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michel123456

I see you, You see me.

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Posted (edited)

This below is an attempt to explain my conception of the mechanism of Time.

The beginning is quite simple: You have two friends A (Alex) and B (Beatrix)  resting next to each other. There is no motion, only time passing by.

Alex looks to Beatrix.

Since light takes time to travel from Beatrix to Alex, Alex observes Beatrix as it was a slight instant ago (slightly in the past).

I can make the following diagram, where the vertical axis is Time, and the horizontal is Space.

At T=0, Alex observes Beatrix at a distance & at T=-1

time-AB0.jpg.0cba712bba1bfc2e4a96370809716e23.jpggraph 1

But at the same time (T=0) Beatrix looks at Alex also at T=-1 (the situation is symmetric). Thus we also have the following graph

time-BA0.jpg.c10226d94aa2b40a51e04be12ebe1850.jpggraph 2

How can that be? The conventional way of thinking is the following: when Alex slides in time, he leaves behind him a path of events (Alex existing in past times)

It goes like this:

accepted1.gif.268249c43759286af11d44e65a1767b4.gifgraph 3

The bold line is the 4D existence of Alex along its life-line. Every point of this line will find Alex at a specific point of his life.

In this way of thinking there is no difficulty to understand that As Alex sees Beatrix, Beatrix sees Alex. It goes like this:

time-AB0-cross.jpg.effb36138bd5438f5bc87707f0fee319.jpggraph 4

Alex looks at Beatrix &reversely Beatrix looks at Alex. They live in the same time frame (T=0) and they see each other at T=-1.

No problem.

But what do I mean when I say "they see each other at t=-1? Do I mean that B & A below on the graph truly "exist" there?

No I don't think so. Here begins the speculation:

The speculation is that when Alex sees Beatrix behind in time, it means that the signal (the photons) send by Beatrix have reach the eye of Alex. In order to travel the distance from B to A, the photons need time. And the bottom B on the previous graph are void, there is nothing there. There is only the image of B in the eye of A. And respectively, at bottom A on the graph, there is nothing. There is only the image of A in the eyes of B.

It goes like this: the following graph shows the sliding of A in Time (to be compared with graph 3)

michA.gif.bc33080d9c619ac1baa8ae2233140e64.gifgraph 5

And below Beatrix sliding in time

MichB.gif.0d4242e9e4479a50a428cacda7574626.gifgraph 6

The graph below shows Alex looking at Beatrix while sliding in time (remember that both observers do not move, they rest in place, simply sliding in time)

MichAB.gif.fa949b4c37ce017e8b267985a3548206.gifgraph 7

And graph 8 is the reverse B observing A

MichBA.gif.113eb54418b11ae81c5534bf69162906.gifgraph 8

Graph 9 represents both observers while sliding in Time

Mich-ABab.gif.d28ba921bbe8e0edff55f7bfa5dff985.gifgraph 9

In the above graph 9, the solid circles represent the real observers A & B sliding in time, the empty circles represent the image as observed: the image in the retina of solid A and solid B. In fact, there is nothing there, following graph 5 & graph 6.

Now let's get a little more complicated: say that Alex throws a ball to Beatrix.

Graph 10 represents a ball (in red) from A to B. In orange, the image of the ball a seen by A.

Mich-Total-A.gif.9fa5a790f8e9a08970b515893478fb0b.gifgraph 10

Graph 11 is the same as graph 10, but as seen by B

Mich-Total-B.gif.d19332a66a88a4b7046b8ffaac5400e4.gifgraph 11

In both graph 10 & 11, the ball makes the same trajectory in the same time. But observers A & B have a different point of view, the image that come to their eyes is different. In both diagrams, the image is the vertical projection of the ball to the diagonal of view. Both observers will agree that the ball was send at T=0, and received at T=4. They may compare their point of views and agree on distance & time. Alex will throw the ball and as it goes away, it goes into the past until reaching B. And B will see the ball coming from the past. It corresponds to reality. And there is no need for small a & b to "exist". In both graphs, a & b are images of A & B in the eyes of each other. There is nothing actually at points a & b.

