Would the one accelerating please stand up?

[David Carroll]

Ok. I'm trying to get this straight in my mind. I am familiar with Einstein's equivalence of acceleration and gravity. But what I am trying to figure out is: who exactly is accelerating? If someone is falling to the Earth and I am looking at his pitiful circumstance, he is clearly - according to my observation and calculations - accumulating distance geometrically...and therefore accelerating. Oh, but wait a minute. He feels weightless. According to him - if he hasn't suffered an infarction already - all the subjective signs of one accelerating are missing: he does not feel any force pressing against him (we'll discount the atmosphere). However, I in my stationary position do feel a force against my feet. This leads me to conclude that I am the one accelerating and not him. After all, when I am in a car, I know when I am accelerating when I am pushed back into my seat.

Furthermore, the time dilation, the Lorentzian contraction, and all that mess...does that apply to people in a free fall toward a gravitational body or only to people that are on the ground (or a platform or what have you) and who are feeling the "acceleration" of the ground toward their feet?

 

[phinds]

Ok. I'm trying to get this straight in my mind. I am familiar with Einstein's equivalence of acceleration and gravity. But what I am trying to figure out is: who exactly is accelerating? If someone is falling to the Earth and I am looking at his pitiful circumstance, he is clearly - according to my observation and calculations - accumulating distance geometrically...and therefore accelerating. Oh, but wait a minute. He feels weightless. According to him - if he hasn't suffered an infarction already - all the subjective signs of one accelerating are missing: he does not feel any force pressing against him (we'll discount the atmosphere). However, I in my stationary position do feel a force against my feet. This leads me to conclude that I am the one accelerating and not him. After all, when I am in a car, I know when I am accelerating when I am pushed back into my seat.

Yep, you got it. Everyone is motionless in their own frame of reference but for you to stay motionless relative to the Earth, the Earth has to push up against you while spacetime curvature wants you to move toward the center of the Earth. The guy in free-fall is happily following spacetime curvature.

Furthermore, the time dilation, the Lorentzian contraction, and all that mess...does that apply to people in a free fall toward a gravitational body or only to people that are on the ground (or a platform or what have you) and who are feeling the "acceleration" of the ground toward their feet?

They apply to ALL objects that are moving when viewed from a specific frame of reference in which they are not at rest, it's just that the effects are SO tiny at human-scale speeds that they might as well not be happening (except where incredible precision is needed, such as in the GPS system). Just to be sure I'm clear, all motion is relative. NOTHING is ever moving in its own frame of reference and so nothing ever "experiences" time dilation, for example. It is something that is observed/calculated by someone in a different frame of reference. You, for example, are MASSIVELY time dilated right now from the frame of reference of an "accelerated" particle at CERN (because it sees itself as motionless and you moving at .999999c)

 

[David Carroll]

Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?

 

[phinds]

Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?

I say again, time dilation is an observation/calculation from a frame of reference in which an object is not moving. It is utterly irrelevant WHY the object is moving in that FoR, it only matters that it is moving. Your guy in free-fall is moving relative to the spaceship (unless it performs acceleration to exactly match, outside the atmosphere, the spacetime path of the guy in free-fall. The space-ship can't free-fall at the same rate at the guy, because it is farther outside the Earth's gravity well.

There are two kinds of time dilation. One is cause by relative motion and the other is caused by gravitational fields. The GPS system, for example, has to compensate for both. The GPS satellites are all moving relative to the surface of the Earth AND they are farther outside the Earth's gravity well than a clock on the surface so they have to compensate for both. If the GPS satellites were geosynchronous (which many people mistakenly believe them to be) the speed issue would not be there.

 

[Dale]

But what I am trying to figure out is: who exactly is accelerating?

There are two separate but related concepts of acceleration. One is called "coordinate acceleration", it is the second time derivative of the position and it depends on the reference frame. The other is called "proper acceleration", it is the acceleration measured by an accelerometer.

In your frame: You have a coordinate acceleration of 0 and a proper acceleration of g upwards. The free fall guy has a coordinate acceleration of g downwards and a proper acceleration of 0.

In the free fall guys frame: You have a coordinate acceleration of g upwards and a proper acceleration of g upwards. He has a coordinate acceleration of 0 and a proper acceleration of 0.

Note that the coordinate acceleration changes depending on which frame you use. Note that the proper acceleration does not change depending on which frame you use.

 

[David Carroll]

So, will I observe the same degree of Lorentz contraction of the falling dude as he does of me? (I'm talking over my own head here.)

 

[Dale]

So, will I observe the same degree of Lorentz contraction of the falling dude as he does of me? (I'm talking over my own head here.)

No. The time dilation formula only applies to inertial frames, which are frames where objects with 0 proper acceleration move in straight lines at constant speed. Your frame and his frame are not equivalent to each other.

 

[collinsmark]

Yeah, I realize about the frame of reference part. What I mean is: if someone flew down from outer space, placed a watch on the falling guy, stole it back during mid-fall and brought it to flying spaghetti monster or what have you and flying spaghetti monster looked at the watch, wouldn't he see that the watch read the same time as his (after correcting for the slight acceleration from and to the earth), whereas if he picked up a clock from the Dead Sea he would see the clock show an earlier time than his own (however slight a difference it would be)? Because when the falling guy was in free-fall, he was not effected by gravitational time/length/mass warps (according to spaghetti monster's frame of reference). Or am I way off?

