Gravitational Time Dilation - Confused

[rosie]

I'm puzzled by gravity - let alone time dilation. Stuck at first base. How is it that I can take two tennis balls - fill the one with rocks and drop them both down a tube, say, so no extraneous forces - then they both land at the same time? Yet picking up the one is more difficult than the other? More confusing yet - the one will bounce the other wont.

If gravity causes the initial drop - what comes into the equation that gives either of these balls the property of weight?

 

[yuiop]

I'm puzzled by gravity - let alone time dilation. Stuck at first base. How is it that I can take two tennis balls - fill the one with rocks and drop them both down a tube, say, so no extraneous forces - then they both land at the same time? Yet picking up the one is more difficult than the other? More confusing yet - the one will bounce the other wont.

If gravity causes the initial drop - what comes into the equation that gives either of these balls the property of weight?

Hi Rosie,

It is an interesting question, but it is slightly sidetracking the theme of this thread. Perhaps it is best if you post your question as new thread and I (and hopefully others) will have shot at answering it ;)

 

[yuiop]

What that page is saying is that if a photon is directed outward in a radial direction by an event exactly on the event horizon, then the photon will hover there (in Kruskal-Szekeres coordinates its diagonal worldline would coincide with the diagonal representing the event horizon in these coordinates). It is certainly not saying that photons in general hover at the horizon, a photon moving inward will cross it just fine.

For any radius > 2m the vertical coordinate speed of light is exactly the same whether the photons are going up or down. Only the direction sign changes but the magnitude is the same. Why is it that at the event horizon an outwardly emitted photon has zero velocity and inwardly emmited photon heads off to the central singularity? Clearly the magnitude of the velocity of the up and down photons is now different. Why is that? Why is it, that if I am below the event horizon and shine a torch downwards the photons travel forward in time to the central singularity and if I shine the torch upwards the photons still head towards the central singularity but this time they go backwards in time? Can I really make a photon go backwards in time just by changing the orientation of my torch? What is about the equation for the coordinate velocity of light that says the light emitted upwards from below the EH goes backwards in time and towards the central singularity, rather than forwards in time and away from the central singularity? I don't see anything there. It seems to be just an arbitary interpretation. When we look at the equation for motion of a massive particle in proper time the direction of motion is clearly indicated by the direction of advancing proper time and the equation clearly shows that a massive particle can leave the central singularity and head upwards towards the event horizon. Why can a particle with mass head up towards the event horizon and a photon can not? I would have put my money on the photon being the most likely to overcome gravity due to its higher velocity.

 

[JesseM]

For any radius > 2m the vertical coordinate speed of light is exactly the same whether the photons are going up or down.

Perhaps this is true in Schwarzschild coordinates, but it wouldn't be true in some other coordinate systems like Eddington-Finkelstein coordinates, for example. In any case, coordinate speeds are not important, the issue being discussed by that page is the actual physical behavior of photons--photons directed outward from events on the event horizon stay on the horizon forever, other photons do not.

Clearly the magnitude of the velocity of the up and down photons is now different. Why is that?

Different in what coordinate system?

Why is it, that if I am below the event horizon and shine a torch downwards the photons travel forward in time to the central singularity and if I shine the torch upwards the photons still head towards the central singularity but this time they go backwards in time?

They don't go "backwards in time" in any physical sense, they just go backwards relative to the time coordinate of Schwarzschild coordinates, which physically is not even timelike inside the horizon. Do you understand that the question of whether a given worldline is timelike or spacelike has a coordinate-independent answer, and that a line of constant t coordinate in Schwarzschild coordinates is spacelike outside the horizon but timelike inside, while a line of constant radial coordinate in Schwarzschild coordinates is timelike outside but spacelike inside? This is why if you want to talk about what happens inside the horizon you'll get yourself much less confused if you use Kruskal-Szekeres coordinates, where the vertical time axis is always timelike, and any timelike worldline will be closer to the vertical than a 45-degree light ray (all light rays are 45 degree straight lines in Kruskal-Szekeres coordinates) at every point, both outside and inside the horizon. Compare the timelike worldline A - A' - A'' in both Schwarzschild and Kruskal-Szekeres coordinates, in this diagram from the Misner-Thorne-Wheeler Gravitation textbook (p. 835):

Can I really make a photon go backwards in time just by changing the orientation of my torch? What is about the equation for the coordinate velocity of light that says the light emitted upwards from below the EH goes backwards in time and towards the central singularity, rather than forwards in time and away from the central singularity? I don't see anything there. It seems to be just an arbitary interpretation.