Now let's get more complicated, here below, introducing Clong (C). Clong is an hypothetical observer out of sync, behind in Time.Mich-total-C.gif.4e3a0d4f9e1a62d687885c1e3ad8af95.gifgraph 12

Intuitively, C should be observable, I have simply to choose another observer D sufficiently far away, see below graph 13

Mich-Total-D-explained.gif.e0018404fd29022330345aa96e020f21.gifgraph 13

Here it gets a bit complicated: what is D actually observing? Is he observing C, or the image of A (labeled a' on the graph)

My answer is that D has in his eyes the image of A. There is no real superposition, on one hand you have a real object C at specific coordinates, on the other hand you have an image on the retina of an observer.

Reversely, A has in his eyes the image of D (labeled d on the graph). Observer A does not see the void at d, there is nothing at d. The only one who will see d is observer C, exactly the same way B sees A.

Any comment appreciated.

 

 

 

 

 

 

 

 

 

MichAB.gif

Edited by michel123456

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1 hour ago, michel123456 said:

This below is an attempt to explain my conception of the mechanism of Time.

The beginning is quite simple: You have two friends A (Alex) and B (Beatrix)  resting next to each other. There is no motion, only time passing by.

Alex looks to Beatrix.

Since light takes time to travel from Beatrix to Alex, Alex observes Beatrix as it was a slight instant ago (slightly in the past).

I can make the following diagram, where the vertical axis is Time, and the horizontal is Space.

At T=0, Alex observes Beatrix at a distance & at T=-1

time-AB0.jpg.0cba712bba1bfc2e4a96370809716e23.jpggraph 1

But at the same time (T=0) Beatrix looks at Alex also at T=-1 (the situation is symmetric). Thus we also have the following graph 

“At the same time” loses meaning. In your scenario, clocks are not synchronized. Everyone has their own time. Unlike Einstein clock synchronization used in relativity.

 

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Posted (edited)
1 hour ago, michel123456 said:

This below is an attempt to explain my conception of the mechanism of Time.

Still? 

Time isn't a thing, anymore than your conception of it is, therefore logically it has no mechanism/meaning; se la vie...😉

Edited by dimreepr

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Good post.
Everything was sensible up to and including graph #4.

Then you went 'off the rails' again.
And you really have to stop with the 'flying', 'sliding', and any other kind of ill-defined terminology of 'movement' through time.
Events don't 'move' through time; our individual perceptions of 'now' do.

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I'm in no position to judge the physics behind michel's speculation, but it reminds me of someone trying model the universe from a geocentric perspective. It might ultimately work, but why complicate something that is (relatively) simple and works just fine?

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Posted (edited)
1 hour ago, zapatos said:

I'm in no position to judge the physics behind michel's speculation, but it reminds me of someone trying model the universe from a geocentric perspective. It might ultimately work, but why complicate something that is (relatively) simple and works just fine?

If you make the comparison between graph 3 (accepted) & graph 5 (speculation), which one is simpler?

accepted1.gif.000736b1f8eb272911fb7d1d1639c578.gifmichA.gif.21fdd040ef0644e333e5f8db21f62f36.gif

If I can show that the multiple A's of graph 3 are redundant, isn't that a good point? Even if it is harder to understand?

2 hours ago, MigL said:

Then you went 'off the rails' again.
And you really have to stop with the 'flying', 'sliding', and any other kind of ill-defined terminology of 'movement' through time.
Events don't 'move' through time; our individual perceptions of 'now' do.

That's what I am disputing. To stop me you have to convince me: show me where I made a mistake.

16 minutes ago, michel123456 said:

and works just fine?

No it doesn't "work just fine". Our theories are missing a lot.  The cosmological measurements (based on theories) are off by 85% of matter and I don't know how much energy.

My explanation opens a window to a lot of matter & energy (see object C in graph 12). In fact, it is the possibility of another multiple unobservable universe.

And I have an extra argument that I keep for later.

Edited by michel123456

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These graphs only illustrate the positions of Alex, B and C which are fixed in space.
It is true that thereafter you can represent their displacement along the horizontal from left to right.
But graphically speaking, should we still be able to simulate the time dilation according to speed.
Do you have a solution with your ball as packets transmitted from one to the other?

This may be your extra argument ... the time dilation following the 'more or less speed movement' of Alex or Beatrix. 

Or is it possible to integrate a Minkowski diagram into yours?

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Posted (edited)
35 minutes ago, Kartazion said:

But graphically speaking, should we still be able to simulate the time dilation according to speed.

