Time dilation for clock thrown up and caught back

[michelcolman]

If you throw up a clock in a field of gravity and it falls back down, will it show the same time as a clock that stayed at the surface? In other words, will the gravitational time dilation of the stationary clock exactly equal the sum of the smaller gravitational time dilation of the other clock (higher up in the field) and the dilation caused by its speed?

Just to make things perfectly clear: clock 1 stays at the surface, clock 2 is thrown up and falls back down, so its time passes more slowly because of its speed, but also more quickly because it is higher up in the field of gravity. The question is: do the two dilations balance each other out exactly?

I know this may look like a homework question, but trust me, I'm a 35 year old airline pilot :)

The reason I'm asking is that I want to understand gravitational time dilation in the expanding universe.

If I understood correctly, you can use two different coordinate systems for the universe: the coordinate system that we would normally use based on special relativity (time for distant (and therefore moving) objects passes more slowly), or the god-like cosmological coordinate system in which the time coordinate at any location in the universe is chosen to be local time as experienced by a local observer who is moving together with the expansion of the universe.

The first view just looks at the universe as a collection of things that happen to be flying apart, while the second view is closely tied to the particular structure of our expanding universe. Both are just a choice of coordinates, nothing more.

The two systems give different, but consistent views of the universe. Just one example: In the first system, very distant objects will never reach our age because time grinds to a halt as they approach (but never exceed) the speed of light. In the second system, they are the same age as us but their light will never reach us because space between us is expanding faster than the speed of light (even the light that is trying to come towards us, is actually retreating away from us due to the expansion of the universe). Both views result in the same observation: we never see them reach our age.

It all seems to make perfect sense, except if I consider a universe that shrinks back together (like people used to think before they found out the universe is actually accellerating, but never mind that). In the event of a Big Crunch, the first coordinate system results in a twin paradox, while the second obviously has no such problem.

I know the Big Crunch will probably never happen, but I just want to figure out why the two models seem to be contradictory in such a theoretically possible case.

If the two clocks from my first question (one thrown up and caught, the other stationary) show the same time, that would solve my paradox since any observer would consider himself to be at the center of the universe's great field of gravity. But I'm not very good at this kind of math (yet) and somebody else can probably show me a calculation in a minute or so, or even better, give an argument why the clocks (hopefully) must show the same time.

 

[A.T.]

If you throw up a clock in a field of gravity and it falls back down, will it show the same time as a clock that stayed at the surface?

This thread is about tossing clocks:
https://www.physicsforums.com/showthread.php?t=249722

 

[Naty1]

If you throw up a clock in a field of gravity and it falls back down, will it show the same time as a clock that stayed at the surface?

no.

nor will such a self powered clock that is launched without any gravity present run at the same rate as a "stationary clock"...for faster things time moves more slowly... as does time in a high gravitational potential.

So when when you skyjockies are airborne your clock runs a tad slower for both reasons and the GPS system you use for navigation must also be corrected for such relativistic time effects. http://en.wikipedia.org/wiki/Global_Positioning_System#Timekeeping

 

[Dale]

Just to make things perfectly clear: clock 1 stays at the surface, clock 2 is thrown up and falls back down, so its time passes more slowly because of its speed, but also more quickly because it is higher up in the field of gravity. The question is: do the two dilations balance each other out exactly?

The tossing clocks thread is a good one to go over for this question. The bottom line is that the tossed clock will record more time than the surface clock. In fact, the tossed clock will always record the maximum time possible between the "toss" and "catch" events. If you put a little rocket booster, or wings, or anything that makes it not purely free-falling, then the clock will record less time than a free-falling clock between the same two events.

 

[diazona]

The tossing clocks thread is a good one to go over for this question. The bottom line is that the tossed clock will record more time than the surface clock. In fact, the tossed clock will always record the maximum time possible between the "toss" and "catch" events. If you put a little rocket booster, or wings, or anything that makes it not purely free-falling, then the clock will record less time than a free-falling clock between the same two events.

Interesting... never thought about it like that. I like that explanation.

 

[michelcolman]

nor will such a self powered clock that is launched without any gravity present run at the same rate as a "stationary clock"...for faster things time moves more slowly... as does time in a high gravitational potential.