We've been through this before, kev--once again, can you tell me yes or no if you accept the standard GR notion that there is a clear distinction between coordinate-independent geometric facts about spacetime and worldlines, as opposed to facts which are only true in a particular choice of coordinate system? Do you agree that only the first type of facts have any objective physical meaning, and that no special holy significance should be attributed to the quirks of the Schwarzschild coordinate system?

When we look at the equation for motion of a massive particle in proper time the direction of motion is clearly indicated by the direction of advancing proper time and the equation clearly shows that a massive particle can leave the central singularity and head upwards towards the event horizon.

What does "advancing proper time" mean? In theory you can parametrize proper time to advance in either direction relative to coordinate time (even in SR nothing is stopping you from having proper time decrease as coordinate time increases, since the laws of physics are time-symmetric it is only thermodynamics that sets a physical arrow of time), but geometrically physicists can show that the worldline A - A' outside the horizon in the linked diagram connects up smoothly to the worldine A' - A'' inside the horizon (so both represent sections of a single continuous worldline), and if you want to have proper time defined in such a way that proper time advances along with coordinate time in Schwarzschild coordinates outside the horizon, then to be consistent you must have proper time decrease with increasing coordinate time inside the horizon (and decrease with increasing radial coordinate, which is really the timelike coordinate inside the horizon). If you didn't do this, you'd have a weird situation where a single particle's proper time went from increasing as it approached the horizon to decreasing as soon as it passed it, which wouldn't make any sense if you wanted proper time to represent elapsed time on an actual physical clock accompanying the infalling object. This is of course much more clear if you look at the A - A' - A'' worldline as represented in Kruskal-Szekeres coordinates, which avoids the misleading aspects of Schwarzschild coordinates.

 

[yuiop]

Do you understand that the question of whether a given worldline is timelike or spacelike has a coordinate-independent answer,

Yes I do, and if the worldline is spacelike in Schwarzschild coordinates, then it is spacelike in Kruskal-Szekeres coordinates too.. precisely because it coordinate-independent.

and that a line of constant t coordinate in Schwarzschild coordinates is spacelike outside the horizon but timelike inside, while a line of constant radial coordinate in Schwarzschild coordinates is timelike outside but spacelike inside?

The proper time of a falling observer is always positive as it falls in Schwarzschild coordinates and this is true above or below the horizon. This means above and below the horizon the invariant interval is timelike, but since timelike become spacelike below the horizon in Schwarzschild coordinates then the motion of the falling observer is spacelike below the horizon and by definition the observer is exceeding the speed of light. Since the invariant interval of an infalling observer is spacelike below the horizon in Schwarzschild coordinates it follows that the motion of a falling observer is spacelike below the horizon in Kruskal-Szekeres coordinates too, because it is an independent invariant quantity as you keep mentioning.

(all light rays are 45 degree straight lines in Kruskal-Szekeres coordinates)

A light ray at 45 degrees in a coordinate system does not indicate which way the light is going and this is difficult to determine because the proper time of a photon is zero. Proving that the path of a falling particle is timelike by comparing it to the supposed path of a photon whose direction you have not proven, proves nothing.

We've been through this before, kev--once again, can you tell me yes or no if you accept the standard GR notion that there is a clear distinction between coordinate-independent geometric facts about spacetime and worldlines, as opposed to facts which are only true in a particular choice of coordinate system? Do you agree that only the first type of facts have any objective physical meaning, and that no special holy significance should be attributed to the quirks of the Schwarzschild coordinate system?