Yes. If it is considered as an argument I can step into this, but I am afraid to escape the important thing.

 

 

35 minutes ago, Kartazion said:

Or is it possible to integrate a Minkowski diagram into yours?

important differences are that a Minkowski diagram is static & there is no ct vertical axis in mine, only T.

Edited by michel123456

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3 hours ago, michel123456 said:

To stop me you have to convince me: show me where I made a mistake.

For one thing you keep insisting on 'motion' through the spatial and temporal dimensions.
If you take relativity seriously, there is no space, and there is no time; only space-time, as the two cannot be separated.

The argument is that you can neither move through time, nor move through space, but you are at ALL locations you occupy in space-time.
In a simplified case, as you used, where we consider only two dimensions, displacement horizontally and duration vertically, you exist as a line through space-time. Vertical if you are standing still, with a slope if you undergo a displacement.
And if your axis are marked in seconds and kilometers, the slope of this line can never be less than 1/300000. IOW you CANNOT move only in space.
In this example, 'now' is a point on the line, and differs for every observer; that point, 'now' is what each of us perceives to be moving, through space-time.
I can't really explain the 4dimensional equivalent, but you exist as a volume, 'extruded' through a fourth orthogonal time dimension, and your 'now' is a hypersurface or 'foliation' of the space-time manifold, and again, none are common to differing observers.

You are trying to understand 'motion' through time, and inventing all these complexities, much like the 'add-ons' to the Ptolemaic System in its later years before Copernicus replaced it, to try and explain things which are not even required.
Take Occam's razor and cut away everything that isn't needed.

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Posted (edited)
8 hours ago, MigL said:

For one thing you keep insisting on 'motion' through the spatial and temporal dimensions.
If you take relativity seriously, there is no space, and there is no time; only space-time, as the two cannot be separated.

The argument is that you can neither move through time, nor move through space, but you are at ALL locations you occupy in space-time.
In a simplified case, as you used, where we consider only two dimensions, displacement horizontally and duration vertically, you exist as a line through space-time. Vertical if you are standing still, with a slope if you undergo a displacement.
And if your axis are marked in seconds and kilometers, the slope of this line can never be less than 1/300000. IOW you CANNOT move only in space.
In this example, 'now' is a point on the line, and differs for every observer; that point, 'now' is what each of us perceives to be moving, through space-time.
I can't really explain the 4dimensional equivalent, but you exist as a volume, 'extruded' through a fourth orthogonal time dimension, and your 'now' is a hypersurface or 'foliation' of the space-time manifold, and again, none are common to differing observers.

Well understood. But I think it does not explain how things are going on.

Look at it back again. It seems you are agreeing till graph 4 (here below)time-AB0-cross.jpg.5545b2790bedbf28ac65262718740b77.jpggraph 4

In this graph we have simply Alex standing still, looking at Beatrix, a few meters away (no relativistic velocity involved). The only thing that is counted for is the gap in time, the  information delay (whatever value it takes). Alex & Beatrix stand next to each other, where are they currently on this diagram? See below:

time-AB0-crossx2.jpg.36e80a7f66f7e54842d60bd2dd37c946.jpg

There they are: they are together in time, next to each other. Say on Monday 9th of March 2020, 11.01 in the morning.

They are not Alex 9th March & Beatrix 1st of January.

(beware before giving rep.points, I have edited this post)

Edited by michel123456

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8 hours ago, MigL said:

 The argument is that you can neither move through time, nor move through space, but you are at ALL locations you occupy in space-time.

Whose argument is this? In relativity, your 4-velocity is c. 

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Posted (edited)
22 hours ago, michel123456 said:

But what do I mean when I say "they see each other at t=-1? Do I mean that B & A below on the graph truly "exist" there?

No I don't think so.

That's the error. A spacetime diagram is a depiction, a map, not reality itself. So your question 'Do I mean that B & A below on the graph truly "exist" there?' makes no sense. 

So what is a spacetime diagram: it is a map of who and what was where. So A and B, assuming they sit still for a while, say from T-4 until T+4. Then your diagram should show that: you should draw A and B on every time sitting at their place. If A was at X0 at T0, then you must draw that point and leave it there. The question 'where was A at T0?' must have an unambiguous answer! A was at X0 at T0, and your diagram should show that. 

