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Old 07-09-2008   #271 (permalink)
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Re: What can we know of reality?

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Originally Posted by modest View Post
I’ve moved the above post..
Thank you. I was having a bad time with some of LaTex expressions last night and I guess I opened the wrong window. I had started to make a post to that thread which I had decided to abort. If you know anybody in power, let them know that this LaTex in quotes is a bear. Other forums don't seem to have the problem so they ought to be able to fix it.

Thank you again -- Dick
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Old 07-09-2008   #272 (permalink)
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Re: What can we know of reality?

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Originally Posted by ldsoftwaresteve View Post
About the rainbow, no, unless there's someone perceiving it. The construct is created by the observer and by the positional relationship of the observer and the light passing through the water droplets. Good example by the way. When we project it into existence as an actual object, we might conclude that we can slide down it.
Actually the rainbow might be a very confusing example as far as explaining what the epistemological analysis is about, because it is sort of a special kind of "thing" in the common physical worldview; an observer-dependent "thing". That is why I said "not having that much to do with the epistemological analysis but with your debate about what we should mean by "real".

Soooo, for this thread forget I ever mentioned it

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Mentally, I suppose we can but not ontologically, if I understand the meaning of the word as you are using it. Could I have substituted the word, 'epistemologically' in place of 'Mentally' in the previous statement?
No I don't think so.

Epistemology commonly refers to the study of knowledge. I.e. what does it mean to "know" or to "understand", and how does one acquire "knowledge". Especially that first bit of knowledge that allows one to interpret any data at all in any sense at all.

Ontology refers to the study of "existence", i.e. what actually exists in reality as oppose to only in our conception of reality (e.g. is "soul" or "mind" or "self" or "spacetime" or "wave function" or "shadow" or "electron" ontologically real entity, or are they rather mental concepts - imaginary entities - that we use as part of our model/conception of reality)

Note that physics is (or should be) different from ontology in that it seeks valid models; valid for making predictions. The entities that are defined as part of a valid model are far too often taken as ontologically real more or less tacitly (and completely unnecessarily), but that's just where a strong sense of ontology (philosophy) comes in handy. (Funny sidenote; often the original author(s) of a new model see the entities they invented as completely imaginary tools, but then many people who interpret and/or validate their work slowly start conceiving them as ontologically real entities just because the model works, i.e. "it explains what they see")

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For example, if the math checks out, it proves unequivocally that it is valid to model reality relativistically, predictionwise. But at the same time it implies strongly that the source of the relativistic description is not relativity of simultaneity or relativistic spacetime construction in any ontological sense, since completely epistemological standpoint already makes that sort of description valid.
Assuming, of course, that the 700 pound gorilla is consistent. Still, just so I don't screw this up, please restate this another way.
Yes... I think I very definitely should (btw, 700 pound gorilla = one's worldview?)

So let me restate with some more detail.

I referred to Doctordick's presentation as "epistemological analysis", in an attempt to underline the fact that it only deals with the constraints regarding our conception of reality, instead of constraints to reality itself. Too many people erroneously interpret the analysis as if it referred to how reality is (and tune out having decided it doesn't make sense).

The analysis is about properties found from our models of reality, and if such an analysis happens to show that;
1. A certain way of classifying reality (=making definitions) into "intelligible entities" leads inevitably to relativistic description being valid
AND
2. That "certain way of defining entities" is forced upon us as long as we don't want to make undefendable assumptions about reality.

Then it means that a (predictionwise) valid relativistic conception of reality can always be built regardless of what the true ontology of that reality is. Let me reinforce that it can always be built because "we can always define entities that way", i.e. we can always tack identity to the patterns of noumena that way. And not only we "can"; devoid of undefendable assumptions, we MUST.

That means there is absolutely no reason to place the concepts associated with relativity up to any ontological status, or to choose between any ontological views of relativity. You still can make all sorts of ontological assumptions, but it would be somewhat amazing co-incidence if you happened to land upon correct ontology (not to mention you would never be able to know)

I.e. if the math checks out, reality is what it is, and time dilation etc is brought through to our models by how we have defined (many) things in our worldviews (from the very bottom of our worldviews).

If that sounds like I'm saying the same thing over and over, it's because you probably understood what I tried to say ;D

And let it be said this is just one small (but interesting) part of the analysis.