I don't atribute special holy significance to Schwarzschild coordinates but I do not think it is an unfortunate choice of coordinates as it often described. If something is coordinate-independent geometric fact in any given coordinate system then it true in any other coordinate system. Kruskal-Szekeres coordinates do not have any higher claim to the truth over the supposedly misguided Schwarzschild coordinates. They have equally valid.

I wonder why you attribute holy significance to the quirks of Kruskal-Szekeres coordinates?

What does "advancing proper time" mean?

It means nothing get younger, not even twins in relativity. In SR and GR, the proper time can slow down relative to another clock, but never reverse. None of the twins ever ends up younger than they started in any thought experiment.

In theory you can parametrize proper time to advance in either direction relative to coordinate time (even in SR nothing is stopping you from having proper time decrease as coordinate time increases,

From now on I am counting birthdays in parametrized proper time. ;)

.. and if you want to have proper time defined in such a way that proper time advances along with coordinate time in Schwarzschild coordinates outside the horizon, then to be consistent you must have proper time decrease with increasing coordinate time inside the horizon (and decrease with increasing radial coordinate, which is really the timelike coordinate inside the horizon). If you didn't do this, you'd have a weird situation where a single particle's proper time went from increasing as it approached the horizon to decreasing as soon as it passed it, which wouldn't make any sense if you wanted proper time to represent elapsed time on an actual physical clock accompanying the infalling object.

Yep, it does not make sense, if objects really do continue to fall below the horizon, but I agree that is the conventional wisdom.

Take a look at the attached world maps in various projections. The top chart of the first diagram shows a view of the globe (Orthographic azimuthal projection) from both sides and the blue lines represent the path of a polar orbiting satellite projected onto the surface of the earth. The lower chart of the first image is a Mercator type projection of the world map that flattens the globe into a flat rectangle. The satellite path is also shown on this projection. In the second image the Mercator projection is squeezed in the middle to produce an unusual but still still valid chart of the Earth. Notice the large white empty triangular spaces. There is no need to fill those empty spaces with a parallel virtual Earth to make the geodesics complete. All the information from any of the first two charts is present on this chart also. Now a further chart on the right is created by rotating the left part of the squeezed chart clockwise to form a Kruskal-like map of the Earth. Note that the blue satellite path which is a true geodesic path does not look smooth and continuous in the Kruskal-like map as it crosses from the East quadrant to the North quadrant and yet we know in reality it is. Note that in this Kruskal-like Earth map "proves" that what we normally think of as North-South in Australia is actually East-West!. "Northlike" has become "Eastlike" just as timelike becomes spacelike in Kruskal-Szekeres coordinates. Because we are familiar with the Earth chart we are not fooled and we know North does not become East by the use of an unusual chart and we demonstrate this by simply checking a compass. However, if we are using an unusually distorted coordinate system to study unfamiliar territory like a black hole, then the absurdities are not so obvious. An additional red line has been added to the Kruskal-like map that represents a smooth and continuous geodesic in the Kruskal-like chart but it does not take a lot of visualisation to see that it does not represent the natural inertial motion of anything real. The moral is that a smooth and continuous continuation of a path as it crosses a boundary in a particular distorted coordinate system is not proof by itself that the path is smooth and continuous in reality.

 

[JesseM]

Yes I do, and if the worldline is spacelike in Schwarzschild coordinates, then it is spacelike in Kruskal-Szekeres coordinates too.. precisely because it coordinate-independent.

And do you agree that the line A'-A'' in the diagram I posted above is timelike, not spacelike? Do you agree that a line of constant radial coordinate in Schwarzschild coordinates is timelike outside the horizon but spacelike inside, whereas a line of constant t-coordinate in Schwarzschild coordinates is spacelike outside the horizon but timelike inside? Finally, do you agree that a horizontal line of constant t coordinate in Kruskal coordinates is spacelike everywhere, and a vertical line in Kruskal coordinates is timelike everywhere? If you disagree with any of these you are simply confused, if you would actually do the math and check whether the line element ds^2 is positive or negative for a small increment on any of these lines, you'd see that all these statements are correct.