So when when you skyjockies are airborne your clock runs a tad slower for both reasons and the GPS system you use for navigation must also be corrected for such relativistic time effects. http://en.wikipedia.org/wiki/Global_Positioning_System#Timekeeping

That is wrong. Time moves more slowly for objects with a lower gravitational potential (closer to a massive object). For example, time actually stops when someone approaches a black hole's event horizon (as seen from outside).

So the two effects, speed and gravitational potential, have opposite effects on proper time. The tossed clock is higher up (further away from the earth) so it's running more quickly, but it's also moving which makes it run more slowly. I was kind of hoping the two would cancel each other out, but it appears they don't...

Anyway, I'll have a good look at the tossing clocks thread and see if I can make any sense of it.

 

[A.T.]

The tossing clocks thread is a good one to go over for this question.

Yes, but at https://www.physicsforums.com/showthread.php?t=249722&p=1840160" it was sill not clear what time clock B (free fall, oscillating in a tunnel) measures. It would be interesting to compare that to clock D (free fall rest at the center) and clock E (free fall orbiting at surface level).

 

[michelcolman]

no.

nor will such a self powered clock that is launched without any gravity present run at the same rate as a "stationary clock"...for faster things time moves more slowly... as does time in a high gravitational potential.

So when when you skyjockies are airborne your clock runs a tad slower for both reasons and the GPS system you use for navigation must also be corrected for such relativistic time effects. http://en.wikipedia.org/wiki/Global_Positioning_System#Timekeeping

From the very Wikipedia article you linked to: (emphasis mine)

A number of sources of error exist due to relativistic effects. For example, the relativistic time slowing due to the speed of the satellite of about 1 part in 10^10, the gravitational time dilation that makes a satellite run about 5 parts in 10^10 faster than an Earth based clock, and the Sagnac effect due to rotation relative to receivers on Earth.

 

[michelcolman]

Yes, but at https://www.physicsforums.com/showthread.php?t=249722&p=1840160" it was sill not clear what time clock B (free fall, oscillating in a tunnel) measures. It would be interesting to compare that to clock D (free fall rest at the center) and clock E (free fall orbiting at surface level).

That's exactly what I was really looking for (while only just realising it now): the comparison between a clock at the center of the earth, and a clock that is thrown up from the center and falls back again (without passing the surface, if that makes a difference).

If they don't match, I'm at a loss to explain why a Big Crunch of the universe would not induce a twin paradox.

 

[Dale]

That's exactly what I was really looking for (while only just realising it now): the comparison between a clock at the center of the earth, and a clock that is thrown up from the center and falls back again (without passing the surface, if that makes a difference).

Well, that is not at all what you asked above. In your original question the metric was the exterior Schwarzschild metric, the tossed clock was traveling on a geodesic (free fall), and the surface clock was not. Since geodesics are always the locally longest path we know immediately that the tossed clock records the most time.

Your revised question now deals with the interior Schwarzschild metric, and is comparing two different geodesics (both clocks are free falling). So you cannot immediately say which is shorter. There is no way around it other than to integrate the metric along the two geodesics.

If they don't match, I'm at a loss to explain why a Big Crunch of the universe would not induce a twin paradox.

The above questions have little to do with any possible Big Crunch. The reason is that the metric is different. Your first question was about the exterior Schwarzschild metric, your second about the interior Schwarzschild metric, and this one is about the FLRW metric.

 

[michelcolman]

The above questions have little to do with any possible Big Crunch. The reason is that the metric is different. Your first question was about the exterior Schwarzschild metric, your second about the interior Schwarzschild metric, and this one is about the FLRW metric.

I think I'm going to repost under a different title then, forgetting about tossed clocks and just asking the question what would happen in a big crunch...

 

[George Jones]

Since geodesics are always the locally longest path

This isn't always true in general relativity.

 

[A.T.]

Your revised question now deals with the interior Schwarzschild metric, and is comparing two different geodesics (both clocks are free falling). So you cannot immediately say which is shorter. There is no way around it other than to integrate the metric along the two geodesics.

