Reversal of proper time for a falling object

[yuiop]

Consider the Schwarzschild line element for radial motion only:
c^2 d\tau^2 = \left(1-\frac{2GM}{rc^2}\right) c^2 dt^2 - \frac{dr^2}{ \left(1-\frac{2GM}{rc^2}\right)}
Solve for d\tau^2/dt^2:
\frac{d\tau^2}{dt^2} = \left(1-\frac{2GM}{rc^2}\right) - \frac{dr^2/dt^2}{ c^2\left(1-\frac{2GM}{rc^2}\right)}
The velocity of a object falling from infinity in Newtonian physics is given by:
dr/dt = -\sqrt{\frac{2GM}{r}}
which is the same magnitude as the Newtonian escape velocity at r. In Schwarzschild coordinates the free fall velocity is scaled by the gravitational gamma factor squared and is given by:
dr/dt = -\sqrt{\frac{2GM}{r}} \left(1-\frac{2GM}{rc^2}\right)
If we substitute this expression into the 2nd equation at the top we obtain:
\frac{d\tau^2}{dt^2} = \left(1-\frac{2GM}{rc^2}\right) - \left(1-\frac{2GM}{rc^2}\right)\frac{2GM}{ rc^2}
which simplifies to:
\frac{d\tau}{dt} = \left(1-\frac{2GM}{rc^2}\right)

For r<2GM/c^2 the proper time runs in the reverse direction relative to Schwarzschild coordinate time. Note that there are no complication due to imaginary or complex numbers.

Note also that many alternative metrics such Gullstrand-Painleve, Eddington-Finkelstein, Kruskal-Szekeres are based on the point of view of an observer using a falling clock as a reference for coordinate time.

 

[PAllen]

Well, you are ignoring the physical nature of the coordinates. Inside the event horizon, dt is not a timelike direction it is a spacelike direction, so the meaning you propose for d τ /dt is absurd (inside the event horizon). And, of course, r<2GM/c^2 means inside the event horizon.

 

[yuiop]

Well, you are ignoring the physical nature of the coordinates. Inside the event horizon, dt is not a timelike direction it is a spacelike direction,...

One way to determine the spacelike or timelike nature is to observe the sign of the spacetime interval dS^2. The radial Schwarzschild metric in terms of the spacetime interval is:
dS^2 = \left(1-\frac{Rs}{r}\right) dt^2 - \frac{dr^2}{ \left(1-\frac{Rs}{r}\right)}

where Rs = 2GM/c^2 and using units of c=1. For a stationary test particle dr/dt=0 so in this case:

dS^2 = \left(1-\frac{Rs}{r}\right) dt^2

For r>Rs the interval is positive (and so timelike) and for r<Rs the interval is negative (so spacelike) for a stationary particle.

For a falling particle:

dS^2 = \left(1-\frac{Rs}{r}\right)^2 dt^2

Note that for a falling particle ds^2 remains positive (and timelike) even when r<Rs.

 

[PAllen]

One way to determine the spacelike or timelike nature is to observe the sign of the spacetime interval dS^2. The radial Schwarzschild metric in terms of the spacetime interval is:
dS^2 = \left(1-\frac{Rs}{r}\right) dt^2 - \frac{dr^2}{ \left(1-\frac{Rs}{r}\right)}

where Rs = 2GM/c^2 and using units of c=1. For a stationary test particle dr/dt=0 so in this case:

dS^2 = \left(1-\frac{Rs}{r}\right) dt^2

For r>Rs the interval is positive (and so timelike) and for r<Rs the interval is negative (so spacelike) for a stationary particle.

For a falling particle:

dS^2 = \left(1-\frac{Rs}{r}\right)^2 dt^2

Note that for a falling particle ds^2 remains positive (and timelike) even when r<Rs.

Sorry, this is wrong. The character of a coordinate direction at a point is determined by the sign of the contraction of a unit tangent vector in that coordinate direction, at that point. As you have shown yourself, that shows the t direction (and dt) is spacelike inside the horizon. A timelike path is a coordinate independent feature, so, of course a freefall path has a timelike tangent, always, everywhere, in any coordinates; but that is completely irrelevant to your OP argument.

Your OP argument is claiming that dτ/dt is measuring time flow along a world line. This is just wrong if dt spacelike, no matter what the character of the world line.

 

[PAllen]

An example in Minkowski space of the OP error:

suppose I have a particle moving in the -x direction. Then dτ/dx is negative. So what?

 

[PeterDonis]

For a falling particle:

dS^2 = \left(1-\frac{Rs}{r}\right)^2 dt^2

Note that for a falling particle ds^2 remains positive (and timelike) even when r<Rs.

I assume this is a mistype--a falling particle has nonzero dr, but what you wrote only has the dt^2 term. Also, to properly assess the sign of ds^2 for a falling particle, you need an equation for dr/dt, i.e., an equation for the particle's trajectory. Assuming that by "falling particle", you meant "freely falling particle", the equation for dr/dt would be derived from the geodesic equation.

 

[yuiop]

Sorry, this is wrong. The character of a coordinate direction at a point is determined by the sign of the contraction of a unit tangent vector in that coordinate direction, at that point. As you have shown yourself, that shows the t direction (and dt) is spacelike inside the horizon. A timelike path is a coordinate independent feature, so, of course a freefall path has a timelike tangent, always, everywhere, in any coordinates; but that is completely irrelevant to your OP argument.

Your language is slightly above my technical ability, but you seem to be saying that that the spacetime interval that determines the timelike nature can only be measured in terms of a stationary observer where dr or dx =0. I would disagree. For the SR spacetime interval, we use dS^2 = dt^2-dx^2 and we do not require that dx=0 because dS^2 is invariant for all values of dx^2. Also you seem to be requiring that dS be measured by a stationary clock in the coordinates, when it generally accepted that there is no such thing as a stationary clock in Schwarzschild coordinates, below the event horizon.

Your OP argument is claiming that dτ/dt is measuring time flow along a world line. This is just wrong if dt spacelike, no matter what the character of the world line.

In my perhaps limited understanding, dt is a short time interval measured by the Schwarzschild observer at infinity. This observer and his clock remain outside the event horizon and therefore dt remains timelike. dτ/dt simply says the ratio of a small change in the proper time of a test clock relative to a small change in the time interval registered by an inertial master clock far away from the gravitational mass and stationary in the gravitational field. As the test clock falls are you saying that the small time interval dτ measured by the test clock ceases to be a measure of proper time? On the other hand, dt measured by the master clock does not change its nature because it is not moving relative to the gravitational field and remains outside the event horizon.

