Does an event horizon ever exist?

[Jonathan Scott]

However, this method breaks down at the horizon, because at the horizon, that symmetry of the spacetime is no longer a time translation symmetry.

I don't understand why what happens "at the horizon" is relevant to the problem.

At any finite distance from the horizon, in a static metric relative to the source, the normal relativity method of assigning simultaneous times by the half-way point of a light-speed signal gives an unambiguous result which matches the coordinate time. It is symmetrical and there seems to be no reason to assume that it would be biased in any special way. This time coordinate however goes to infinity as the object gets closer to the horizon.

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

It is also theoretically possible (with an appropriate hypothetical method of propulsion such as a relativistic bungee cord) for the falling object to reverse course and return along a path with similar velocity to its original path, in which case it can return to its starting point, and that too can follow a symmetric path and hence show that the static time coordinate is a physically valid way to describe its path from the static point of view.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

This seems to violate standard physical assumptions about sequence and causality so strongly that I can't see how the maths could be considered to "prove" anything.

As far as I'm concerned, it seems perfectly physical reasonable for the maths to show a continuous solution from the point of view of the falling observer, but for it to be impossible for that solution to be reached because the falling observer's time rate relative to the normal universe approaches zero, as in my "stasis box" example. I consider that any statement that the crossing "actually happens" is not necessarily disproved by this model, but rather that such an extraordinary idea requires extraordinary evidence, and the continuity of the maths doesn't constitute such evidence.

 

[Nugatory]

If I drop a stone from my hand to the ground, I can calculate within some small error margin, what the time on my clock will be when the stone hits the ground.

No, actually, you cannot. What you know is the time on your watch when the light from the stone hitting the ground reaches your eyes. You can then subtract the light travel time between the ground and your eyes and say that that is the time on your watch when the stone hit the ground: "My wristwatch read exactly 12:00 noon when the stone hit the ground; I know this because the light hit my eyes at 6 nanoseconds past noon and my eyes are 180 centimeters off the ground". But notice that you've slipped a claim about simultaneity in there: you're saying that the stone hitting the ground and the watch reading 12:00 noon are simultaneous. An observer moving relative to you will not see these two events as simultaneous; so you have not actually calculated "when" the stone hits the ground in any absolute sense.

I agree that you what have calculated is when the stone hits the ground using the only definition of simultaneity that makes any sense to you: If at time T you observe something happening at distance X away from you, you say that it happened at time T-(X/c). It would be totally perverse to use any other definition of simultaneity. So...

What's so special about the event horizon?

The method I described above, the one for which there is no non-perverse alternative, doesn't work with an event horizon.

 

[Passionflower]

At any finite distance from the horizon, in a static metric relative to the source, the normal relativity method of assigning simultaneous times by the half-way point of a light-speed signal gives an unambiguous result which matches the coordinate time. It is symmetrical and there seems to be no reason to assume that it would be biased in any special way. This time coordinate however goes to infinity as the object gets closer to the horizon.

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

I disagree with that, the term 'after' here does not make any sense.

The fact that it takes an (near) infinite time for light to reach the outside observer informing him that a probe crossed the event horizon does not contradict the fact that the probe simply crosses the event horizon more or less as a non-event. Why would you want to call it 'after' for light from inside the event horizon to reach the outside observer as this light would never be able to reach him?

By the way the internal Schwarzschild solution is not static, unlike the external solution.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

It is simply because it takes light an (near) infinite time to reach the observer, he simply does not see it or at least very late. Think of sound when the wind goes in opposite direction, it would take longer to reach you, taken that to the extreme would you claim someone did not shout 'hello' because the wind blew the sound waves the wrong way and they never reached your ears?

 

[Jonathan Scott]

...

It is simply because it takes light an (near) infinite time to reach the observer, he simply does not see it or at least very late. Think of sound when the wind goes in opposite direction, it would take longer to reach you, taken that to the extreme would you claim someone did not shout 'hello' because the wind blew the sound waves the wrong way and they never reached your ears?

The problem with that analogy is that, as I mentioned earlier, you can bounce light off a mirror at a lower potential and the situation is completely symmetrical with respect to time.

