# Does the time exist?

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25 minutes ago, Luc Turpin said:

Will do!

This will help. As Susskind says in the General Relativity: The Theoretical Minimum,

Quote

it is very important to grasp perfectly the fundamental diagram of space-time near a black hole, shown in figure 5, in order to understand gravitational fields created by massive bodies, and eventually the theory of general relativity.

This is the figure 5, for reference:

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5 hours ago, joigus said:

IMO, this would be a "no"

The diagram is necessarily correct (it shows a valid solution to the EFE), and I think so is Genady’s interpretation of it.

The thing is that, at the event horizon, something very unintuitive happens - the physical meaning of the coordinates we use is no longer the way we are accustomed to. Imagine an astronaut in free fall, just as he crosses the horizon - let’s for simplicity’s sake say his feet emit light. Once his feet have crossed the horizon, and always assuming free fall, this light signal is now no longer “below” the eyes, but in their future. Light below the horizon is perfectly free to propagate in all spatial directions, yet it can still never leave the BH, because the singularity is in the future, and the horizon is in the past. It is no longer a question of up, down, above or below, once you’re past the horizon.

Thus, for the astronaut, the light leaves his feet, and his eyes will necessarily “meet” it, because he’s in free fall. Both age towards the singularity, their relative velocity remains c (so everything is locally Minkowskian), yet their geodesics must intersect, just as the diagram shows. Thus he sees his feet like normal, perhaps slightly redshifted and dimmed. He will otherwise never notice anything special at the horizon. And he can’t, because locally everything must look Minkowskian.

This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings.

13 hours ago, MigL said:

as there are no geodesics for light to follow in the outwards direction.

It’s the other way around, see also above - light past the horizon remains past the horizon, but the eyes which see that light are falling inwards, and can intersect that light in the future.

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Ok. If everything boils down to "there cannot be a backward photon" I think we all agree. But I have no time to think about it now.

I still think Pinoccio cannot see the tip of his nose if he's facing towards the horizon, but he can if he's facing the other way, which is what didn't seem to me compatible with what Genady said in an extremely cursory way.

It seems I misunderstood. I'll get back to it probably tomorrow.

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The primed observer, falling through the horizon at event C', sees the unprimed observer, which is falling through the horizon at event C.

Based on the diagram and reading, unprimed c will see primed c fade away and disappear as he gets closer and closer to EH. This is because light will emit ever more slowly, get dimmer until completely fading away as one reaches EH. Nothing comes out of a black hole, not even light. Correct?

As for primed c seeing unprimed c falling through the horizon, I can only speculate that primed c will see the incoming light without impediment, but I am not so sure. Common logic tells me that if light fades away on one side, it should not be present on the other side. But this is of course a black hole, which abides by its own internal laws.

7 hours ago, Markus Hanke said:

This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings.

So, space-time is inverted? space (down) becomes time (future)?

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13 minutes ago, Luc Turpin said:

unprimed c will see primed c

Here the green arrow shows that the unprimed observer being at event D sees the primed observer being at event M' (approximately):

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Here the green arrow shows that the unprimed observer being at event D sees the primed observer being at event M' (approximately):

I was visually seeing it on the diagram, but cannot grasp the concept conceptually!

Am-I asking a silly question if I ask if there is a difference in seing at EH as opposed to inside the BH?

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3 minutes ago, Luc Turpin said:

I was visually seeing it on the diagram, but cannot grasp the concept conceptually!

Am-I asking a silly question if I ask if there is a difference in seing at EH as opposed to inside the BH?

The 45o line that goes through the origin and the events C and C', is a worldline of EH. The greyish area above it and to the left, with the events D, E, D', E', is inside the BH. The white area to the right and below - outside the BH. The hyperbola separating the light grey and the dark grey, with event F on it - singularity.

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The 45o line that goes through the origin and the events C and C', is a worldline of EH. The greyish area above it and to the left, with the events D, E, D', E', is inside the BH. The white area to the right and below - outside the BH. The hyperbola separating the light grey and the dark grey, with event F on it - singularity.