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or refusing to believe that (initially) completely unknown data stream can be predicted meaningfully
I assume you mean the future, when you refer to completely unknown data stream.
No I mean, ~when you are born. When you don't yet know jack about how reality is, and you are supposed to start building a worldview. Imagine a brain that has not got a clue what the alien patterns mean, e.g. how to interpret the impulses so to conceive it in a form of a 3D environment, or to "hear sounds" or even "words" and "language". Before that brain has even formed a conception of such a thing as "self".

I suppose the reason some people think such data cannot be predicted is because they think that the meaning of some small part of the data must be first known before we can use that knowledge to interpret other parts of data and figure out new information. Think of a book in alien language. That is how our own learning of anything is often seen; we must first know some information, and then use that to interpret new data and recover new information from it. (Think about how we build physics models, make experiments, interpret the experiments according to those models, etc.)

But those people fail to recognize that one can always build a valid model about that data (for example making assumptions until they yield meaningful interpretation) without truly and explicitly knowing the meaning of any single pattern in that data. Without ever becoming completely certain about the meaning of anything in that data! But still being able to make reasonable predictions. That is how you know your native tongue; no one explained you what the words mean when you were a baby, you made assumptions that yielded reasonable interpretation of what you heard (like in the duck example Doctordick just gave)

And still one more important point; people often think that "immediate perception" is not part of "interpreting sensory data according to one's worldview". When I refer to interpretation, I don't necessarily refer to a conscious process. And when we perceive anything at all, that is always a pattern being recognized as something that is defined in our worldview. I.e. what makes a perception is some pattern having been "understood" as "something".

Also it makes no difference - as far as this analysis goes - how much of that interpretation is hardwired in us at birth and how much is result of our building of a worldview. For all intents and purposes, we can here consider a learning process that starts with sensory data patterns whose meaning is initially completely unknown, and yet it manages to start building a worldview (~form ideas of what such and such data patterns might mean), and start interpreting further data accordingly (and recover further information from that data, so to build a larger worldview). All that "modeling of reality" is done so that predictions can be made about the data, without ever knowing the explicit meaning of any of it.

Interesting property of such worldview is semantics. There's no information about the "real meaning" of any of the data to be found from that worldview at all, there is only a large self-coherent set of definitions or "set of assumptions that support one another and yield sensical interpretation of incoming data". When some learning system uses such a "circle of beliefs" to "understand" its sensory data and to further work on its worldview, it amounts to semantical understanding, even if all that further learning and interpretation is done in completely mechanical manner. That right there is a solution (in principle) to a very perplexing problem of AI, how could a machine understand semantics? Short answer; by not allowing it to "understand" any bit of reality explicitly, but force it build a model about the data from complete scratch.

Notice the parallels to how we don't know the reality itself, but instead our awareness is our mental model of reality; it is how we conceive reality.

...aand that's what I meant with "initially completely unknown data stream", i.e. data stream whose meaning is completely unknown.

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Also note that people tend to base their query of reality on their perceptions, as if there is something ontologically given in the meanings that have been assigned to some otherwise unintelligible patterns. Let us not forget that any sort of perception of anything, no matter how simple, entails that the perceived thing/pattern/noumena had already been defined. I.e. some sort of identity had been assigned to some spatial/temporal pattern. When you make a definition, allowing you to tack identity on some pattern/noumena, it does not mean that suddenly ontological reality exists accordingly.
Agree. You say, 'rainbow' and I know what you mean.
Now I regret I ever said rainbow
I hope with the above you have a, let's say "more detailed" idea of what I meant.

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Does the analysis still hold water if the worldview is weak? Or do you mean if it is self-coherent, it isn't weak?
Well any actual worldview that any actual person carries, is probably quite incoherent. It is impossible to keep such a huge amount of associations or assumptions coherently together, especially when the worldview is changing all the time. That is, we often make logical mistakes. Btw, that's a solution to another perplexing problem of AI; how does a machine make logical mistakes? Well, its logical problem solving does not come about the behaviour of logic gates (transistors) like is the case with calculators. Instead it has built a model of logical relationships; it's understanding of logic is part of its huge semantical worldview.

About whether the analysis "holds water", I can only point out that it does not even attempt to uncover properties that would be found from "all possible worldviews", but let's say properties that are to be expected from a hypothetical self-coherent worldview.

The purpose being that if such properties are found, and if those properties just happen to look exactly like some things or relationships that are commonly taken as "ontologically real", then one might want to re-check their ontological perspective on reality, I would say.