 

[yuiop]

An example in Minkowski space of the OP error:

suppose I have a particle moving in the -x direction. Then dτ/dx is negative. So what?

So what indeed. I am talking about dτ/dt being negative. That argument only holds if you can demonstrate that dt becomes dx in Schwarzschild coordinates.

 

[Passionflower]

Inside the event horizon, dt is not a timelike direction it is a spacelike direction, so the meaning you propose for d τ /dt is absurd (inside the event horizon). And, of course, r<2GM/c^2 means inside the event horizon.

Right!

 

[yuiop]

For a <EDIT>freely</EDIT> falling particle:
dS^2 = \left(1-\frac{Rs}{r}\right)^2 dt^2
Note that for a falling particle ds^2 remains positive (and timelike) even when r<Rs.

I assume this is a mistype--a falling particle has nonzero dr, but what you wrote only has the dt^2 term.

I have taken that into account. In the OP I entered the freefall velocity of a clock falling from infinity to obtain the above equation.

Also, to properly assess the sign of ds^2 for a falling particle, you need an equation for dr/dt, i.e., an equation for the particle's trajectory. Assuming that by "falling particle", you meant "freely falling particle", the equation for dr/dt would be derived from the geodesic equation.

I used:

dr/dt = -\sqrt{\frac{2GM}{r}} \left(1-\frac{2GM}{rc^2}\right)

and when substituted into the metric, the dr/dt term disappears. The full explanation is in the OP, but I have not derived the free falling velocity, but just used a generally recognised result. e.g see http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates#Speeds_of_raindrop for the speed of rain drop in Schwarzschild coordinates.

Slightly more directly:

Consider the Schwarzschild line element for radial motion only:
dS^2 = \left(1-\frac{2GM}{rc^2}\right) dt^2 - \frac{dr^2}{ c^2\left(1-\frac{2GM}{rc^2}\right)}
In Schwarzschild coordinates the free fall velocity is given by:
dr/dt = -\sqrt{\frac{2GM}{r}} \left(1-\frac{2GM}{rc^2}\right)
Substitute this expression into the equation at the top we obtain:
dS^2 = \left(1-\frac{2GM}{rc^2}\right) dt^2 - \left(1-\frac{2GM}{rc^2}\right)\frac{2GM}{ rc^2} dt^2
which simplifies to:
dS^2 = \left(1-\frac{2GM}{rc^2}\right)^2 dt^2

 

[PAllen]

Your language is slightly above my technical ability, but you seem to be saying that that the spacetime interval that determines the timelike nature can only be measured in terms of a stationary observer where dr or dx =0. I would disagree. For the SR spacetime interval, we use dS^2 = dt^2-dx^2 and we do not require that dx=0 because dS^2 is invariant for all values of dx^2. Also you seem to be requiring that dS be measured by a stationary clock in the coordinates, when it generally accepted that there is no such thing as a stationary clock in Schwarzschild coordinates, below the event horizon.

Good questions for focusing on a common confusion. Does the letter x or t have any physical meaning? No. Coordinates form a possibly non-orthogonal basis at every point - a unit vector in each coordinate direction. The metric, expressed in given coordinates then tells you the nature of the coordinate directions at each point. This question has to be answered before you can even pose whether your coordinate differential at a given point represents a clock measurement or a ruler measurement. Your core error remains you are trying to interpret a ruler measurement as a clock measurement due to misleading letter used for coordinates inside the event horizon.

In my perhaps limited understanding, dt is a short time interval measured by the Schwarzschild observer at infinity. This observer and his clock remain outside the event horizon and therefore dt remains timelike.

This statement is only true for the coordinates outside the event horizon. The crux of the strictly coordinate singularity at the event horizon is change in nature of the t and r coordinate. Note, the event horizon has a coordinate free definition; the fact that SC coordinates are singular here is a 'defect' of SC coordinates related to the change on nature of t and r across the horizon.

 

[PAllen]

So what indeed. I am talking about dτ/dt being negative. That argument only holds if you can demonstrate that dt becomes dx in Schwarzschild coordinates.

I have. You are disputing the mathematical definition of the timelike/spacelike character of a coordinate differential (or direction).

 

[PeterDonis]

I have taken that into account. In the OP I entered the freefall velocity of a clock falling from infinity to obtain the above equation.

Ah, I see -- I had missed the squared superscript in the equation for the freely falling particle. Yes, that looks right as far as the equation for the proper time of a freely falling particle goes. However, since this equation describes an interval with both nonzero dt and nonzero dr (you are eliminating dr by using the equation for dr/dt, but that doesn't make dr zero), you can't use it to deduce anything about the physical interpretation of the dt coordinate differential taken by itself.

 

[PeterDonis]

Your language is slightly above my technical ability, but you seem to be saying that that the spacetime interval that determines the timelike nature can only be measured in terms of a stationary observer where dr or dx =0.

No, what he is saying is that, to determine the nature (spacelike, timelike, or null) of a given coordinate, you have to look at intervals where *only* that coordinate changes. In the case of the t coordinate, that means looking at intervals where only dt is nonzero. Yes, that means that dr is zero (and also dtheta and dphi), and that, in this particular case, equates to a "stationary" observer (at least outside the horizon). But that's a side effect. The key point is that *all* coordinate differentials other than dt have to be zero in order to test whether a nonzero dt corresponds to a spacelike, timelike, or null interval.

So what we really need to look at is this:

c^{2} d\tau^{2} = \left( 1 - \frac{2GM}{rc^2} \right) c^{2} dt^{2}

i.e., the Schwarzschild line element with dr = dtheta = dphi = 0. This describes a "pure dt" line element, one that "points" solely in the t direction. And as you can see, this line element gives a positive ds^2 outside the horizon, a zero ds^2 at the horizon, and a negative ds^2 inside the horizon. That means the t coordinate is timelike outside, null on, and spacelike inside the horizon.

 

[Passionflower]

Coordinates form a possibly non-orthogonal basis at every point - a unit vector in each coordinate direction. The metric, expressed in given coordinates then tells you the nature of the coordinate directions at each point.

Exactly, well spoken!

{running the risk of making a simplification too simple to be right} here is an analogy:

Coordinates are like colored flag posts that do not necessarily stand perpendicular to the direction they measure. Imagine a set of red and blue flag posts. The red ones measure a horizontal property and start perpendicular to the ground while the blue ones measure a vertical property and start flat on the ground. Gradually they angle towards each other, e.g. the red ones towards the ground and the blue ones towards the vertical axis. At one point the red flag post is flat on the ground and the blue one is perpendicular to the ground and now they continue their angling.

Before the flag posts measured a resp. horizontal and vertical property but now it is the other way around.
The world stays the same however at one point the red and blue flag posts start to measure something different.