Much better visual comprehension now than before. It was generally what I thought.

So unprimed c cannot been seen as he has passed EH and faded away, while unprimed c has not passed EH and then can still be seen! correct? Or is this a too simplistic explanation?

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13 minutes ago, Luc Turpin said:

unprimed c cannot been seen ... while unprimed c ...

You said "unprimed c" twice, so I don't know what you mean. Anyway, event C is seen at event C'.

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You said "unprimed c" twice, so I don't know what you mean. Anyway, event C is seen at event C'.

So primed c cannot been seen as he has passed EH and faded away, while unprimed c has not passed EH and then can still be seen! correct? Or is this a too simplistic explanation?

Anyway, event C is seen at event C'.

Got that

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5 minutes ago, Luc Turpin said:

So primed c cannot been seen as he has passed EH and faded away, while unprimed c has not passed EH and then can still be seen

It depends on a "seeing" event. Draw two 45o light rays up from any event: it can be seen at any event on these rays.

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My mistake.
I realized both observer and the light source ahead of him are falling towards the future, but forgot to take into account  the light source is ahead of the observer temporally, not spatially.

I guess a barefooted observer, falling into a small enough BH, would notice his toenails need trimming, as they may be months ahead of him, in the future.

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It depends on a "seeing" event. Draw two 45o light rays up from any event: it can be seen at any event on these rays.

Understood!

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15 hours ago, Markus Hanke said:

The diagram is necessarily correct (it shows a valid solution to the EFE), and I think so is Genady’s interpretation of it.

The thing is that, at the event horizon, something very unintuitive happens - the physical meaning of the coordinates we use is no longer the way we are accustomed to. Imagine an astronaut in free fall, just as he crosses the horizon - let’s for simplicity’s sake say his feet emit light. Once his feet have crossed the horizon, and always assuming free fall, this light signal is now no longer “below” the eyes, but in their future. Light below the horizon is perfectly free to propagate in all spatial directions, yet it can still never leave the BH, because the singularity is in the future, and the horizon is in the past. It is no longer a question of up, down, above or below, once you’re past the horizon.

Thus, for the astronaut, the light leaves his feet, and his eyes will necessarily “meet” it, because he’s in free fall. Both age towards the singularity, their relative velocity remains c (so everything is locally Minkowskian), yet their geodesics must intersect, just as the diagram shows. Thus he sees his feet like normal, perhaps slightly redshifted and dimmed. He will otherwise never notice anything special at the horizon. And he can’t, because locally everything must look Minkowskian.

This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings.

It’s the other way around, see also above - light past the horizon remains past the horizon, but the eyes which see that light are falling inwards, and can intersect that light in the future.

8 hours ago, Luc Turpin said:

So, space-time is inverted? space (down) becomes time (future)?

I was asking because of this:

"After jumping into a black hole and passing the horizon, time and space change roles as now the singularity is not longer a point in space, but a moment in time."

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19 hours ago, Markus Hanke said:

The diagram is necessarily correct (it shows a valid solution to the EFE), and I think so is Genady’s interpretation of it.

The thing is that, at the event horizon, something very unintuitive happens - the physical meaning of the coordinates we use is no longer the way we are accustomed to. Imagine an astronaut in free fall, just as he crosses the horizon - let’s for simplicity’s sake say his feet emit light. Once his feet have crossed the horizon, and always assuming free fall, this light signal is now no longer “below” the eyes, but in their future. Light below the horizon is perfectly free to propagate in all spatial directions, yet it can still never leave the BH, because the singularity is in the future, and the horizon is in the past. It is no longer a question of up, down, above or below, once you’re past the horizon.

Thus, for the astronaut, the light leaves his feet, and his eyes will necessarily “meet” it, because he’s in free fall. Both age towards the singularity, their relative velocity remains c (so everything is locally Minkowskian), yet their geodesics must intersect, just as the diagram shows. Thus he sees his feet like normal, perhaps slightly redshifted and dimmed. He will otherwise never notice anything special at the horizon. And he can’t, because locally everything must look Minkowskian.

