# Black Hole Riddle 3

1. Oct 26, 2005

### Zanket

This is an analysis involving general relativity (GR). Let a spherically symmetric, uncharged, nonrotating star (not a black hole—that comes later) have an escape velocity = infinitesimally less than c at its surface. Then its r-coordinate radius (also known as the reduced circumference and hereafter just R) is infinitesimally above its Schwarzschild radius. Let an observer having negligible mass free-fall from rest at infinity to traverse the star via a tunnel drilled diametrally through it. Escape velocity is the velocity needed by an object to rise freely from a given altitude to rest at infinity. An object’s fall reverses its rise, so an object free-falling from rest at infinity passes each altitude at the escape velocity for that altitude. Then the observer enters the tunnel at infinitesimally less than c, accelerates to reach a maximum velocity still infinitesimally less than c at the star’s center, and then decelerates to exit the tunnel at infinitesimally less than c, where these velocities are all directly measurable.

The equivalence principle tells us that special relativity (SR) applies in each infinitesimal region of spacetime that the observer passes. Then the Lorentz factor of SR tells us that, because the observer passes each and every infinitesimal region of spacetime within the star at infinitesimally close to c, the observer passes each and every infinitesimal region of spacetime within the star in an infinitesimal proper time (time elapsed on the observer’s watch). The sum of any number of infinitesimal times is an infinitesimal time, so the observer traverses the whole star in an infinitesimal proper time, regardless of the magnitude of R. Were the R of the star even a million light years, the observer would still traverse it in an infinitesimal proper time.

Now consider a black hole. A GR equation for the proper time of the same observer (that is, an observer having free-fallen from rest at infinity) to fall from the black hole’s event horizon to its central singularity is proportionate to the black hole’s mass (equation provide upon request). The more mass, the more proper time elapses for the observer. The R of a black hole is proportionate to its mass. So the greater the R of the black hole, the more proper time elapses for the observer.

Here is the riddle: How do you explain that the observer could take millions of proper years to fall from the black hole’s event horizon to its central singularity, for a black hole with a certain R, whereas the observer could traverse the whole of a star having the same R in an infinitesimal proper time? After all, the only relevant difference between these objects is that the black hole has an infinitesimally greater mass.

(Don’t ask me; I don’t have a mainstream answer. I don’t see how it can be a “time/space coordinates are reversed below the event horizon” thing—that doesn’t seem to explain how an infinitesimal time for the star becomes x number of years when a gram of material is added to the star to make it a black hole.)

2. Oct 26, 2005

### pervect

Staff Emeritus
Your first analysis is probably wrong.

Do a general relatvistic analysis of both cases, up to the point where the object reaches the stars surface.

To make the problem easy, we need to make an approximation. The easy approximation is that the added acceleration while the object continues to fall to the stars center doesn't significantly reduce the trip time. I can't rigorously justify this approximation at this point, but it seems pretty reasonable, at least to sastisfy the finite/infinte question.

If we make this approximation, we can apply the formula that I derived in
https://www.physicsforums.com/showpost.php?p=602558&postcount=29
George Jones has done a similar analysis and come to a similar conclusion later in this same thread. (This is good, because I screwed up this analysis the first time I tried it :-(, but this time around it looks right).

While this formula is for falling into a black hole, it will also give us the velocity and rapidity (dr/dtau) for an object falling into an almost-black-hole, we just need to say that r is almost, but not quite, 2M

The result is that $dr/d\tau = \sqrt{E^2 - 1 + 2M/r}$

Here E=1 when the object is at rest at infinity, E>1 if the object is initally falling into the black hole, and E<1 if the object free-falls from some distance that's less than infinity.

It's fairly trivial to say that from this result that the proper transit time should be proportional to the mass M of the black hole, not zero, because dr/dtau is nearly 1, and r is nearly 2M, and we are assuming that dr/dtau doesn't change "much" in the interior of the star.

Note that the velocity of the object near the "almost horizon" is NOT given by this formula, at least not by the usual definitions - dr/dtau is a different quantity altogether. Numerically they turn out to be closely related, though.

If you want to compare the velocity of the object to the velocity of light, you need to use the usual definitions which assumes a locally Lorentzian metric, which means that g_00 = 1, and that g_rr = 1. But near the almost-black-hole, g_00 is nearly zero (nearly infinite gravitational time dilation), and g_rr is nearly infinite.