[Edit: On further consideration, the contents of this post are wrong (more than just the 42 vs 84 corrections; rather the whole thing is wrong). Sorry for the confusion.]

Allow me to replace the scenario with a new one that hopefully demonstrates the important concepts, without the need to bring in the flying spaghetti monster and its delivery person.

Consider a simple, non-rotating Earth. Drill a hole through the Earth from one side to the other. (If you wish, you can also remove the atmosphere from the Earth so it doesn't leak into the hole. If not, just neglect air resistance).

Now drop a clock down the hole. Also keep another clock (initially synchronized) on the surface of the Earth near one of the ends of the hole. After around 42 minutes [Edit: should be 84 minutes, not 42], the clock that you dropped down will come back up the same hole (the clock will oscillate back and forth through the Earth, very similar to a mass on a spring, with a period of roughly 42 minutes [Edit: should be 84 minutes, not 42]). At that moment, you can easily compare the two clocks.

Which clock has ticked slower?

Before you answer, here are a couple of things to note:

• At the moment you dropped the clock, both clocks were synchronized, and existed in the same place (well close enough) at the same time and had zero relative velocity.
• When the clock comes back up through the hole (after about 42 [Edit: should be 84, not 42] minutes or so), once again both clocks exist in the same place and time and have zero relative velocity; although now they are no longer synchronized. (The clocks are easy to compare though.)

The important point with the above notes is that both clocks share a single point in spacetime at the beginning, and they also share another point in spacetime at the end.

So which path through spacetime extremalizes proper time, and which does not?

Be careful here. A little knowledge about relativity can actually lead you to the wrong answer. Many would follow the following logic and reach the wrong conclusion:

• Well, the clock falling through the Earth is moving back and forth, and thus, like the standard twin paradox, it would tick slower, right?
• Also, the falling clock goes deeper into Earth's gravity well, so that's another reason why the falling clock should tick slower, due to gravitational time dilation, right?

With that, someone might conclude that the clock falling through Earth would tick slower, for multiple reasons. But that someone would be wrong.

The correct answer is that the falling clock will, overall, tick faster! The clock sitting there at rest by the end of the hole will experience more time dilation than the falling clock.

There is a little truth to the logic about the gravity-well deepness part, but the comparison to the standard twin paradox is very wrong here.

The falling clock is in freefall, meaning it is tracing a path through spacetime on a geodesic. And that causes its proper time to be extremal. And it will always tick faster than any other clock that went though spacetime taking a different path. (Recall that both clocks share the same beginning and ending points in spacetime.)

 

[David Carroll]

Thank you, collinsmark. That was very helpful, enlightening,...and somewhat humorous. What you said is basically what I thought was the case (though my bumbling language above might have indicated otherwise): That someone accelerating, or as Dale Spam named it, accelerating in the proper way, would have a slower clock than someone in free-fall.

 

[collinsmark]

Thank you, collinsmark. That was very helpful, enlightening,...and somewhat humorous. What you said is basically what I thought was the case (though my bumbling language above might have indicated otherwise): That someone accelerating, or as Dale Spam named it, accelerating in the proper way, would have a slower clock than someone in free-fall.

Yes, you are correct. That is always the case.

If lazy Bob synchronizes his clock with Alice when the floating freely in space (together, at the same place and time, and same relative velocities), and then Alice goes off and explores the galaxy, leaving Bob effortlessly floating the whole while (Bob never has any proper acceleration), then later comes back to Bob, such that they again share the same space, time and velocity, Bob will always age more than Alice. It doesn't matter what Alice does -- travels the various stars, rests on the surface of heavy planets -- anything, she will never age more than Bob during her exciting trip.

[Edit: this not necessarily true. There are situations where it is decidedly false. Sorry for the confusion. ]

To be fair though, in order to compare Alice's and Bob's aging, they need to be in the same place at the same time at the time of comparison. Making the comparison half way through Alice's trip doesn't make sense.

 

[collinsmark]

[Once again I'm temporarily deleting this post until I can double check something about a claim I made.]

 

[collinsmark]

I'll leave my last post deleted. I had made a claim about very low-Earth-Orbit satellite clocks (lower than GPS -- so low they're just skimming the Earth's surface) still ticking faster.

My claim was incorrect. Although the effects of being in freefall would tend to speed up the orbiting clocks (relative to a terrestrial clock), it's not enough to compensate for the ticking slowdown due to the changing of reference frames during the initial acceleration.

My mistake was that even though the satellite and terrestrial clock share the same points in spacetime at the comparison events, they still do not share the same frame of reference (their instantaneous, relative velocities are different). And that's enough to keep the very low orbit satellite clocks ticking slower.

(The hole through the Earth scenario is still correct though, as-is.)

 

[David Carroll]

Is free-fall measurably continuous? Can something be in "more of a free-fall" than another thing? Or is it a go/no-go sort of thing?