I'm bad at calculations. In the https://www.physicsforums.com/showthread.php?t=249722&p=1840160" kev didin't do clock B (free fall, oscillating in a tunnel) and said you might need a certain mass distribution to make clock B meet clock E (free fall orbiting at surface level) again. But B and D (free fall rest at the center) meet in any case. So you could assume interior Schwarzschild metric (which already assumes a certain mass distribution).

The interior Schwarzschild metric, or at least the spatial part of it, is that of a hypersphere. It would make sense intuitively, if two geodesics on it had the same lengths between intersections, like on a normal sphere. I have conflicting statements on this. Maybe some day I will learn how to do these numeric integrations
.

 

[Dale]

This isn't always true in general relativity.

I am pretty sure that it is if we limit ourselves to timelike geodesics. Null geodesics, of course, all have 0 length, and spacelike geodesics are always the locally shortest path.

Note that the word "locally" is very important here!

EDIT: did you mean that there are some exotic geometries where a timelike geodesic is a locally minimal path or a "saddle point"? If so, I would be surprised, but I could understand.

 

[Frame Dragger]

Am I the only person who, in appreciating alternate (UK/European) definitions of "toss" finds this dialogue a bit amusing? Is a tossed clock actually tossed, or is it a clock used to time the tossing?

On a serious note, I'm with DaleSpam in being a bit confused by what you (George Jones) meant. It seems... contradictory unless Dale is on the right track.

 

[George Jones]

On a serious note, I'm with DaleSpam in being a bit confused by what you (George Jones) meant. It seems... contradictory unless Dale is on the right track.

I've always meant to get back to this thread. Okay, I'll try to type up some explicit calculations, and to talk about about what's happenintg.

 

[Frame Dragger]

I've always meant to get back to this thread. Okay, I'll try to type up some explicit calculations, and to talk about about what's happenintg.

Thanks, I don't doubt that you have an interesting point to make. You have at least two of us on tenterhooks. ;)

 

[yuiop]

Thanks, I don't doubt that you have an interesting point to make. You have at least two of us on tenterhooks. ;)

Make that 3

 

[bcrowell]

In both this thread and the earlier one on tossed clocks, I don't see anyone making what seems to me to be the simplest argument. By the equivalence principle, we can consider everything in the tossed clock's frame. Imagining ourselves in this frame, and adopting the coordinates that are natural there, there is zero gravitational field, and the metric is not only flat but also has the usual Minkowski form when written in our coordinates. In this frame, we observe a desk and a clock go zooming by. The desk exerts a force on the clock, causing it to stop and then come back. The whole thing is simply the standard twin paradox (for constant acceleration). The clock being manipulated by the desk reads a shorter elapsed time.

One way in which this description is simpler is that you don't have to worry about whether geodesics are minima or maxima (which actually depends on issues like +--- versus -+++ metrics), whether they're global or local extrema, etc. The metric is purely Minkowskian when written in our coordinates, geodesics are linear equations written in our coordinates, and we just recycle the usual result of the twin paradox.

Viewed in this frame, the example also doesn't require reasoning about two competing effects, one from gravitational time dilation and one from time dilation due to motion.

 

[George Jones]

In both this thread and the earlier one on tossed clocks, I don't see anyone making what seems to me to be the simplest argument. By the equivalence principle, we can consider everything in the tossed clock's frame. Imagining ourselves in this frame, and adopting the coordinates that are natural there, there is zero gravitational field, and the metric is not only flat but also has the usual Minkowski form when written in our coordinates. In this frame, we observe a desk and a clock go zooming by. The desk exerts a force on the clock, causing it to stop and then come back. The whole thing is simply the standard twin paradox (for constant acceleration). The clock being manipulated by the desk reads a shorter elapsed time.

In the quote below, clocks E and B are freely falling and thus are both in local inertial frames, and clock C (sitting on a desk) is accelerated. Doesn't your argument above lead to T_C < T_E and T_C < T_B? But this isn't true, as, actually, T_C > T_E and T_C > T_B. This is the whole point: this type of reasoning breaks down.

In this post, I will summarize the results, and the I will gives an explanation of the results in another post.