This is probably the biggest mistake people make when trying to visualise black holes - they think that, past the horizon, the singularity is “down”; but it’s not, it’s in the future. Likewise, the horizon isn’t “up”, but in the past. This is extremely important to understand, or else there’ll be all sorts of misunderstandings.

It’s the other way around, see also above - light past the horizon remains past the horizon, but the eyes which see that light are falling inwards, and can intersect that light in the future.

Very far from understanding this but can I ask ,if the up and down coordinates that apply outside the EH are replaced with future and past inside the EH  what are the future and past coordinates that applied  outside the EH replaced with inside the EH?

Are there any spatial coordinates that apply inside the EH and  can they be visualized?

They wouldn't be flattened on the EH itself ,would they? (that's a wild ,desperate guess)

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16 hours ago, Luc Turpin said:

So, space-time is inverted? space (down) becomes time (future)?

In some sense, yes. But it isn’t really a physical change that one would notice - spacetime remains smooth, regular, and locally Minkowskian everywhere (outside the singularity). What changes is mostly the physical meaning of the coordinates we use, relative to the exterior region.

4 hours ago, geordief said:

what are the future and past coordinates that applied  outside the EH replaced with inside the EH?

They now become spatial in nature, along the radial direction. Future means going “down” radially, past means going “up”. So the physical meaning of the r and t coordinates trade places.

But again, it’s not something you would notice; spacetime looks just the same below as above the horizon. The only difference is its causal structure - below the horizon, all physically possible world lines (whether geodesics or not) terminate at the singularity.

4 hours ago, geordief said:

Are there any spatial coordinates that apply inside the EH and  can they be visualized?

Locally, everything looks perfectly normal there, at least up to the point where tidal forces become noticeable.

4 hours ago, geordief said:

They wouldn't be flattened on the EH itself ,would they?

No, nothing special happens at the horizon at all - spacetime is perfectly regular there, and if you were to fall through it, you wouldn’t locally notice anything out of the ordinary. It’s really just a mathematical concept, not a physical entity.

18 hours ago, joigus said:

I still think Pinoccio cannot see the tip of his nose if he's facing towards the horizon

He can, because he’s in free fall. Visualise it like this (though it’s not really correct) - a photon emitted radially outwards very close to the horizon has a very slow radial (!) velocity wrt to the event horizon. On the other hand though, Pinocchio falls through the horizon at nearly the speed of light (wrt some outside reference), and thus meets the photon on the way. So it’s not like the photon necessarily propagates to his eyes, but rather that his eyes fall right to where the photon is. It’s kind of like jumping upwards in an elevator - you can put relative motion between yourself and the elevator floor, but both you and the elevator continue to move down regardless (maybe a stupid example, but you get my drift hopefully).

Note that the situation is different if you’re not in free-fall. If Pinocchio, after the tip of his nose crosses the horizon, somehow fires magical thrusters that arrest his fall before he reaches the EH, then his nose will visually disappear for him (and get ripped away).

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8 hours ago, geordief said:

Are there any spatial coordinates that apply inside the EH and  can they be visualized?

I want to add to the @Markus Hanke's explanation above, that there are two more spatial coordinates, the "latitude" and the "longitude", which apply in the same way inside the EH as outside it. They are not shown on the Kruskal diagrams.

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5 hours ago, Markus Hanke said:

He can, because he’s in free fall. Visualise it like this (though it’s not really correct) - a photon emitted radially outwards very close to the horizon has a very slow radial (!) velocity wrt to the event horizon. On the other hand though, Pinocchio falls through the horizon at nearly the speed of light (wrt some outside reference), and thus meets the photon on the way. So it’s not like the photon necessarily propagates to his eyes, but rather that his eyes fall right to where the photon is. It’s kind of like jumping upwards in an elevator - you can put relative motion between yourself and the elevator floor, but both you and the elevator continue to move down regardless (maybe a stupid example, but you get my drift hopefully).