 

[collinsmark]

Is free-fall measurably continuous? Can something be in "more of a free-fall" than another thing? Or is it a go/no-go sort of thing?

Interesting question. :)

One thing you can say for sure: Check your accelerometer. If your accelerometer reads 0, you are in freefall.

But what happens if you experience a little air resistance when falling out of a tree? You're still accelerating toward Earth. But your proper acceleration is not quite 0, but it's almost 0. Is that "almost freefall"?

I won't comment on the terminology, but I can say that the effects of time dilation are continuous, in this respect.

 

[PeterDonis]

The correct answer is that the falling clock will, overall, tick faster!

Are you sure? I strongly suggest checking your intuition here with math.

The falling clock is in freefall, meaning it is tracing a path through spacetime on a geodesic. And that causes its proper time to be extremal.

In flat spacetime, you would be correct without qualification. In curved spacetime, however, there are complications. A free-falling worldline is always locally extremal; that is, it always has more elapsed proper time between two given events than non-free-falling worldlines that are "close" to the free-falling one. But the worldline of the person who stays at one end of the hole is not "close" to the free-falling one in this sense.

Globally, however, a given free-falling worldline is not necessarily the one with the absolutely longest proper time between two given events; there may be other worldlines, not free-falling, which have more elapsed proper time. You have, I believe, (apparently unwittingly) given one example of such a case.

It is true that, given a pair of events in curved spacetime, the worldline between them which does have the globally maximal proper time will be a freely falling worldline. But the statement that any free-falling worldline must always have more elapsed proper time than any non-free-falling worldline between the same two events, is much stronger than that (and is false).

[Edit: I see you've retreated from that very strong version, since you agree that the free-falling clock in the "skimming the surface" orbit has less elapsed proper time than the clock at rest on the surface. So I've edited my statement above to no longer attribute that strong version to you; but I think it's still worth stressing that it's false.]

(The hole through the Earth scenario is still correct though, as-is.)

Have you checked the math? If so, can you show it?

 

[Dale]

An example which is closely related to the hole-in-the-earth scenario would be the "thrown clock" scenario.

If you have two clocks, one at rest on the table, and one thrown briefly upwards so that it passes by the table clock twice, then the thrown clock will accumulate more time. This is basically just a twins scenario with the thrown clock being like the stay-at-home twin.

Here "briefly" means that the time and distance are small enough that tidal effects are negligible.

 

[PeterDonis]

An example which is closely related to the hole-in-the-earth scenario would be the "thrown clock" scenario.

"Closely related" in the sense that it illustrates different proper times for different worldlines, yes. But in this scenario, as you say, the free-falling clock (the thrown one) has greater elapsed time. (In fact, the thrown clock's worldline is the globally maximal one among all worldlines between the same start and end events.)

However, if we set up the scenario so that the thrown clock goes up and comes back down in 84 minutes, the same time it takes for the clock in the hole to come back and for the clock in the "skimming the surface" orbit to come back, the proper time elapsed on the clock in the hole will not be the same as the proper time elapsed on the clock thrown upwards; it will be less. If I've done my calculations correctly, the ordering of proper times will be:

clock in hole < clock in orbit < clock at rest on surface < clock thrown upwards

 

[collinsmark]

Have you checked the math? If so, can you show it?

Here is a link to an pdf document, compete with the mathematics, that explains the principle of extremal aging, that (as far as I can tell) does a better job than I could do (particularly in short order):

http://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=0CCAQFjAA&url=http://ocw.mit.edu/courses/physics/8-224-exploring-black-holes-general-relativity-astrophysics-spring-2003/assignments/gravitationalforcesnotes3a.pdf&ei=x5TVVJftL4irogTz9oHwCw&usg=AFQjCNG0njnplKtWXXZcKeHwTNNOA_S-ew&sig2=lR5FIGqQk-iflesM0G8w6Q&bvm=bv.85464276,d.cGU

Here is a relevant quote:

"There is no way to distinguish between the effects on an observer of a uniform gravitational field and of constant acceleration.

"The meaning of this statement is shown by a famous thought experiment proposed by Einstein. Consider a person in an elevator whose cable has broken and is accelerating at 9.8 ms^-2 toward the center of the earth. The person is weightless and floats in the middle of the elevator with no acceleration relative to the elevator. (This follows from the Newtonian equivalence of gravitational and inertial mass discussed in the notes Coordinates and Proper Time -- recall that all bodies fall the same way in a gravitational field.) Locally, that is over sufficiently small intervals of time and space, there is no way for the person in the elevator to determine whether she is in an elevator falling in a gravitational field or floating freely in an unaccelerated spaceship far from all sources of gravity.

"We know already that, in the absence of gravity, unaccelerated bodies take the path that maximizes proper time. Arguing from the Equivalence Principle, Einstein proposed that the same "extremal aging" principle holds in curved spacetime. Locally there is no difference between free-fall and unaccelerated motion. Locally, bodies must therefore follow paths that maximize proper time. If a long path is divided into many short segments, each of which maximizes proper time, then the total path maximizes proper time. The effects of gravity must be represented not by local acceleration but rather by global curvature."
​

This applies to the "hole-in-the-Earth" scenario, by the way.