Consider a spherical planet of uniform density and five clocks (changing notation slightly):

clock A is thrown straight up from the surface and returns to the surface;
clock B is dropped from rest through a tunnel that goes through the centre of the planet;
Clock C remains on the surface;
clock D remains at the centre of the planet;
clock E orbits the body right at the surface.

Assume that A is thrown at the same time that B is dropped, and that the initial velocity of A is such that A and B arrive simultaneously back at the starting point. The times elapsed on the clocks A, B, and C between when they are all are together at the start and when they are all together at the end satisfy $t_A > t_C > t_B$.

Since A and B are freely falling and C is accelerated, it might be expected that $t_A > t_C$ and $t_B > t_C$, so $t_C > t_B$ seems strange.

Assume that clock E is coincident with clocks A, B, and C when A and B start out. As Fredrik has noted, unless the density of the planet has a specific value, E will not be coincident with with A, B, and C when A and B arrive back, but E will be coincident again with C at some other event. The elapsed times between coincidence events of E and C satisfy $T_C > T_E$. Again, since E is freely falling and C is accelerated, this seems strange.

 

[bcrowell]

In the quote below, clocks E and B are freely falling and thus are both in local inertial frames, and clock C (sitting on a desk) is accelerated. Doesn't your argument above lead to T_C < T_E and T_C < T_B?

No. My argument is based on the equivalence principle, which is local. A and C occupy the same local piece of spacetime. You can't use the equivalence principle in a nonlocal way to compare with B, D, and E. (I assume that E is going to make a complete orbit and then be compared with C again when it gets back -- otherwise it's just like A.)

So I would agree that my argument is not the way to go if you also want to analyze B, D, and E, but it still seems to me like the most direct way to compare A and C.

 

[George Jones]

By the equivalence principle, we can consider everything in the tossed clock's frame. Imagining ourselves in this frame, and adopting the coordinates that are natural there, there is zero gravitational field, and the metric is not only flat but also has the usual Minkowski form when written in our coordinates.

The metric is not flat, because, even with respect to a local inertial coordinate system, the second derivatives of the components of the metric are non-zero, and, consequently, the curvature tensor in non-zero at all points.

No. My argument is based on the equivalence principle, which is local. A and C occupy the same local piece of spacetime. You can't use the equivalence principle in a nonlocal way to compare with B, D, and E. (I assume that E is going to make a complete orbit and then be compared with C again when it gets back -- otherwise it's just like A.)

So I would agree that my argument is not the way to go if you also want to analyze B, D, and E, but it still seems to me like the most direct way to compare A and C.

For a clock that actually is tossed physically, the situation involving A and C is just as non-local as the other situations. In particular, all the situations are non-local because definite integration (over a non-infinitesimal interval for which a single inertial coordinate system is not possible) is required to calculate times in all the situations.

The difference is that the situations involving C and E and C and B involve conjugate points, while the situation involving C and A doesn't involve conjugate points. I hope to say more about this in later posts.

 

[atyy]

The difference is that the situations involving C and E and C and B involve conjugate points, while the situation involving C and A doesn't involve conjugate points. I hope to say more about this in later posts.

So may I guess that your objection to time-like geodesics being "locally longest" is that "length" is intrinsically non-local. Locally, a geodesic has zero acceleration, which only translates into the longest distance between two points if there is one time-like geodesic connecting those two points.

 

[Frame Dragger]

So may I guess that your objection to time-like geodesics being "locally longest" is that "length" is intrinsically non-local. Locally, a geodesic has zero acceleration, which only translates into the longest distance between two points if there is one time-like geodesic connecting those two points.

OK, now that makes sense in the context of George's answer.

 

[bcrowell]

The metric is not flat, because, even with respect to a local inertial coordinate system, the second derivatives of the components of the metric are non-zero, and, consequently, the curvature tensor in non-zero at all points.

Hmm...I don't think so. You can start from a Minkowski metric and apply a coordinate transformation into the frame of an observer with constant proper acceleration, which gives a metric $ds^2=(1+ax)^2dt^2-dx^2$. The curvature tensor is zero at all points, because curvature is intrinsic, so you can't get a nonvanishing curvature by doing a coordinate transformation on a Minkowski spacetime. Its time-time component has a second derivative that's nonzero, but that isn't sufficient to prove that it has a nonvanishing curvature. You get some Christoffel symbols that are nonzero, but the Riemann tensor vanishes.