 

[yuiop]

The key point is that *all* coordinate differentials other than dt have to be zero in order to test whether a nonzero dt corresponds to a spacelike, timelike, or null interval.

I am afraid I have to beg to differ here. Let us consider an example in Minkowski spacetime using the (+---) sign convention, where we observe two events with Δx=0.8, Δy=0, Δz=0 and Δt=1 then:

ΔS² = Δt² - Δx² = 1- 0.64 = 0.36

This is positive and suggests a time like interval, but can we be sure if Δx≠0 as in this case?

Lets transform to a different reference frame where Δx=0 and so Δt=0.6 and recalculate the spacetime interval:

ΔS² = Δt² - Δx² = 0.36 - 0 = 0.36

The above shows that not only does the sign (and therefore the timelike nature) of ΔS² stay the same for any value of Δx² but the magnitude is identical too. Determining the timelike nature of ΔS² does not appear to require Δx = Δy = Δz = 0 as you assert.

One definition of a timelike interval (and I am quoting from wikipedia here) is that

There exists a reference frame such that the two events are observed to occur in the same spatial location, but there is no reference frame in which the two events can occur at the same time.

.
The spacetime interval I calculated for a falling particle meets that requirement. There does exist a reference frame in which the events occur at the same spatial location, and that is the reference frame of the falling observer and there also is no reference frame in which the two events can occur at the same time.

All the maths supports my assertion that the spacetime interval for the proper time measured by the free falling clock remains timelike in Schwarzschild coordinates even below the event horizon. Everything else here seems to be waving of hands, blue flags and red herrings.

 

[PAllen]

I am afraid I have to beg to differ here. Let us consider an example in Minkowski spacetime using the (+---) sign convention, where we observe two events with Δx=0.8, Δy=0, Δz=0 and Δt=1 then:

ΔS² = Δt² - Δx² = 1- 0.64 = 0.36

This is positive and suggests a time like interval, but can we be sure if Δx≠0 as in this case?

Lets transform to a different reference frame where Δx=0 and so Δt=0.6 and recalculate the spacetime interval:

ΔS² = Δt² - Δx² = 0.36 - 0 = 0.36

The above shows that not only does the sign (and therefore the timelike nature) of ΔS² stay the same for any value of Δx² but the magnitude is identical too. Determining the timelike nature of ΔS² does not appear to require Δx = Δy = Δz = 0 as you assert.

One definition of a timelike interval (and I am quoting from wikipedia here) is that .
The spacetime interval I calculated for a falling particle meets that requirement. There does exist a reference frame in which the events occur at the same spatial location, and that is the reference frame of the falling observer and there also is no reference frame in which the two events can occur at the same time.

All the maths supports my assertion that the spacetime interval for the proper time measured by the free falling clock remains timelike in Schwarzschild coordinates even below the event horizon. Everything else here seems to be waving of hands, blue flags and red herrings.

You are confusing two different things here. The interval between two events (in GR, also need to specify path) is invariant. Further, if it represents the world line of a material body, it is timelike everywhere.

This has nothing to do with the timelike/spacelike nature of a coordinate differential. Don't know why you are having so much trouble with this.

I'll try a another approaches to get this across. Suppose I write the metric:

dτ^2 = f(a,b)^2 da^2 -g(a,b)^2 db^2

and also tell you that I am using time like signature for my metric. How do you know which coordinate is space like and which is time like (note, for generality, in GR it is perfectly permissible to have coordinate systems in which all coordinates are spacelike, or all timelike, or all null - the latter are quite useful sometimes)? [Actually, all null coordinates are useful in SR too, and Dirac and Born were fond of them].

Well, you set da=0; the db^2 coefficient is negative, so b is spacelike. Similarly, a is timelike. Given this metric, it happens that this feature will be true everywhere. For a different metric, it could vary point to point. Independent of the character of each coordinate nature at each point, you can always have paths through a point that are timelike, spacelike or null. However, to determine the meaning of dτ/da you must determine the nature of da. That dτ is invariant is a given. But the nature of da is not, and is not determined by the nature of the path - it is determined by the metric expression at this point.

 

[pervect]

For r<2GM/c^2 the proper time runs in the reverse direction relative to Schwarzschild coordinate time. Note that there are no complication due to imaginary or complex numbers.

For R<2M, the coordinate "t" doesn't measure time at all. The coordinate "t" is spacelike, in the sense that the lorentz interval between any point P and any point P+dt is a spacelike interval, while the coordinate "r" is timelike, in the sense that the Lorent interval between P and P+dr is timelike for an point P inside the event horizon.

The direction of time is reversed for the r coordinate. Positive proper time is always r decreasing.

 

[PAllen]

All the maths supports my assertion that the spacetime interval for the proper time measured by the free falling clock remains timelike in Schwarzschild coordinates even below the event horizon.

This is key to the confusion. Believe it or not, everyone agrees with you on this point! The disagreement is not over what interval between two points on a free fall world line means - that is trivially timelike everywhere in all coordinate systems. The disagreement is specifically over the meaning of dτ/dt in a particular coordinate system. Further, the disagreement is over dt not dτ.

 

[PeterDonis]

I am afraid I have to beg to differ here. Let us consider an example in Minkowski spacetime

In this example you have turned an interval with dx nonzero into an interval with dx = 0 by switching coordinate charts. The equivalent in your example of an observer falling into the black hole would be to switch from global Schwarzschild coordinates (in which dr is nonzero) to the local inertial frame of the freely falling observer at some particular event on his worldline (and inside the horizon). In *that* chart, yes, "dr" would be zero and the "t" coordinate would be timelike, but it's a different chart, so a different t coordinate than the Schwarzschild t coordinate, which will still be spacelike.

 

[Dale]

Hi yuiop, maybe this will help.

Suppose that I gave you a list of coordinates (t,x,y,z), and asked you which one represented time, you would confidently say the first. But there is no reason that I couldn't give you the list of coordinates (x,y,t,z), and ask which represents time, you would probably say the third, but you may not be as confident. Now say I gave you the list of coordinates (μ,κ,ε,β) and ask which represents time, now you have no idea.

All of these are valid coordinate systems. Because the first follows a standard convention, the meaning is clear, but as I get further away from the standard convention it becomes less clear which coordinate represents time. These other coordinate systems are not bad or wrong, just different.

So we need a way to distinguish which coordinate is the time coordinate which doesn't depend on the convention, and that is given by the metric:

c²dτ² = +dt² -dx² -dy² -dz² let's me know that t is the time coordinate
c²dτ² = -dx² -dy² +dt² -dz² let's me know that t is still the time coordinate despite the reordering
c²dτ² = -dμ² -dκ² -dε² +dβ² let's me know that β is the time coordinate without any doubt

Do you understand that? If so, then here is a question:

What is the time coordinate given the metric c²dτ² = -dt² +dx² -dy² -dz² ?