 

[Passionflower]

The problem with that analogy is that, as I mentioned earlier, you can bounce light off a mirror at a lower potential and the situation is completely symmetrical with respect to time.

It is not.

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

 

[Dale]

This seems to imply that although we don't have a way to describe the event of crossing the horizon in the static coordinate system, we can at least say that it definitely occurs AFTER our coordinate time becomes infinite.

Yes, there are coordinate systems which can assign a definite coordinate time to the crossing event. In all such coordinate systems that I am aware of the event crossing does occur after infinite time according to the Schwarzschild coordinates.

In the same way, in flat spacetime the inertial coordinates claim that a horizon crossing occurs after an infinite amount of Rindler time.

Note, as Passionflower noted you cannot use either Schwarzshild coordinates nor Rindler coordinates to make such a statement, you have to use a different coordinate chart, one which actually covers the event horizon.

I am therefore quite puzzled that people seems prepared to assert that something which clearly does not happen from a normal point of view until after an infinite time has elapsed - not even the end of the universe, but an INFINITE time - "actually happens".

Again, by that same logic anything in flat spacetime occurs after some Rindler horizon and therefore occurs after an INFINITE time by some observer's radar coordinates. Your logic would therefore say that no event can ever be considered to have "actually happened", even if we ourselves observed it, simply because some other observer might not be able to assign a finite time to it using radar coordinates.

 

[A.T.]

It is not.

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

You mean the time elapsed for sender and final receiver?

 

[Jonathan Scott]

If you take two static observers A and B at different r-coordinate values then light going from ABA will show a different elapsed time then going from BAB.

What do you mean here?

If you mean that observers at different potentials see different proper times for the round trip from their point of view, that's perfectly normal, as that depends on the time dilation at the observer potential. However, the coordinate times for the light times for the trip AB and BA are equal as plotted by any observer, regardless of which end they are on.

Anyway, that seems irrelevant here. What I'm pointing out is that from the point of view of ANY static observer, the assumption that the turnaround point occurs at a time which is "simultaneous" with the half way time at the start and end location (as in SR) gives a consistent definition of "simultaneous" for all static observers at all potentials. This means that any static observer can naturally extend their time coordinate in the usual way down to any point arbitrarily close to the horizon.

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

 

[Passionflower]

What I'm pointing out is that from the point of view of ANY static observer, the assumption that the turnaround point occurs at a time which is "simultaneous" with the half way time at the start and end location (as in SR) gives a consistent definition of "simultaneous" for all static observers at all potentials.

You can make that assumption if you pick the appropriate coordinates but that does not say anything about the physics.

Do you really think it is a natural view to assume it takes light just as much time to go down into a gravitational well as it takes to come back out? Remember the speed of light is always c locally, once you look at the speed of light over an extended curved region you can show just about anything with the appropriate coordinate chart.

 

[PAllen]

Would you consider the simultaneity of a thrown baseball compared to your own unphysical, or implausible? Then consider a static observer choosing to consider, moment to moment, the simultaneity of fastball thrown toward the BH. Whether you use Fermi-Normal simultaneity for the baseball at that moment, or computed radar (assuming extension of inertial trajectory of baseball the moment it leaved your hand, extended to the distant past and future) simultaneity, you would find by this slightly different point of view:

- at every point of the static world line, there are simultaneous events at and inside the horizon
- in the case of SC geometry, even the pure vacuum case, there is an earliest moment on your world line where thrown baseball simultaneity reaches the singularity (before this, it (Normal simultaneity) goes inside the horizon to some minimum r>0 then r grows (on the way to alternate KS sheet we are not considering here). Thus you have a sense of 'when' the singularity formed.
- all later history of static world line, your baseball 'now', includes events all the way to the singularity.

Whether one believes such predictions of classical GR apply to our universe, the above is unambiguously true of classical GR.

 

[Dale]

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

The Rindler horizon involves exactly the same infinities as the Schwarzschild horizon. What is your justification for dismissing the analogy, besides the obvious fact that it is damaging to your argument? You have asserted the physical significance of Schwarzschild time through the use of radar coordinates, just like those used for Rindler time.

 

[PeterDonis]

The argument about a Rindler horizon seems unhelpful, as that is a theoretical concept which also involves infinities.