The proper time of a falling observer is always positive as it falls in Schwarzschild coordinates and this is true above or below the horizon.

What do you mean by "always positive"? You can set the zero point of proper time anywhere you like. Perhaps you mean that proper time is always increasing with the t coordinate in Schwarzschild coordinates, but in this case you are mistaken, as long as we adopt the convention that there are no "local maxima" of proper time along a given worldline (meaning no points on a worldline where if you go in either direction away from a given point, the proper time decreases from its value at that point) then if the proper time is increasing with the t coordinate outside the horizon, it must be decreasing with the t coordinate outside the horizon. The only way to have the proper time increase with t coordinate both inside and outside is to have such a "local maxima" on the object's worldline at the point it crosses the horizon (which is not represented in Schwarzschild coordinates, but geometrically it's a perfectly real point in the spacetime manifold, and it is represented in Kruskal-Szekeres coordinates).

This means above and below the horizon the invariant interval is timelike but since timelike become spacelike below the horizon in Schwarzschild coordinates then the motion of the falling observer is spacelike below the horizon and by definition the observer is exceeding the speed of light.

No, you're completely confused here. First of all, just talking about the "invariant interval" ds^2 is meaningless without specifying what path you're calculating the invariant interval along--are you talking about a short interval along the worldline of the infalling particle? If so, then if you find it's timelike then it is timelike, period, that's a coordinate-independent geometric fact and it has absolutely nothing to do with whether the t-coordinate of your chosen coordinate system is timelike or spacelike. You calculate the value of the ds^2 along a path by integrating the line element along that path; the line element at every point is in turn determined by the metric, which is what encapsulates the actual geometric curvature of the spacetime. You can use any arbitrary coordinate system you like, including one where the t coordinate becomes spacelike or whatever, but the metric is specifically adjusted to each possible coordinate system in such a way that the integral of ds^2 between two events along a given worldline will have the same value in every coordinate system, so that "path length" is a geometric fact which is independent of your choice of coordinate system. Since the integral of ds^2 along a given path must be the same in ever possible coordinate system, then whether ds^2 is positive or negative must also be the same in every possible coordinate system, and that is what determines whether the path is spacelike or timelike.

So the worldline is still timelike in the horizon, as I'm sure could be checked if you parametrized a path like A'-A'' in the diagram and then checked the value of the line element ds^2 for an infinitesimal variation in the parameter--it would have the same sign as the line element for an infinitesimal variation in the parameter for the path A-A' of the same particle outside the horizon. It is the line element, based on the metric, that tells you in a real geometric terms whether a worldline is spacelike or timelike--do you understand that this is the sole determinant of the geometry, that all other features of particular coordinate systems are irrelevant to the meaning of "spacelike" and "timelike"?

Here's the line element in Schwarzschild coordinates (copied from here):

$$ds^2 = (1 - \frac{2m}{r}) c^2 dt^2 - \frac{1}{(1 - \frac{2m}{r})} dr^2 - r^2 ( d\theta^2 + r^2 sin^2 \theta d\phi^2 )$$

If you like, free feel to verify that for an infinitesimal increment along a line parallel to the t axis (meaning dr, d-theta and d-phi are all zero but dt is nonzero), if the line is at a value of r where r>2m (outside the event horizon), ds^2 will be positive (meaning the increment is timelike in a geometric sense), but if the line is at a value of r where r<2m (inside the event horizon), ds^2 will be negative (meaning it's spacelike). Likewise you could look at an infinitesimal increment along a line parallel to the r axis and verify that outside the horizon it's negative but inside the horizon it's positive.

Incidentally, if you look at the diagram below, also from MTW's Gravitation, you can clearly see that the worldline of a particle lies inside the future light cone of events on its worldline both inside and outside the horizon, which is another way of thinking about the physical meaning of what it means for a worldline to be "timelike" (a physicist or mathematician could show that this definition is equivalent to the positive ds^2 notion, which is handy because in relativity light cones are very basic to the understanding of causality).

