What can we know of reality?

[ughaibu]

If you are not interested why do you bother?

Because I thought the article might be of interest to you. Is there something difficult about post 247?

[AnssiH]

You are right on the money except for one simple point. The probability P(set #2) has been expressed as the squared magnitude of a vector ([imath]\vec{\Psi}_2[/imath]) in an abstract vector space. That vector space has nothing to do with either the vector space of x and tau nor the abstract vector space used to express P(set #1) as the squared magnitude of [imath]\vec{\Psi}_1[/imath].

 

Ah, of course. Guess I should have realized that. So there will never appear a dot product between [imath]\vec{\Psi}_2[/imath] and [imath]\vec{\Psi}_1[/imath]

 

Also Modest pointed out how to use LaTex in quotes. The problem is that the backslashes get stripped out from quotes when the post is being parsed. But if you add two, only one of them gets stripped out and LaTex gets displayed correctly.

 

http://hypography.com/forums/tutorials-how-s/6457-how-use-latex-equations-2.html#post216498

 

Additionally looks like in quotes you cannot have spaces inside LaTex tags.

 

Of course reviewing your posts becomes little bit of a *****; prolly a good idea to copy the whole text to a clipboard before clicking review, and then pasting it back to editing box immediately.

 

Perhaps the issue can be clarified with a simple set of parenthesis. Instead of writing [imath]\vec{\Psi}_2^\dagger\cdot\vec{\alpha}_i\cdot\vec{\nabla}_i\vec{\Psi}_1\vec{\Psi}_2[/imath] as I did, suppose we instead write it as [imath]\vec{Psi}_2^\dagger\cdot\left\{\vec{\alpha}_i\cdot\vec{\nabla}_i\vec{\Psi}_1\right\}vec{\Psi}_2[/imath]. As far as the vector [imath]\vec{\Psi}_2^\dagger[/imath] is concerned, what is enclosed in the parenthesis is nothing more than an algebraic operator which operates on each vector component of [imath]\vec{\Psi}_2[/imath].

 

Okay, that seems pretty clear.

 

Essentially, it is very important that the factor [imath]\vec{\alpha}_i\cdot\vec{\nabla}_i\vec{\Psi}_1[/imath] does not have the same effect on every component of [imath]\vec{\Psi}_2[/imath].

 

Hmmm, actually I don't understand why that is important... *scratching head*

 

It is integration over all arguments of set #2 which is used to eliminate those arguments from the problem. In order to do that, we must uncover the relationships from the fundamental equation which can be factored from the original relationship. This means we need to examine the consequences of the differential operator (which I have explicitly given) and the consequences of the Dirac delta function (which only comes into play with the integration itself). The final purpose of this endeavor will be to obtain a functional relationship which explains the behavior of one element. This first step was to elucidate the analytical procedure which, through integration (essentially a sum over all possibilities), is capable of removing a set of variables (in this case, set #2) from my fundamental equation.

 

Right, I'm starting to have some idea about how that set #2 gets removed.

 

I don't understand what are the "consequences of the Dirac delta function which come into play with the integration", but then I still have large parts of #246 to digest.

 

One thing anyone familiar with differential equations will realize is that it is the boundary conditions which set the solutions of a differential equation to specific functions. A differential equation tells one about relationships but seldom limits a solution to one possibility. From an analytical perspective, boundary conditions are conditions imposed on the differential equation by, “the rest of the universe”. That is to say, it is exactly what impact the rest of the universe has on the behavior of a body which actually sets that behavior. These integrals (which remove the specific arguments from my fundamental equation) represent exactly what the impact of the rest of the universe is (at least, exactly what our expectations of that impact is; a subtle but important point).

 

Very important, especially since so many people tend to assume quickly that you are talking about ontological reality and I suspect that is what makes many people tune out so quickly. I have to say that what I find so healthy about this analysis - and I suspect I'm not alone - is exactly that it is dealing with epistemological consequences rather than posing them as consequences to ontological reality.

 

What I am getting at is the fact that the issues I am closing in on here are issues which modern physics totally ignores (and that includes Stenger's publication ughaibu brought up).

 

I don't have the expertise to follow Stenger's logical steps closely, but from what I gather, I do like the fact that his paper also has to do with the unavoidable consequences of certain symmetries in our worldviews. Certainly it is fair to say there are many similarities between your ideas, even if Stenger has not established these symmetries as absolutely necessary for "any self-coherent worldview that does not contain undefendable assumptions"... After all there are number of issues in your presentation that can be dealt with more or less separately if one so wishes. E.g. you can come to the conclusion that such and such symmetry always leads to a relativistic description of reality without knowing whether that symmetry is always necessary or not.

 

Also usually having couple of different perspectives to the same issue is helpful, since one perspective tends to emphasize certain features of the issue while another perspective may make different features clearer.

 

I can't help but wonder if Stenger was able to review your presentation; at least he should be able to pick up what the presentation is about quite readily :)

 

Later

-Anssi

[Doctordick]

Hi Anssi, sorry you are the last to receive my attention. I got the others first because answering them required me to think; answering you is easy because you almost always pick up on what I am saying almost no matter how I put it.

Also Modest pointed out how to use LaTex in quotes.

That is nice to know. Actually I am quite surprised that Hypography is the only forum using LaTex which still has that particular problem. You would think they would have it fixed by now.

So there will never appear a dot product between [imath]\vec{\Psi}_2[/imath] and [imath]\vec{\Psi}_1[/imath]

Absolutely correct!

Additionally looks like in quotes you cannot have spaces inside LaTex tags.

Spaces are obtainable through “backslash semi-colon”: i.e., [imath]\vec{\Psi}_1\;\;\;\vec{\Psi}_2 [/imath] as opposed to.[imath]\vec{\Psi}_1\vec{\Psi}_2 [/imath].

Hmmm, actually I don't understand why that is important... *scratching head*

The factor [imath]\vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1[/imath] contains a differential operator (the expression is going to become, when expanded, a sum of terms containing the factors [imath]\frac{\partial}{\partial x_i}[/imath] and [imath]\frac{\partial}{\partial \tau_i}[/imath]). What you have to remember is that [imath]\vec{\Psi}_2[/imath] is a vector function expressible in a abstract space: i.e., [imath]\vec{\Psi}_2=\Psi_{d_1} \hat{d}_1+\Psi_{d_2} \hat{d}_2+\cdots+\Psi_{d_k} \hat{d}_k[/imath]. There need be no relationship between those “k” [imath]\Psi[/imath] functions. Each and every one of those k functions could be entirely different relationships; for example [imath]\Psi_{d_2}[/imath] might be proportional to [imath]x_3^2[/imath] while [imath]\Psi_{d_7}[/imath] could, at the same time be proportional to sin[imath](x_3)[/imath]. Certainly you should understand that the differential of [imath]\Psi_{d_2}[/imath] need bear no resemblance whatsoever to the differential of [imath]\Psi_{d_7}[/imath]. It follows, as the night the day, that the [imath]\vec{\nabla}_i[/imath] operator certainly cannot be factored out of between the [imath]\vec{\Psi}_2^\dagger[/imath] and the [imath]\vec{\Psi}_2[/imath] vector functions. Each component of that dot product must be proportional to whatever that differential yields on that particular component.