If the Rindler horizon is a "theoretical concept", then so is the black hole event horizon. Rindler coordinates are the natural coordinates for observers with constant proper acceleration in Minkowski spacetime in the same way as Schwarzschild coordinates are the natural coordinates for observers with constant proper acceleration in Schwarzschild spacetime. So Rindler coordinates, and their associated horizon, are only "theoretical" to the same extent that Schwarzschild coordinates, and their associated horizon, are.

Yes, Rindler coordinates become singular (which is how I would rephrase your "involves infinities") at the Rindler horizon, in the same way as Schwarzschild coordinates become singular at the event horizon. That's the point: any argument that events can't happen "after" an infinite Schwarzschild time coordinate would also prove, if it were valid, that events can't happen "after" an infinite Rindler time coordinate. Which would prove, if valid, that a whole region of Minkowski spacetime can't exist. Which is why the argument isn't valid to begin with, for Rindler coordinates or for Schwarzschild coordinates.

 

[Jonathan Scott]

Do you really think it is a natural view to assume it takes light just as much time to go down into a gravitational well as it takes to come back out? Remember the speed of light is always c locally, once you look at the speed of light over an extended curved region you can show just about anything with the appropriate coordinate chart.

I would consider it very unnatural to assume anything else, as outside the event horizon of a black holes, classical mechanics is entirely reversible with respect to time.

In this case, I'm specifically referring to a static metric for a spherically symmetrical situation. That means that the coordinate speed of light inwards at a given point in the metric is the same as the speed outwards.

 

[Passionflower]

In this case, I'm specifically referring to a static metric for a spherically symmetrical situation. That means that the coordinate speed of light inwards at a given point in the metric is the same as the speed outwards.

No it does not.

We have the metric, the one you are talking about is the external Schwarzschild metric and we have a coordinate chart for a particular metric. Just because a particular coordinate chart of a given metric shows that the coordinate speed of light is isotropic does not mean it is a physical statement about the speed of light over a larger distance.

By analogy, on a Mercator projection it shows that Greenland is larger than Australia, however in reality it is much smaller than Australia.

 

[photonkid]

Does Montgomery, Alabama "exist"? How do you know it exists?

I'm sure one could write buckets of philosophical prose about this question :-(

However, it's perfectly possible to go visit Montgomery Alabama in a finite amount of time. The major differences between Montomery Alabama and a black hole event horizon , according to current theory and experiment, are as follows:

1) The nearest event horizon, at the center of our galaxy, is a lot further away than Montgomery. So the "finite" amount of time is a lot longer, and the vehicles one would need to use aren't technologically feasible.

But the time needed to get there is still finite according to theory.

2) The more troubling question is that if you do reach the event horizon, you won't be able to report your findings back to the people on Earth. But you'll still reach there in a finite time - according to current theory.

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

 

[PAllen]

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

These are good questions without good answers. Nobody has observed either Hawking radiation or Unruh radiation. There is strong and increasing observational evidence of horizons, which used to be entirely theoretical (like Hawking and Unruh radiation). If the black hole firewall proposal is true, then a well established horizon would be very detectible as you passed through it. However, classically, there is no way to detect it locally as you pass through it. Indirectly, as you observe from a distance, it is detectible, thus you can compute when you've passed it. (The firewall proposal is just that: some big names associated with it, but also prominent QG researchers who believe it is refuted).

 

[pervect]

But the person who goes to Alabama can tell when he has arrived at Alabama. The person who goes to visit an event horizon has no way to know when he has reached it. Is there some way to know when you've reached an event horizon? Does the absence of radiation tell you? Can you calculate size versus mass reliably if you're close enough?

You certainly won't be able to tell when you've passed the absolute horizon by any local measurement. It would also require an infinite amount of time to determine even a global measurement, as it's defined by the fact that light rays emitted never reach infinity.

This may be an issue, it becomes most important when a black hole is evaporating.

I *think* you can detect when you've passed an apparent horizon. This is defined by having the property that the expansion parameter \theta is negative for both ingoing and outgoing null geodesics (light rays).

I don't know specificially what apparatus you'd need to measure it, but I think the expansion parameter should be locally measurable. It might be good to get independent confirmation of this if it's really an important point.