For a clock that actually is tossed physically, the situation involving A and C is just as non-local as the other situations. In particular, all the situations are non-local because definite integration (over a non-infinitesimal interval for which a single inertial coordinate system is not possible) is required to calculate times in all the situations.

No, I don't think that's right. If you were going to be that strict about interpreting the locality of the e.p., then it would be meaningless, because it would never apply to anything. It's true that this is a subtle point. The reason the e.p. is not a totally well defined mathematical law is that it's very hard to provide a perfectly rigorous definition of how local is local. But I'm convinced that A and C are local enough to qualify, for the following reason. The e.p. does not prevent you from measuring accelerations. That is, the guy in the elevator can't tell whether his elevator is accelerating with acceleration $a$ or whether it's in a gravitational field $a$, but he *can* tell what the numerical value of $a$ is. A typical experiment he would do to determine that would be to release a rock above the floor and time how long it takes to reach the floor. That's a measurement of a certain world-line which is finite, but not so extended that it falls outside the kind of local measurement that the e.p. allows. If you compare it with the observation of the clocks A and C, it should be clear that those observations are no less local than the rock in the elevator; they're simply whole parabolas rather than half parabolas, but a factor of 2 doesn't make the e.p. suddenly not apply. The kind of measurement that the e.p. doesn't apply to is something like measuring tidal forces, e.g., determining whether the rock's acceleration is different near the floor than near the ceiling. Basically you can determine $a$, but you can't determine $a$'s derivatives.

The difference is that the situations involving C and E and C and B involve conjugate points, while the situation involving C and A doesn't involve conjugate points. I hope to say more about this in later posts.

The reason that the e.p. can't be applied to compare, say, C sitting at the surface and E orbiting the planet is that E's world-line extends so far through space that it clearly demonstrates a tidal effect: its acceleration vector differs measurably from point to point along its trajectory. By measuring it, you're measuring an acceleration's derivatives.

 

[atyy]

I thought we were in the Schwarzschild metric?

 

[bcrowell]

In particular, all the situations are non-local because definite integration (over a non-infinitesimal interval for which a single inertial coordinate system is not possible) is required to calculate times in all the situations.

Sorry for the double reply, but I had some further thoughts about this portion of your post.

Okay, there are two elements to this statement: (1) integration over a non-infinitesimal interval, and (2) needing more than a single inertial coordinate system.

I think #2 is the easy one to handle. The reason you need more than one Minkowski frame to cover a certain region of spacetime is that it's curved. This is the same as the issue of not being able to cover the entire curved surface of the Earth with a single map. I'm assuming my #25 will convince you that we can compare A and C in a spacetime with zero intrinsic curvature, so it should be clear that it is not necessary to use more than one Minkowski frame to cover A and C. (It *is* necessary to use more than one Minkowski frame to cover E's motion.)

Re #1, a good way to think about infinitesimals, in GR or elsewhere, is that they're a shorthand for talking about different degrees of smallness. E.g., there's the practice of neglecting the square of an infinitesimal compared to the infinitesimal itself, which has sometimes been derided as non-rigorous, but which can actually be made rigorous using Robinson's non-standard analysis, or smooth infinitesimal analysis. The important point is that there aren't just two categories of sizes, the reals and the infinitesimals. There are infinitesimals like $dx^2$ that are just as small compared to dx as dx is compared to 1. What this is mocking up about physical reality is the distinction between the ability to measure things at different scales. You look at a spacetime at a certain scale, and you say it's a Schwarzschild metric. Zoom in some more, and you may only be able to determine things like the Weyl tensor. Zoom in some more, and it looks indistinguishable from Minkowski space. The Minkowskian picture is "local," and the Weyl tensor is "local," but they're not equally local. One is more local than the other.

A typical way of stating the strong equivalence principle is that local experiments can't detect the curvature of space. To make any sense of this, you have to chose a specific level to call "local." The lab has to be small enough so you can't see tidal effects, but big enough so that you can lay out some apparatus on a lab bench that occupies a certain amount of space, and perform an experiment that takes a certain amount of time. You have to be able to have apparatus like clocks lying on the lab bench. Clocks perform a definite integral of dt. So just because something involves a definite integral, that doesn't mean it's not local for the purposes of the e.p.

