Reversal of proper time for a falling object

1. Apr 18, 2012

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.

2. Apr 18, 2012

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.

3. Apr 18, 2012

yuiop

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.

4. Apr 18, 2012

PAllen

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 irrelevent 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.

5. Apr 18, 2012

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?

Last edited: Apr 18, 2012
6. Apr 18, 2012

Staff: Mentor

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.

7. Apr 18, 2012

yuiop

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.

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.

8. Apr 18, 2012

yuiop

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.

9. Apr 18, 2012

Passionflower

Right!

10. Apr 18, 2012

yuiop

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.

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:
$$dS^2 = \left(1-\frac{2GM}{rc^2}\right)^2 dt^2$$

Last edited: Apr 18, 2012
11. Apr 18, 2012

PAllen

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.
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.

12. Apr 18, 2012

PAllen

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

13. Apr 18, 2012

Staff: Mentor

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.

14. Apr 18, 2012

Staff: Mentor

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.

15. Apr 18, 2012

Passionflower

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.

Last edited: Apr 18, 2012
16. Apr 18, 2012

yuiop

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.

17. Apr 18, 2012

PAllen

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.

Last edited: Apr 18, 2012
18. Apr 18, 2012

pervect

Staff Emeritus
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.

19. Apr 18, 2012

PAllen

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.

20. Apr 18, 2012

Staff: Mentor

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.

21. Apr 18, 2012

Staff: Mentor

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 lets me know that t is the time coordinate
c²dτ² = -dx² -dy² +dt² -dz² lets me know that t is still the time coordinate despite the reordering
c²dτ² = -dμ² -dκ² -dε² +dβ² lets 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 ?

22. Apr 18, 2012

yuiop

Hi Dale. It would not be the first time I learnt 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.

23. Apr 18, 2012

Passionflower

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.

24. Apr 18, 2012

Staff: Mentor

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.

25. Apr 18, 2012

Passionflower

That does not make any sense to me.

You cannot have the cake and eat it.