 

[photonkid]

There is a theorem (originally due to Einstein) that says that an object in a static equilibrium, like a neutron star (or a white dwarf or anything else that holds itself up statically against its own gravity) can't have a radius smaller than 9/8 of the Schwarzschild radius for its mass. The time dilation factor at that radius is only 3 (i.e., time at the surface of such an object "flows" 1/3 as fast as it does at infinity); it certainly isn't approaching infinity.

One of the key things about black hole candidates is that at least some of them appear to be confined within a radius smaller than the above limit. (At least, that's my understanding.) So whatever is in there, it can't be something in static equilibrium like a neutron star.

So I'm guessing that the time dilation factor approaches infinity as you approach the Schwarzschild radius - because on the relativity faq website it says
<quote On my worldline as I fall into the black hole, it turns out that the Schwarzschild coordinate called t goes to infinity when I go through the event horizon.</>
and
<quote> In any case, this doesn't indicate that I take forever to fall in, since the proper time involved is actually finite.</>

I have to confess I have no understanding whatsoever of why a "finite proper time" means it won't take an infinite amount of "earth time" for something to reach an event horizon - even if the collapsing star was the only thing in the universe. However, it's also difficult to imagine matter pancaking indefinitely at the edge of the event horizon. The whole situation is very bizarre.

 

[Jonathan Scott]

No it does not.

We have the metric, the one you are talking about is the external Schwarzschild metric and we have a coordinate chart for a particular metric. Just because a particular coordinate chart of a given metric shows that the coordinate speed of light is isotropic does not mean it is a physical statement about the speed of light over a larger distance.

By analogy, on a Mercator projection it shows that Greenland is larger than Australia, however in reality it is much smaller than Australia.

I still really don't know what you're going on about here. For any spherically symmetrical static metric, isn't it obvious that the coordinate radial speed of light is the same both ways, and the coordinate time taken for light to travel between two points in a radial direction is the same regardless of direction?

 

[PAllen]

Would you consider the simultaneity of a thrown baseball compared to your own unphysical, or implausible? Then consider a static observer choosing to consider, moment to moment, the simultaneity of fastball thrown toward the BH. Whether you use Fermi-Normal simultaneity for the baseball at that moment, or computed radar (assuming extension of inertial trajectory of baseball the moment it leaved your hand, extended to the distant past and future) simultaneity, you would find by this slightly different point of view:

- at every point of the static world line, there are simultaneous events at and inside the horizon
- in the case of SC geometry, even the pure vacuum case, there is an earliest moment on your world line where thrown baseball simultaneity reaches the singularity (before this, it (Normal simultaneity) goes inside the horizon to some minimum r>0 then r grows (on the way to alternate KS sheet we are not considering here). Thus you have a sense of 'when' the singularity formed.
- all later history of static world line, your baseball 'now', includes events all the way to the singularity.

Whether one believes such predictions of classical GR apply to our universe, the above is unambiguously true of classical GR.

A correction and addition to above:

- Using radar simultaneity for the baseball thrown at the BH at the moment it leaves your (static observer) hand (with its then inertial trajectory extended to the distant past), there will be events at and inside the horizon simultaneous with this event (of leaving your hand). However, the singularity will never be radar simultaneous to any event on a free fall trajectory before it reaches the singularity.

- I had an error in my analysis of the Normal simultaneity for this event. Corrected, it becomes too complex for me to be willing to solve in detail. All that I can reliably show is that (as for the radar case) there exist events at and inside the horizon simultaneous to the moment it leaves your hand. My guess is that the qualitative features would be the same as radar, meaning that the singularity would never be Normal simultaneous to the thrown baseball.

 

[PeterDonis]

I have no understanding whatsoever of why a "finite proper time" means it won't take an infinite amount of "earth time" for something to reach an event horizon

Because they are two different kinds of questions: one is about an invariant, the other is about a coordinate-dependent simultaneity convention.

The first question, how much proper time does it take, is simple to answer because it's an invariant: just integrate the proper time along the infalling object's worldline. This will give a finite answer that is independent of the coordinates you choose.