This is the basic thing that your analsyis does not take into account - the distortion of space-time geometry near the almost-black-hole.

The calculation of velocity in the locally Lorentzian frame is done in the thread I mentioned, and is given by the quantity drr/dtt, which turns out to be equal to be

drr/dtt = (dr/dtau) / E

Since E=1, we wind up with dr/dtau AND drr/dtt both nearly 1. This means that the object is nearly moving at C, but when you look at the space-time geometry, the object falling into the black hole still sees the black hole as having a finite "thickness". This is possible only because the geometry near the black hole is not nearly Minkowskian, but is almost singular in Schwarzschild coordinates (g_00 is nearly zero, and g_rr tends to infinity).

Last edited: Oct 26, 2005
3. Oct 26, 2005

### Zanket

I’m having a tough time understanding your response. For the black hole an analysis isn’t needed since the formula is available for that (I can dig it up). For the star it seems clear that it must be traversed in an infinitesimal proper time. What do you think the proper time would be to traverse the star (that is, the proper time through the tunnel)? I don’t see that in your response. Can you put it in layman's terms?

Last edited: Oct 26, 2005
4. Oct 26, 2005

### pervect

Staff Emeritus
I'm saying that the time to traverse the star is roughly 2M (in geometric units). That's 2GM/c^2 in standard units. M is the mass of the star.

The rationale is this: |dr/dtau| is nearly equal to 1 when the object passes the event horizon. I am making an assumption that the acceleration in the interior of the star does not shorten the trip time appreciably, to avoid a messy calculation. The actual time will be a little shorter. Using this assumption, r must change from r=2M to r=0 with dr/dtau = -1, so dtau = dr / (dr/dtau) = 2M / 1 = 2M.

I'm using geometric units here, so a proper time of 2M corresponds to 2GM/c^2 in standard units. In geometric units, c=1 and G=1, so G/c^2=1, and G/c^2 has MKS units of meters/ kg.

The point I'm making, ultimately, is that the clocks and rods of an observer hovering near the almost-event-horizon are the clocks and rods we use to measure the velocity of the infalling object, and that these clocks and rods are not the same as the Schwarzschild coordinates r and t. We have to distinguish between local distance and time standards, and the Schwarzschild coordinates.

The speed of light is equal to 'c' only in a _local_ coordinate system, one with local clocks and local rulers - i.e. we set up a cesium clock near r=2M which defines our time standard, and we set up a standard meter based on the same cesium clock standard, and that's how we measure distances and times to compute the velocity of the infalling object nearly equal to 'c'.

Note that an outside obsever at infinity, who thinks that our cesium clock near r=2M is nearly stopped, will also think that the infalling object has a velocity of zero, not 'c'.

5. Oct 26, 2005

### Zanket

I understood that better, thanks, and I understand geometric units. I don’t see how you are supporting a claim that the proper time for the observer to traverse the star is roughly 2M. My analysis used only local measurements, where SR applies. Do you agree or disagree with this statement: Within in a local region of spacetime where SR applies, an object passed at a directly-measurable velocity of infinitesimally less than c is passed in an infinitesimal proper time? If you disagree, then why? If you agree, then do agree or disagree that a sum of any number of infinitesimal proper times is an infinitesimal proper time? If you disagree, then why? If you agree, then have you not agreed that the proper time for the observer to traverse the star is an infinitesimal proper time? If not, then why?

6. Oct 26, 2005

### JesseM

Phrasing this in terms of "infinitesimals" seems a little confusing, can you rephrase it into a statement about limits? For example, it is true that in the limit as your velocity approaches c in my frame, the proper time as measured in your frame to cross any fixed finite distance (as measured in my frame) would approach zero, because Lorentz contraction is causing the distance in your frame to approach zero. On the other hand, if you allow the distance to increase along with your velocity, you can't say that in the limit as your velocity approaches c the time to cross any distance approaches 0. In this problem you seem to be taking a double limit, both the limit as the star gets closer and closer to being a black hole, and the limit as the escape velocity at its surface approaches c. Maybe the conceptual answer is that as the velocity is approaching c in the second limit, the length of the path through the star as seen by moving object is not approaching zero, because you can't just apply the normal Lorentz contraction formula to the diameter of the star as seen by a distant observer, you also have to take into account curvature of spacetime which is increasing in the first limit, and which might somehow be increasing the apparent distance beyond what the distant observer would measure.