I don't understand what are the "consequences of the Dirac delta function which come into play with the integration", but then I still have large parts of #246 to digest.

It is a simple issue of deciding what can and can not be factored out of that integration. The Dirac delta function in the fundamental equation is a function of the difference between two variables, usually denoted by [imath](\vec{x}_i -\vec{x}_j)[/imath]. The integration is over the arguments of set #2. If either of those two variables are taken from set #2, the function cannot be factored from the integration. You might look at the definition of the Dirac delta function as part of its definition is a statement of the integration. I will answer any questions you have after you look at the definition of its integration.

Very important, especially since so many people tend to assume quickly that you are talking about ontological reality and I suspect that is what makes many people tune out so quickly. I have to say that what I find so healthy about this analysis - and I suspect I'm not alone - is exactly that it is dealing with epistemological consequences rather than posing them as consequences to ontological reality.

I sure did warm up to that word “healthy” in there; I took it as quite a complement. Thank you. I think you and I see things from a very similar perspective.

I do like the fact that his paper also has to do with the unavoidable consequences of certain symmetries in our worldviews.

I am not upset with him on that point at all. It is a rather common issue expressed quite a bit in the theoretical physics community. What I didn't like was his presentation which is very much from the authoritarian physicists perspective, a perspective I think hurts more than it helps. His perspective is as embedded in the standard perspective as is any perspective put forth by the academy. For example, he talks about what he calls, “Point-of-View Invariance (POVI)”; which seems to be the central theme of his presentation. To quote his definition of “POVI”, “The models of physics cannot depend on the point of view of a particular observer. This implies that they should not single out any particular position or direction in space-time.” He clearly considers “position”, “direction” and/or “space-time” to be fundamentally well understood concepts.

 

These are exactly what the standard authorities consider to be the “fundamentals” of physics. My position is that none of these concepts can be supported a-priori. Essentially, these are some of the very concepts which I find presumptuous. He has already stepped out on an unsupportable branch. What I want the reader to do is to back off and consider the actual problem facing them; which is, they must start from a position that they have no idea what these concepts constitute. The very problem I am having with Qfwfq is that he wants things to be expressed in terms of these concepts and the whole issue I am talking about is lost if one starts there.

.. After all there are number of issues in your presentation that can be dealt with more or less separately if one so wishes. E.g. you can come to the conclusion that such and such symmetry always leads to a relativistic description of reality without knowing whether that symmetry is always necessary or not.

Yes, but doing that is side-stepping the entire issue I am trying to present. Most all decent physicists out there are quite familiar with what Stenger is presenting and most of them jump to the conclusion that I am trying to present that kind of argument.

I can't help but wonder if Stenger was able to review your presentation; at least he should be able to pick up what the presentation is about quite readily :)

That is because you are beginning to comprehend the physics implied by my attack. You understand where I am coming from because it is exactly the question which has been bothering you from before you first talked to me. Stenger already knows all the physics and how it relates to symmetry principals but I am quite sure that the question you and I have been talking about is absolutely beyond his comprehension. To him, space, time, position, mass and dynamics are all well defined concepts with which he has no problems. He is explaining how physics arises from these concepts and has no concern at all how these concepts managed to arise.

 

Sorry for the rant but Stenger almost symbolizes exactly the problem I have with the modern physicists perspective.

 

Have fun -- Dick

[AnssiH]

Hi Anssi, sorry you are the last to receive my attention. I got the others first because answering them required me to think; answering you is easy because you almost always pick up on what I am saying almost no matter how I put it.

 

 

No worries, I'm probably also often the slowest to reply back. :/ Though my summer holidays are coming up soon, should have little bit more time at my hands then :)

 

 

Spaces are obtainable through “backslash semi-colon”: i.e., [imath]\vec{\Psi}_1\;\;\;\vec{\Psi}_2 [/imath] as opposed to.[imath]\vec{\Psi}_1\vec{\Psi}_2 [/imath].

 

 

That's good to know. Though, I meant if you have spaces inside your LaTex code - which normally does not have any effect to the end result - in quotes such code often results in some error. I noticed that the hard way :)

 

 

The factor [imath]\vec{\alpha}_i\cdot\vec{\nabla}_i\vec{\Psi}_1[/imath] contains a differential operator (the expression is going to become, when expanded, a sum of terms containing the factors [imath]\frac{\partial}{\partial x_i}[/imath] and [imath]\frac{\partial}{\partial\tau_i}[/imath]). What you have to remember is that [imath]\vec{\Psi}_2[/imath] is a vector function expressible in a abstract space: i.e., [imath]\vec{\Psi}_2=\Psi_{d_1}\hat{d}_1+\Psi_{d_2}\hat{d}_2+\cdots+\Psi_{d_k}\hat{d}_k[/imath]. There need be no relationship between those “k” [imath]\Psi[/imath] functions. Each and every one of those k functions could be entirely different relationships; for example [imath]\Psi_{d_2}[/imath] might be proportional to [imath]x_3^2[/imath] while [imath]\Psi_{d_7}[/imath] could, at the same time be proportional to sin[imath](x_3)[/imath]. Certainly you should understand that the differential of [imath]\Psi_{d_2}[/imath] need bear no resemblance whatsoever to the differential of [imath]\Psi_{d_7}[/imath]. It follows, as the night the day, that the [imath]\vec{\nabla}_i[/imath] operator certainly cannot be factored out of between the [imath]\vec{\Psi}_2^\dagger[/imath] and the [imath]\vec{\Psi}_2[/imath] vector functions. Each component of that dot product must be proportional to whatever that differential yields on that particular component.

 

 

I think I understand what you are saying above, but I'm not sure what it says about my original question... So, when you said "Essentially, it is very important that the factor [imath]\vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1[/imath] does not have the same effect on every component of [imath]\vec{\Psi}_2[/imath].", you meant that otherwise it could be factored out directly? I do not know how important this bit of information is to understanding the equations so I don't know if I should dwell on it...

 

 

But one thing that I would like to ask right now; what effect does it have to [imath]\vec{\Psi}_2[/imath] that there's that [imath]\vec{\Psi}_1[/imath] as part of that algebraic operator. I.e. what's the difference between expressions:

 

 

[imath] \vec{\Psi}_2^\dagger \cdot \left\{ \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1\right\}\vec{\Psi}_2[/imath]

 

 

and

 

 

[imath] \vec{\Psi}_2^\dagger \cdot \left\{ \vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_2[/imath]

 

 

I remember the [imath]\vec{\Psi}_1[/imath] had to be there because it was potentially having an effect to [imath]\vec{\Psi}_2[/imath], but I am somewhat confused about that expression and how it would be carried out... I am not entirely sure if these questions even make much sense :P

 

 

It is a simple issue of deciding what can and can not be factored out of that integration. The Dirac delta function in the fundamental equation is a function of the difference between two variables, usually denoted by [imath](vec{x}_i -vec{x}_j)[/imath]. The integration is over the arguments of set #2. If either of those two variables are taken from set #2, the function cannot be factored from the integration. You might look at the definition of the Dirac delta function as part of its definition is a statement of the integration. I will answer any questions you have after you look at the definition of its integration.