The second question, how much "earth time" does it take, doesn't have a well-defined answer because it depends on which coordinates you adopt and the simultaneity convention that goes with them. That's true of *any* question about how to assign a "time" for some observer to events that are not on that observer's worldline. The infalling object is not on the Earth's worldline once it starts to fall, so any assignment of "earth time" to events on it is coordinate-dependent.

If you adopt the most natural simultaneity convention for "earth time", the one that goes with Schwarzschild coordinates, then you can't assign *any* finite "earth time" to events on or inside the horizon. Saying that "it takes infinite Earth time" for an infalling object to reach the horizon is a sloppy way of putting it because it assumes that assigning "earth time" to events on the horizon is even well-defined in the first place, and it isn't.

Also, the Schwarzschild coordinate simultaneity convention is not the only one you can adopt for assigning "earth time" to events; there are others, as PAllen has noted, which allow a finite "earth time" to be assigned to events on or inside the horizon.

 

[photonkid]

ok, yeah, I pretty much gathered that that's what the rationale was but...

Because they are two different kinds of questions: one is about an invariant, the other is about a coordinate-dependent simultaneity convention.

The first question, how much proper time does it take, is simple to answer because it's an invariant: just integrate the proper time along the infalling object's worldline. This will give a finite answer that is independent of the coordinates you choose.

...there seems to be a problem with this. If I move towards a gravitational field, time runs progressively more slowly (relative to where I was) and I cannot tell that it's doing so. If time actually stopped or ran infinitely slowly I would never know. If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?
It's too strange for me to grasp. Equally strange is that there are supposedly three time coordinates inside the event horizon. Why can't we just say that there is nothing happening inside an event horizon. If something is happening inside the event horizon (events are occurring) then time does exist inside the event horizon.

The second question, how much "earth time" does it take, doesn't have a well-defined answer because it depends on which coordinates you adopt and the simultaneity convention that goes with them.

If we had a way of determining whether an event horizon is present (e.g. size / mass) and that it is growing, we could say that the time taken on our clock for at least some of the matter to cross the event horizon is between zero and 13 billion years.

 

[PAllen]

...there seems to be a problem with this. If I move towards a gravitational field, time runs progressively more slowly (relative to where I was) and I cannot tell that it's doing so. If time actually stopped or ran infinitely slowly I would never know. If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?

For what observer a light year away? For someone a light year away moving sufficiently rapidly toward the BH, your clock would appear to be going at the same rate as theirs.

It's too strange for me to grasp. Equally strange is that there are supposedly three time coordinates inside the event horizon. Why can't we just say that there is nothing happening inside an event horizon. If something is happening inside the event horizon (events are occurring) then time does exist inside the event horizon.

Where did you get the idea there are 3 time coordinates inside the EH? There is one, just like outside. What really happens inside an EH we don't know (and many believe GR breaks down somewhere inside the EH; some believe it breaks in some ways at the EH). However, what GR predicts about the interior is unambiguous - sufficiently local physics is the same as anywhere else.

 

[Nugatory]

Equally strange is that there are supposedly three time coordinates inside the event horizon

I think that you may be misunderstanding the common (and correct) statement that inside the event horizon the Schwarzschild t coordinate becomes spacelike and the Schwarzschild r coordinate becomes timelike. This does indeed happen, but the other two spatial coordinates (θ and ψ) remain spacelike so we still have three space and one time coordinate.

You can see this if you look carefully at the Schwarzschild metric:
<br /> ds^2 = -(1-\frac{R}{r})dt^2+(1-\frac{R}{r})^{-1}dr^2+r^2(d\theta^2+sin^2{\theta}cos^2\phi)<br />
For r>R, outside the horizon, the coefficient of dt^2 is negative and the coefficient of dr^2 is positive. For r<R these signs flip; but the coefficients of the d\theta^2 and d\phi^2 terms remain positive.

 

[PeterDonis]

If I move towards a gravitational field, time runs progressively more slowly (relative to where I was)

You keep talking as if this statement has a physical meaning, when the whole point is that it doesn't; it's only a statement about the coordinate convention you pick. There is *no* way of assigning a physical meaning to the statement "time runs more slowly relative to where I was". You have to pick a simultaneity convention to assign a time "relative to where I was", and a simultaneity convention is just a convention; it doesn't tell you anything about the physics.