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This conversation is good on many levels. I think that mankind has made a habit of projecting into existence things which exist only in our minds.
Yes, very annoying habit. "Naive realism" it's called. Almost no one thinks they abide to naive realism, yet most insist certain definitions in their worldview are also ontological things by themselves.

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And I suspect that somehow that mistake is the cause of a lot of pain and misery. The one big example I can think of besides 'time' is 'God'.
Yes, what I'm saying above is also reflected in issues of religious faith and on many things. This mail is so lengthy because this issue has so much to do with just about any human activity one way or another, and I just keep pointing them out, hoping people can interpret me more properly.

It is easy to misunderstand a lot of things I'm saying above, especially as I am being rather careless in some assertions (don't have time to try and remove ambiguities more). If something seems incredibly odd, ask me and I'll try and clarify it for you.

I hope this has been helpful for anyone trying to figure out what this thread is all about. It would be good to have more people to really understand the conversation.

Doctordick, I'll have to get to your post tomorrow, it is once again very late

-Anssi
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Old 07-10-2008   #273 (permalink)
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Re: What can we know of reality?

Jesus H. Christ.

Thanks Ansii. I appreciate the time it took to put it all down here. I have some shifting to do to get on the same page and that'll take some time. And I don't want to distract you guys.

I am very happy that someone else is talking to DoctorDick who can converse intelligently with him. I had the impression from day 1 that he had something important to say. And I felt absolutely horrible that I couldn't understand him. It wasn't his fault, it was mine. I had so many blindspots (and still do) in my worldview that conversation on this level wasn't and probably still isn't possible.

I may comment on some things but only to spur or trigger more clarity in a given area. Sometimes blindspots are a bitch to get rid of. And my brain always walks slowly and in random directions.

One more thing. The 700 pound gorilla to me means 'now', the bleeding edge of experience, the current state of existence. Perhaps it would be more accurate to say that it is the invisible gorilla since we do not perceive the underlying activity except indirectly.

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Old 07-10-2008   #274 (permalink)
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Lightbulb Re: What can we know of reality?

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AnssiH
So let me restate with some more detail.
Good Post Anssi,
Thanks for taking the time to spell it out.
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AnssiH
Yes, very annoying habit. "Naive realism" it's called. Almost no one thinks they abide to naive realism, yet most insist certain definitions in their worldview are also ontological things by themselves.
IMHO I think, you got this right.


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Old 07-12-2008   #275 (permalink)
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Hi, sorry I haven't been able to use much time on this the past couple of days...

Quote:
Originally Posted by Doctordick View Post
Notice that there are only two operators there. The issue is that those two operators are embedded in a collection of operators so what I really want to know is the exact form of the factor when the two are commuted. That is why I put in that term “(some junk B)”; in order to remind you that there were additional factors to the right of the two terms of interest.
I think it finally dawned on me what I was missing (while doing something else actually)...

It's that;

\left[\frac{\partial}{\partial x_i}f(\vec{x}) + \left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](B)

is the same as:

\left\{\frac{\partial}{\partial x_i}f(\vec{x})\right\}(B) + \left[\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](B)

...which is somewhat what I was expecting to see... Yeah, it's my unfamiliarity with math that makes me miss the most obvious things, especially when I'm not sure what (and where) I'm supposed to look for So, did I get that right this time?

Still, about how \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2 relates to \left\{(\frac{\partial}{\partial x})(f(x))\right\} (or about me trying to figure out how does the "product rule of differentiation" help with the commutation of \vec{\Psi}_2^\dagger)

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The first step here is to examine carefully exactly what the differential term looks like. The dot product, \vec{\alpha}_i \cdot \vec{\nabla}_i was defined to be

\left[\alpha_{xi}\hat{x}_i+\alpha_{\tau i}\hat{\tau}_i\right] \cdot \left[\frac{\partial}{\partial x_i}\hat{x}_i + \frac{\partial}{\partial \tau_i}\hat{\tau}_i\right].

Now, if you remember the dot products between unit vectors (the answer is unity if they point in the same direction and zero if they are orthogonal) you will see that every term of this product is an alpha operator times a differential with respect to a specific argument.
I think I understand everything up to this point, but about this I have a question.

So I suppose the dot product essentially turns like this:
(\alpha_{x1}\hat{x}_1)(\frac{\partial}{\partial x_1}\hat{x}_1) + (\alpha_{\tau 1}\hat{\tau}_1)(\frac{\partial}{\partial \tau_1}\hat{\tau}_1) + ... etc?