 

[yuiop]

Do you understand that? If so, then here is a question:

What is the time coordinate given the metric c²dτ² = -dt² +dx² -dy² -dz² ?

Hi Dale. It would not be the first time I learned something from you so I will play along and give the obvious answer, that in that case +dx² would appear to be the time coordinate.

I am very aware that it is generally accepted that timelike becomes spacelike and vice versa below the event horizon in Schwarzschild coordinates, but I have yet to see a convincing proof as to why that should be so. Maybe you can do it.

 

[Passionflower]

I am very aware that it is generally accepted that timelike becomes spacelike and vice versa below the event horizon in Schwarzschild coordinates.

When you say "timelike becomes spacelike" it may hold the crux of your misunderstanding.

Remember nothing physically changes passed the event horizon, all that changes is that the Schwarzschild coordinate that mapped coordinate time now maps the reduced circumference and the r coordinate now maps coordinate time.

Coordinates only map things, they may, but not necessarily must, represent a physical measurable.

 

[PeterDonis]

the Schwarzschild coordinate that mapped coordinate time now maps the reduced circumference

This is not quite correct; the Schwarzschild r coordinate still corresponds to the "reduced circumference" inside the horizon, even though it is timelike. That is, the area of a 2-sphere at radial coordinate r is still 4 pi r^2 inside the horizon; and the area of a 2-sphere at a given "t" coordinate is not a function of t. The Schwarzschild r coordinate being timelike inside the horizon refers to the *differential* dr, i.e., to the *direction* that the unit vector associated with "r", \partial / \partial r (with t, theta, phi held constant) points; it does not affect the labeling of the 2-spheres by the r coordinate.

 

[Passionflower]

This is not quite correct; the Schwarzschild r coordinate still corresponds to the "reduced circumference" inside the horizon, even though it is timelike.

That does not make any sense to me.

You cannot have the cake and eat it.

 

[PeterDonis]

That does not make any sense to me.

You cannot have the cake and eat it.

I know it seems that way, but it isn't. Consider:

(1) In Painleve coordinates, r is spacelike inside the horizon, not timelike, and therefore r is unproblematically equal to the "reduced circumference" (i.e., a 2-sphere at r has area 4 pi r^2) all the way down to r = 0.

(2) The radial *coordinate* r is defined the *same* for Schwarzschild coordinates as for Painleve coordinates. In other words, if I look at the spacetime itself, the geometric object, and focus on any particular 2-sphere in it, that 2-sphere will be labeled by a "Painleve r" and a "Schwarzschild r", and *both* r's will be the *same*. (A given 2-sphere will be labeled, in general, by different *time* coordinates--its Painleve T will be different than its Schwarzschild t--because the two coordinate charts take spacelike slices out of the spacetime in different ways. But both sets of spacelike slices label the 2-spheres within them by the *same* r coordinates, based on the physical area of the 2-sphere, which is invariant.)

(3) Therefore, the Schwarzschild r coordinate inside the horizon must be equal to the "reduced circumference" of the 2-sphere it labels, just as it is outside the horizon, even though the Schwarzschild r is timelike inside the horizon. Basically, on a given 2-sphere there are two unit vectors--the Schwarzschild \partial / \partial r and the Painleve \partial / \partial r. The former points in a timelike direction inside the horizon; the latter still points in a spacelike direction. But if both originate from the same 2-sphere, they will both be associated with the same r coordinate.

(4) So the apparent paradox is resolved by carefully distinguishing between the labeling of individual 2-spheres by the r coordinate, which is how r gets defined as the "reduced circumference", and the direction in which the vector \partial / \partial r originating from a given 2-sphere at a given r points. The latter can change (spacelike to timelike) by changing coordinate charts, without changing the former at all.

 

[PAllen]

Hi Dale. It would not be the first time I learned something from you so I will play along and give the obvious answer, that in that case +dx² would appear to be the time coordinate.

I am very aware that it is generally accepted that timelike becomes spacelike and vice versa below the event horizon in Schwarzschild coordinates, but I have yet to see a convincing proof as to why that should be so. Maybe you can do it.

Before there can be a proof, there needs to be agreement on axioms and definitions. Further, I disagree with your summary above. Timelike is timelike spacelike is spacelike, period. What changes across the event horizon in SC coordinates (and not necessarliy in other coordinates) is the nature of the coordinate labeled 't'. If we can't agree on definitions and axioms, there is no basis for logical argument.

Every single contributor to this thread except you accepts that any of the following equivalent definitions determine the character of a coordinate called 'q' at a point:

1) Contract the unit vector (q,f,g,h)=(1,0,0,0), repeated twice, with the metric. The sign of this determines the character of the coordinate.

2) In the line element convention, set df=dg=dh=0, and determine the sign of dτ^2

3) In the the limit as Δq goes to zero, for the point P, what is the sign of the interval between P and P+Δq?

Using any of these formulations, I hope it is obvious that t is spacelike inside the event horizon. Note that these definitions are independent of any particular world line. This is important, because we change coordinates so t in one is timelike and t in another is spacelike, but this will not change the interval between two events on a given world line.

Looking at all coordinates in various schemes, using the above, we find (inside the horizon):

1) In SC interior coordinates, t is spacelike, r is timelike,\theta and \phi are spacelike.

2) In Gullstrand–Painlevé coordinates, all are spacelike inside the horizon.

3) In Kruskal–Szekeres coordinates, U is spacelike (everywhere), V is timelike (everywhere),
\theta and \phi are spacelike (everywhere).

 

[yuiop]

This as I understand it, is the textbook demonstration that the proper time of a falling particle advances smoothly and in finite time as a particle falls from outside an event horizon to the centre of a black hole.

The radial velocity of an object falling from infinity, in terms of the proper time of the falling object is:


\frac{dr}{d\tau} = -\sqrt{\frac{2GM}{r}}


To find the elapsed proper time between two radial coordinates we can invert the above expression and integrate with respect to r to obtain:

\tau = \int^{r=r_2}_{r=r_1} -\sqrt{\frac{r}{2GM}} dr = \frac{2}{3}\sqrt{\frac{r_1^3-r_2^3}{2GM}}

For r1=2GM and r2=0 we obtain the textbook result that:

\tau = \frac{2}{3}\sqrt{8G^2M^2} = \frac{4}{3}GM

Does that seem about right so far?