 

 

You mean [math]\int_{-\infty}^{\infty} f(x)\delta(x-a)dx = f(a) [/math] ?

 

 

I am not sure how to interpret that expression properly... I do remember that the total integral of the dirac delta function is 1 by definition, maybe that is exactly what is being implied there?

 

 

If so, I'm not sure what it means regarding "what can and can not be factored out of that integration"

 

 

Hmmm, I think next, when I have time, I should try and go through #246 again as you've clarified few issues that did not make sense to me before, maybe that will clear some things up to me...

 

 

I sure did warm up to that word “healthy” in there; I took it as quite a complement. Thank you. I think you and I see things from a very similar perspective.

 

 

I think so too. Yeah people tend to see their perceptions as ontological facts far too readily, and it was good to see that Stenger's paper leaning at least little bit towards an epistemological perspective. Perhaps his analysis is still starting from undefendable assumptions (or "arbitrary definitions") regarding space and time, but still I'd say "one step to the right direction".

 

 

Talk about space and time, it is actually somewhat curious that people tend to take their immediate perceptions as "ontological facts" rather than as a (useful) perspective on reality, when the same people also believe in relativistic description of spacetime, and recognize that relativistic logic also leads to a multitude of logically equal ontological interpretations, without any hope of ever recognizing which one is correct. After that, it is strange how reluctant people are to consider the possibility that reality can always be seen from many different but equally valid perspectives just by defining things differently, and to consider that any definition of reality might just be what is done in order to draw predictions about reality, instead of being how reality has been somehow built.

 

 

I'd expect any "healthy" scientific analysis of reality to somehow recognize that issue. Just find it odd if people take your stance as some sort of idealistic scheme...

 

 

Sorry for the rant but Stenger almost symbolizes exactly the problem I have with the modern physicists perspective.

 

 

I understand your frustration. (maybe you should not have vented it on Ughaibu though :)

 

 

-Anssi

[Doctordick]

So, when you said "Essentially, it is very important that the factor [imath]\vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1[/imath] does not have the same effect on every component of [imath]\vec{\Psi}_2[/imath].", you meant that otherwise it could be factored out directly?

Yes!

I do not know how important this bit of information is to understanding the equations so I don't know if I should dwell on it...

What is under examination at the moment is, how does one obtain Schrödinger’s equation from my fundamental equation. This is essentially a straight forward problem in algebra so “understanding the equations” means understanding the structure and how it can be handled algebraically. There is really no deeper meaning in there; my whole presentation constitutes deducing alternate relationships embedded in the original expression.

 

We are talking about “expectations” being given via the squared magnitude of [imath]\vec{\Psi}[/imath] when the arguments of that function are numerical labels. The fact is that “physics” is all about the calculations of expectation when things being expected are numbers so physics must be a subfield of this analysis.

But one thing that I would like to ask right now; what effect does it have to [imath]\vec{\Psi}_2[/imath] that there's that [imath]\vec{\Psi}_1[/imath] as part of that algebraic operator. I.e. what's the difference between expressions:

 

 

[imath] \vec{\Psi}_2^\dagger \cdot \left\{ \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1\right\}\vec{\Psi}_2[/imath]

 

 

and

 

 

[imath] \vec{\Psi}_2^\dagger \cdot \left\{ \vec{\alpha}_i \cdot \vec{\nabla}_i \right\}\vec{\Psi}_2[/imath]

Right multiply the second by [imath]\vec{\Psi}_1[/imath] and they are exactly the same thing.

I remember the [imath]\vec{\Psi}_1[/imath] had to be there because it was potentially having an effect to [imath]\vec{\Psi}_2[/imath], but I am somewhat confused about that expression and how it would be carried out... I am not entirely sure if these questions even make much sense :P.

The “effect” is simple multiplication; expand all these expressions into their explicit meaning and then multiply those expanded factors times one another per the instructions of the representation: i.e., “dot” products require knowing that orthogonal unit vectors multiply to zero and parallel unit vectors multiply to unity and the differential operators yield the derivative of the factor they are multiplying plus the commutated result of the original product. The rule can be explicitly written as

[math]\left\{(\frac{\partial}{\partial x})(f(x))\right\}=\left\{f'(x)+(f(x))(\frac{\partial}{\partial x})\right\}[/math]

 

where what is contained within the curly brackets is embedded in a string of products. By the way, both [imath]\vec{\Psi}_1[/imath] and [imath]\vec{\Psi}_2[/imath] have to be there because the fundamental equation is an equation on [imath]\vec{\Psi}[/imath] (which is a function of all numerical arguments) and we have divided that set of arguments into two: set #1 and set #2 which have led to the definition of [imath]\vec{\Psi}_1[/imath] and [imath]\vec{\Psi}_2[/imath]. The problem in our factorization is that one of the two must include the arguments of the other to be absolutely correct. Remember, only the arguments from set number two will be removed by integration.

 

You mean [math]\int_{-\infty}^{\infty} f(x)\delta(x-a)dx = f(a) [/math] ?

Yes; essentially the integral over x is the function f(x) evaluated when x is equal to a.

I am not sure how to interpret that expression properly... I do remember that the total integral of the dirac delta function is 1 by definition, maybe that is exactly what is being implied there?

Just add in the fact that the Dirac delta function explicitly vanishes everywhere except when its argument (in this case, “x-a”) is exactly zero and the result of the integral should be obvious to you.

If so, I'm not sure what it means regarding "what can and can not be factored out of that integration"

In our case, we are essentially integrating over [imath]\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2[/imath] (which is effectively the f(x) in the above expression) times a Dirac delta function of the difference between an argument from set #1 and one from set #2. Now [imath]\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2[/imath] is defined to be the probability you expect the specific set of arguments represented by set #2 so the result of the integration is no more than that probability evaluated for that specific set of arguments (that is set #2) when they are identical to set #1. That is to say, the Dirac delta function when integrated over set #2 results in an interaction field for set #1 which is the probability of finding set #2 at the points represented by set #1. That simple fact requires all first order interaction fields in this interpretation of the universe to be boson fields; quite a reasonable requirement considering modern physics explanations.

… it was good to see that Stenger's paper leaning at least little bit towards an epistemological perspective. Perhaps his analysis is still starting from undefendable assumptions (or "arbitrary definitions") regarding space and time, but still I'd say "one step to the right direction".