One could say that your animations are useless. Spacetime diagrams depict space and time in relation to each other, but then, by animating them, you introduce a second time dimension!

Edited by Eise

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2 hours ago, michel123456 said:

There they are: they are together in time, next to each other. Say on Monday 9th of March 2020, 11.01 in the morning.

They are not Alex 9th March & Beatrix 1st of January.

But they could be, using your protocol, if they were 68 (or 69) light-days apart. 

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1 hour ago, swansont said:

But they could be, using your protocol, if they were 68 (or 69) light-days apart. 

Using my protocol, yes A & B can be very far apart. As long as one observes the other, the reverse applies. I see you, you see me. Except if you disappeared in some explosion. It works for the entire Observable Universe. IOW this planet on this galaxy billion light years away is not billion light years in the past. Its image took billion years to travel to us. At this moment, this planet is on the horizontal line of the graph at T=0. Exactly as Beatrix, if I am Alex the observer.

2 hours ago, Eise said:

That's the error. A spacetime diagram is a depiction, a map, not reality itself. So your question 'Do I mean that B & A below on the graph truly "exist" there?' makes no sense. 

So what is a spacetime diagram: it is a map of who and what was where. So A and B, assuming they sit still for a while, say from T-4 until T+4. Then your diagram should show that: you should draw A and B on every time sitting at their place. If A was at X0 at T0, then you must draw that point and leave it there. The question 'where was A at T0?' must have an unambiguous answer! A was at X0 at T0, and your diagram should show that. 

One could say that your animations are useless. Spacetime diagrams depict space and time in relation to each other, but then, by animating them, you introduce a second time dimension!

A spacetime diagram is a depiction of events: it is the story of A & B as seen by A or B, or by any other observer that can observe A & B. My graphs show that there may be more than that: observers that cannot observe A & B & reversely, observers that A & B cannot observe.

Now, you are correct that by animating them I am introducing a second time dimension.

But a moving slot upon a block Universe is also adding a time dimension.

 

It is the same as a standing slot & a movable BU, like an old fashioned camera lens & a movie film.

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19 minutes ago, michel123456 said:

As long as one observes the other, the reverse applies. I see you, you see me.

How do you know?

Your graphics/observations are essentially assumptions, so any conclusion is just a guess; you know, metaphysical (not real).

You seem to be trying to equate, I think therefore I am, with I think therefore you are.

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3 hours ago, swansont said:

Whose argument is this? In relativity, your 4-velocity is c. 

Sorry Swansont, poorly worded on my part.
Here is a quote from Wiki on four-velocity which may explain better...

"Physical events correspond to mathematical points in time and space, the set of all of them together forming a mathematical model of physical four-dimensional spacetime. The history of an object traces a curve in spacetime, called its world line. If the object has mass, so that its speed is less than the speed of light, the world lie may be parametrized by the proper time of the object. The four-velocity is the rate of change of four-position with respect to the proper time along the curve. The velocity, in contrast, is the rate of change of the position in (three-dimensional) space of the object, as seen by an observer, with respect to the observer's time."

https://en.wikipedia.org/wiki/Four-velocity

IOW things can't actually move in space, UNLESS there is a change in time also.
( they would be moving at infinite speed )

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Posted (edited)
1 hour ago, dimreepr said:

How do you know?

 

I know. I don't think I have introduced anything new. If you & me sit on a chair at 1 meter interval, we are sharing the same time. The delay of the graph will be ridiculously small but still exists. If we increase the distance between our chairs, the time gap will increase but we will still share the same T. You can increase the distance infinitely, there is no reason for the situation to change, as long as we are both sitting still on a chair. There is no scale on the graph. If I can observe a far away Galaxy, then theoretically an alien on this Galaxy can also observe me(1). Like Alex & Beatrix. The alien is not in the past, he has traveled in time as we did.

And the alien is not viewed by someone in his future. As we are not currently under the scrutiny of a future alien.

(1) I know it hasn't left the Milky Way but Voyager can look at us.https://www.nasa.gov/mission_pages/voyager/multimedia/pia00452.html

Edited by michel123456
(1)

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50 minutes ago, MigL said:

Sorry Swansont, poorly worded on my part.
Here is a quote from Wiki on four-velocity which may explain better...