 

[DrGreg]

yuiop, (in response to post #22)

maybe your problem is that you are not recognising that "external" and "internal" Schwarzschild coordinates are two different coordinate systems. It just so happens that, by convention, we use the same symbols, and the same metric equation, for both systems. It might be less confusing if we said that the external metric was<br /> <br /> ds^2 = \left(1-\frac{2GM}{rc^2}\right) c^2 dt^2 - \frac{dr^2}{ \left(1-\frac{2GM}{rc^2}\right)} - R^2 \left(d\theta^2 + \sin^2 \theta \, d\phi^2 \right)<br /> <br />for r > 2GM/c², and the internal metric was<br /> <br /> ds^2 = \frac{c^2 dT^2}{ \left(\frac{2GM}{Tc^3}-1\right)} - \left(\frac{2GM}{Tc^3}-1\right) dR^2 - c^2 T^2 \left(d\Theta^2 + \sin^2 \Theta \,\, d\Phi^2 \right)<br /> <br />for cT < 2GM/c².

 

[Passionflower]

Technically you are correct as PG coordinates are a mixture or a free falling frame and a stationary observer infinitly far removed.

But if we consider a rain frame which is a true free falling observer then r is not the same as r in Schwarzschild coordinates passed the EH. To go from one to the other we need a LT before we integrate, try a LT with Schwarzschild coordinates with the assumption the r is a measure of space passed the EH. Let me know what you get when you do that.

 

[PeterDonis]

This as I understand it, is the textbook demonstration that the proper time of a falling particle advances smoothly and in finite time as a particle falls from outside an event horizon to the centre of a black hole.

Nobody is disputing that. But the Schwarzschild coordinate "t" is not the same as the proper time of a falling particle, so arguments about that proper time do not tell us anything about whether "t" is timelike or spacelike inside the horizon.

 

[PeterDonis]

But if we consider a rain frame which is a true free falling observer then r is not the same as r in Schwarzschild coordinates passed the EH.

I assume you mean the "rain frame" described here:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

In this frame, "r" is not the same as the Schwarzschild r outside the EH either. This frame rescales "r" so that the metric coefficient g_rr is locally 1. So the "r" in this chart is *not* the same as the "reduced circumference".

Edit: That Wiki page actually has a frame it calls the "rain frame" *and* a frame it calls the "free-float frame". I'm not sure which one you're referring to, but I think it's the former.

 

[Passionflower]

In this frame, "r" is not the same as the Schwarzschild r outside the EH either. This frame rescales "r" so that the metric coefficient g_rr is locally 1. So the "r" in this chart is *not* the same as the "reduced circumference".

Why do you think it is not the reduced circumference?

In this frame r2-r1 is actually the physical distance and thus the space foliation for this observer is actually Euclidean.

 

[PeterDonis]

Why do you think it is not the reduced circumference?

Because if you look closely at the line element for the frame (I'm assuming you mean the "rain frame" on the Wiki page, not the "free-float" frame), it has dr_{r}^{2} (the "rain" radial coordinate differential) in the second term, but just r (the "reduced circumference") in the last two, the ones with the angular differentials.

So r_{r} does directly measure radial distance (at that particular radius--note that the coordinate transformation for r_{r} is radius-dependent, so it's not a global chart--there is a *different* definition of r_{r} for each 2-sphere in the spacetime), but it does *not* correctly measure "reduced circumference" because the area of a 2-sphere is 4 \pi r^{2}, *not* 4 \pi r_{r}^{2}.

 

[yuiop]

yuiop, (in response to post #22)

maybe your problem is that you are not recognising that "external" and "internal" Schwarzschild coordinates are two different coordinate systems.

Hi DrGreg, I appreciate what you are getting at, but my argument would be this. When we want to know what happens in a region we have never tested, we can only take the laws we have tested and confirmed in our region and extend them to the unknown region. If we says the laws change in the unknown region, we can say anything we like about the unknown region. By way of example let us say we had a hypothetical universe where proper time τ of a moving clock related to coordinate time t by the relation dτ = dt(1-v^2/c^2) and have never actually tested what happens when v>c. We could predict using the laws we have tested that for v>c that dτ runs in the opposite direction to dt. Now if we are uncomfortable with proper time running in reverse, we can say that v>c that are different laws and the relation becomes dτ = -dt(1-v^2/c^2) and that it is the dt that changes it nature rather than the proper time of the moving clock. Let's say sometime later we actually (in the hypothetical universe) get the clock exceed c and observe that the moving clock is now running backwards relative to all our clocks and natural processes. Should we still cling to the notion that the proper time of the moving clock cannot run backwards and therefore all our clocks and natural processes must have become spacelike instead of timelike? Why should our clocks and rulers change their nature, when they have not undergone any acceleration or other measurable change, just because one clock in the (hypothetical) universe is exceeding c? Of course if there was an observer riding with the accelerated clock, they would say the clock is still advancing in time while we say it is going backwards.

For another example, still in a hypothetical universe but a bit more like ours, we have determined that for speeds less than c that dτ = dt√(1-v^2/c^2). For some reasons physicists are of a mindset that the light speed barrier can be broken. They notice that the equation predicts that proper time becomes imaginary for v>c but this does not fit comfortably with their belief that v can exceed c, so they conjecture that v<c there is one metric dτ = dt√(1-v^2/c^2) and for V>c there is another metric dτ = i*dt√(1-v^2/c^2) and by making a patchwork quilt of these metrics they can demonstrate that the proper time of a particle passing form v<c to v>c smoothly increases and remains real. They are of course only kidding themselves to make the universe fit their preconceptions.

 

[PAllen]

Hi DrGreg, I appreciate what you are getting at, but my argument would be this. When we want to know what happens in a region we have never tested, we can only take the laws we have tested and confirmed in our region and extend them to the unknown region. If we says the laws change in the unknown region, we can say anything we like about the unknown region. By way of example let us say we had a hypothetical universe where proper time τ of a moving clock related to coordinate time t by the relation dτ = dt(1-v^2/c^2) and have never actually tested what happens when v>c. We could predict using the laws we have tested that for v>c that dτ runs in the opposite direction to dt. Now if we are uncomfortable with proper time running in reverse, we can say that v>c that are different laws and the relation becomes dτ = -dt(1-v^2/c^2) and that it is the dt that changes it nature rather than the proper time of the moving clock. Let's say sometime later we actually (in the hypothetical universe) get the clock exceed c and observe that the moving clock is now running backwards relative to all our clocks and natural processes. Should we still cling to the notion that the proper time of the moving clock cannot run backwards and therefore all our clocks and natural processes must have become spacelike instead of timelike? Why should our clocks and rulers change their nature, when they have not undergone any acceleration or other measurable change, just because one clock in the (hypothetical) universe is exceeding c? Of course if there was an observer riding with the accelerated clock, they would say the clock is still advancing in time while we say it is going backwards.