My complaint arises almost directly from the fact that most all physicists see Stenger’s approach as being a very reasonable attack (notice that he has managed to get published). My difficulty is that I have solved a problem they “know” cannot be solved. The professionals who read my stuff are all wasting their time trying to figure out exactly what problem I have actually solved because they know it cannot be the one I claim it to be. They try and interpret it as some version of what Stenger has done but they can’t figure out what my starting assumptions are; they think I am just omitting to set them forth. So, in effect, they don’t even look at my arguments.

… when the same people also believe in relativistic description of spacetime, and recognize that relativistic logic also leads to a multitude of logically equal ontological interpretations, without any hope of ever recognizing which one is correct. After that, it is strange how reluctant people are to consider the possibility that reality can always be seen from many different but equally valid perspectives just by defining things differently, and to consider that any definition of reality might just be what is done in order to draw predictions about reality, instead of being how reality has been somehow built.

I think the majority of human beings find it very difficult to live with the idea that the basis of their beliefs on any level might be wrong especially if they think the consequences are important. I am of the opinion that this is the real source of their failure to consider possible alternatives. Most people take knowing the answer to a question to be more important than being correct: i.e., they operate on the assumption that what they “know” is correct. It’s a pretty overwhelming urge. I, on the other hand, decided many years ago that I didn’t know anything for sure and decided to just go with my gut: i.e., I always do whatever seems like the right thing to do and don’t worry about being wrong. My mother told me that I can always be forgiven if I do what I think is right but there is no forgiveness for doing what you feel is wrong no matter how reasonable it might seem to be.

I understand your frustration. (maybe you should not have vented it on Ughaibu though :)

It is possible you are right. Ughaibu, if you are reading this, I apologize; I didn’t mean to take my frustration out on you.

 

Have fun -- Dick

[ughaibu]

Ughaibu, if you are reading this, I apologize; I didn’t mean to take my frustration out on you.

No problem, I wasn't upset. A now deceased friend used to say, "dont say sorry, say thank you".

[AnssiH]

Hello again, sorry I'm slow but now I should have some more time on my hands.

 

I've been going through this Schrödinger equation deduction today and yesterday, and while I did figure some things out, I feel I'm quite a bit lost still...

 

Part of the problem is that it's hard to follow all the issues spread over multiple posts when I have so shaky understanding of these mathematical concepts. I'll have questions from the earlier posts too but I need to scratch my head with them little bit more still before I even know what to ask, but until then there are questions about your latest reply;

 

[imath]\vec{\Psi}_2^\dagger\cdot\left\{\vec{\alpha}_i\cdot\vec{\nabla}_i\vec{\Psi}_1\right\}\vec{\Psi}_2[/imath]

 

and

 

[imath]\vec{\Psi}_2^\dagger\cdot\left\{\vec{\alpha}_i\cdot\vec{\nabla}_i\right\}\vec{\Psi}_2[/imath]

Right multiply the second by [imath]\\vec{\\Psi}_1[/imath] and they are exactly the same thing.

 

What confuses me is that I'm not sure what does it mean that there's a multiplication between [math]\vec{\Psi}_1[/math] and [math]\vec{\Psi}_2[/math]. "Forming a dot product" has been referred to as "multiplication" so I've assumed "multiplication" and "dot product" are the same thing, but then I also understood that [math]\vec{\Psi}_1[/math] and [math]\vec{\Psi}_2[/math] exist in different vector spaces and cannot have a dot product between them...? Clearly I am missing something here, and its so easy to make wrong assumptions that it kind of feels like I'm walking on a mine field... :P

 

I apologize if you have already explained that part to me, I have either forgotten and cannot find it now, or I did not even understand the explanation properly in the first place...

 

The “effect” is simple multiplication; expand all these expressions into their explicit meaning and then multiply those expanded factors times one another per the instructions of the representation: i.e., “dot” products require knowing that orthogonal unit vectors multiply to zero and parallel unit vectors multiply to unity and the differential operators yield the derivative of the factor they are multiplying plus the commutated result of the original product. The rule can be explicitly written as

 

[math]\left\{(\frac{\partial}{\partialx})(f(x))\right\}=\left\{f’(x)+(f(x))(\frac{\partial}{\partialx})\right\}[/math]

 

where what is contained within the curly brackets is embedded in a string of products.

 

That might be an explanation for what I just asked but I don't really understand it. I understand that "orthogonal unit vectors multiply to zero and parallel unit vectors multiply to unity", but I do not understand what that says about what I asked, and I do not understand what "...plus the commutated result of the original product" means, what the math expression says and what it means that the contents of the curly brackets are "embedded in a string of products", and how does this relate to something in post #194...

 

Yup, fair to say I am slightly lost right now :/

 

I also am not sure I understand what you are saying about the integral of dirac delta function, but right now I feel I am so confused with the above that I let it rest for a min...

 

-Anssi

[Doctordick]

Sorry I have been slow in my response. We have been in Denver for our granddaughters second birthday and I have come down with a head cold which won't quit. There were about ten kids at the birthday party and I think kids carry more virulent diseases than any other crowd. At any rate, I really got sick the day before we left Denver. It has been almost a week now and I am starting to become human again.

Yup, fair to say I am slightly lost right now :/

Yes, I think you have a grasp of the problem; you are indeed “slightly lost”. The problem revolves around your concept of “mathematics”. I believe you see mathematics as a defined set of operations whereas I see mathematics as the invention and study of internally consistent systems. There is a subtle but very important difference in our perspectives. You are confused because you think an expression such as “AB” has to be defined before it can be used. Two symbols placed one after the other are generally presumed to imply multiplication and we can use the term “multiplication” to refer to such an expression even if multiplication between the two symbols has not been defined. If the operation is not defined, it simply means that “it cannot be done”: i.e., every time you come across such an expression, it must be left in that form. That is to say, writing down “AB” does not mean that A and B have been defined nor does it mean that the outcome of multiplication between the two has been defined. It is no more than a logical expression. If, further on down the road, the outcome of the operation is defined then we can use that definition; however, until such a thing is defined, what is important is that our manipulations are consistent with all possible definitions of that operation.

 

A good example of the sort of thing I am talking about are the [imath]\vec{\Psi}_a[/imath] functions which were defined by the fact that [imath]\vec{\Psi}_a^\dagger\cdot\vec{\Psi}_a[/imath] is the probability that your expectations for the arguments [imath](\vec{x}_1,\vec{x}_2 \cdots)[/imath] is the specific set for which that function is to be evaluated. That dot product is completely defined and results in an ordinary number bounded by zero and one. Against this, the expression [imath]\vec{\Psi}_a \vec{\Psi}_b[/imath] is simply not defined. It is rather, “an irreducible expression” analogous to the “AB” expression mentioned above where neither A nor B have been defined; that is, what one gets under multiplication is undefined.