That is also what I would expect to see when looking at the dot product definition at Wikipedia: Dot product - Wikipedia, the free encyclopedia
(and your explanation of the same issue at #246)

But I feel a bit uneasy about this because when looking at that Wikipedia explanation, I do not understand why you mention the orthogonal unit vectors, as I don't understand when would orthogonal unit vectors ever get multiplied by each others... Or is that dot product "definition" at Wikipedia omitting the dot product of orthogonal vectors exactly because they'd amount to 0 anyway? I am really just guessing :I

Also I'm not quite sure why the unit vectors are indexed there... i.e. what's the difference between \hat{x}_1 and \hat{x}_2. Just asking to make sure I'm not interpreting something wrong again.

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So, what we really want to know is what is the consequence of that differential operator (it is the only operator which makes any difference).

So that is how I came to concern myself with commutation of that differential with the various functions. You must understand that the central issue is the integral over all arguments of set #2. In that respect there is no difference between commuting \vec{\Psi}_2 to the left and commuting \vec{\Psi}_2^\dagger\cdot to the right. What is important is that we bring \vec{\Psi}_2^\dagger\cdot \vec{\Psi}_2 against one another because that is a well defined expression.
So, if I'm getting this right, the f(\vec{x}) represents whatever the function that the index argument(s) have on... um, \vec{\Psi}_2? (That's what f means in post #194 then too?)

So is the issue here that you are effectively commuting \vec{\Psi}_2 to the left? No?

I'm really guessing here, the biggest hole in my understanding right now has to do with how does the product rule of differentiation actually aid in our goal of bringing the \vec{\Psi}_2^\dagger\cdot and \vec{\Psi}_2 together... I mean I think I understand how the product rule is used as far as that example with \left\{(\frac{\partial}{\partial x})(f(x))\right\} goes, but then I'm still not sure what the fundamental equation looks like by the time all the \vec{\Psi}_2^\dagger's are against \vec{\Psi}_2's. I don't think it was explicitly stated anywhere.

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What I am getting at here is that the single most important factor is commutation of the differential operator with a function which is dependent upon the argument with respect of which that differential is being made.
This makes it sound like I might be on the right track with my assumptions on this post...

-Anssi
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Old 07-12-2008   #276 (permalink)
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Re: What can we know of reality?

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Originally Posted by Doctordick View Post
If you know anybody in power, let them know that this LaTex in quotes is a bear. Other forums don't seem to have the problem so they ought to be able to fix it.
It's been passed up the chain-o'-command and it's in the works.

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Old 07-13-2008   #277 (permalink)
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Re: What can we know of reality?

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Originally Posted by modest View Post
It's been passed up the chain-o'-command and it's in the works.
Thank you!

Hi Anssi,

You make me ashamed of myself with all the effort you are putting into this. After reading your post, I realize I need to put more effort into looking for errors in my posts.
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Originally Posted by AnssiH View Post
Hi, sorry I haven't been able to use much time on this the past couple of days...
You owe me no appologies.
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Originally Posted by AnssiH View Post
\left[\frac{\partial}{\partial x_i}f(\vec{x}) + \left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](B)

is the same as:

\left\{\frac{\partial}{\partial x_i}f(\vec{x})\right\}(B) + \left[\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](B)
Exactly correct; as long as you remember that the left hand partial derivative with respect to x_i operates only on f(\vec{x}) and not on (B) while the right hand partial derivative does operate on (B). The notation here is not very clear; that is why I keep harping on the issue.

Yes, I think you are getting it quite right. The real issue here is when and where we can commute these factors. As I have said a number of times the algebra being done in post # 194 in order to get to Schroedinger's equation is actually quite simple though you have to understand how these element commute.
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Originally Posted by Doctordick View Post
If we left multiply the above equation by \vec{\Psi}_2^\dagger (forming the inner or dot product with the algebraically modified \vec{\Psi}_2) and integrate over the entire set of arguments referred to as set #2, we will obtain the following result:

\left\{\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\right\}\vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right. +
\left.\int \vec{\Psi}_2^\dagger \cdot \left[\sum_{\#2} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1

This is the expression we are trying to algebraically derive from my fundamental equation. Notice that there is essentially no doubt as to what is being differentiated here; it is the intermediate steps which are difficult to express. Post #194 is the one we need to be getting through “one line at a time”.
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Originally Posted by AnssiH View Post
Still, about how \vec{\nabla}_i \vec{\Psi}_1\vec{Psi}_2 relates to \left\{(\frac{\partial}{\partial x})(f(x))\right\} (or about me trying to figure out how does the "product rule of differentiation" help with the commutation of \vec{\Psi}_2^\dagger)
What you are missing is the vanishing of both \vec{\Psi}_2 and \vec{\Psi}_2^\dagger in the first term of the equation I just quoted from post #194. You can see it either way: commute the \vec{\Psi}_2^\dagger to the right or \vec{\Psi}_2 to the left. They are essentially the terms I refer to as (some junk A) and (some junk B). Notice that our product rule has produced a term (the differential of f(x)) where B is not differentiated.