The "thrown clock" scenario is different, and I don't think it necessarily applies. The thrown clock undergoes acceleration before it's thrown and after it lands. That needs to be considered before comparing it to a clock just sitting on the surface. [Edit: Oh, wait. I didn't read that "passing the tabletop twice" part. It still might not apply though since the two clocks have different velocities at the comparison events.]

On the other hand, the "hole-in-the Earth" scenario presents the principle of extremal aging because

• The falling clock is always in a state of freefall -- from beginning to end
• At the comparison events, not only do the two clocks share the same points in spacetime, but also their relative velocity is zero.

 

[PeterDonis]

This applies to the "hole-in-the-Earth" scenario, by the way.

Once again, I strongly suggest that you actually do the math. You will find that the principle of extremal aging doesn't mean quite what you appear to think it does. The word "locally" is very important in the quote you gave: a free-falling worldline has locally maximal proper time, but as I said in an earlier post, that does not necessarily mean it has globally maximal proper time. But the way you are applying the principle to the hole in the Earth scenario assumes that the proper time must be globally maximal, not just locally maximal.

It still might not apply though since the two clocks have different velocities at the comparison events.

That doesn't matter. The clock at rest on the surface and the clock in the "skimming the surface" free-fall orbit also have different velocities at the comparison events; but you can still compare their proper times between successive events where they meet. Comparison of proper times does not require that relative velocities are zero at the start and end events. It only requires that those events are common to both worldlines.

 

[PeterDonis]

an pdf document, compete with the mathematics

I forgot to comment on this. The math given in that document only applies in vacuum; it does not apply in the interior of the Earth, because the interior of the Earth is not vacuum and the metric there is not the metric given in the pdf. The general principle as given in equation (2) of the pdf is correct, but to apply it to the worldline of the clock falling through the hole in the Earth, you have to know the metric in the Earth's interior (which is not the one given later on in the pdf).

However, you don't have to know the full metric in the Earth's interior to prove the statement I made in a previous post about the ordering of the proper times. You only have to know certain general features of it, as compared to the metric in the vacuum region outside the Earth.

(Note that even in vacuum, there can still be free-falling worldlines between two events which are not globally maximal; the worldline of the clock in the "skimming the surface" orbit is an example.)

 

[Orodruin]

clock in hole < clock in orbit < clock at rest on surface < clock thrown upwards

I checked the clock in hole vs rest on surface and I agree with this. The clock in the hole has the least elapsed time.

 

[PeterDonis]

I checked the clock in hole vs rest on surface

Have you tried to calculate the clock in the hole vs. the clock in the "skimming the surface" orbit? That to me is the most interesting comparison. ;)

 

[collinsmark]

Once again, I strongly suggest that you actually do the math. You will find that the principle of extremal aging doesn't mean quite what you appear to think it does.

Yes, you are right. I realize my error now.

Sorry for the confusion.

 

[collinsmark]

Again, I'm sorry about my hole-through-the-Earth giving the falling clock maximal proper time claim. That claim is wrong. I'm not sure where I originally got that idea. But it's true that I was misunderstanding the principle of extremal aging. Thank you PeterDonis for pointing me in the right direction.

I had forgotten that undergoing/changing one's proper acceleration doesn't just change one's path through a curved, vast, spacetime landscape; it can change the whole landscape (so to speak -- in certain qualitative and quantitative ways as observed by the traveler). And sometimes that can increase the overall proper time between two events (beyond what was previously the "maximum" according to previous observations of the then-"landscape").

Well, the good news is that I learned something exciting and useful today. :)

 

[Dale]

This applies to the "hole-in-the-Earth" scenario, by the way.

The "thrown clock" scenario is different, and I don't think it necessarily applies.

You have the above backwards. It applies to the thrown clock scenario, but not the hole through the Earth scenario. The reason is that the thrown clock scenario was specifically constructed to satisfy the "local" requirement.

The hole in the Earth scenario was not and tidal effects cannot be ignored. There is no accelerating elevator in flat spacetime where a dropped clock will fall down and then fall back up.

Edit: oops, never mind I see you already figured that out.

 

[Orodruin]

Have you tried to calculate the clock in the hole vs. the clock in the "skimming the surface" orbit? That to me is the most interesting comparison. ;)

Did now, interesting indeed.

 

[PeterDonis]

It applies to the thrown clock scenario,

Just to clarify, the local requirement applies to your version of the thrown clock scenario. My version (where the clock stays in flight for 84 minutes) violates the "local" requirement.

 

[PeterDonis]

it can change the whole landscape (so to speak -- in certain qualitative and quantitative ways as observed by the traveler). And sometimes that can increase the overall proper time between two events (beyond what was previously the "maximum" according to previous observations of the then-"landscape").

I'm not sure I understand what you mean by this. Changing your own proper acceleration does not change the geometry of spacetime. It only changes your particular worldline. So on the most obvious interpretation I can find of your statement above (that changing the "landscape" means changing the geometry of spacetime), it is false. If you meant something else, I think you need to clarify what you meant. (There are also issues with constructing the "reference frame" of an accelerated observer, even in flat spacetime, which I suspect you may not be fully taking into account.)