It really seems like you are not reading what people are writing. We keep saying clocks and rulers don't change their nature across the horizon, neither do any laws don't change. Neither do any intervals change nature. Further, as I noted, in some coordinate systems (e.g. Kruskal), no coordinates change their nature. All that is true is that using some particular schemes for labeling events, the label called 't' has one meaning in one region and another meaning in another region. What determines its meaning? The metric. The metric is the only thing in GR that gives meaning to coordinate labels.

Let me ask you this: try to phrase your argument in Kruskal coordinates. If you are saying something physical, it should not matter what coordinates you use. Only, there you will find that along a free fall trajectory that crosses the horizon dτ/dV remains positive. Why? Because in these coordinates V is timelike coordinate everywhere.

 

[PeterDonis]

For r<2GM/c^2 the proper time runs in the reverse direction relative to Schwarzschild coordinate time.

Let me go back to the OP and ask the following question: what do you think the above quoted observation means, physically?

 

[yuiop]

Let me go back to the OP and ask the following question: what do you think the above quoted observation means, physically?

This is a big question and I will come back to it later and there are many other replies here that deserve to be digested and responded to as time allows. Thanks all. For now I would like to leave you all with an analogy and ask what you make of it.

Let us say we have a rectangular piece of tracing paper with x-axis along the long edge of the rectangle and y-axis on a short edge. On a wall we mark off a similar coordinate system with the x' axis horizontal and y' axis vertical. We lay the tracing paper on the wall and ensure the x-axis of the tracing paper is parallel with the x' axis of the wall grid and the y and y' axes are similarly parallel. We glue the left edge of the paper to the wall. Now we put a diagonal fold about a third of the way along the tracing paper so that that the last two thirds of the long rectangle are now vertical instead of horizontal. On the paper that has been moved position, the x lines are now parallel to the y' lines on the wall and the y lines are now parallel to the x' lines. Can we now say the x coordinates of the paper have become ylike or should we say that that the y' coordinates of the wall have become x'like? We now put another diagonal fold in the last third of the paper so that in this section x is antiparallel to x' so that increasing x on the paper is related to decreasing x' on the wall. Can we argue that nothing has changed on the paper and that x' coordinates on the wall have now changed their nature and no longer measure horizontal distances on the wall correctly? Imagine an a two dimensional creature restricted to living on the tracing paper. As far as it is concerned, the paper is flat and the wall must be somehow distorted in space. Who is correct. The 2D creature on the paper who says the paper is flat and the wall is folded in space or another observer who says the wall is flat and the paper is folded? As far as I can see, the x' coordinates on the wall always measure an x'like or horizontal property of the wall whatever we do with the paper. Folding the paper changes the relation of the paper to the wall, but does not change anything about the quantity that x' coordinates are defined to measure, namely horizontal distances along the wall. Why is it that in Schwarzschild coordinates that we seem to take the position that when the paper folds, it is actually the wall that folds and refuse to acknowledge the possibility that maybe the paper folded?

 

[PeterDonis]

I would like to leave you all with an analogy and ask what you make of it.

I'm not sure I understand the analogy you are trying to draw. Which coordinates on a black hole spacetime do you see as corresponding to the paper coordinates? Which ones do you see as corresponding to the wall coordinates?

 

[A.T.]

What about this from an old thread:

Note: The negative proper time for clock D when R=2.13 is not a typo. The proper time of a clock at the centre of a gravitational mass is negative always negative for radii less than 2.25m or less than \frac{9}{8} R_{s}

If true, you would have proper time reversal without any event horizon.

Question to yuiop, if you still remeber: What interior Schwarzshild solution did you use to derive this? Was is the exact one, or just a weak field approximation? The latter might explain the strange result because R = 9/8 R_s is definitely not a weak field.

 

[yuiop]

What about this from an old thread:

If true, you would have proper time reversal without any event horizon.

Question to yuiop, if you still remeber: What interior Schwarzshild solution did you use to derive this? Was is the exact one, or just a weak field approximation? The latter might explain the strange result because R = 9/8 R_s is definitely not a weak field.

At Rs < R < 9/8 Rs there is an apparent horizon where gravitational time dilation goes to zero located below Rs. The result came from an exact solution. This was discussed in a recent thread and we agreed there are certain limitations to the validity of the result. See https://www.physicsforums.com/showpost.php?p=3857054&postcount=24 and following posts.

The interior solution assumes that the massive body has even density and is static. The negative proper time of the clock at the centre probably suggests that a massive body cannot have even density or be static when the surface radius is less than 9/8 Rs.

 

[yuiop]

Let me go back to the OP and ask the following question: what do you think the above quoted observation means, physically?

The observation that the free falling clock goes in the reverse direction to coordinate time suggests to me that we may be wrong in some of our assumptions about what happens inside black holes such as the assumption that an object passes straight through the event horizon and continues to the singularity at the centre. See my previous post which is related. Reversal of proper time implies that the thermodynamic arrow of time is reversed which is unlikely so if we obtain a negative time result we should revisit our assumptions. I understand that most people here take the view that the free falling clock is not measuring time any more below the event horizon.

I'm not sure I understand the analogy you are trying to draw. Which coordinates on a black hole spacetime do you see as corresponding to the paper coordinates? Which ones do you see as corresponding to the wall coordinates?

To me the wall represents Schwarzschild coordinates, because that solution is static and unchanging and the falling observer does not materially change anything about the gravitational field is the massive body is large relative to the observer. The paper coordinates represent the reference frame of the free falling observer. The location of the free falling observer is continually changing relative to the gravitational field and he can probably even notice changes in his own coordinate system over time due to tidal effects and limitations of the equivalence principle, if his coordinate system is extended over a large distance. The rate of the clock that the falling observer uses as a reference, is continually changing over time as he falls, relative to the clock of the Schwarzschild master clock. In SR we can argue about whose clock is really going slower, but in GR it is indisputable that the falling clock is ticking slower than the stationary SC clock and continues to get even slower as it falls.