Last edited: Oct 26, 2005
7. Oct 26, 2005

### Zanket

Agreed.

The velocity is infinitesimally close to c through the whole tunnel of the star. Of two stars, one with an R = 1 light year, and the other with an R = 2 light years, it is true that the second star would take about twice as much proper time for the observer to traverse. But what is twice an infinitesimal proper time? It is an infinitesimal proper time. Any number times zero is zero. Any number times an infinitesimal is an infinitesimal. An infinitesimal is a number arbitrarily close to zero; it isn’t a finite number. For any small finite positive number you can think of, there exists a smaller finite positive number, any multiple of which can be arbitrarily close to zero. So I can say that when my velocity is infinitesimally close to c the proper time to cross any distance is infinitesimal (arbitrarily close to zero). In the limit, at c, the proper time is zero. At infinitesimally close to c, the proper length of anything I measure is infinitesimal (as you point out), and a sum of any number of infinitesimal proper lengths is an infinitesimal proper length, which takes an infinitesimal proper time to cross.

Those are synonymous to me. A black hole is defined as an object with an escape velocity >= c. As a star gets closer and closer to being a black hole, by definition the escape velocity at its surface approaches c.

The free-falling observer in the riddle makes measurements within local regions of spacetime within which SR applies. The observer applies the Lorentz contraction formula only locally, inputting into the formula direct measurements of velocity relative to the star. Because all the measurements are direct, there is no curvature for the observer to contend with. Each local region of spacetime is flat. The curvature along the whole tunnel does not affect the finding that the observer’s proper time to traverse the tunnel is infinitesimal, for the tunnel is an unbroken series of locally flat spacetime regions, and the calculation is straightforward: you simply sum the proper times to traverse each local region. The effect of the curvature of the star as regards the riddle is nothing more than to vary the observer’s proper time to traverse each locally flat region, but the proper time to traverse each and every locally flat region is still infinitesimal.

Last edited: Oct 26, 2005
8. Oct 26, 2005

### pervect

Staff Emeritus
Let's use limits, and not infinitesimals, I'm more comfortable with them.
In the limit as v->c, an object of finite size in its own rest frame will be passed in zero proper time.

I would agree that in SR the sum of any number of objects of finite width in their own rest frame would also be passed in zero proper time.

Note however that the limit of $f(v) \tau$ is not guaranteed to be zero for arbitrary f(v) when the limit of f(v) as v->c is not finite - it is quite possible for it to be a constant non-zero number if f(v) has the right form.

However, I would say that these statements about SR do not prove anything about the problem of what happens to an object that is falling into a black hole.

You think that your analysis "uses only SR", but in fact it is not possible to analyze the problem of an object falling into a black hole only using SR. The SR approach works only when the metric coefficients are nearly constant. Your metric coefficients are not only not constant for an object falling into a black hole - they are not even finite! (At least not in the Schwarzschild coordinate system being used).

Some results using Finklestein coordinates r,v,tau for an object falling into a black hole of mass 1 from rest at r=infinity, t=-infinty might (or might not) help you.

The Finkelstein r coordinate is the same as the Schwarzschild r coordinate. The Finkelstein v coordinate is a bit tough to explain, so I'll just not explain it :-(.

The result from my notes (again, the mass of the black hole is 1)

$$r(\tau) = \frac{1}{2} \sqrt[3]{36 \tau^2}$$

$$dr/d\tau = \frac{1}{3} \sqrt[3]{\frac{36}{\tau}}$$

Thus the schwarzschild radius, r=2 is reached at tau=-4/3, and the singularity is reached at tau=0. (Thus my fall-in time of 2M was in fact too large).

Note that dr/dtau is always GREATER than 1 inside the Schwarzschild radius (i.e for -4/3 < tau < 0). But the velocity of light is not being exceeded.

Last edited: Oct 26, 2005
9. Oct 26, 2005

### Zanket

OK, then I imply from that that you agree that the observer would traverse the star in zero proper time in the limit.