 

[Dale]

Just to clarify, the local requirement applies to your version of the thrown clock scenario. My version (where the clock stays in flight for 84 minutes) violates the "local" requirement.

Yes, I agree. I was referring to my version.

 

[PeterDonis]

Did now, interesting indeed.

Did you find that, for the idealized case of a constant density spherical mass, the two proper times are actually the same?

 

[Dale]

The geometric concept is this: in Euclidean space there is a single straight line which connects two points and it is the shortest path connecting those two points.

In curved space there can be multiple straight lines (geodesics) which connect two points, and they may be different lengths. Consider a sphere, if you pick two nearby points you can go "straight there" as you usually think of it, but you can also walk straight in the exact opposite direction all the way around the globe to get there.

 

[collinsmark]

I'm not sure I understand what you mean by this. Changing your own proper acceleration does not change the geometry of spacetime. It only changes your particular worldline. So on the most obvious interpretation I can find of your statement above (that changing the "landscape" means changing the geometry of spacetime), it is false.

Right; perhaps it was a poor choice of words on my part. Of course spacetime itself is not significantly altered by a single observer's acceleration, but it can alter the observer's observed spatial components and time components separating distinct spacetime events. Given sufficient timelike separation of two spacetime events, and given sufficient curvature between and around the events, there might exist worldlines that ultimately result in greater proper time than a purely freefall worldline between the same events (assuming such a purely freefall worldline exists).

 

[PeterDonis]

it can alter the observer's observed spatial components and time components separating distinct spacetime events

The word "observed" is somewhat misleading here; the "spacial and time components" are coordinates, not observables. In flat spacetime, there is no unique way to construct coordinates for an accelerated observer, but there is, at least arguably, a unique way to do it for a free-falling observer, namely, the standard Minkowski chart. So the temptation is there to view the space and time coordinates in this chart as being "directly observable".

However, in curved spacetime, there is no longer a unique way to construct coordinates even for a free-falling observer, if they are to cover more than a small patch of spacetime. (In a small enough patch of spacetime, you can always construct local inertial coordinates in which a given free-falling observer is at rest.) So there are no unique "space and time components" for a given observer between events outside a single local inertial frame, even if that observer is free-falling. This is an important reason why the geometric viewpoint--viewing everything in terms of geometric invariants instead of coordinates--is desirable in general relativity.

Given the above, viewing proper acceleration as "altering the space and time components" is not a good way to view what's going on; those components aren't uniquely determined even without the acceleration. I'm also not sure that it's a good idea to single out proper acceleration as what "changes your view of spacetime" so that worldlines with more proper time than a given free-fall worldline are possible. See further comments below.

Given sufficient timelike separation of two spacetime events, and given sufficient curvature between and around the events, there might exist worldlines that ultimately result in greater proper time than a purely freefall worldline between the same events (assuming such a purely freefall worldline exists).

I'm not sure this is the best way to state what spacetime curvature can do. Here's how I would state it: in curved spacetime, there can be multiple free-falling worldlines between the same two events (given sufficient timelike separation between the events and sufficient curvature between and around the events). These different free-falling worldlines can have different lengths (proper times); at least one of these must be the globally maximal length for all worldlines between those two events, but other lengths can be shorter.

Adding accelerated worldlines to the mix does not change anything about the above; it just draws attention to the fact that, if some of the multiple free-falling worldlines between two given events have lengths shorter than the globally maximal length, there will, in general, be accelerated worldlines between those two events that are longer than some of the free-falling worldlines between those events. But that will only be the case if there are multiple free-falling worldlines between those events, and if some of them are shorter than the globally maximal length. So the effects of spacetime curvature can be described entirely in terms of free-falling worldlines; you don't need to consider accelerated worldlines.

In the example under discussion, there are three free-falling worldlines between the same pair of events: the worldline of the clock falling through the hole, the worldline of the clock in the "skimming the surface" orbit, and the worldline of the clock thrown upward that comes back after exactly one orbit of the other clocks (i.e., after 84 min). The last of these is the one whose length is globally maximal (for all worldlines between that pair of events). The others are shorter. (The fourth worldline, that of the clock at rest at one end of the hole, is accelerated, and is longer than the two free-falling worldlines whose lengths are not globally maximal.)

 

[Orodruin]

Did you find that, for the idealized case of a constant density spherical mass, the two proper times are actually the same?

To first order in the gravitational potential, yes. I did not have time to pursue it further than that.

I'm not sure this is the best way to state what spacetime curvature can do. Here's how I would state it: in curved spacetime, there can be multiple free-falling worldlines between the same two events (given sufficient timelike separation between the events and sufficient curvature between and around the events). These different free-falling worldlines can have different lengths (proper times); at least one of these must be the globally maximal length for all worldlines between those two events, but other lengths can be shorter.

Perhaps a comparison to a Riemannian case can help a bit here. If one looks at a sphere, the shortest distance between two points which are not antipodal is along a great circle. However, if you follow the same great circle in the other direction, you will still be following a geodesic curve, but it will not be a global minimum of the length between the points (it is a saddle point).

 

[PeterDonis]

To first order in the gravitational potential, yes.

