What is the Relationship Between Proper Time and Coordinate Time in Black Holes?

[George Jones]

I don't think that's right, George. How could the local observer ever experience redshift? The ground is preventing his free fall, it isn't being "pulled out from under him". The observer on the surface would be experiencing incredible acceleration as the star radius approached the Schwarzschild radius and, analogous to the distant observer seeing the local one being redshifted into nothingness, I believe the local one would see the outside world blueshifted towards infinity, wouldn't it?

No.

Consider freely falling observer B (that is about to splat on the surface) coincident with an observer C that is on the surface of the collapsing star. Just before splat, B is moving towards C with some local speed that is strictly less than the speed of light, and, consequently, there is a finite (Doppler) time shift between B and C. In my previous post, I pointed that B sees a finite redshift of the light emitted by A, an observer who hovers far from the collapsing star. The composition of two finite shifts is always finite, i.e., is never infinite.

If the relative speed between B and C is small, then C will see the light emitted by A to be redshifted by a somewhat smaller amount than does B. If the relative velocity between B and C is large, then C will see the light emitted by A to be blushifted by a finite amount.

 

[rjbeery]

George: thanks for explaining. I have more questions though:

strictly less than the speed of light

Doesn't this presume that B and C (and the surface) are all above the horizon? I thought that from A's perspective the velocity of the free falling beer can is c at the horizon, and dealing with mass traveling at c is the main source of all of these strange behaviors and peculiar explanations.

If the relative speed between B and C is small, then C will see the light emitted by A to be redshifted by a somewhat smaller amount than does B.

Whether B exists as an intermediary or not I do not believe that C will ever see redshifting of A's light. I'm using intuition here so I am obviously liable to be proven wrong mathematically.

 

[rjbeery]

The Schwarzschild line element describing the geometry outside a static black hole is:

ds^2 = - \left( {1-{2M \over r}} \right) dt^2 + \left( {1 - {2M \over r}} \right) ^{-1} dr^2 + r^2 d \Omega ^2

where d\Omega^2 = d\theta^2 + sin^2\theta\phi^2 and (t, r, \theta, \phi) the Schwarzschild coordinates.

A body free-falling from the far distance takes a finite proper time \tau of about 0.3 milliseconds to go from r0=100 km to the horizon of a 10-solar masses black hole at r=29km:

\delta\tau = {2 \over 3} {1 \over \sqrt{2M} } \left[ r_0^{3/2} - r^{3/2} \right] = 0.000348 s

On the other side, if we express the infall in terms of coordinate time by means of the following differential equation, it can be seen that the same body takes infinite coordinate time t to reach the limit of the horizon r=2M.

{dt \over dr} = {dt \over d\tau} {d\tau \over dr} = - \sqrt {r \over 2M} \left( 1- 2M \over r \right) ^{-1}

Xantox: I have a degree in Comp Sci and took Calc and Diff EQ but that was many moons ago. Could you hold my hand a bit in understanding the transition from the Schwarzschild metric to calculating the coordinate times for the local vs the distant (Minkowski metric, right?) frames? I'm back in school for Physics but it would be nice to get a good grasp on this before classes start (on Monday!)

 

[xantox]

Could you hold my hand a bit in understanding the transition from the Schwarzschild metric to calculating the coordinate times for the local vs the distant (Minkowski metric, right?) frames?

The first part is to understand that finite proper time intervals for the infalling body are equivalent to infinite coordinate time intervals for the distant observer. We may proceed by expressing the Schwarzschild metric in timelike form:

d\tau^2 = - \left( {1-{2M \over r}} \right) dt^2 - \left( {1 - {2M \over r}} \right) ^{-1} dr^2 - r^2 d \Omega ^2

At fixed r and \Omega, dr=d\Omega=0, and the relation between proper time and Schwarzschild time is:

{d\tau \over dt} = \sqrt{{1-{2M \over r}}}

This shows that at infinite r, proper time intervals are equal to coordinate time intervals. The Schwarzschild metric is minkowskian and flat at infinity, ie Schwarzschild time has a special relativistic meaning in that limit. But this also shows that when the infalling particle approaches the horizon at r=2M, its wristwatch runs infinitely fast when expressed in Schwarzschild time coordinate units. This is the consequence of the curvature of the time dimension – signals emitted near the horizon at regular intervals get delayed longer and longer when they are projected onto curved away regions. Thus we can equivalently say that in this case the particle is both crossing a finite proper time and an infinite coordinate time.

The second part is to understand that the proper time elapsed on the free-falling particle worldline to reach the horizon from a finite distance is indeed finite. For this it is needed to determine its equation of motion, solve it and integrate on the worldline. You may find the full derivation in any general relativity textbook, see for example in Frolov and Novikov here.