built into the metric, it could still be argued that one effect is due to relativistic gravity (GR), while the other is an effect of relativistic velocity (SR).

I will response in more detail as clearly you can do the same trick with dr/dt for a free-falling object. The maths seems to show that there is a solution for [dr/dt] that involves (1-Rs/r) * SQRT(Rs/r). This implies that to a distant observer, a free-falling object starts at zero velocity at infinite radius [r] and ends up at zero velocity at the Schwarzschild radius [r=Rs]. Weird, but this makes sense to me, but I'm really interested in understanding dr/d-tau case, which I am assuming tells me the distance travel against time as perceived by an onboard observer, but I haven't balanced the maths, as yet.
Again, many thanks. keithh

Last edited: Feb 7, 2008
10. Feb 9, 2008

keithh

Based on the helpful feedback to my original question about the effects of both general & special relativity, I wanted to try and outline my current understanding of the implications of Schwarzschild metric. Of course, I would still welcome any clarification or corrections on any issue.

1) What the metric tells you depends on the frame of reference being considered, i.e. distant observer [dt] or onboard observer $$[d\tau]$$.

2) Basic Schwarzschild metric for an equatorial orbit:

$$c^2 d\tau^2 = (1-Rs/r) c^2 dt^2-[(1-Rs/r)^-^1 dr^2 + r^2 d\phi^2]$$

3) If we only consider a radial path, the last term goes to zero:

$$c^2 d\tau^2=(1-{Rs/r}) c^2 dt^2- (1-{Rs/r})^-^1 dr^2$$

4) If divide through by $$dt^2$$ we relate the observation to the distant observer:

$$c^2 (d\tau/dt)^2 = (1-{Rs/r}) c^2 - (1-{Rs/r})^-^1 (dr/dt)^2$$

5) The equation in (4) reflects the ratio of proper time to distant time as a function of [dr/dt]. If we just position an object at radius [r] with no velocity, i.e. [dr/dt=0], we could simplify the ratio as follows:

$$d\tau/dt =(1-{Rs/r})^-^{1/2}$$

In this context, this would appear to be time dilation due to gravity only, i.e. GR, as there is no associated velocity, radial or orbital.

6) If [dr/dt] is not zero, we have to consider a different solution of (4). While I have not worked through the derivation, various texts define energy as a constant of motion, which makes sense for a free-falling object where potential energy is converted to kinetic energy without any change in the total energy. The net result leads to an equation that differs from (5) as follows:

$$E/m =(1-{Rs/r}) dt/d\tau = 1$$
$$d\tau/dt=(1-{Rs/r})$$

Question: Is it reasonable to interpret this additional time dilation being the result of gravity and velocity, i.e. GR & SR. Substituting for $$[v^2=2GM/r]$$ for [Rs] leads to an equivalence such that:

$$(1-{Rs/r}) = (1-({v/c})^2)$$

As such, combining GR + SR time dilation would lead to the same result:

$$d\tau/dt = GR + SR = (1-{Rs/r})^{1/2} * (1-({v/c})^2)^{1/2} = (1-{Rs/r})$$

7) Dealing with a case where [dr/dt] is not zero, then we substitute (6) into (4) and re-arrange:

$$c^2 (1-{Rs/r}) [ (1-{Rs/r})-1] = -(1-{Rs/r})^-^1 (dr/dt)^2$$
$$(dr/dt)^2 =-( c^2 (1-{Rs/r})^2 (Rs/r) )$$
$$dr/dt = -c (1-{Rs/r}) \sqrt(Rs/r)$$

8) The implication of the final equation in (7) suggests that a distant observer sees the velocity of an object free-falling towards the event horizon, which starts at an infinite radius [r] with zero velocity and ends with zero velocity at Rs. From a sanity perspective, nobody could ever see this effect, as the light returning from the free-falling object would disappear due to gravitational redshift. As such, this perspective might be considered conceptual.

9) So what is the perspective of the onboard observer? If we start again at (3), but now divide by $$d\tau$$:

$$c^2 = (1-{Rs/r})\ c^2 \ (dt/d\tau)^2\ -\ (1-{Rs/r})^-^1 (dr/d\tau)^2$$

9) Again substitute the logic of (6)

$$c^2 = (1-{Rs/r})\ c^2 \ (1-{Rs/r})^-^2\ -\ (1-{Rs/r})^-^1 (dr/dt)^2$$
$$c^2 - c^2(1-{Rs/r})^-^1 = -(1-{Rs/r})^-^1 (dr/d\tau)^2$$
$$(1-{Rs/r}) \ * \ c^2[ 1\ -\ (1-{Rs/r})^-^1 ]\ = \ -(dr/d\tau)2$$
$$(dr/d\tau)^2 \ = \ -c^2[(1-{Rs/r})\ -\ 1 ]$$
$$dr/d\tau \ =\ -c\sqrt(Rs/r)$$

10) The implication of (9) suggests that an onboard observer would perceive the velocity at infinite radius to be zero, as per the previous case, but the perception of velocity, as the radius approached [Rs], would increase to light speed [c]. Therefore, this result seems to completely contradict the previous perception.

11) If the arguments of (5) & (6) are correct, both GR & SR are integrated into the Schwarzschild metric and both suggest that time onboard our free-falling object is slowing to zero due to both gravity and velocity.

12) These various outcomes clearly raise questions. Based on the assumptions of equation (5) & (6), the proper time $$[d\tau]$$ of an object appears to stop at the event horizon, irrespective of its velocity, due to GR alone. As such, even if the velocity [v] equals [c], the implication is that forward motion [x=ct] stops due to time going to zero. As such, it might appear that an onboard observer would never perceive the moment of crossing the event horizon, no matter what his velocity.

13) Despite the apparent implications of the metric, standard text seem to explain away the infinities as a false perception of the coordinate system, which they say does not apply to the actual spacetime geometry. As such, an onboard observer will simply sail through the event horizon on a one-way trip into the black hole. While I have a few questions on this aspect, I will raise it as a separate thread entitled Climbing out of a Black Hole`.

Last edited: Feb 9, 2008
11. Feb 12, 2008

Ich

I observed your progress in the other threads you opened. I agree with the moderators that you seem to prefer giant leaps to small steps, which causes more confusion than necessary.

To get back to the roots, let me comment some of your points.
It is essential that you get the concepts straight: Tau is the proper time of the infalling observer, exactly what his clock measures. It's rate with respect to itself is, of course, 1 second per second. You can't make it "stop", no matter what you do. In this example, literally, it keep running through the end of the universe.
What you calculated is the ratio $$\frac{d\tau}{dt}$$. This ratio becomes zero because dt is ill-behaved at the horizon, not dtau. The observer will reach r=0 in finite proper time. He will not even notice that he crossed the horizon, no matter what "t" an external observer might attribute to this event. tau is real, t is a coordinate.
Same thing for your twin paradox: Don't bother about time dilation. Assign consistent inertial coordinates (i.e. pick one frame an stick to it) to all events, and calculate the real proper times of all involved using the especially simple metric $$d\tau^2=dt^2-dx^2$$. Prolem resolved.
In the case of black holes, don't draw any conclusions from Newtonian calculations, which are simply not applicable. You can calculate the acceleration from the geodesic equation, and you'll find that it diverges at the horizon. In every case you mentioned, you find consistently that nothing (except light) can keep its static position at the horizon, no matter how hard they accelerate. That's what prevents somebody from climbing up the rope. It even prevents the rope from being there in the first place.