The black hole analysis isn’t in question, since for that there’s already a formula available that returns the proper time for the observer to travel from the event horizon to the singularity. That formula shows that the proper time is proportionate to the black hole’s mass. My analysis used SR only for the star. I didn’t analyze the black hole scenario except to report the existing formula and compare its results to the results for the star.

So the riddle remains, why the potentially arbitrarily large difference between the proper time for the star and the proper time for the black hole, when the only relevant difference between the two could be as small as a gram of material?

10. Oct 26, 2005

### JesseM

It might make sense to say the sum of any finite number of infinitesimal proper lengths is zero, but I don't see why that has to be the case here. As an analogy, suppose we are using the parameter d to describe a series of trials where you cross a ruler of length of $$L/\sqrt{1 - (dc)^2/c^2}$$ travelling at velocity dc. What proper time does it take you to cross the ruler as d approaches 1? Well, in your own frame the length will always be L regardless of the value of d, so as your velocity approaches c the proper time will approach L/c. But according to your argument, wouldn't you conclude that the time must be zero, since when d is infinitesimally close to c, the distance between successive 1 cm marks on the ruler must be infinitesimal, and "a sum of any number of infinitesimal proper lengths is an infinitesimal proper length which takes an infinitesimal proper time to cross"? Of course this would be wrong, and the reason is that the although the apparent distance between 1 cm marks is approaching zero as you approach c, the total number of 1 cm marks is approaching infinity since the ruler is growing without bound as the parameter d approaches 1.

So perhaps something similar would be true of marks along the inside of the tunnel in a series of trials where the star gets closer and closer to the critical mass needed to create a black hole, and your initial speed at the surface gets closer and closer to the speed of light? Suppose as you travel through the tunnel, markings have been placed along the inside with the property that if you measure how far apart two nearby markings are in your own local rest frame, and you use the Lorentz contraction to figure out how far apart they'd be in their own local rest frame, you'll always find that they're some fixed small distance apart (like 1 cm). If we take an infinite series of stars that get closer and closer to being a black hole without quite reaching it, and in each star's tunnel markings have been placed with this property, could it be that the number of markings you pass will approach infinity? I don't know enough about GR to say if this would be true, I'm just thinking out loud here...in any case I don't think your reasoning about infinitesimals is very rigorous, and it has the potential to be pretty misleading in simpler cases like the one I mentioned.

11. Oct 26, 2005

### Zanket

Yes, I would conclude that.

I don’t see how the ruler is growing without bound. I see that the total number of 1 cm marks (the marks imprinted on the ruler) remains constant.

Agreed, you could calculate that.

Well, for an infinite series of stars, the number of markings you pass will be infinite. But I get your point. You could pass any number of markings. But again, all of any number of markings, no matter the distance they indicate in their own local rest frame, will be passed by you in zero proper time in the limit (at c). Even all of an infinite number of markings are passed in zero proper time in the limit. GR itself has that feature, when it says that the directly measured radius of a black hole is infinite above its event horizon. Then an observer falling across an event horizon passes an infinite number of potential markings some finite distance apart in their own local rest frame. In the limit, at the horizon, at an in-falling velocity of c, the observer just passed an infinite number of potential markings.

12. Oct 26, 2005

### JesseM

I'm talking about an infinite series of different trials, each one using a different ruler. Each trial is associated with a particular value of the parameter d, and the parameter d is approaching 1, with the ruler's rest length on each trial being $$L/\sqrt{1 - (dc)^2/c^2}$$, which you can see will approach infinity as d approaches 1.
I'm not talking about passing an infinite number of stars, I'm talking about taking an infinite series of trials just as above, except this time with the stars getting closer and closer to the critical mass needed for a black hole. Then if you look at the number N of marks the traveler passes on a single trial, with the marks drawn according to the rule I described above, I'm saying N may approach infinity in the limit of later and later trials. On one trial the traveler may pass a million marks as he goes through his star, on a later trial a different traveler may pass a billion marks as he goes through his own star, and so on, even though to outside observers the stars all appear to have the same radius. Again, I don't know if this is what would happen, but it could be. Just think of those "rubber sheet" embedding diagrams, which I realize can be misleading--a black hole is a depression of infinite depth, so a series of stars approaching a black hole would look like deeper and deeper finite depressions with the same rim width. The length of the path of an observer who starts at the rim, goes to the bottom and climbs back out to the opposite side of the rim will increase without bound, even though the width of the rim always looks the same to distant observers. So it seems plausible the number of markings in the tunnel could increase without bound, although this is just a guess.
No, this is definitely not true as you can see if you examine my ruler example, and I'm on firmer ground there than I am with the star example because I know more about SR than GR. If the length of the ruler in its own rest frame is $$L/\sqrt{1 - (dc)^2/c^2}$$ in a given trial, and the velocity of the observer is dc in that trial, you can see that the observer will observe the ruler moving relative to him at velocity dc so he will see it shrunk by $$\sqrt{1 - (dc)^2/c^2}$$, meaning on every single trial the observer will observe the ruler's length in his own rest frame to be L. So, as the relative velocity between the observer and the ruler approaches c on successive trials, the proper time for a given observer to cross his ruler will approach L/c.
Wait, are you saying you could fit an infinite number of small test masses at constant distances along a radial line from the horizon out to some finite distance (in Schwarzschild coordinates), such that each test mass locally observed the same distance between itself and its nearest neighbors?