If I'm one millimeter from the event horizon and my clock tells me it takes one microsecond to move one micron, how many billion years would pass on a clock one light year away while I move one micron?

Same problem here. To answer this question you have to pick a simultaneity convention, and that's just a convention; it doesn't tell you anything about the physics.

If we had a way of determining whether an event horizon is present (e.g. size / mass) and that it is growing, we could say that the time taken on our clock for at least some of the matter to cross the event horizon is between zero and 13 billion years.

Once again, to do this you would need to pick a simultaneity convention, and that wouldn't tell you anything about the physics.

The problem you appear to be having is that you keep wanting to ask a question that doesn't have a well-defined answer. I know it seems to you like it should, but it doesn't. Repeating the question in different ways won't change that.

 

[PeterDonis]

Where did you get the idea there are 3 time coordinates inside the EH? There is one, just like outside.

Not necessarily; this also depends on the coordinate system you pick. There are coordinate charts in which all four coordinates are spacelike inside the horizon--for example, Painleve coordinates. (AFAIK it's not possible to have *more* than one timelike coordinate, but you can have less than one.)

This just makes it even more important to understand that coordinates don't tell you about the physics; you have to compute invariants to understand the physics.

 

[PAllen]

Not necessarily; this also depends on the coordinate system you pick. There are coordinate charts in which all four coordinates are spacelike inside the horizon--for example, Painleve coordinates. (AFAIK it's not possible to have *more* than one timelike coordinate, but you can have less than one.)

This just makes it even more important to understand that coordinates don't tell you about the physics; you have to compute invariants to understand the physics.

True, as I often stress. You could also have 2 spacelike and two light like coordinates. What I should have said is that inside the EH you can set up normal coordinates at any event the look locally like Minkowski coordinates.

 

[photonkid]

You keep talking as if this statement has a physical meaning, when the whole point is that it doesn't; it's only a statement about the coordinate convention you pick. There is *no* way of assigning a physical meaning to the statement "time runs more slowly relative to where I was".

Yes there is. If I have two identical clocks on top of a mountain and take one down to sea level for a while, then back to the top of the mountain, it will be behind the clock that stayed on top of the mountain.

 

[photonkid]

For what observer a light year away? For someone a light year away moving sufficiently rapidly toward the BH, your clock would appear to be going at the same rate as theirs.

For an observer not moving towards the black hole.

Where did you get the idea there are 3 time coordinates inside the EH?

I read it somewhere on the internet but I can't find where just at the moment.

There's a kind of allusion to it here but this is not where I read it.
http://www.einstein-online.info/spotlights/changing_places

[Edit] Found it here
http://www.astronomycafe.net/qadir/ask/a11339.html

 

[Nugatory]

Yes there is. If I have two identical clocks on top of a mountain and take one down to sea level for a while, then back to the top of the mountain, it will be behind the clock that stayed on top of the mountain.

That only works when the two clocks begin and end at the same location so there is no simultaneity problem at the beginning and end of the roundtrip. Take a look at my long answer reply to opti_mus back in post #81 of this thread - you're describing case #1 where we're measuring the proper time between two events (clocks separate, clocks rejoin). Unless and until they've rejoined, we're in case #2 of that post - no universal notion of "at the same time".

Here's a simpler example from special relativity:
Two spaceships traveling in opposite directions both set their clocks to the same time as they fly past one another, heading off to infinity never to rejoin. Because of time dilation and relativity of simultaneity, both will say that the other one has the slow clock, and they'll both be right. In this situation, the statement "this clock is really the slow one, that one is running normally" has no absolute physical significance - it can't because they're both right and they can't both be the slow one.
But if instead of flying off to infinity, one of the ships turns around so that eventually the two meet again, then we have an unambiguous and observer-indepedent "at the same time" at the ends of the journey, so we can compare the time elapsed. That's analogous to your example of the clock going down the mountain and coming back up and turns the mutual time dilation into the classic twin paradox.

(I seem to remember saying somewhere back in this thread that you (or someone?) are are making things harder for yourself by jumping into a fairly subtle problem of GR before you've solidly nailed down the basics of special relativity and classical mechanics. If I didn't say it already, I'm saying it now)