michel123456

Understanding common displacement

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57 minutes ago, swansont said:

What does that have to do with changing past events? Not knowing is not the same thing as changing.

Fair. But not knowing may lead to "not existing".

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Posted (edited)
On 5/23/2019 at 3:10 AM, michel123456 said:

In this simple diagram I am representing the displacement of an object from point a to point b (see below very short video explaining the concept)

 

 

record_000010.avi

Seriously, I'm not sure which is going to take me longer. Figuring out what the apparent riddle is here, or figuring out how to open an.avi file? Let me see first I have to get out the lap top wait for the backlog of updates to finish before I jam the works up by trying to do something else at the same time. Not to mention that I read that there is a major update that MS says is crucial for all MS system users to download and install and conventional wisdom suggests that said crucial download will not be included in the update backlog. Meaning that about the time I think everything is caught up and ready to go, I will go to SFN to play the .avi file, thenI will run into some serious wait time only to be informed when I run out of patience an begin to investigate that a critical update needs to downloaded, Do I want to do it now? To which I will click yes, only to come back an hour or more later to a new notification asking if I want to install the update now? I will sigh and shake my head then click yes, but now I'm afraid to leave, so I will sit there for what seems like forever for a little bar to fill in across the screen, and that is only if I'm lucky cause sometimes there is just a busy indicator without any sign of progress to encourage me... Really? An .avi file?

I'm just joking :):) my laptop needs to be caught up anyway. Lol, :(, oh God an .avi file... Truthfully I've always enjoyed your illustrations...

Edited by jajrussel

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Well that was painless. I'm going to have to read everything again just to make sure I understand what you are asking. But from the hip the illustration looks like an illusion, but then you never say that a,c is equal to a,b. However the diagram seems to suggest it. As you roll it posistion b would have to drop below the plain because if it rolls true then a,c would have to be equal to a,b. What you have shown is that posistion b is transitioned out of bounds as you roll the diagram. We would never consider a,c as a legitimate path because initially posistion c is out of bounds. Thus beyond consideration.

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Posted (edited)

Okay my original observation is flawed. The position of c was indicated on the original diagram. You have folded? a rectangle along the diagonal? To show what? Okay maybe I see it? a,c is actually the T axis, and c on the original diagram merely marks the intersect on the diagonal? The reality is that a,c is equal to b,c, but I chose to think the path was a,b, so I mentally associated a, b to a, c.The question is why? Because in the original diagram the placement of c suggest an association with b. Then I suspected a trick then shooting from the hip I devised a mental illusion to explain what I thought the original diagram was suggesting. It only gets worse if I'm still wrong...

Edited by jajrussel

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Okay I am completely out if phase in this conversation. It's not a riddle, but in my defense my contributions didn't do anything to make the conversation any more convoluted then it already was/is. It's actually seems to be starting to make some sense to me so I'm hoping Michel 123456 will continue by explaining why Q is is placed where it is and how that placement implies ancient. I'd ask him to draw me a picture, but... Well, he already did... Of course if anyone knows why Q is placed where it is, and why it implies ancient and wouldn't mind explaining it to me it would probably only make me feel more foolish, but at this point that seems to be the only direction left to me, and I'm strangly okay with it.

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On 6/6/2019 at 7:56 AM, jajrussel said:

I'm hoping Michel 123456 will continue by explaining why Q is is placed where it is and how that placement implies ancient.

Q is placed at point a (t=0) at the beginning of the path. The observer at point c has moved from there (t=1) and would consider Q in his past.

The common understanding say that c would consider that Q is the same himself in the past 1 minute ago (or 1 year ago : Q and c are the same observer that "travel in time").

What I say (unconventionally) is that since c cannot observe Q (you cannot directly see yourself 1 minute ago) it means that Q could be a different object than c.

The condition for Q to exist as a different object is that the spacetime coordinates at point a are free.

Because if c is a continuing entity that is somehow extruded through time between a and c, then it means that the coordinates at point a are occupied and thus Q cannot be a different object (and that is the convention).

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49 minutes ago, michel123456 said:

Q is placed at point a (t=0) at the beginning of the path. The observer at point c has moved from there (t=1) and would consider Q in his past.

The common understanding say that c would consider that Q is the same himself in the past 1 minute ago (or 1 year ago : Q and c are the same observer that "travel in time").

What I say (unconventionally) is that since c cannot observe Q (you cannot directly see yourself 1 minute ago) it means that Q could be a different object than c.

The condition for Q to exist as a different object is that the spacetime coordinates at point a are free.

Because if c is a continuing entity that is somehow extruded through time between a and c, then it means that the coordinates at point a are occupied and thus Q cannot be a different object (and that is the convention).

But that point is already occupied. That’s the problem. The original object can (in principle) observe it, because it was at the same coordinates: 0,0,0

You can’t have a second object with identical coordinates as the first. You said so.

 

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Posted (edited)
2 hours ago, swansont said:

But that point is already occupied. That’s the problem. The original object can (in principle) observe it, because it was at the same coordinates: 0,0,0

You can’t have a second object with identical coordinates as the first. You said so.

In this scenario, objects are somehow extruded through the time dimension. I existed yesterday & I exist today: that makes 2 "me" along the time line. Which means that the Time dimension does not behave like the Space dimensions. When an object is displaced in space it does not "extrude", it changes coordinates.

But if you consider that there is only one single "me" sliding along the Time line, then the back & forth coordinates are empty.

 

Edited by michel123456

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1 hour ago, michel123456 said:

In this scenario, objects are somehow extruded through the time dimension. I existed yesterday & I exist today: that makes 2 "me" along the time line. Which means that the Time dimension does not behave like the Space dimensions. When an object is displaced in space it does not "extrude", it changes coordinates.

But if you consider that there is only one single "me" sliding along the Time line, then the back & forth coordinates are empty.

 

It doesn’t matter. The spatial coordinate (0,0) is occupied at t=0

You can’t have another object occupying that bit of spacetime

It’s important to understand and enforce the rules of relativity and clock synchronization to be able to communicate about this

And the “back and forth” coordinates are not “empty” in the sense that you were at 0,0,0 and that cannot change

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On 6/7/2019 at 8:09 PM, swansont said:

You can’t have another object occupying that bit of spacetime

Can you prove it? By which means?

For example, can you communicate with an alien (or a spaceship) that is currently observing the Earth a year (or a month) ago?

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Just now, michel123456 said:

Can you prove it? By which means?

This was a condition you described earlier in the thread.

For fermions it’s the Pauli exclusion principle 

Just now, michel123456 said:

For example, can you communicate with an alien (or a spaceship) that is currently observing the Earth a year (or a month) ago?

I can send them a signal if they are  a distance greater than ct from me

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