 

The logical rules in such a circumstance essentially require that the mathematical operations available in your algebraic maneuvers are any operations consistent with the ordinary concept of multiplication. For example, (A+B)(C+D)=A(C+D)+B(C+D)=AC+AD+BC+BD. The only difficulty one need be aware of when performing algebraic manipulations with undefined entities is that of commutation. That issue concerns the order with which the multiplication is done (an issue I am referring to whenever I use the terms “left multiply” or “right multiply”): i.e., what is the relationship between AB and BA. Ordinary numbers yield exactly the same result; however, there exist definitions of multiplications which are not commutative so there is a need to be careful.

 

It turns out that commutation can not be a problem in the algebraic manipulation of our [imath]\vec{\Psi}[/imath] functions because our definitions require that [imath]\vec{\Psi}_a^\dagger\cdot\vec{\Psi}_b[/imath] be an ordinary number. That is to say, [imath]\vec{\Psi}_a^\dagger\cdot\vec{\Psi}_a\vec{\Psi}_b^\dagger\cdot\vec{\Psi}_b=\vec{\Psi}_b^\dagger\cdot\vec{\Psi}_b\vec{\Psi}_a^\dagger\cdot\vec{\Psi}_a[/imath]. The point of all this is the fact that our equations may be manipulated with simple straight forward high school algebraic operations.

 

I have already commented earlier that the “dot” products are defined only between certain expressions. Just for convenience, I tend to place the “dot” immediately after the first component of the product so that you can then easily locate the pair of expressions which are implied in that product.

What confuses me is that I'm not sure what does it mean that there's a multiplication between [math]\vec{\Psi}_1[/math] and [math]\vec{\Psi}_2[/math].

It is an abstract multiplication which has no meaning until after the expressions are completed by performing the dot product required to obtain the probability implied by the specific [imath]\vec{\Psi}[/imath] of interest. The purpose of the algebra is not to solve for these expressions but rather to discover constraints upon those functions implied by my fundamental equation.

"Forming a dot product" has been referred to as "multiplication" so I've assumed "multiplication" and "dot product" are the same thing, but then I also understood that [math]\vec{\Psi}_1[/math] and [math]\vec{\Psi}_2[/math] exist in different vector spaces and cannot have a dot product between them...? Clearly I am missing something here, and its so easy to make wrong assumptions that it kind of feels like I'm walking on a mine field... :P

The issue is that some of these products are defined and some are not; the elements between which a dot product has been defined are defined operations. The other multiplications are not defined and simply cannot be preformed until they are defined.

 

You can look at the problem as one of simple High School algebraic manipulation of undefined symbols. When you come across a pair which has been defined, you can perform the defined operation but that doesn't mean you have to (in most cases, the actual base functions involved will never be defined, we are only concerned with their existence); all you really have to do is recognize the kind of thing the defined operation will yield and I will let you know what and why that result is achieved. This is exactly what makes mathematics so powerful: you don't need to know what the things being represented actually are, not if you have defined meanings for all operations you actually perform.

 

Just as an aside, the central issue of algebra is the fact that if you have two things which are equal (that is references on opposite sides of an equal sign) then, so long as you do exactly the same thing to both sides, the equality remains valid. What you started with or what you did with it is unimportant so long as you avoid some critical issues like dividing by zero or commuting a differential operator with a differential-able function. (Two issues of significance in this work which need to be addressed whenever they may happen.)

I do not understand what "...plus the commutated result of the original product" means

Commutation of a differential operator has significant consequences, [imath]\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}[/imath] means that the partial differential is to be taken of everything to the right of the operator (including those functions which have been omitted) and not of the functions to the left of the partial (including those functions which have been omitted. By that definition, it should be quite clear to you that [imath]\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}[/imath] is certainly not equal to [imath]\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}[/imath] (the commutated expression.

 

Note the the original fundamental equation is made up of many products of things which are (or can be) sums of many terms. When fully written out the end result will be a sum of ineducable product of operator expressions far too complex to write down in this presentation. But that isn't really our concern. What we need to do is to express that end result as a valid product of factors which are, for the most part, exactly the same sums of many terms we started out with. That process involves knowing when and where the commutation of the various terms is a valid operation and what results when commutation is not valid. Note that the following two expressions yield identical results:

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

 

That is to say, the commuted term fails to include differentiation of the expressed [imath]f(\vec{x})[/imath] so it is necessary that we expressly include that term.

..., what the math expression says and what it means that the contents of the curly brackets are "embedded in a string of products", and how does this relate to something in post #194...

In post #194, I continually switch from writing out the entire sum (in terms of the expressed factors) to writing out individual terms in order to show what happens during the commutation of [imath]\vec{\Psi}_2^\dagger \cdot[/imath] as I try to commute that over against [imath]\vec{\Psi}_2[/imath] because that product is a defined entity.

I also am not sure I understand what you are saying about the integral of dirac delta function, but right now I feel I am so confused with the above that I let it rest for a min...

The expression [imath]\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2[/imath] is defined to be the probability of finding the set of arguments “set #2” to be explicitly as they are expressed in the function [imath]\vec{\Psi}_2[/imath]. When we integrate that expression over all possibilities for set #2 we get “unity” if everything, except [imath]\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2[/imath], can be factored from the integral before the integral is done. (That fact is used a couple of times.)

 

When we cannot factor a term such as [imath]\delta(\vec{x}_i -\vec{x}_j)[/imath] out of that integral, we can either leave it as a term yet to be integrated (which we must do if both i and j are chosen from set #2; however, when i is chosen from set #1 and j is chosen from set #2, we have a very unique case: in that case the integral over set #2 becomes exactly value of [imath]\vec{\Psi}_2^\dagger \cdot \vec{\Psi}_2[/imath] when that function is evaluated at the positions defined by the actual values of interest for set #1 (see the definition of an integral over the Dirac delta function). It follows that the probability density of set #2 generates a potential well for the elements of set #1. The sign of that potential well is quite interesting.

 

What we really need to do is examine the algebraic steps one at a time.

 

Have fun -- Dick

[AnssiH]

Sorry I have been slow in my response. We have been in Denver for our granddaughters second birthday and I have come down with a head cold which won't quit. There were about ten kids at the birthday party and I think kids carry more virulent diseases than any other crowd. At any rate, I really got sick the day before we left Denver. It has been almost a week now and I am starting to become human again.

 

Heh, yeah, good thing you are starting to get over it then.

I'm reverting back to the old way of quoting in this post because I can't figure out how to make some expressions work inside quotes (I think it's those \left\ and \right\ tags that break it, or the space in \partial x_i... Hypography please, everyone's waiting :D)

 

Yes, I think you have a grasp of the problem; you are indeed “slightly lost”.

 

"slightly" lost :)

 

Thank you for the helpful explanation about undefined expressions.

 

Related to that ->

===== QUOTE #263 =====

Commutation of a differential operator has significant consequences, [imath]\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}[/imath] means that the partial differential is to be taken of everything to the right of the operator (including those functions which have been omitted) and not of the functions to the left of the partial (including those functions which have been omitted. By that definition, it should be quite clear to you that [imath]\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}[/imath] is certainly not equal to [imath]\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}[/imath] (the commutated expression.

===================

 

Yup.