For example in these forums it is often stated that the speed of light near a black hole is the same as it is at infinity and that local measurements trump distant measurements. Let's examine these beliefs more closely. Let us say that a observer, (call him Adam) near the black hole but at constant radius claims that he measures the speed of light to be c locally. Eve, who is far away from the black hole also measures the local speed of light to be c but additionally claims that that the speed of light near Adam is slower than the speed of light at her radius. Adam counter claims that he measures the local speed of light to be c and the same as the local speed of light measured by Eve so the speed of light does not vary with distance from the black hole, because local measurements trump distant measurements. Eve points out Adam measures the speed of light using a clock that is ticking slower than her own clock so his measurements are distorted. Adam says his his clock is ticking at a rate of one second per second so it must be running at the same rate as Eve's clock. Eve then gets Adam to agree to a series of experiments transporting clocks and bringing them back together and eventually gets Adam to agree that clocks lower down do actually physically run slower than clocks higher up. He also agrees that the time dilation affects are not reciprocal as in SR. Once he concedes those points, he has to agree that the speed of light lower down must be relatively slower than the speed of light higher up and that local measurements can be distorted by the effects of gravity on his measuring tools.

If we time a race and find that the everyone in the race has broken the world record by a significant margin and later find the the race clock had a fault and was running slow, do we still make a new entry in the record books or declare that result invalid? At the very least we should try and determine by how much the race clock was running slow and correct for that error.

 

[Dale]

Hi Dale. It would not be the first time I learned something from you so I will play along and give the obvious answer, that in that case +dx² would appear to be the time coordinate.

I am very aware that it is generally accepted that timelike becomes spacelike and vice versa below the event horizon in Schwarzschild coordinates, but I have yet to see a convincing proof as to why that should be so. Maybe you can do it.

So, the rule is that the time coordinate is the one with the + sign in the metric, irrespective of the symbol used. Therefore, we simply apply this rule to the Schwarzschild metric. Let's just use units where c=G=1 and M=1/2 and let's compute the metric at r=2 and r=1/2 with θ=90º.

At r=2 we have:
dτ² = +0.5dt² -2dr² -4dθ² -4dφ² (t is the time coordinate)

At r=1/2 we have:
dτ² = -dt² +dr² -0.25dθ² -0.25dφ² (r is the time coordinate)

It follows directly from the rule for identifying the time coordinate. Do you disagree with the rule for identifying the time coordinate? If so, then what rule would you propose instead?

 

[PAllen]

The observation that the free falling clock goes in the reverse direction to coordinate time suggests to me that we may be wrong in some of our assumptions about what happens inside black holes such as the assumption that an object passes straight through the event horizon and continues to the singularity at the centre. See my previous post which is related. Reversal of proper time implies that the thermodynamic arrow of time is reversed which is unlikely so if we obtain a negative time result we should revisit our assumptions. I understand that most people here take the view that the free falling clock is not measuring time any more below the event horizon.

How can you you so persistently misread statements that have been repeated and clarified dozens of times? Everyone here (except possibly you) has stated multiple times that the free fall clock continues to measure time, perfectly normally, in the forward direction. Yet another way to state the flaw in your interpretation is the dτ/dt is a coordinate dependent quantity, and per universal interpretation of GR cannot be taken to mean anything observable without a computation based on invariant values. There is the rub: dτ/dτ is obviously 1; dτ/dt referred to the local frame of the free fall world line is also 1 straight through. So this is what I call the 'fundamental mistake of newbie GR interpretation' - why I rail against relating observables to coordinate quantities without analysis in terms of invariants.

Note something observable that is quite easy to compute: as the free faller passes through and beyond the EH, they can continue to get signals from outside the EH (it is only the reverse that is prevented by the EH). Suppose these signals are clock readings from an exterior clock. Then the free faller notes that the exterior clock continues to advance in the same direction as theirs, with no discontinuity or behavior change at the EH, right down to the singularity. (Note: the relative rates of the two clocks will vary depending on the particular free fall trajectory and the location of the exterior clock; but they will both be advancing steadily, in the same direction, as observed by the free faller).

[Edit: The only thing we say changes at the EH in this discussion is the nature of dt in one particular set of coordinates; and that this coordinate feature has no particular physical significance, because it doesn't affect observables such as the scenario described in the prior paragraph.]

 

[PAllen]

To me the wall represents Schwarzschild coordinates, because that solution is static and unchanging and the falling observer does not materially change anything about the gravitational field is the massive body is large relative to the observer.

Here is yet another key misunderstanding. The SC metric inside the EH is not static at all. This would be clear if you analyzed in terms of tensors and invariants rather than coordinates. In particular, there are no timelike killing vectors inside the EH, so it is not static. This, of course, is related to your error of insisting that (P, P+Δt) has a nature determined by the letter t rather than by observing the sign of its contraction with the metric.

 

[yuiop]

...
At r=2 we have:
dτ² = +0.5dt² -2dr² -4dθ² -4dφ² (t is the time coordinate)

At r=1/2 we have:
dτ² = -dt² +dr² -0.25dθ² -0.25dφ² (r is the time coordinate)

It follows directly from the rule for identifying the time coordinate. Do you disagree with the rule for identifying the time coordinate? If so, then what rule would you propose instead?

Two alternative proposals:

1)Time coordinate is that measurement made using a clock.
2)Time coordinate is that unique coordinate that can only increase while all other coordinates can increase or decrease.

I am trying to get at an understanding of the actual physical meaning of time coordinate changing from being timelike to being spacelike. I am sure it is not the most simplistic casual conclusion that clocks become rulers and vice versa.

Using your rule for identifying the time coordinate of the timelike coordinate as being the coordinate with the opposite sign to all the others in the metric we can note that above the event horizon the coordinate with the + sign can only increase for increasing dτ. By logical extension, below the event horizon we can conclude that dr can only increase and dt can run in either direction for increasing dτ.

 

[PAllen]

Two alternative proposals:

1)Time coordinate is that measurement made using a clock.
2)Time coordinate is that unique coordinate that can only increase while all other coordinates can increase or decrease.

I am trying to get at an understanding of the actual physical meaning of time coordinate changing from being timelike to being spacelike. I am sure it is not the most simplistic casual conclusion that clocks become rulers and vice versa.

Using your rule for identifying the time coordinate of the timelike coordinate as being the coordinate with the opposite sign to all the others in the metric we can note that above the event horizon the coordinate with the + sign can only increase for increasing dτ. By logical extension, below the event horizon we can conclude that dr can only increase and dt can run in either direction for increasing dτ.

Well let's apply your alternatives to interior SC coordinates:

1) A curve of t constant, r varying, inside the EH can be computed to be a timelike curve, and the the proper time along it will be a monotonic function of r. Thus, this criterion says r is timelike, not t.

2) For this criterion, we must clarify 'along a timelike path'. For, trivially, a spacelike path can go back in time. Then, using the interior metric it is easy to see that any timelike path must have r changing monotonically towards 0, while t can vary in either direction along a timelike path.

[The only peculiarity is that r decreasing is 'forward in time' along free fall trajectory continued across the EH.]

So, your own criteria agree with what we've all been saying.