A light ray at 45 degrees in a coordinate system does not indicate which way the light is going and this is difficult to determine because the proper time of a photon is zero. Proving that the path of a falling particle is timelike by comparing it to the supposed path of a photon whose direction you have not proven, proves nothing.

Huh? I wasn't talking about the "direction" of a photon, and timelike vs. spacelike has nothing to do with the direction that proper time is increasing on a given path, it's only about the set of points in spacetime that the path passes through. Again, in GR it's always true that a path which remains inside the light cones (future or past) of every point along the path is a timelike one.

I don't atribute special holy significance to Schwarzschild coordinates but I do not think it is an unfortunate choice of coordinates as it often described. If something is coordinate-independent geometric fact in any given coordinate system then it true in any other coordinate system. Kruskal-Szekeres coordinates do not have any higher claim to the truth over the supposedly misguided Schwarzschild coordinates. They have equally valid.

I wonder why you attribute holy significance to the quirks of Kruskal-Szekeres coordinates?

Er, I don't. You don't seem to understand that timelike vs. spacelike is a geometric fact which is the same in all coordinate systems (once you've figured out the correct form of the metric in that coordinate system, you can use it to calculate the integral of ds^2 along a given path and your answer will be a coordinate-independent one), and that physicists have checked this already and verified that Kruskal-Szekeres happen to have the nice property that worldlines which are timelike in the coordinate-independent sense always have a slope closer to the vertical than 45 degrees when represented in these coordinates. Of course it's the geometry that's fundamental, but because of this "nice" property of Kruskal-Szekeres coordinates they are less likely to mislead you about the basic geometrical issues, as you have clearly been mislead by thinking about the "less nice" Schwarzschild coordinates.

It means nothing get younger, not even twins in relativity. In SR and GR, the proper time can slow down relative to another clock, but never reverse. None of the twins ever ends up younger than started in any thought experiment.

Then like I said, if you want to avoid a "local maxima" in proper time which would correspond to time decreasing in both directions from the maxima, you must define proper time in such a way that if proper time is increasing with Schwarzschild t coordinate on a given worldline outside the horizon, then it is decreasing with Schwarzschild t coordinate on the same worldline outside the horizon.

Yep, it does not make sense, if objects really do continue to fall below the horizon but I agree that is that conventional wisdom.

You're being a crackpot if you keep talking as though these issues are a matter of "conventional wisdom" as opposed to objectively right or wrong and probably not hard to check mathematically for anyone well-versed in GR.

Take a look at the attached world maps in various projections. The top chart of the first diagram shows a view of the globe (Orthographic azimuthal projection) from both sides and the blue lines represent the path of a polar orbiting satellite projected onto the surface of the earth. The lower chart of the first image is a Mercator type projection of the world map that flattens the globe into a flat rectangle. The satellite path is also shown on this projection. In the second image the Mercator projection is squeezed in the middle to produce an unusual but still still valid chart of the Earth. Notice the large white empty triangular spaces. There is no need to fill those empty spaces with a parallel virtual Earth to make the geodesics complete.

As on a previous thread, you talk as though it's a matter of arbitrary fancy whether we think of different points in a coordinate map as being "the same" or "different", ignoring this thing called the metric which tells you how distances between points actually work in a geometric sense. The globe analogy is actually helpful in thinking about the meaning of the metric, though. Do you agree that on a globe, there is a geometric notion of distance along a given path, and that path with the shortest distance between any two points is a section of the great circle that contains both points? Do you agree that the apparent length of a given path in a given coordinate representation will not in general match up well with the actual geometric length? For example, if you have two horizontal lines of apparent equal length in a Mercator projection, one at the equator and one near the pole, they will not actually have equal geometric length if you map the corresponding points on the globe, the one at the equator will have a greater geometric length. Well, the metric in Mercator coordinates would tell you the line element at every point, and it would tell you that a given increment of the horizontal coordinate near the equator corresponds to a larger increment of ds^2 then the same sized increment of the horizontal coordinate near the pole, and if you integrated ds^2 along a path using the right formula for the line element in Mercator coordinates, you'd get back the actual geometric length of that path on the surface of the globe. And if you can calculate the coordinate-independent distance along all possible paths, then you can define how "nearby" an arbitrary pair of points are by looking at the length of the shortest possible path between them.