 

Is [imath]\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}[/imath] a defined operation at all then? At first I would have assumed it isn't, but when you say...

 

===== QUOTE #263 =====

Note that the following two expressions yield identical results:

 

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

===================

 

...I'm getting little bit confused.

 

If you were to remove everything after the plus sign, would you then just have exactly the same partial differentiation on both sides? (I assume the curved parentheses are there for clarity but otherwise meaningless)

 

But then something can be added to the other side and the equal sign still holds? Or is that precisely because the commutated expression is undefined?

 

Anyway, as I'm starting to get over the frustration I had about not understanding above, suddenly I understand what you are saying about integration with dirac delta function. I re-read your explanation from #260 and understood everything up to the point you make an interpretation of it, that I didn't quite get;

 

That is to say, the Dirac delta function when integrated over set #2 results in an interaction field for set #1 which is the probability of finding set #2 at the points represented by set #1. That simple fact requires all first order interaction fields in this interpretation of the universe to be boson fields; quite a reasonable requirement considering modern physics explanations.

 

Likewise I didn't get the interpretation from the last post: It follows that the probability density of set #2 generates a potential well for the elements of set #1. The sign of that potential well is quite interesting.

 

Heh, well I don't know what's meant with a potential well and what would make its sign interesting :)

 

Anyway, one step forward. Thanks.

 

-Anssi

[Doctordick]

Heh, yeah, good thing you are starting to get over it then.

Not near as well as I would like. Still hacking up flem like an old man. And we have to go to Atlanta this next weekend; a nephew is getting married. Maybe four days will fix things up a bit.

I'm reverting back to the old way of quoting in this post because I can't figure out how to make some expressions work inside quotes (I think it's those \left\ and \right\ tags that break it, or the space in \partial x_i... Hypography please, everyone's waiting :D)

I suspect you are just failing to double up all the left slashes. That “\left\” is not a valid tag; the back slash is an escape character which gives meaning to the following text. “\left” is the tag which creates a parenthesis which grows to cover the size of the expression after it. That means it has to have a closing tag “\right” or it doesn't know the extent of the expression which determines its size. “\{“ lets the system know you want to use a curly parenthesis which is ordinarily used to enclose LaTex expressions and does not create a printable character.

 

I have used your suggestion for getting around the stripping problem. I presently edit all my posts in “openoffice” and just paste it into the hypography edit box. I don't really think it takes much additional time and I can save partial work when something comes up (like my wife wanting some attention).

Is [imath]\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}[/imath] a defined operation at all then?

Certainly [imath]f(\vec{x})[/imath] has not been defined (I am using it as stand in for any function of [imath]\vec{x}[/imath] but the partial differential operator is well defined (though the outcome of differentiating [imath]f(\vec{x})[/imath] can't be).

At first I would have assumed it isn't, but when you say...

 

===== QUOTE #263 =====

Note that the following two expressions yield identical results:

 

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

===================

 

...I'm getting little bit confused.

You are failing to consider the consequences of things which might or might not be surrounding the expressions explicitly displayed. Consider the expression

[math](some\;junk\; A)\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}(some\;junk\; B)=(some\;junk\; A)\left[\frac{\partial}{\partial x_i}f(\vec{x})+\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](some\;junk\; B)[/math]

 

The differential operator operates on every function to the right of its position and does not operate on any functions to the left of its position. You need to understand the product rule of differentiation. Examine the differentiation of [imath]u\cdot v[/imath] as shown in that post. (Please note that the dot in that expression stands for general multiplication.) With regard to our problem, u is [imath]f(\vec{x})[/imath] and v is “some junk B”.

If you were to remove everything after the plus sign, would you then just have exactly the same partial differentiation on both sides? (I assume the curved parentheses are there for clarity but otherwise meaningless)

Sometimes, in spite of what some authorities state, mathematical notation is somewhat ambiguous. The parentheses were there to encourage you to think of the operation within the parenthesis to be an operator unto itself (the term without the parentheses is simply a term: i.e., the differential operation has already been performed).

But then something can be added to the other side and the equal sign still holds? Or is that precisely because the commutated expression is undefined?

The term on the right of the plus sign does not differentiate [imath]f(\vec{x})[/imath] but only ends up differentiating the omitted term, “some junk B”. The term to the left of the plus sign is taken to have operated only on [imath]f(\vec{x})[/imath] and to have resulted in a new function as apposed to the left side of the equal sign where the partial is an operator standing on its own which operates on the product of [imath]f(\vec{x})[/imath] and “some junk B”.

 

The whole purpose of this representation is to allow one to derive the algebraic result without being concerned with the whole of the total representation: i.e., exactly how can we handle the commutation of expressions without worrying about other terms in the finished expression. The fact that such results may be deduced piece meal is what makes algebra so powerful. If you had to hold the whole thing in your head while you performed the operations, the process would soon become unwieldly.

I re-read your explanation from #260 and understood everything up to the point you make an interpretation of it, that I didn't quite get;

 

Likewise I didn't get the interpretation from the last post: It follows that the probability density of set #2 generates a potential well for the elements of set #1. The sign of that potential well is quite interesting.

These comments have to do with the physics and don't actually have anything to do with the deduction itself. After you are sure you understand the algebra used to derive the Schroedinger's equation, we can worry about the pertinent physics. For the moment don't worry about it.

 

Have fun -- Dick

[AnssiH]

Not near as well as I would like. Still hacking up flem like an old man. And we have to go to Atlanta this next weekend; a nephew is getting married. Maybe four days will fix things up a bit.

 

Oh that could turn out nasty if it involves airplanes... Good luck!

 

I suspect you are just failing to double up all the left slashes.

 

Could be that too, but have you been able to quote [math]\partial x_i[/math] yet?

 

[math]partial x_i[/math]

 

:/

 

======= QUOTE #265 =======

You are failing to consider the consequences of things which might or might not be surrounding the expressions explicitly displayed. Consider the expression

 

[math](some\;junk\; A)\left\{\frac{\partial}{\partial x_i}\right\}\left\{f(\vec{x})\right\}(some\;junk\; B)=(some\;junk\; A)\left[\frac{\partial}{\partial x_i}f(\vec{x})+\left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](some\;junk\; B)[/math]

 

The differential operator operates on every function to the right of its position and does not operate on any functions to the left of its position. You need to understand the product rule of differentiation. Examine the differentiation of [imath]u\cdot v[/imath] as shown in that post. (Please note that the dot in that expression stands for general multiplication.) With regard to our problem, u is [imath]f(\vec{x})[/imath] and v is “some junk B”.

==============

 

I studied that wikipedia page a bit and I think I now have a superficial understanding of the product rule of differentiation. At some point I did make exactly the assumption dubbed as "common error" in that web page :P

 

But still I've got something little bit wrong apparently... Let me know where I get it wrong;

 

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

 

I assume that the above means essentially the same as

 

[math] d(u \cdot v) [/math]

 

from the wikipedia page.