I wonder about that. The integral I get for the clock falling through the hole is

$$
\tau = \int_{0}^{T} \sqrt{1 - \frac{3M}{R} - \frac{M}{R} \left[ \sin^2 ( 2 \pi t / T ) - \cos^2 ( 2 \pi t / T ) \right]} dt
$$

The average value of the expression in brackets above is zero over the interval of integration. I thought I remembered some theorem or other that says the contribution of that term to the integral vanishes in that case (which would mean we wouldn't have to expand out the square root to show that $\tau$ is the same here as for the clock in the circular orbit), but I haven't been able to find a statement of the theorem.

 

[Gary Feierbach]

A practical problem is 2 rocket probes, clocks synchronized, take off from high Earth orbit. One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

 

[PeterDonis]

One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

I strongly doubt this scenario is even possible in our solar system (at least not without one or both probes having to accelerate for a large part of their trip, not just at the start and end).

However, if we assume that we're in some solar system with planets arranged just right to make it possible, the general answer is that you would have to do the detailed computation; there's no simple rule of thumb that will give you the right answer for a scenario like this.

 

[Gary Feierbach]

We've been going to Mars this way for quite some time. The exception being ion rockets and mid course corrections of conventional rockets. Of course you have to decellerate when you get to Mars unless it is okay to crash into Mars. let's assume both probes crash into Mars but both have hardened blue boxes containing the clocks. The point of this is to determine if acceleration due to gravity alters the clock relative to the direct probe. I'm also assuming that the planets are aligned in such a way that this is possible.

 

[PeterDonis]

We've been going to Mars this way for quite some time.

I didn't mean that the general idea of sending spaceships to other planets this way is impossible; I meant that the specific scenario you describe, where spaceships traveling on the two routes you describe leave Earth orbit at the same time and arrive at Mars at the same time, is most likely impossible.

The point of this is to determine if acceleration due to gravity alters the clock relative to the direct probe.

Attributing any difference in elapsed time (on the assumption that the two trips take the same amount of Earth time) to "acceleration due to gravity" during the Venus flyby is much too simplistic. If elapsed proper time along the two routes is different, it's not any particular point in either journey that makes the difference; it's the entire journey, as compared with the other.

I'm also assuming that the planets are aligned in such a way that this is possible.

And that's the assumption that I think is extremely unlikely in our actual solar system; I don't think there is any possible alignment of the three planets, given their actual orbits, that would make the scenario possible.

 

[Gary Feierbach]

Well I can visualize it but can't prove it without a NASA solar navigation model.

 

[A.T.]

Have you tried to calculate the clock in the hole vs. the clock in the "skimming the surface" orbit? That to me is the most interesting comparison.

Also interesting: Clock oscillating within the hole vs. clock at the center of the sphere.

Old thread on this:
https://www.physicsforums.com/threads/tossing-a-clock.249722

 

[Dale]

A practical problem is 2 rocket probes, clocks synchronized, take off from high Earth orbit. One will loop venus for a gravitational velocity assist to go to Mars while the other takes a more direct approach to Mars with the same initial acceleration. They both arrive at Mars at the same Earth time. How do their internal clocks differ.

I strongly doubt this scenario is even possible in our solar system (at least not without one or both probes having to accelerate for a large part of their trip, not just at the start and end).

I agree that, as stated, it is probably not possible. However, if you relax the "same initial acceleration" part and just have them launch from the same event at different initial velocities, then I think it would be possible. You would calculate the time required for the gravitational assist path for the first rocket and then choose the initial velocity of the other rocket to arrive at Mars at that time. I am quite certain that the initial accelerations will be different, but for any arrival time there should be some initial velocity which will get you there.

 

[PeterDonis]

if you relax the "same initial acceleration" part and just have them launch from the same event at different initial velocities, then I think it would be possible. You would calculate the time required for the gravitational assist path for the first rocket and then choose the initial velocity of the other rocket to arrive at Mars at that time.

Hm, yes, you're right, if the initial velocities are unconstrained (except for the obvious speed of light limit), then there should always be some geodesic on the direct path that matches the endpoints of the gravity assist geodesic.

 

[PeterDonis]

Also interesting: Clock oscillating within the hole vs. clock at the center of the sphere.

Yes, that case is interesting too. Since I need to correct the equations I posted previously anyway , I'll go ahead and post one for that case too.

First, the general expression for $g_{tt}$ in the interior of a spherically symmetric mass of constant density is:

$$
g_{tt} = \left( \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2M r^2}{R^3}} \right)^2
$$

where $R$ is the radius at the surface and $r$ is the radial coordinate. I had misremembered this before, which led to the incorrect formula I gave in post #35 for the clock falling through the hole (though, as we'll see, that formula would still lead to the same approximation for small $M / R$). The general formula for time dilation (relative to a clock at rest at infinity) is

$$
\tau = \int \sqrt{g_{tt} dt^2 - g_{rr} dr^2 - g_{\phi \phi} d\phi^2} = \int \sqrt{g_{tt} - g_{rr} v_r^2 - r^2 \omega^2} dt
$$

where I have used the fact that $g_{\phi \phi} = r^2$ in the "equatorial plane" ($\theta = \pi / 2$) of Schwarzschild coordinates, and $\omega$ is the angular velocity (again, relative to an observer at rest at infinity).