Your last summary paragraph is correct except for the inverted convention on r. The metric only contains squared differentials, thus the timelike coordinate is constrained to change monotonically, but can follow either directional convention.

Noting that the exterior and interiors SC solutions are actually two separate coordinate patches, following a free fall world line through the EH, the coordinate description of the path (within the problematic SC coordinates) has the following character:

coordinate value of timelike coordinate just above EH goes to infinity. Then, the time along the world line just below the EH resumes with time like coordinate r=Rs-ε (coordinate discontinuity at EH itself), and continues to r=0 at time of reaching singularity.

 

[Dale]

1)Time coordinate is that measurement made using a clock.

This rule doesn't work. A clock measures proper time, not coordinate time. In the metric, that is the quantity on the left hand side, not any of the quantities on the right hand side.

2)Time coordinate is that unique coordinate that can only increase while all other coordinates can increase or decrease.

This rule might work, but it needs a little bit of clarification. First, if you are just considering a chart on the manifold then all coordinates can vary over their complete range. So, you must be restricting it somehow.

As much as possible, we don't want to rely on other theories, like thermodynamics, so I would think that the best way to do that would be to identify some event and then to identify all nearby events that are inside the future light cone from that event. If we do that we will usually find that for three of the coordinates the "inside" events will have larger and smaller coordinate values compared to the reference event and for one there will only be coordinates which are larger than the reference event.

I think this clarifies your intention without distorting it in any way.

Using your rule for identifying the time coordinate of the timelike coordinate as being the coordinate with the opposite sign to all the others in the metric we can note that above the event horizon the coordinate with the + sign can only increase for increasing dτ. By logical extension, below the event horizon we can conclude that dr can only increase and dt can run in either direction for increasing dτ.

Yes. Although for a black hole r can only decrease, so the time coordinate would be -r.

 

[PeterDonis]

The observation that the free falling clock goes in the reverse direction to coordinate time suggests to me that we may be wrong in some of our assumptions about what happens inside black holes such as the assumption that an object passes straight through the event horizon and continues to the singularity at the centre.

So you're interpreting the reversal to imply that *coordinate* time is the "fixed point" whose direction is unchanging, and *proper* time is what has "reversed direction". That clarifies things. It's also backwards; actually, Schwarzschild coordinate time is more like the "paper", and the underlying spacetime geometry is the "wall", as we'll see below.

I understand that most people here take the view that the free falling clock is not measuring time any more below the event horizon.

As PAllen said, this is a serious misunderstanding of what all of us have said. You yourself showed that the worldline of a freely falling object remains timelike all the way down to r = 0. That means that a clock falling along such a worldline continues to "measure time". We all agree that it does.

To me the wall represents Schwarzschild coordinates, because that solution is static and unchanging

Only outside the horizon! This may be a key source of the misunderstanding. Inside the horizon, Schwarzschild spacetime is *not* static. That means that you cannot use the "static" faraway observer as a point of reference for events inside the horizon.

Try this exercise: look at a diagram of Schwarzschild spacetime in Kruskal coordinates. Look at what the curves r = constant and t = constant look like. Outside the horizon, in "region I", the r = constant curves are hyperbolas running up the diagram that gradually get flatter as you move outward. The t = constant curves are straight lines whose slope varies with t (the t = 0 line is the "X-axis" of the diagram, t < 0 lines slope down and to the right, t > 0 lines slope up and to the right). So the r = constant curves are timelike and the t = constant curves are spacelike. (Note that the horizon, r = 2M, the 45 degree line up and to the right, is not included in the Schwarzschild exterior chart, except for the "center point" at the origin of the diagram, where t = plus infinity; this is where all of the t = constant lines *cross*, which is why there is a coordinate singularity in the Schwarzschild chart.)

What does this tell us about the character of the coordinates? (Remember that this diagram leaves out the angular coordinates altogether, so we are only considering r and t.) Since the r = constant lines are timelike, that means any small line element with dr = 0 and dt nonzero will be timelike. So t is a timelike coordinate. Similarly, since the t = constant lines are spacelike, any small line element with dt = 0 and dr nonzero will be spacelike. So r is a spacelike coordinate.

Now look at "region II", inside the horizon. Here the r = constant curves are hyperbolas running across the diagram, and the t = constant curves are lines running up the diagram from the "center point". The t = 0 line goes straight up; t > 0 lines go up and to the right, t < 0 lines go up and to the left. So here, r = constant lines are spacelike, and t = constant lines are timelike. This tells us, using the same reasoning as above, that inside the horizon, r is timelike and t is spacelike. This also tells us why the spacetime inside the horizon is not static: since the r = constant lines are now spacelike, there cannot be any static observers that stay at the same r.

But it also tells us something else. What will the worldline of a freely falling observer look like in this diagram? It will start out somewhere in the lower right corner, at very large r and very large negative t. It will gradually move up the diagram, and to the left (but staying at less than 45 degrees to the vertical, since it is always timelike), until it reaches the horizon line. At that point, it is at r = 2M, t = plus infinity. Now continue following the worldline into region II, the interior. It will keep moving up, and somewhat to the left. That means it will have to continue decreasing its r coordinate, but it will also now be *decreasing* its t coordinate! This should be obvious from looking at how the t = constant lines are laid out in region II; the ones closer to the horizon line have higher values of t, so as the freely falling worldline moves away from the horizon, it moves to lower values of t.

So what has actually happened? What has happened is that the Schwarzschild *coordinate* time is like the "paper" in your scenario; it flips around inside the horizon so that it runs in the opposite direction, relative to the "wall", the actual underlying spacetime geometry as diagrammed in the Kruskal diagram. (Actually, it would be more correct to say that the "paper" has been flipped by 45 degrees, sort of like the horizon line being the "fold".)

For example in these forums it is often stated that the speed of light near a black hole is the same as it is at infinity and that local measurements trump distant measurements.

No, that's not what is often stated. What is often stated is that an observer anywhere in spacetime will measure the *local* speed of light to be c. That is *not* the same as saying the speed of light is "the same as it is at infinity" without qualification--in a curved spacetime there is not a unique way to compare speeds between distant locations. You discuss different ways the comparison can be made, and rightly point out that they give different answers; but there is no way of showing that any of those answers is the "right" one. One observer says that his local measurements stay the same, so the speed of light stays the same; another says that gravity distorts the measuring tools, so the speed of light changes with location. In a sense, both are right, because they are talking about different interpretations of the term "speed of light".

 

[PeterDonis]

(Actually, it would be more correct to say that the "paper" has been flipped by 45 degrees, sort of like the horizon line being the "fold".)

Correction, I should have said "flipped by 90 degrees".