Commuting the two terms together and integrating over all possibilities for set #2 then yields an answer of unity and the first term in that quoted equation arises. The other terms still have integrals to be done.
Quote:
Originally Posted by AnssiH View Post
I think I understand everything up to this point, but about this I have a question.
And indeed you should! First of all you are absolutely correct, there should be no index on \hat{x} or \hat{\tau}. It is easy to get sloppy with LaTex, particularly when one is impatient and I get that way easily.

The dot product actually turns out like this:

(\alpha_{x1}\hat{x})\cdot(\frac{\partial}{\partial x_1}\hat{x})+(\alpha_{x1}\hat{x})\cdot(\frac{\partial}{\partial \tau_1}\hat{\tau}) + (\alpha_{\tau 1}\hat{\tau})\cdot(\frac{\partial}{\partial x_1}\hat{x})+ (\alpha_{\tau}\hat{\tau})\cdot (\frac{\partial}{\partial \tau_1}\hat{\tau}) + ... etc?

If you then add to this the fact that \hat{x}\cdot\hat{x}=\hat{\tau}\cdot\hat{\tau}= 1 and that \hat{x}\cdot\hat{\tau}=\hat{\tau}\cdot\hat{x}= 0 then you get very close to what you wrote down: i.e.,

\alpha_{x1}\frac{\partial}{\partial x_1} + \alpha_{\tau 1}\frac{\partial}{\partial  \tau_1} + ... etc?

Orthogonal unit vectors yield a dot product of zero and parallel unit vectors yield a dot product of unity. I suspect Wikipedia is omitting the step I just laid out for you because it is usually taken to be obvious but, when you are doing involved algebra with many terms, leaving out steps can be dangerous.
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Originally Posted by AnssiH View Post
...but then I'm still not sure what the fundamental equation looks like by the time all the \vec{\Psi}_2^\dagger's are against \vec{\Psi}_2's. I don't think it was explicitly stated anywhere.
It is explicitly laid out in post #194; the post we should be examining term by term.

I think you are very much on the right track. In order to make things a little easier, I am going to post a new thread called “Deriving Modern Physics from my Fundamental Equation”. I am going to open the thread with a copy of the essence of post #194. Hopefully the powers that be will allow such a thing. This thread is beginning to get a bit cumbersome.

To all those who are worried about a completion to this thread, I will answer the question, “What can we know of reality?” Modern physics is currently the best answer to that question.

Just so you understand where all this leads you should understand that by deriving Schroedinger's equation from basic principals, I have, in effect, shown that all conceivable universes may be seen as a three dimensional space occupied by objects which are required by definition to obey classical mechanics in the classical limit. What I have shown can be taken in two different ways. One can see the result as demonstrating that our classical view of the universe (a three dimensional space occupied by objects which obey classical mechanics) is entirely general and capable of representing any conceivable universe or one can view my results as demonstrating that the fact that classical mechanics is true by definition and that no classical experiment tells us anything about the universe except perhaps that our definitions are self consistent.

With regard to the second viewpoint, if one takes the position that the job of a research scientist is to search out the rules which separate the "true" universe from all possible universes, then no classical experiment can provide any guidance on the subject whatsoever. Classical mechanics is itself a tautology.

And this is just the opening salvo of my work; there is considerably more to come.

Wish me luck -- Dick
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Old 07-15-2008   #278 (permalink)
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Re: What can we know of reality?

