- #1
Doctordick
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Hi Russell,
New thread because I think it deserves a new thread!
I think Chapter I of "Foundations" is very difficult for people to follow. Hopefully, you now understand the essence of what I was trying to communicate there. I think that, under the definitions I listed for the concepts convenient to talking about my equation, that equation is both true by definition and totally capable of representing any possible circumstance. I would really appreciate serious criticism of that logic and, until I get some serious rational complaints, I have no option but to assume I am correct.
So, let us look at the solutions. To begin with, a direct solution is simply impossible, at least for any worthwhile problem, as the number of arguments in the equation is almost beyond comprehension. The argument [itex]\vec{x}[/itex] represents the entire set [itex](\vec{x}_1,\vec{x}_2,\vec{x}_3\cdots,\vec{x}_n),[/itex] where n runs over the entire universe, knowables and unknowables combined (remember, NO information outside that referred to by [itex]\vec{x}[/itex] can be available to any examination of the problem)!
For your convenience, Chapter II is at:
http://home.jam.rr.com/dicksfiles/reality/CHAP_II.htm
My first step is to separate the equation into two parts. In my book, I make this first separation between "knowables" and "unknowables". I will follow that exact attack as I have the equations in the book to refer to (saves me a lot of latex problems). Note that, in equation 2.3, although the arguments of [itex]\vec{\Psi}_1[/itex] are only the finite set called "knowables", the arguments of [itex]\vec{\Psi}_2[/itex] span the entire set [itex]\vec{x}[/itex]. If you do not understand why this must be so let me know.
There is one strange notation I use which needs to be understood to avoid confusion. When I want to indicate an integral over all the [itex]\vec{x}_i[/itex] in a defined set, instead of writing down
[tex]\int_{x_a=-\infty}^{x_a=+\infty}\int_{\tau_a=-\infty}^{\tau_a=+\infty}\int_{x_{a+1}=-\infty}^{x_{a+1}=+\infty}}\int_{\tau_{a+1}=-\infty}^{\tau_{a+1}=+\infty}\cdots\int_{x_{a+n}=-\infty}^{x_{a+n}=+\infty}}\int_{\tau_{a+n}=-\infty}^{\tau_{a+n}=+\infty}\cdots [/tex]
I use a single integral sign with a circle in the middle (the sign commonly used for a "line integral"). I use the circle to imply the closed completeness of the integration. I also use the symbol dv to indicate the rest of the integration notation which should have been written
[tex]dx_a d\tau_a dx_{a+1} d\tau_{a+1}\cdots dx_{a+n} d\tau_{a+n} \cdots[/tex]
I do this simply to save space as all integrals are over exactly the same ranges and the complete notation serves no purpose. Secondly, I do not use any "line integrals" so confusion should not arise.
The only serious issue between equation 2.4 and equation 2.5 is the need for [itex]\vec{\Psi}_1[/itex] to be asymmetric with respect to exchange of variables. The [itex]\tau[/itex] axis was introduced to make sure that the fact of multiple occurrences of the same reference in B was not lost when those references were mapped into points on the x axis. If [itex]\vec{x}_i[/itex] can equal [itex]\vec{x}_j[/itex] then that purpose is defeated. That necessary constraint does not appear in any of the mathematical constraints I have written down.
Asymmetric with respect to exchange is a phrase which means that the result of the algorithm changes sign if two arguments are exchanged. If you wish to make an algorithm asymmetric, all you have to do is add to that algorithm the exactly the same algorithm with two arguments exchanged and multiplied by -1. That sum will be an algorithm asymmetric with respect to exchange of those two arguments which is still a solution to the same differential equation as the original. Now choose a second pair and do the same thing. Eventually you will use up all the arguments (their number is finite) and the result will be asymmetric with respect to any exchange. Notice that I will often propose doing things which are fundamentally impossible to do in fact, the real issue is, are they doable in principal!
This leads to a very important consequence. If the two arguments being exchanged are identical, then the result of the algorithm can not change. There exists but one number which does not change when it's sign changes: that is zero! It follows that [itex]\vec{\Psi}_1[/itex] must vanish if any two arguments are the same and the constraint is now implicit in the representation. Likewise the second term can not contribute as the argument of the Dirac delta function is never zero.
If you do not agree with "theoretical possibility" of actually calculating [itex]\frac{\longleftrightarrow}{f}[/itex] let me know and we can discuss it. The last point is the equivalence of the explicit set defined by the collection of [itex]\vec{x}_i[/itex] and the explicit set defined by the collection of arguments ([itex]\vec{x}_i -\vec{x}_j[/itex]). Again, if you have any question about proving the collections are equivalent, let me know and I will explain it in detail.
Thus we arrive at equation 2.5 which displays explicitly the form of equation any collections of knowables must obey. Now I haven't said I can tell you what [itex]\frac{\longleftrightarrow}{f}[/itex] looks like; I have merely said that such a thing could be known if [itex]\vec{\Psi}_2[/itex] were known.
Think about all that and let me know if you have any questions -- Dick
PS Sorry about the [itex]\frac{\longleftrightarrow}{f}[/itex] notation, it's the best latex notation I can come up with at the moment.