Last edited: Oct 26, 2005
13. Oct 26, 2005

### pervect

Staff Emeritus
Sorry, no. I'm trying to explain why not, I guess I haven't entirely succeeded. But if you at least realize that I don't agree, it will be a first step :-).
The fact that it doesn't match up with the black hole analysis should serve as a "red flag" to you that there is something wrong with your SR analysis.
This may help.

The proper time it takes an object to move down a trajectory is just the intergal of $d\tau$.

In the Lorentzian metric of SR $d\tau = \sqrt{dt^2 - dr^2}$ assuming $\theta = \phi =$ constant

So we could summarize your argument - Look, if the local velocity is 'c', then dt=dr, and $d\tau = 0$.

But in GR, this is *NOT* the case. The Lorentz interval is given by the equation.

$d\tau = \sqrt{g_{tt} dt^2 - g_{rr} dr^2}$

For definitenes, in the Schwarzschild metric
$$g_{tt} = (1-r_s/r) \hspace{.5 in} g_{rr} = \frac{1}{1-r_s/r}$$

$r_s$ being the Schwarzschild radius, 2M in geometric units.

Note that at r=r_s, g_tt=0 and g_rr is infinite!

At the event horizion, dr/dt is zero, $d\tau/dt$ is infinte, and the system just isn't well behaved mathematically. dt for an object transversing a geodesic path turns out to be infinite.

There's no way around this bad behavior without using well behaved coordinates like Kruskal coordiantes or Finklestein coordinates that are not pathological. The problem coordinate is not r, but is the Schwarzschild 't' coordinate, Finklestein coordinates can use r and v to avoid the problem.

Inside the event horizon, though, the coordinates are at least not singular. Here we can at least see how the Lorentz interval $d\tau$ is non-zero and real. The positive term that is timelike is due to g_rr dr^2, because 'r' is a timelike direction. The negative term that is space-like is the g_tt dt^2, because 't' is spacelike inside the event horizon. As long as g_rr dr^2 > g_tt dt^2, the Lorentz interval is real and greater than zero.

Let me go over that again.

We are adding up an infinite number of very small time intervals $d\tau$ when we integrate. As long as each piece is real and positive, the sum will be real and positive.

The size of each piece is sqrt(g_tt dt^2 - g_rr dr^2), however both g_tt and g_rr are negative numbers. Thus the size of each piece is

sqrt(|g_rr| dr^2 - |g_tt| dt^2)

and as long as the quantity inside the sqrt above is positive and non-zero, we will have a positive intergal for the proper time which is positive (and nonzero).

I'm afraid a really convincing mathematical analysis for what happens exactly at the event horizion (which is where you are focused right now) would require that you use non-singular coordinates, you are basically beating your head against a mathematical singularity due to a poor coordinate choice that does not represent a physical singularity. The only way around this is to adopt a set of coordinates that is not singular.

Last edited: Oct 27, 2005
14. Oct 27, 2005

### pervect

Staff Emeritus
There's something I realized. If you assume the star has all its mass at its surface, my previous approximate solution actually is exact.

You haven't specified the mass distribution of the star, so assuming a hollow star for ease of calculation is perfectly fair.

The basic point is that moving mass away from the center will always increase the proper time it takes to traverse the star. You get the minimum of 4/3M when the mass is all at the center, because the mass keeps accelerating and gaining speed as long as possible, and you get the maximum time of 2M when the mass is in a spherical shell, because the acceleration stops sooner.