Even if the coordinate system does something weird like stretching out a single geometric point on the globe into a line in the coordinate representation (which is exactly what happens with the set of events at the event horizon in Schwarzschild coordinates), you could figure out that this set of different coordinate points is "really" a single geometric point by nothing that the shortest path between such points would have a geometric length of zero. Similarly, in the case of your triangular Earth diagram with empty white spaces, if you define the metric on each triangular slice and then identify points on the diagonals of each slice as in the right-hand version of your diagram, you could show geometrically using the metric on either slice that all the points along the diagonal are really a single geometric point (the North Pole). In GR I think defining what it means for different coordinate points to be the "same" geometrically would be a bit more complicated, since you can have lightlike paths between distinct points which have ds^2 = 0 along the path, but just from reading physicists talking about coordinate representations vs. geometry I'm sure that physicists and mathematicians do have some well-defined notion of what it means for points in spacetime with distinct coordinates to be the "same point" geometrically. And from reading physicists talk about the Kruskal diagram, they make clear that different points along the diagonal representing the event horizon actually are geometrically distinct points in the spacetime manifold, so this is unlike your triangular mapping of the Earth where all the points on the diagonal are really just the single geometric point of the North Pole, and thus the blue line can really be a single continuous line geometrically even though it appears to cross the diagonal at two different places.

 

[Max™]

They don't go "backwards in time" in any physical sense, they just go backwards relative to the time coordinate of Schwarzschild coordinates, which physically is not even timelike inside the horizon. Do you understand that the question of whether a given worldline is timelike or spacelike has a coordinate-independent answer, and that a line of constant t coordinate in Schwarzschild coordinates is spacelike outside the horizon but timelike inside, while a line of constant radial coordinate in Schwarzschild coordinates is timelike outside but spacelike inside? This is why if you want to talk about what happens inside the horizon you'll get yourself much less confused if you use Kruskal-Szekeres coordinates, where the vertical time axis is always timelike, and any timelike worldline will be closer to the vertical than a 45-degree light ray (all light rays are 45 degree straight lines in Kruskal-Szekeres coordinates) at every point, both outside and inside the horizon. Compare the timelike worldline A - A' - A'' in both Schwarzschild and Kruskal-Szekeres coordinates, in this diagram from the Misner-Thorne-Wheeler Gravitation textbook (p. 835):

Jesse is a smart guy, I was reading and thinking that something very obvious was missing, but couldn't figure out what it was, then it hit me that no one was considering the nature of the worldlines directly.

If you don't consider the worldlines themselves, you can do silly things like decree arbitrarily simultaneous timeslices to be universal for all observers, when they're obviously unique to each individual observer when compared against the worldlines themselves.


I always liked to say the degrees of freedom within a Schwarzschild hole are inwardly directed after being reversed. That isn't as clear as the above explanation though.

 

[A-wal]

I'm puzzled by gravity - let alone time dilation. Stuck at first base. How is it that I can take two tennis balls - fill the one with rocks and drop them both down a tube, say, so no extraneous forces - then they both land at the same time? Yet picking up the one is more difficult than the other? More confusing yet - the one will bounce the other wont.

If gravity causes the initial drop - what comes into the equation that gives either of these balls the property of weight?

I've said it before and I'll say it again:

1). It takes more energy to move an object which weighs more?

2). Because if it didn't work that way then if two objects were to touch in mid air then they would fall faster. If touching isn't enough then what difference would it make if they were properly joined?

3). Because objects in free fall aren't really falling?

The answer is 3. When an object is in free fall, it is in fact at rest. It's the ground that's moving up, which pushes anything on the surface down. Length contraction stops things from actually moving outwards.

Read my blogs. Then tell me what you thought. In fact I think they would be good to put in this section and made sticky, providing there's no errors of course. Technically they're not close to what real physicists could write but I think they'll be good for all the beginners that frequently come here.