Is time just an illusion?

AnssiH

I suppose Doctordick is going to be back soon, so time to reply...

Quote by Doctordick

I tried to correspond with Chalmers for a while a number of years ago. I am sure he wouldn't remember me as he insisted on pushing the idea that he knew a lot more about the problem than I did (I took it to be pretty well an emotional reaction sans thought). I could not get it across to him that the problem he was trying to solve was actually a consequence of presuming his world-view was a valid representation of reality: i.e., he was presuming his question had meaning when, if fact, it was “completely up to how one wants to understand the situation”.

Exactly.
Seems you've met a lot of famous people btw :)

But back to your last post on this thread; all in all I think you pretty well understand what I am doing. I think your only confusion seems to be from trying to extend what I am saying beyond what I am actually saying. The sole purpose of the current discussion is to convince you that my equation,

[tex]\left{\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j} \beta_{ij} \delta(\vec{x}_i - \vec{x}_j) \right}\vec{\Psi}= K \frac{\partial}{\partial t}\vec{\Psi} = iKm \vec{\Psi},[/tex]

does indeed follow directly from my definitions. Trying to interpret what those definitions mean (beyond the meanings required by the deduction itself) isn't really an important issue. Not now anyway and I think my being drawn into such a discussion really isn't beneficial though such things will arise later. The meanings required by the deduction will become an important issue when it comes to interpreting the solutions to that equation. It is at that time that I will start defining other things. What is important is that these additional definitions must not conflict with those definitions already established.
A wise choice; however, you are making it clear that you will not be satisfied until you understand the solutions so we do have some work ahead of us.

Yes, I think there is. It may take a while since I don't always have a lot of time to dedicate on the issue. So, thank you for your patience :)

I have just reread both #477 and #478 (the two are actually one post) and there is quite a bit of the important issues covered there. If you do manage to figure out the details of those two posts, I think you will be close to being convinced that my equation is a valid representation of the most important logical constraints on a flaw-free explanation.

I am sure you will have some questions so I will figure on ironing those out when I get back.

Yeah, I've been spending some time with the #477/#478 post, and I still have some questions I couldn't figure out.

To summarize where I'm at;
I understand the addition of invalid ontological elements, in order to:

- make the number of elements the same in each "present".
- make sure there are no identical present (to make "t" retrievable).
- make sure there are no identical presents even if any given single ontological element was removed from any given present (to make sure a missing element could be retrieved from the table if we were given all but one element)

(Btw, since each tau was associated with a specific X, they together constituted a "single element", i.e. a "missing element" always refers to a "x & tau"-pair... Right? Thought I'd say it out loud since this can cause confusion)
---

Then, a function "f" was defined, as the function which outputs the missing element when input with any given "present" missing any given element. Seen as a vector function:

[tex]\vec{(x,\tau)}_n= x_n\hat{x}+\tau_n\hat{\tau} = \vec{f}((x,\tau)_1,(x,\tau)_2, \cdots, (x,\tau)_{n-1})[/tex]

(I don't know what "X hat" and "tau hat" mean, and so I couldn't figure out what the middle part of that equation says)
---

Then, a function F was defined, as the difference between the missing index and the result of the function "f". I.e;

[tex]F((x,\tau)_1,(x,\tau)_2, \cdots, (x,\tau)_n})= \vec{(x,\tau)}_n - \vec{f}((x,\tau)_1,(x,\tau)_2, \cdots, (x,\tau)_{n-1})\equiv 0[/tex]

(Perhaps I asked this before but forgot; what does [tex]\equiv[/tex] mean there exactly?)

Would it be correct to say that F is a function which "tests" every element of a "present" with some function f? Or is it more proper to just say it is "any function which returns 0 when its input with a full present"?
---

And here I start to struggle little bit more. I'm note sure how does the above turn into:

[tex]\sum_{i \neq j}\delta(x_i -x_j)\delta(\tau_i -\tau_j) = 0[/tex]

I understood this is just one of many functions that satisfies the requirements for F, and since we have chosen it as F, it will have an effect on what labels we put on the ontological elements. To quote you:
"It is thus a fact that the equation will constrain all labels to be different and any specific collection of labels can be reproduced by the simple act of adding “invalid ontological elements” until all the wrong answers are eliminated."

What I don't understand is the latter part of that sentence. Perhaps I have misundestood something, but maybe you could just explain in more detail, how is it that it reproduces a specific collection?
---

Then we get to the propability function. Thank you for the helpful information about "complex conjugate". I understand the need of squaring psi, but I don't know what the psi itself was to accomplish. I went back to the old posts, but all I can find is the idea of seeing any result of any function as a vector. (I do remember the stuff about squaring and re-normalizing)

Consequently, I cannot figure out what you mean by;

Quote by Doctordick

With the “invalid ontological elements” I introduced to make that sum over Dirac's delta function become the F function I needed, I know that, whenever I have the correct set of numerical references to my ontological elements,

[tex]\sum_{i \neq j}\delta(x_i -x_j)\delta(\tau_i -\tau_j) = 0, [/tex]

If I don't, then that sum is infinite! Against this, I also know that, if I have an incorrect set,

[tex]\vec{\Psi}(x_1,\tau_1,x_2,\tau_2,\cdots,x_n,\tau_n,t)=0[/tex]

as the probability of seeing that particular set of references must be zero and the probability is the sum of the positive definite squares of the components of Psi (that means that every one of those components must be zero and Psi must totally vanish). This means that no matter what arguments are inserted as numerical references to that collection of ontological elements, the product of those two above must be zero (if one isn't zero, the other is). It follows that

[tex]\sum_{i \neq j}\delta(x_i -x_j)\delta(\tau_i -\tau_j)\vec{\Psi}(x_1,\tau_1,x_2,\tau_2,\cdots,x_n,\tau_n,t) = 0, [/tex]

without exception.

Perhaps you could explain that to me in more detail?

Thank you
-Anssi

Evo

I'm afraid that this thread has been allowed to remain in violation of the guidelines on personal theories for too long.

It is against our Posting Guidelines to discuss, in most of the PF forums, new or non-mainstream theories or ideas that have not been published in professional peer-reviewed journals or are not part of current professional mainstream scientific discussion.

http://physicsforums.com/showthread.php?t=5374

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