 

Then following that

 

[math] d(u \cdot v) = v \cdot du + u \cdot dv [/math]

 

I would expect our case to turn out like this:

 

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

 

Basically I just added that one extra [math](B)[/math] there in the beginning to be multiplied by the partial derivative of f.

 

These seem like rather minor hiccups, will prolly be cleared quickly. Next, I can start working backwards in the posts towards that old #194 :)

 

-Anssi

[Doctordick]

Oh that could turn out nasty if it involves airplanes... Good luck!

No airplanes; we're driving!

Could be that too, but have you been able to quote [math]\partial x_i[/math] yet?

I think I just have!

But still I've got something little bit wrong apparently... Let me know where I get it wrong;

You really don't “get it wrong” you are merely misinterpreting what I am saying. As I have said before, mathematical notation has a few ambiguities which one occasionally runs into.

I would expect our case to turn out like this:

 

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

The problem here is that you have commuted (B} to the left in one of the two terms we are interested in commuting. You have done that in order to make it look just like the referenced example. The referenced example is somewhat simpler than what we are dealing with. Remember, we are concerned with commuting two operators in the center of a string of undefined operators and we cannot throw those undefined operators about willy nilly.

 

In the referenced example, notice that they have commuted u and v. This makes the express assumption that u and v commute. The reason the author did that was because of the standard convention that differential operators operate on all functions to the right: i.e., had he written [imath]d(u\cdot v)= du \cdot v+ u\cdot dv[/imath] the reader might have interpreted the first differential operator as operating on the whole expression, [imath]u \cdot v[/imath] which would be the same as the starting expression.

 

Actually, I have somewhat the same problem. When I wrote

[math]\left[\frac{\partial}{\partial x_i}f(\vec{x}) + \left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](some\;junk\;B)[/math]

 

that first expression [imath]\frac{\partial}{\partial x_i}f(\vec{x})[/imath] could erroneously have been interpreted as implying the differential was to be taken on all elements to its right. I tried to avoid that interpretation with my comment that all operators were enclosed in curly brackets: i.e., that expression standing all alone as it did meant that the expression was the differential of f and not the differential operator operating on f. Subtleties Subtleties Subtleties; can we really avoid them completely?

 

Actually, I had hoped you would bring that up as it is a problem not easily explained to someone who has not been bothered by it. If I had tried to explain the issue before doing it I was afraid I would just confuse you. You are doing very well.

 

Have fun -- Dick

[AnssiH]

I think I just have!

 

Doh! Let's see now...

 

[math]\partial x_i[/math]

 

Hmm, I must have gotten the backslashes stripped out again by hitting preview too much :I (I use a text editor to edit the post too but sometimes I forget to re-paste stuff back after preview)

 

Actually, I had hoped you would bring that up as it is a problem not easily explained to someone who has not been bothered by it. If I had tried to explain the issue before doing it I was afraid I would just confuse you.

 

That was probably good thinking, because even now I'm still very much confused :P

 

In the referenced example, notice that they have commuted u and v. This makes the express assumption that u and v commute. The reason the author did that was because of the standard convention that differential operators operate on all functions to the right: i.e., had he written [imath]d(u\cdot v)= du \cdot v+ u\cdot dv[/imath] the reader might have interpreted the first differential operator as operating on the whole expression, [imath]u \cdot v[/imath] which would be the same as the starting expression.

 

Actually, I have somewhat the same problem. When I wrote

[math]\left[\frac{\partial}{\partial x_i}f(\vec{x}) + \left\{f(\vec{x})\right\}\left\{\frac{\partial}{\partial x_i}\right\}\right](some\;junk\;B)[/math]

 

that first expression [imath]\frac{\partial}{\partial x_i}f(\vec{x})[/imath] could erroneously have been interpreted as implying the differential was to be taken on all elements to its right. I tried to avoid that interpretation with my comment that all operators were enclosed in curly brackets: i.e., that expression standing all alone as it did meant that the expression was the differential of f and not the differential operator operating on f.

 

So when you say "that expression standing all alone as it did...", by "that expression" you refer to [imath]\frac{\partial}{\partial x_i}[/imath] or [imath]\frac{\partial}{\partial x_i}f(\vec{x})[/imath]

 

Assuming it's the former, then I understand how that looks a lot like [imath]du \cdot v+ u\cdot dv[/imath].

 

I say "a lot" because I still don't understand why in our case "v" is [imath]f(\vec{x})[/imath] while "dv" is [imath]\frac{\partial}{\partial x_i}B[/imath]

 

The problem here is that you have commuted (B} to the left in one of the two terms we are interested in commuting. You have done that in order to make it look just like the referenced example. The referenced example is somewhat simpler than what we are dealing with. Remember, we are concerned with commuting two operators in the center of a string of undefined operators and we cannot throw those undefined operators about willy nilly.

 

I think I need to take a step back to make sure I get this right. Let's review;

 

I was confused about the meaning of the expression:

 

[math] \vec{\Psi}_2^\dagger \cdot \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2[/math]

 

Or more specifically the end part;

 

[math] \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2 [/math]

 

In your reply (#260) you brought up the product rule of differentation as:

 

[math]\left\{(\frac{\partial}{\partial x})(f(x))\right\}=\left\{f';(x)+(f(x))(\frac{\partial}{\partial x})\right\}[/math]

 

I see there you use notation [math]f';(x)[/math]. I don't know why there's a semicolon but if it wasn't a mistake but means something like multiplication, then the expression once again looks a lot like [imath]du \cdot v+ u\cdot dv[/imath], except I don't know what the (x) alone would mean.

 

And last but not least, I am not at all sure anymore how [imath] \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2 [/imath] relates to [imath]\left\{(\frac{\partial}{\partial x})(f(x))\right\}[/imath]

 

Subtleties Subtleties Subtleties; can we really avoid them completely?

 

Nope :P

 

As I'm trying to learn these math concepts, it's easy for me to make just a slightly wrong assumption about something, and have it pass unnoticed because it still works quite far along the way. Just the additional assumption that build on the wrong become slightly askew as well. Until I come to a point where my little theory doesn't work at all and comes down like a house of cards.

 

It can sometimes be hard to find where that original false assumption is... For this reply, I worked out a few new theories about what your explanations might mean (going back to post #260), and I'm expressing the one(s) that seem like a best fit, with hopes you can spot where I'm going wrong :/

 

I'm just stating that because it is, amusingly, exactly how we normally build our worldviews regarding any subject, be it a physics model, or how to use the VCR :) It is easy and very common to have many underlying assumptions slightly askew, making everything on top more or less askew as well (and often unnecessarily complicated) to accommodate the situation... And this all reminds me of Thomas Kuhn's The Structure of Scientific Revolution... ...which I still have not read myself

 

ps. sorry I messed up the order of your paragraphs when quoting, I just thought it's clearer to ask the questions in this order.