Within a static spherical shell, an object will experience no acceleration in GR (just as it will not experience any acceleration in Newtonian gravity).

Last edited: Oct 27, 2005
15. Oct 27, 2005

### Zanket

It’s hard for me to get my mind around the concept of a ruler that grows with each subsequent trial. I’m assuming you’re talking about a ruler that is fixed in place in the tunnel. Why not just keep these rulers the same rest length every time? Like have them have a rest length of a meter, say, in every trial. Then have the star have an every higher escape velocity in each trial. In each successive trial more metersticks can be fixed end-to-end in the tunnel, measuring the star’s physically-measured diameter.

Agreed.

It is my understanding that it could.

I’ve read this a couple times but don’t get it. When you say “the observer will observe the ruler's length in his own rest frame to be L”, is “the observer” the free-falling one? Is the ruler the one fixed in place in the tunnel? Is “his own rest frame” the observer’s rest frame? When you say “for a given observer to cross his ruler” is “a given observer” the free-falling one? Is “his ruler” the observer’s ruler in his own rest frame? If so, how does he cross his own ruler?

You need not answer those questions if you can agree that I see what you’re getting at. What you are getting at, I think, is that as the free-fall velocity approaches c with each successive trial, the number of metersticks (rulers that are a proper meter in length) laid end-to-end throughout the tunnel can be increased. So the free-falling observer finds that the proper (measured by the observer) length of each passing meterstick is shorter than the last trial, but now there’s more of them. So the observer won’t necessarily (at first glance) get through the tunnel in less proper time with each successive trial. Did I see what you are getting at? Let me know and then I’ll respond.

Yes. Taylor and Wheeler discuss this in the book Exploring Black Holes, in response to a student’s question (paraphrased), “How can a free-falling observer cross an event horizon in a finite time, when an infinite number of metersticks laid end-to-end must be passed, since for metersticks having any finite proper (measured by the observer) length it would take an infinite proper time to cross an infinite number of them?” The authors explain how this infinite number of metersticks can be passed by the observer in a finite time.

16. Oct 27, 2005

### JesseM

No, forget the tunnel, I'm not talking about a star at all here. This is just supposed to be an analogy involving only SR and no curved spacetime, just to illustrate the principle that your reasoning about infinitesimals is flawed. And the ruler isn't growing, I'm just talking about the limit of a series of separate cases--you can imagine separate rulers and separate travelers in each trial if you like, it doesn't matter. Maybe it will also help if we imagine the trials are numbered, and on trial n you set d = 1 - 1/2^n, so on trial 1, d=1/2, on trial 2, d=3/4, on trial 3, d=7/8, and so forth.
Again, there's no tunnel and no free-falling here, just an observer in flat spacetime traversing a ruler which has rest length $$L/\sqrt{1 - (dc)^2/c^2}$$ at constant velocity dc. It's not too hard to see that for any given trial with d<1, the traveller will see the proper time it takes to pass from one end of the ruler to the other as exactly L/dc, so naturally the limit as d approaches 1 is L/c.
Interesting. Well, that would lend support to my guess that as the mass of the star approaches the critical mass needed to form a black hole, the number of marks the free-falling observer will pass inside the star would also approach infinity. After all, you could imagine extending the tunnel out beyond the star's surface to twice the star's radius, and what you're saying is that the number of marks the free-falling observer will pass between the entrance to the tunnel and the surface of the star will approach infinity as the star's mass approaches that of a black hole (at which point the surface becomes the event horizon), so it would be pretty weird if the number of marks he'd pass while inside the star didn't also approach infinity.

17. Oct 27, 2005

### Zanket

I haven't forgotten about you, prevect.

OK, I’m with you on this paragraph.

You’re calculating this in a different way of the way I normally think of this. I can’t picture the rest length of the ruler being passed depending on the observer’s velocity. That’s just too weird for me to grasp.

Can we approach this from the way I normally see it done? For simplicity, I set c = 1. The d is the observer’s velocity, a percentage of c, so I denote it with v. The rest length of the ruler is L, and the proper (as measured by the observer) length is L * sqrt(1 – v^2). The proper time to cross the ruler is (proper length) / v. The proper length approaches zero as v approaches 1, so the proper time to cross the ruler approaches zero as v approaches 1.