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Originally Posted by Doctordick View Post
You make me ashamed of myself with all the effort you are putting into this.
Well I've made sure to spend plenty of time enjoying the (incredibly short) summer too Pretty soon it's going to become all cold and dark over here again :P

Quote:
What you are missing is the vanishing of both vec{Psi}_2 and vec{Psi}_2^dagger in the first term of the equation I just quoted from post #194. You can see it either way: commute the vec{Psi}_2^dagger to the right or vec{Psi}_2 to the left. They are essentially the terms I refer to as (some junk A) and (some junk B). Notice that our product rule has produced a term (the differential of f(x)) where B is not differentiated.
Hmmm, well there are two things that confuse me at the above. The terms "some junk A" and "some junk B" did not end up next to each others in the example you gave in #265. Another thing is, am I supposed to see that term that the "product rule produced" (the differential of f(x)) somewhere in:


\left\{\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\right\}\vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right. +
\left.\int \vec{\Psi}_2^\dagger \cdot \left[\sum_{\#2} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j (\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1

?

Because if I am, I don't recognize it...

I think I understand the rest of the post. And I'll be replying over to the new thread soon then...

-Anssi

EDIT: Oh, and good luck
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Old 07-17-2008   #279 (permalink)
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Re: What can we know of reality?

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Originally Posted by AnssiH View Post
Well I've made sure to spend plenty of time enjoying the (incredibly short) summer too Pretty soon it's going to become all cold and dark over here again :P
Yes, one of the problems with living in the far north. I was born and raised in rural Illinois outside Chicago; we had plenty of snow come winter time. I now live in Mississippi and to tell you the truth, I don't miss the snow at all. When I was a child, I had an uncle who lived in South Dakota which we used to visit occasionally so I know a little of their weather. I really couldn't understand how anyone could possibly want to live in North Dakota and the idea that there was a major city in Canada north of North Dakota (Winnipeg) was simply beyond me. Yeah, I know Helsinki is six degrees further north but at least you guys have mountains between you and the north pole. Canada is more like Siberia: just a flat plane between them and the Arctic ocean. Here in Mississippi, we can often barbeque a steak outside in a tee shirt on New Year's day.
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Originally Posted by AnssiH View Post
Hmmm, well there are two things that confuse me at the above. The terms "some junk A" and "some junk B" did not end up next to each others in the example you gave in #265.
The answer to that difficulty is quite simple. You just need to recognize the fact that \frac{\partial}{\partial x}f(\vec{x}) is just another function of \vec{x} and, as a simple mathematical function (you can think of it as a table where the value of the function is listed for every argument of interest) it commutes with any other such function. In the proof of the product rule (as I presented it) “some junk A” and “some junk B” can be anything (any kind of mathematical operators); however, in the case of interest (that first term in my explicit expansion) A and B are essentially the functions \vec{\Psi}_2^\dagger (plus appropriate alpha operators) and \vec{\Psi}_2 so commutation is no problem (the order, right to left, of these terms, within the first term of my expansion, is immaterial). When I integrate over set #2, A and B, so defined, yield unity (plus the appropriate alpha operators) and vanish from the expression.
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Originally Posted by AnssiH View Post
Another thing is, am I supposed to see that term that the "product rule produced" (the differential of f(x)) somewhere in:

\left\{\sum_{\#1} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i\neq j (\#1)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\right\}\vec{\Psi}_1 + \left\{2 \sum_{i=\#1 j=\#2}\int \vec{\Psi}_2^\dagger \cdot \beta_{ij}\delta(\vec{x}_i -\vec{x}_j)\vec{\Psi}_2 dV_2 \right. +
\left.\int \vec{\Psi}_2^\dagger \cdot \left[\sum_{\#2} \vec{\alpha}_i \cdot \vec{\nabla}_i + \sum_{i \neq j\#2)}\beta_{ij}\delta(\vec{x}_i -\vec{x}_j) \right]\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1 = K\frac{\partial}{\partial t}\vec{\Psi}_1+K \left\{\int \vec{\Psi}_2^\dagger \cdot \frac{\partial}{\partial t}\vec{\Psi}_2 dV_2 \right\}\vec{\Psi}_1
Not exactly, but it will reappear if the first term of that expansion is worked out in detail. The differential operator is contained in \vec{\nabla}_i and, since there is but one function to the right of it in that term, the only function available to play the role of f(x) is \vec{\Psi}_1.
Quote:
Originally Posted by AnssiH View Post
Oh, and good luck
I think we have good luck. It appears the powers that be are going to allow the new thread. Or, if you are referring to my cold, it's almost totally gone now.

Have fun – Dick
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Old 07-17-2008   #280 (permalink)
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Re: What can we know of reality?

This has been one of the most fascinating threads I have ever tried to follow. I'm glad the major points were explained in english!

Thanks!

Last edited by Overdog; 07-17-2008 at 04:48 AM.. Reason: clarification
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