 

-Anssi

[Doctordick]

Hi Anssi, when I read your post, I had to go look at my post #260. Indeed, there was that semicolon you quoted. Believe me it was an error. The first thing I did was to edit the post to fix it and discovered that there was no semicolon in the actual post. At that point I figured the accent sign must have a special purpose which produced the semicolon so I put a backslash in front of it. When I saved it, that equation showed up as a LaTex error. I also discovered that all the backslashes in the quotes were stripped in spite of the fact that I had not previewed the changes. So I put back double backslashes in the quotes and removed the backslash by the accent sign figuring I would just tell you about the unexplainable error. When I saved the new edit, the semicolon had vanished. ????? Who knows what happened. I have run into that kind of problem before (where the LaTex interpretation was just wrong) and I suspect unprintable characters sometimes get in via a computer error. At any rate, the semicolon does not belong in there.

That was probably good thinking, because even now I'm still very much confused :P

I am quite sorry about confusing you. In the expression [imath]d(u \cdot v)[/imath] shown in the example of the differentiation of a product, you have three “operators” (to go into the terminology I am using). The “d” stands for a differential operator, the u and v stand for functions to be differentiated. The product differentiation rule (without the commutation in referenced example) is [imath]d(u \cdot v)=du \cdot v + u \cdot dv[/imath]

 

To associate that expression with my expression, you must understand that I am talking about the single commutation of the two operators [imath]\frac{\partial}{\partial x_i}[/imath] and the function of the collection of index arguments [imath]f(\vec{x})[/imath]. Note that I could have explicitly displayed the t index also but I left it out because it is immaterial to the commutation being examined.

 

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. Now, to connect those terms to the product differential in the reference, the operator [imath]\frac{\partial}{\partial x_i}[/imath] should be seen as represented by the “d” (the differential operator). The function [imath]f(\vec{x})[/imath] should be seen as represented by the “u”. And, lastly, that factor I added “(some junk B)” is represented by the “v”.

So when you say "that expression standing all alone as it did...",

By "that expression" I was referring to the expression [imath]\frac{\partial}{\partial x_i}f(\vec{x})[/imath], the only one not enclosed in curly brackets. I would have used f'(x) except for the fact that the differential was with respect of a single x argument whereas the function was of the whole set.

 

I hope that clears things up a bit.

And last but not least, I am not at all sure anymore how [imath] \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2 [/imath] relates to [imath]\left\{(\frac{\partial}{\partial x})(f(x))\right\}[/imath]

As I have stated several times, I am talking about the commutation of two product operators within a string of various operators. In order to see these operators explicitly, you have to work them out. First, you should be aware of the fact that [imath]\vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2[/imath] has been abstracted out of [imath]\sum_i \vec{\alpha}_i \cdot \vec{\nabla}_i \vec{\Psi}_1\vec{\Psi}_2[/imath] which is one specific term in my fundamental equation. If you remember, the process was first, the fundamental equation was left multiplied (multiplied from the left) by [imath]\vec{\Psi}_2^\dagger\cdot[/imath] and then the whole fundamental equation was integrated (term by term) over all arguments from set #2. This is a very complex process and we need to seriously simplify it so that we can be sure we make no errors.

 

I want to commute [imath]\vec{\Psi}_2^\dagger\cdot[/imath] through as many factors as I can so that I can simplify the final integration. Obviously I can carry it inside the sum sign (commute it with the sum operator) as left multiplying the sum is the same as left multiplying every term in the sum. I can also commute it with the [imath]\vec{\alpha}_i[/imath] term as that term has no impact on [imath]\vec{\Psi}_2^\dagger\cdot[/imath] (essentially they have nothing to do with one another as [imath]\vec{\alpha}_i[/imath] is nothing more than an abstract operator operating on every term of our sum.

 

Ah, but I can not be so caviler when it comes to the nabla operator because that operator is a complex differential operator and [imath]\vec{\Psi}_2^\dagger\cdot[/imath] is a function of arguments that operator might be operating on. The expressions [imath]\vec{\Psi}_2^\dagger\cdot\vec{\nabla}_i[/imath] can not be assumed to be the same as [imath]\vec{\nabla}_i \vec{\Psi}_2^\dagger\cdot[/imath]. Remember the differential operator is defined to be operating on all elements to the right of its position and the original expression had no differentials of [imath]\vec{\Psi}_2^\dagger\cdot[/imath].

 

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

[math] \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][/math].

 

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. 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 [imath]\vec{\Psi}_2[/imath] to the left and commuting [imath]\vec{\Psi}_2^\dagger\cdot[/imath] to the right. What is important is that we bring [imath]\vec{\Psi}_2^\dagger\cdot[/imath] [imath]\vec{\Psi}_2[/imath] against one another because that is a well defined expression.

 

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.

 

I think I have given you enough information to realize why things turn out the way they do. Let me know if you can follow this post.

As I'm trying to learn these math concepts, it's easy for me to make just a slightly wrong assumption about something, and have it pass unnoticed because it still works quite far along the way. Just the additional assumption that build on the wrong become slightly askew as well. Until I come to a point where my little theory doesn't work at all and comes down like a house of cards.

That is exactly the way learning progresses. I don't know if I told you about my son and the birds but I will repeat it here. He was born in February so the first summer of his life he was a babe in arms. His second summer he was a toddler not yet communicating with us. At the time I owned a Cris-Craft cruiser which I spent a lot of time trying to fix up to usable status. At the time, my wife and he spent time feeding the ducks popcorn at the marina while I worked on the boat.

 

One day the next spring, when he was speaking pretty clear English, he came running into the living room yelling, “daddy, daddy, there are ducks in the front yard!” I went to look and there were no ducks; only a few robins. I told him they were robins and not ducks but he insisted they were ducks. The next week or so I spent a lot of time teaching him about the various birds we could find. He learned the names of all the birds we found but when I asked him to name the birds he would name them with no problems but when he was finished he would always say, under his breath, that they were all “ducks”. A couple of weeks later, a bunch of ducks showed up at the park so I took him to the park and pointed out the ducks. To enforce the thing, I also bought some popcorn so he could feed the ducks. He would not talk about birds for several months after that. I think “I broke his bird” so to speak.

 

We all make assumptions when we learn anything and the most serious fault in the learning process is recognizing exactly where the critical assumption was made. In many respects, my work is based on exactly that problem.

ps. sorry I messed up the order of your paragraphs when quoting, I just thought it's clearer to ask the questions in this order.

No problem! If you don't express your difficulties, I have no way of knowing where you went wrong.

 

Have fun -- Dick

[modest]

Doctordick,

 

I’ve moved the above post from ‘What is time’ where I believe you accidentally posted it into this thread where I believe you intended to post it. If I’ve misunderstood your intentions, please let me know.

 

~modest

[Doctordick]

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

[AnssiH]

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 :eek:

 

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")

 

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. :hyper:

 

Yes... I think I very definitely should :cheer: (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.

 

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.

 

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 :D

I hope with the above you have a, let's say "more detailed" idea of what I meant.

 

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.

 

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.

 

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 :D

 

-Anssi