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$$<br /> ds^2 = \left(1-\frac{2GM}{rc^2}\right) c^2 dt^2 - \frac{dr^2}{ \left(1-\frac{2GM}{rc^2}\right)} - R^2 \left(d\theta^2 + \sin^2 \theta \, d\phi^2 \right)<br />$$for r > 2GM/c², and the internal metric was$$<br /> ds^2 = \frac{c^2 dT^2}{ \left(\frac{2GM}{Tc^3}-1\right)} - \left(\frac{2GM}{Tc^3}-1\right) dR^2 - c^2 T^2 \left(d\Theta^2 + \sin^2 \Theta \,\, d\Phi^2 \right)<br />$$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.