 

[bcrowell]

I thought we were in the Schwarzschild metric?

Let's say physicist Alice actually carries out the tossed clock experiment (clocks A and C) on the surface of the earth. She's in a Schwarzschild metric.

Now physicist Bob is onboard a spaceship with a constant proper acceleration of one gee. He carries out the same experiment. His metric is $ds^2=(1+ax)^2dt^2-dx^2$, which is a flat spacetime expressed in funny coordinates.

Clearly Alice and Bob get the same results, and the reason they do is that the e.p. doesn't allow you to distinguish between a Schwarzschild metric and flat spacetime using local experiments.

 

[atyy]

The difference is that the situations involving C and E and C and B involve conjugate points, while the situation involving C and A doesn't involve conjugate points. I hope to say more about this in later posts.

Let's say physicist Alice actually carries out the tossed clock experiment (clocks A and C) on the surface of the earth. She's in a Schwarzschild metric.

Now physicist Bob is onboard a spaceship with a constant proper acceleration of one gee. He carries out the same experiment. His metric is $ds^2=(1+ax)^2dt^2-dx^2$, which is a flat spacetime expressed in funny coordinates.

Clearly Alice and Bob get the same results, and the reason they do is that the e.p. doesn't allow you to distinguish between a Schwarzschild metric and flat spacetime using local experiments.

Hmmm, interesting. So are we safe if we say "local" means no conjugate points (which I think means we only consider points where there is only one time-like geodesic connecting them)? I'm aware that this is not the traditional definition of "local", which usually means at a single point, or no second derivatives at the point.

Edit: Well, it is true that if there is only one time-like geodesic connecting two points, then that geodesic has the longest length. But the condition of a unique geodesic between two points does go beyond the EP, so maybe local in the no conjugate point sense, but not local in the EP sense.

Edit again: I can't make up my mind about EP for clocks on the surface of the earth. On the one hand, it does seem nonlocal. On the other hand, it doesn't different from the red shift experiment, which is usually stated to be a prediction of the EP.

Edit yet again: EP is dangerous

 

[Frame Dragger]

Hmmm, interesting. So are we safe if we say "local" means no conjugate points (which I think means we only consider points where there is only one time-like geodesic connecting them)? I'm aware that this is not the traditional definition of "local", which usually means at a single point, or no second derivatives at the point.

Edit: Well, it is true that if there is only one time-like geodesic connecting two points, then that geodesic has the longest length. But the condition of a unique geodesic between two points does go beyond the EP, so maybe local in the no conjugate point sense, but not local in the EP sense.

Edit again: I can't make up my mind about EP for clocks on the surface of the earth. On the one hand, it does seem nonlocal. On the other hand, it doesn't different from the red shift experiment, which is usually stated to be a prediction of the EP.

Edit yet again: EP is dangerous

I feel your Edit... I mean pain. This to me is all the more reason to improve the terminology across physics so issues like "locality" are split into specific terms of art.

 

[George Jones]

I feel your Edit... I mean pain. This to me is all the more reason to improve the terminology across physics so issues like "locality" are split into specific terms of art.

I'm on my way out with my family, so I can only make a brief comment. To add to the confusion, physicists and mathematicians often mean completely different things by "local"!

 

[Dale]

I just want to make a few comments about my earlier posts in this thread that sparked this current discussion. This is all according to my own limited knowledge:

A geodesic is a locally extremal path. The path itself may be very long, the word "local" in this context refers to the calculus of variations idea of entire functions that differ from one another by an infinitesimal amount. It in no way implies that the domain of those functions need be small. Specifically, this "local" is in no way related to the "local" of the equivalence principle. It is more closely related to the concept of "local" in optimization where a minimum may be a local minimum, but not a global minimum.

In the case of clocks A and B, they are both geodesics, so they are guaranteed to each be local maxima. The paths differ from each other by a finite amount, so there is no contradiction in one being longer than the other. Both are local maxima, but at most one could be the global maximum. If one is the global maximum then no non-geodesic clock can possibly record more proper time between the two events.