"Physical events correspond to mathematical points in time and space, the set of all of them together forming a mathematical model of physical four-dimensional spacetime. The history of an object traces a curve in spacetime, called its world line. If the object has mass, so that its speed is less than the speed of light, the world lie may be parametrized by the proper time of the object. The four-velocity is the rate of change of four-position with respect to the proper time along the curve. The velocity, in contrast, is the rate of change of the position in (three-dimensional) space of the object, as seen by an observer, with respect to the observer's time."

https://en.wikipedia.org/wiki/Four-velocity

IOW things can't actually move in space, UNLESS there is a change in time also.
( they would be moving at infinite speed )

I agree with that last bit.

But I would say that that everything is moving in spacetime. The faster you move is space, the slower you move in time (from an external observer's standpoint), such that the 4-velocity remains constant.

10 minutes ago, michel123456 said:

I know. I don't think I have introduced anything new. If you & me sit on a chair at 1 meter interval, we are sharing the same time.

You don't seem to be arguing that in the rest of your presentation. Everyone has their own time, and everyone else has a time that is in the past, by L/c.

 

 

 

In physics, the utility of time is that we can make measurements, and those measurements may be for events not co-located, so it's important that we agree on what time it is, and recognize that we will not agree when there is relative motion (or certain position differences in the presence of gravity). We have a clock synchronization protocol to facilitate this.

If you don't follow this protocol, I think you need a new one in order to make sense of any conversations.

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31 minutes ago, michel123456 said:

The delay of the graph will be ridiculously small but still exists.

We can't cross the same river twice... Complications are generally not needed to explain reality...

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5 hours ago, dimreepr said:
5 hours ago, michel123456 said:

The delay of the graph will be ridiculously small but still exists.

We can't cross the same river twice... Complications are generally not needed to explain reality...

Then say that the horizontal segment from A to B is the entire Observable Universe.

 

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12 minutes ago, michel123456 said:

Then say that the horizontal segment from A to B is the entire Observable Universe.

Do you take the expansion of universe into account in that case? Or do you use some other model?

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18 minutes ago, Ghideon said:

Do you take the expansion of universe into account in that case? Or do you use some other model?

No. and no.

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14 hours ago, michel123456 said:

No. and no.

Thanks! Just so I understand*: By "the entire Observable Universe" you mean "a really large distance", and this distance separates two observers in same frame of reference? 

 

*) "Observable universe", to me, implies that GR and other applicable models have to be taken into account.

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On 3/9/2020 at 2:01 PM, michel123456 said:

A spacetime diagram is a depiction of events: it is the story of A & B as seen by A or B, or by any other observer that can observe A & B.

No, it is not: it is a God-like view of the universe, in which God sees past, present and future. If the location T-1 was occupied by A, then this is an unchanging fact, represented by a dot in the spacetime diagram. That dot will stay there, otherwise you are giving a false depiction of what happened at T-1 at X-0. And for such a 'Lord of the Ages' there is no now. 'Now' is only an expression referring to the same time when an utterance was made, and that presupposes that the one who is uttering the expression is in spacetime. 'God' lives outside the spacetime diagram.

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Posted (edited)
2 hours ago, Ghideon said:

Thanks! Just so I understand*: By "the entire Observable Universe" you mean "a really large distance", and this distance separates two observers in same frame of reference? 

 

*) "Observable universe", to me, implies that GR and other applicable models have to be taken into account.

Sort of. The only thing I take from GR so far is constant c (the delay). I am not yet interested to introduce cosmological measurements. When I wrote :

"Then say that the horizontal segment from A to B is the entire Observable Universe." the important word is IS. To me there is a split between what information we get (the image) and what is. In graph 12 the red dot is what is. The orange dot is what is observed.

 

1 hour ago, Eise said:

No, it is not: it is a God-like view of the universe, in which God sees past, present and future. If the location T-1 was occupied by A, then this is an unchanging fact, represented by a dot in the spacetime diagram. That dot will stay there, otherwise you are giving a false depiction of what happened at T-1 at X-0. And for such a 'Lord of the Ages' there is no now. 'Now' is only an expression referring to the same time when an utterance was made, and that presupposes that the one who is uttering the expression is in spacetime. 'God' lives outside the spacetime diagram.

It is a map. A spacetime diagram is a map. A way for us to depict reality. Can we agree on this?

Edited by michel123456

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