Now we can specialize these expressions for each particular case: clocks A (clock tossed up and falling back), B (at rest on the surface), C (in circular orbit skimming the surface), D (falling through the hole), and E (at rest at the center of the Earth). The simplest is clock B, which gives:

$$
\frac{d\tau_B}{dt} = \sqrt{1 - \frac{2M}{R}} \approx 1 - \frac{M}{R}
$$

where for each case the last expression will be what is obtained to first order in $M / R$ in the approxmation that $M / R << 1$.

The next simplest is clock C. We know that $r^2 \omega^2 = M / R$ for this clock, and $g_{tt}$ is the same as for the clock at rest on the surface, so we have

$$
\frac{d\tau_C}{dt} = \sqrt{1 - \frac{3M}{R}} \approx 1 - \frac{3}{2} \frac{M}{R}
$$

Clock A is a bit more complicated because both $g_{tt}$ and $g_{rr}$ vary over its trajectory. We have:

$$
\frac{d\tau_A}{dt} = \sqrt{1 - \frac{2M}{r} - \frac{v_r^2}{1 - 2M / r}} \approx 1 - \frac{M}{r} - \frac{1}{2} v_r^2
$$

where the factor $1/ \left( 1 - 2M / r \right)$ in the last term drops out at this order.

Clock E is next, since it is simpler than clock D because it's at rest:

$$
\frac{d\tau_E}{dt} = \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \approx 1 - \frac{3}{2} \frac{M}{R}
$$

Finally, we have clock D:

$$
\frac{d\tau_D}{dt} = \sqrt{ \left( \frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} \sqrt{1 - \frac{2Mr^2}{R^3}} \right)^2 - \frac{v_r^2}{1 - 2Mr^2 / R^3} } \approx 1 - \frac{3}{2} \frac{M}{R} + \frac{1}{2} \frac{M r^2}{R^3} - \frac{1}{2} v_r^2
$$
$$\approx 1 - \frac{3}{2} \frac{M}{R} - \frac{1}{2} \frac{M}{R} \left[ \sin^2 \left( 2 \pi t / T \right) - \cos^2 \left( 2 \pi t / T \right) \right]
$$

where the second approximate expression is obtained by plugging in the sinusoidal solution for $r$ and $v$. As I noted above, this last expression is the same as what would be obtained from the formula I gave in post #35 by expanding the square root to first order in $M / R$.

So to first order in $M / R$, we have $\tau_C < \tau_B < \tau_A$, as expected; but we also have $\tau_E = \tau_C$, and if we assume that the sinusoidal contributions drop out of the integral for $\tau_D$ since they average to zero over the range of integration, we have $\tau_E = \tau_D = \tau_C$. The differences between these three are in the next order; I'll defer comment on that to a separate post.

 

[PeterDonis]

Ok, so looking at the formulas for $\tau_C$, $\tau_D$, and $\tau_E$ from my previous posts, the exact relationship between them is interesting.

It is evident that, contrary to intuition (at least mine), $\tau_E > \tau_C$, i.e., the clock at rest at the center of the spherical mass runs faster than the clock in the circular orbit skimming the surface. This is easy to show; we just have to verify that

$$
\frac{3}{2} \sqrt{1 - \frac{2M}{R}} - \frac{1}{2} > \sqrt{1 - \frac{3M}{R}}
$$

which can be done by noting that both expressions go to 1 as $M / R \rightarrow 0$, and showing that the derivative with respect to $M / R$ of the left-hand expression is always less negative than the derivative with respect to $M / R$ of the right-hand expression. This works out to

$$
- \frac{3}{2} \frac{1}{\sqrt{1 - 2M / R}} > - \frac{3}{2} \frac{1}{\sqrt{1 - 3M / R}}
$$

which confirms that, as $M / R$ gets larger (i.e., as the object gets more compact), $\tau_E$ decreases more slowly than $\tau_C$, so we will always have $\tau_E > \tau_C$.

The formula for $\tau_D$ is more complicated, but we can gain insight by looking at its behavior at some useful "test" values: $R = 3M$ and $R = 6M$. I will also define the dimensionless radial coordinate $\rho = r / R$ to simplify the formulas.

For $R = 6M$, we have $\tau_C = 1 / \sqrt{2} \approx 0.7071$ and $\tau_E = \left( \sqrt{6} - 1 \right)/2 \approx 0.7247$. The formula for $\tau_D$ works out to

$$
\tau_D = \sqrt{\frac{7}{4} - \sqrt{\frac{3 - \rho^2}{2}} - \frac{1}{12} \rho^2 - \frac{3 v^2}{3 - \rho^2}}
$$

This is easy to evaluate at the endpoints, $\rho = 1$ (where $v = 0$) and $\rho = 0$ (where $v$ has its maximum value, which we'll leave as $v$ in the formula). For $\rho = 1$, we end up with $\tau_D = \sqrt{2/3} \approx 0.8165$ (note that this is equal to $\tau_B$, the clock rate for the observer at rest on the surface at this radius), and for $\rho = 0$, we end up with $\tau_D = \sqrt{7/4 - \sqrt{3/2} - v^2} \approx \sqrt{0.5253 - v^2}$. If we assume that the maximum value of $v$ is approximately the same as the circular orbit velocity for $R = 6M$, which is $1/2$, then for $\rho = 0$ we obtain $\tau_D \approx 0.5246$. This strongly suggests that a full numerical integration would show $\tau_D < \tau_C < \tau_E$ (but see further comments below on the possible behavior of the maximum value of $v$).