I agree with your assessment in this paragraph.

Last edited: Oct 27, 2005
18. Oct 27, 2005

### JesseM

The rest length doesn't physically depend on the observer's velocity, again, you are setting up a bunch of trials, and on each trial you're choosing both the rest length of the ruler and the observer's velocity based on the value of the parameter d. Imagine you have a bunch of pilots, and you say something like "OK Sam, I'm assigning you to run #2, you'll be flying past a ruler whose rest length is 15.12 meters, and I want you to fly your ship by it at 0.75c...now Archibald, I'm assigning you to run #3, you'll be flying past a ruler with a rest length of 20.66 meters, and I want you to fly by it at 0.875c...Suzanne, I'm assigning you to run #4, you'll be flying past a ruler with a rest length of 28.74 meters, I want you to fly by at 0.9375c" etc. etc. Again, it's a bunch of separate rulers and separate pilots flying by their assigned rulers at different speeds...the ruler doesn't physically depend on the pilot's speeds, it's just you who's choosing the length of each pilot's ruler based on a certain formula (In this case, a pilot on run n is assigned to fly at (1 - 1/2^n)*c past a ruler that is constructed to have a rest length of 10 meters / sqrt[1 - (1 - 1/2^2)^2].) You can see that if you assign both pilot speeds and ruler rest length based on the parameter d in the way I described, each pilot will experience a time of L/dc to cross his assigned ruler, right?

19. Oct 28, 2005

### Zanket

Agreed, the rest length doesn't physically depend on the observer's velocity, but your formula does enforce a relationship between the two, which seems overly complicated to me, with all due respect. I don’t think an experiment is needed to determine the proper time required to cross a ruler. Regardless of velocity or rest length of a ruler, the equation is always (proper length) / velocity, which is your L/dc.

Yes. I understand that L is the proper (as measured by the observer) length, and dc is what I call v, the percentage of c where c = 1 (I like to use geometric units). We now have a formula that we agree on for the proper time to cross a ruler, in terms of proper length and v. Is there anything else that you are trying to show by having the rest length of the ruler be related to the velocity by the observer? When the ruler has the same rest length in each pass, I came to the conclusion above that “the proper length approaches zero as v approaches 1, so the proper time to cross the ruler approaches zero as v approaches 1.”

20. Nov 4, 2005

### JesseM

Sorry to take so long to reply, I had some computer troubles...
Even if it seems overcomplicated, you agree that in the limit as the trial number n approaches infinity (remember, in the nth trial I assumed the parameter d would take the value d=(1 - 1/2^n) ), the proper time observed by the pilot in that trial does not approach 0, it approaches L/c, right?

If so, this is the point of the analogy, to show that your reasoning about infinitesimals can lead to bad conclusions when two simultaneous things are happening in the limit--the speed of the pilot is approaching c, and the number of marks on the ruler is approaching infinity. Because of this, the proper time in the limit can be a finite number, despite the fact that the distance between two successive markings is approaching 0 from the perspective of the pilot.

So, exactly the same thing seems to be going on in the limit as the star approaches the critical mass needed to form a black hole. Imagine we again take a series of numbered trials, except that instead of having each pilot cross a ruler on his trial, we have each pilot drop in from infinity towards a star that has a mass slightly less than that needed to form a black hole, with a tunnel through the star that has markings along the wall as I specified. Let's make the following rule--all the stars have the same radius R, and if M is the mass of a black hole with radius R, then on the nth trial the pilot on that trial will fall into a star of mass M*(1 - 1/2^n). I think you will agree that as n approaches infinity, the speed of the pilot as he passes the surface and enters the tunnel will approach c (by your argument about escape velocities), and we also seem to have agreed it's likely that the number of markings the pilot will pass in the tunnel approaches infinity as the star's mass approaches that of a black hole. So this situation is directly analogous to the one with the ruler--again, two simultaneous things are happening in the limit, the pilot's speed is approaching c so the distance between successive markings as seen by the pilot approaches 0, but the total number of markings the pilot passes also approaches infinity. Therefore it's perfectly plausible that, as with the ruler, the limit of the proper time seen by the pilot to cross all those markings is some finite number, it doesn't have to approach zero.