New thread because I think it deserves a new thread!
I think Chapter I of "Foundations" is very difficult for people to follow. Hopefully, you now understand the essence of what I was trying to communicate there. I think that, under the definitions I listed for the concepts convenient to talking about my equation, that equation is both true by definition and totally capable of representing any possible circumstance. I would really appreciate serious criticism of that logic and, until I get some serious rational complaints, I have no option but to assume I am correct.
So, let us look at the solutions. To begin with, a direct solution is simply impossible, at least for any worthwhile problem, as the number of arguments in the equation is almost beyond comprehension. The argument [itex]\vec{x}[/itex] represents the entire set [itex](\vec{x}_1,\vec{x}_2,\vec{x}_3\cdots,\vec{x}_n),[/itex] where n runs over the entire universe, knowables and unknowables combined (remember, NO information outside that referred to by [itex]\vec{x}[/itex] can be available to any examination of the problem)!
For your convenience, Chapter II is at:
http://home.jam.rr.com/dicksfiles/reality/CHAP_II.htm
My first step is to separate the equation into two parts. In my book, I make this first separation between "knowables" and "unknowables". I will follow that exact attack as I have the equations in the book to refer to (saves me a lot of latex problems). Note that, in equation 2.3, although the arguments of [itex]\vec{\Psi}_1[/itex] are only the finite set called "knowables", the arguments of [itex]\vec{\Psi}_2[/itex] span the entire set [itex]\vec{x}[/itex]. If you do not understand why this must be so let me know.
There is one strange notation I use which needs to be understood to avoid confusion. When I want to indicate an integral over all the [itex]\vec{x}_i[/itex] in a defined set, instead of writing down
[tex]\int_{x_a=-\infty}^{x_a=+\infty}\int_{\tau_a=-\infty}^{\tau_a=+\infty}\int_{x_{a+1}=-\infty}^{x_{a+1}=+\infty}}\int_{\tau_{a+1}=-\infty}^{\tau_{a+1}=+\infty}\cdots\int_{x_{a+n}=-\infty}^{x_{a+n}=+\infty}}\int_{\tau_{a+n}=-\infty}^{\tau_{a+n}=+\infty}\cdots [/tex]
I use a single integral sign with a circle in the middle (the sign commonly used for a "line integral"). I use the circle to imply the closed completeness of the integration. I also use the symbol dv to indicate the rest of the integration notation which should have been written
[tex]dx_a d\tau_a dx_{a+1} d\tau_{a+1}\cdots dx_{a+n} d\tau_{a+n} \cdots[/tex]
I do this simply to save space as all integrals are over exactly the same ranges and the complete notation serves no purpose. Secondly, I do not use any "line integrals" so confusion should not arise.
The only serious issue between equation 2.4 and equation 2.5 is the need for [itex]\vec{\Psi}_1[/itex] to be asymmetric with respect to exchange of variables. The [itex]\tau[/itex] axis was introduced to make sure that the fact of multiple occurrences of the same reference in B was not lost when those references were mapped into points on the x axis. If [itex]\vec{x}_i[/itex] can equal [itex]\vec{x}_j[/itex] then that purpose is defeated. That necessary constraint does not appear in any of the mathematical constraints I have written down.
Asymmetric with respect to exchange is a phrase which means that the result of the algorithm changes sign if two arguments are exchanged. If you wish to make an algorithm asymmetric, all you have to do is add to that algorithm the exactly the same algorithm with two arguments exchanged and multiplied by -1. That sum will be an algorithm asymmetric with respect to exchange of those two arguments which is still a solution to the same differential equation as the original. Now choose a second pair and do the same thing. Eventually you will use up all the arguments (their number is finite) and the result will be asymmetric with respect to any exchange. Notice that I will often propose doing things which are fundamentally impossible to do in fact, the real issue is, are they doable in principal!
This leads to a very important consequence. If the two arguments being exchanged are identical, then the result of the algorithm can not change. There exists but one number which does not change when it's sign changes: that is zero! It follows that [itex]\vec{\Psi}_1[/itex] must vanish if any two arguments are the same and the constraint is now implicit in the representation. Likewise the second term can not contribute as the argument of the Dirac delta function is never zero.
If you do not agree with "theoretical possibility" of actually calculating [itex]\frac{\longleftrightarrow}{f}[/itex] let me know and we can discuss it. The last point is the equivalence of the explicit set defined by the collection of [itex]\vec{x}_i[/itex] and the explicit set defined by the collection of arguments ([itex]\vec{x}_i -\vec{x}_j[/itex]). Again, if you have any question about proving the collections are equivalent, let me know and I will explain it in detail.
Thus we arrive at equation 2.5 which displays explicitly the form of equation any collections of knowables must obey. Now I haven't said I can tell you what [itex]\frac{\longleftrightarrow}{f}[/itex] looks like; I have merely said that such a thing could be known if [itex]\vec{\Psi}_2[/itex] were known.
Think about all that and let me know if you have any questions -- Dick
PS Sorry about the [itex]\frac{\longleftrightarrow}{f}[/itex] notation, it's the best latex notation I can come up with at the moment.
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