Note that the situation is different if you’re not in free-fall. If Pinocchio, after the tip of his nose crosses the horizon, somehow fires magical thrusters that arrest his fall before he reaches the EH, then his nose will visually disappear for him (and get ripped away).

Yes, your argument sounds right. I'll have to think about it over the weekend. Big workload now.

Depending on how much Pinoccio has been lying, tidal forces would come into effect though.

Thanks for the careful explanation. +1

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20 hours ago, joigus said:

Depending on how much Pinoccio has been lying, tidal forces would come into effect though.

Yes, true - I’ve been neglecting these, to avoid that extra level of complexity.

20 hours ago, joigus said:

Thanks for the careful explanation. +1

This really is kind of tricky to get one’s head around. That’s one of the reasons why spacetime diagrams, like the one Genady has posted, are so useful - you can see immediately whether geodesics intersect or not.

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3 hours ago, Markus Hanke said:

Yes, true - I’ve been neglecting these, to avoid that extra level of complexity.

This really is kind of tricky to get one’s head around. That’s one of the reasons why spacetime diagrams, like the one Genady has posted, are so useful - you can see immediately whether geodesics intersect or not.

Looking at it physically ,the light from P's nose recedes from him at c   .He follows ,whilst outside the EH at <c

You say he "catches up" with it .Is that because the light from his erstwhile  nose  radiates in all directions and so he "catches up" or "falls into"  that  part of the light that is not receding directly away from him  but with a component back towards him?(which is red shifted?)

If there was just one photon receding directly along  his "line of site" he wouldn't even see it red shifted?

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Forgive my stupidity, but isn't time the 'tick' that we all experience at the same <insert word>???

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1 hour ago, dimreepr said:

Forgive my stupidity, but isn't time the 'tick' that we all experience at the same <insert word>???

Your tick and my tick are not the same. We're moving at different speeds. We are at different elevations with different gravitational potentials. When I'm on a plane and you're on the ground, our ticks differ. When you're walking and I'm still, our ticks are different.

The difference tends to be small at human speeds, but it is not zero.

There is also the issue of human awareness and how our awareness comes after the tick already happened, and when that awareness comes depends on our levels of fatigue, electrolyte saturation, and copious other neurological variables off topic here.

As in my signature, "time is one of those concepts that is profoundly resistant to a simple definition."  ~C. Sagan

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19 hours ago, geordief said:

the light from P's nose recedes from him at c

It goes in all directions, including towards his eyes (roughly the radial direction). So it’s both that the light radiates “upwards” to the degree that it can, given the presence of the horizon, and that he himself falls “down” towards it. The local laws of physics guarantee that the relative velocity remains always exactly c. The crucial bit is that these geodesics (light & eyes) intersect someplace, meaning Pinocchio can see the tip of his nose. If the light was emitted below the horizon, then that’s also where the intersection takes place - unless he somehow manages to arrest his fall before reaching the horizon, then his nose tip of course disappears from sight.

19 hours ago, geordief said:

Is that because the light from his erstwhile  nose  radiates in all directions and so he "catches up" or "falls into"  that  part of the light that is not receding directly away from him  but with a component back towards him?(which is red shifted?)

Yes, pretty much.

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16 hours ago, iNow said:

Your tick and my tick are not the same.

My very badly explained point is, yes they are, the tick frequency remains they same wherever we find ourselves in space and whatever speed we're going there.

It's like the latice/weave of the fabric of space.

In some respects the only meaningful definition of time is a wristwatch, preferably a digital...

“Why are people born? Why do they die? Why do they want to spend so much of the intervening time wearing digital watches?” ― Douglas Adams

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2 hours ago, dimreepr said:

My very badly explained point is, yes they are, the tick frequency remains they same wherever we find ourselves in space and whatever speed we're going there.

Remains the same relative to what/whom? Your point isn’t badly explained. It’s wrong.

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