For $R = 3M$, it is even more interesting. We have $\tau_C = 0$ at this radius, corresponding to the fact that this radius is the photon sphere. We have $\tau_E = \left( \sqrt{3} - 1 \right) / 2 \approx 0.3660$. The formula for $\tau_D$ becomes

$$
\tau_D = \sqrt{1 - \frac{1}{2} \sqrt{3 - 2 \rho^2} - \frac{1}{6} \rho^2 - \frac{3 v^2}{3 - 2 \rho^2}}
$$

This gives $\tau_D = \sqrt{1/3} \approx 0.5774$ for $\rho = 1$ (again, the same as $\tau_B$ for this radius). For $\rho = 0$ it gives $\tau_D = \sqrt{1 - \sqrt{3} / 2 - v^2} \approx \sqrt{0.1340 - v^2}$. If $v \rightarrow 1$ for an object this compact, this obviously gives a negative quantity under the square root; but even if $v$ is anything larger than $\sqrt{1 - \sqrt{3} / 2} \approx 0.3660$, the quantity under the square root will be negative (since $v^2$ is subtracted there). This indicates that, for an object this compact, either the "dropping the clock through the hole" experiment can't actually be done (unlikely since the value of $\tau_E$ shows us that a clock can sit at rest at the center of the object perfectly well), or the maximum velocity achieved in such an experiment, for objects this compact, is much less than the corresponding circular orbit velocity.

If the latter is the case, then we have to re-evaluate the $R = 6M$ case. If the maximum velocity achieved for that case is anything less than $\sqrt{5/4 - \sqrt{3/2}} \approx 0.1589$, then the minimum value of $\tau_D$ (at $\rho = 0$) will be greater than $\tau_C$. Even a somewhat larger maximum velocity will still give an average value of $\tau_D$ greater than $\tau_C$. This would be nice since we know we must have $\tau_C < \tau_D$ for the $R = 3M$ case (since $\tau_C = 0$ for this case), and we would expect the same inequality between the magnitudes to hold for all values of $M / R$ (at least, it seems to me that that ought to be the case).

That still leaves open the question of the relationship between $\tau_D$ and $\tau_E$. Given the values above, I am inclined to think that $\tau_D > \tau_E$ holds; however, checking this would involve doing the full numerical integration, which is more than I want to tackle.

So, bottom line, it looks to me like the full set of inequalities is:

$$
\tau_C < \tau_E < \tau_D < \tau_B < \tau_A
$$

But numerical integration would be required to confirm whether $\tau_D$ belongs where it is, or between $\tau_C$ and $\tau_E$ (and to confirm that the same inequalities hold for all allowed values of $M / R$).

 

[MikeGomez]

... There is no accelerating elevator in flat spacetime where a dropped clock will fall down and then fall back up.

Are you sure?

 

[A.T.]

There is no accelerating elevator in flat spacetime where a dropped clock will fall down and then fall back up.

Are you sure?

I think he means a linearly uniformly accelerating elevator.

 

[MikeGomez]

I think he means a linearly uniformly accelerating elevator.

I see. My thinking was that for the "hole in the earth" scenario, there might be an equivalent situation with a clock in free-fall relative to non-linearly accelerating elevator. Of course the "elevator" wouldn't look much like your everyday elevator.

[Edit] Changed "non-uniformly" to "non-linearly"

 

[Dale]

Are you sure?

Yes, I am sure, but I guess I wasn't clear. I meant an elevator with constant proper acceleration.

This was intended to show that the equivalence principle cannot be applied here. By the equivalence principle an elevator at rest in a gravitational field is locally equivalent to one with constant proper acceleration in flat spacetime.

There is no "equivalence principle" elevator for the hole-in-the-earth scenario.

 

[MikeGomez]

Yes, I am sure, but I guess I wasn't clear. I meant an elevator with constant proper acceleration.

This was intended to show that the equivalence principle cannot be applied here. By the equivalence principle an elevator at rest in a gravitational field is locally equivalent to one with constant proper acceleration in flat spacetime.

There is no "equivalence principle" elevator for the hole-in-the-earth scenario.

Ok. I think you are taking the equivalence principle to mean strictly a linearly accelerating elevator and the static gravitational field at the surface of the earth. I am thinking of it in a more general sense, in that any gravitational effect has an equivalent inertial effect, not just the comparison with gravitation at the surface of the Earth and an elevator accelerating at 1g.

To me it's just semantics to say that there is no "equivalence principle" elevator for the hole-in-the-earth scenario, and therefore the equivalence principle does not apply. The clock falling in the hole and then falling back up has a equivalent inertial counterpart.