Hi Wilhelmus,
This may be in part because English is not your native language, but I find it difficult to understand your comments.
Nevertheless, I will attempt once more to go through my argument step by step, highlighting where my best guess is as to where there might have been a misunderstanding. I know you said that you are not a mathematician but I'd like to ask you to *please* take the time and make the effort to work through the following simple algebraic argument. I will make it easy and go very slow.
Consider a ball, call it ball 1, of radius 1 meter, or
[math]r_1=1m[/math]
where the subscript 1 of r means that it is the radius of ball 1.
What is its radius in terms of light years? 1 light year, abbreviated by 1 ly , is the distance traveled by light in one year, and that is equal to 299792458 m which to keep it simple I will round up to 30000000 meters or 3 X10^8 m.
So the radius of ball 1 in terms of light years is
[math]r_1= 1m \times \frac{1 ly}{3\times10^8 m}\approx 3.33\times 10^{-9}ly[/math]
So far so good? Now, what is the radius in terms of millimeters? 1 meter is 1000 mm or 10^3 millimeters, therefore
[math]r_1=1m \times \frac{1000mm}{1m}=10^3mm[/math]
Now, the A/V ratio of ball 1, denoted by A_1/V_1, in terms of light years is:
[math]\frac{A_1}{V_1}=\frac{4 \pi r_1^2}{4/3 \pi r_1^3}=\frac{3}{r_1}=\frac{3}{3.33\times 10^{-9}ly}[/math]
When A_1/V_1 is expressed in terms of millimeters, it is
[math]\frac{A_1}{V_1}=\frac{4 \pi r_1^2}{4/3 \pi r_1^3}=\frac{3}{r_1}=\frac{3}{1000mm}[/math]
one possible interpretation of what you wrote is that you seem to think that I'm claiming that A_1/V-1 when expressed in terms of light years or in terms of millimeters is different because the units of measure are different.
If this is what you thought, then you have completely misunderstood my argument.
I totally agree that whether you express A_1/V-1 in ly or in mm it is one and the same. There is no difference because in what you units you express a quantity of length does not change that quantity of length in and of itself. By itself, A_1/V-1 doesn't really tell you anything.
So then, what is my argument?
It is based on this:
You have to compare the A/V ratio of ball 1 with the A/V ratio of another ball with a different radius.
This is the point I get the impression you've been missing.
Let me work through the details:
Consider a second ball, call it ball 2, of radius 10^{-11) (this is 10 picometers). Denoting the radius of ball 2 by r_2, this means
[math]r_2=10^{-11}m[/math]
In terms of light years, r-2 is
[math]r_2= 10^{-11}m \times \frac{1 ly}{3\times10^8 m}\approx 3.33\times 10^{-20}ly[/math]
In terms of millimeters, r_2 is
[math]r_2=10^{-11}m \times \frac{1000mm}{1m}=10^{-8}mm[/math]
The A/V ratio of ball 2, denoted by A_2/V_2, in terms of light years is:
[math]\frac{A_2}{V_2}=\frac{4 \pi r_2^2}{4/3 \pi r_2^3}=\frac{3}{r_2}=\frac{3}{3.33\times 10^{-20}ly}[/math]
A-2/V-2 expressed in terms of millimeters is
[math]\frac{A_2}{V_2}=\frac{4 \pi r_2^2}{4/3 \pi r_2^3}=\frac{3}{r_2}=\frac{3}{10^{-8}mm}[/math]
Do you follow so far? Again, I agree with you that this ratio, whether expressed in light years or in millimeters is exactly the same quantity.
But now we finally get to my argument:
Take the ratio of A_1/V-1 and A_2/V-2 in terms of light years:
[math]\frac{\frac{A_1}{V_1}}{\frac{A_2}{V_2}}=\frac{\frac{3}{r_1}}{\frac{3}{r_2}}=\frac{r_2}{r_1}=\frac{3.33\times 10^{-20} ly}{3.33\times 10^{-9}ly}=10^{-11}[/math]
Notice that this is a dimensionless number, the units cancel out.
Expressing the ratio of A_1/V-1 and A_2/V-2 in terms of millimeters:
[math]\frac{\frac{A_1}{V_1}}{\frac{A_2}{V_2}}=\frac{\frac{3}{r_1}}{\frac{3}{r_2}}=\frac{r_2}{r_1}=\frac{10^{-8} mm}{10^3 mm}=10^{-11}[/math]
Did you notice that this dimensionless number is the same as before? It did not matter whether we expressed the radii in terms of light years or in millimeters.
What does the dimensionless number 10^-11 tell us? It tells us that ball_1 has 10^-11 (or hundred billion times) fewer units of volume per unit of area than ball_2, ***regardless*** of what units you choose. You can choose light years, millimeters, inches, yards etc. the number 10^-11 will not change.
Also note that while to keep it simple I used the specific example of balls, the general fact that smaller objects have less volume per area than larger objects of same shape holds for any shape, as long as you use objects that have the same shape and you are consistent in your use of units.
I have attempted here to explain in painstaking detail what my argument is. If you still have trouble with it, I strongly advise you to carefully work through this argument again. I have it laid out in front of you as clearly as I could. Beyond that, I'm afraid I can't help you with understanding the A/V argument.
But let me now move on to the other parts of your comments. Does the A/V argument imply that there has to be some limit in which volume vanishes, since the smaller an object, the less volume per area it has? Absolutely not, this is why I call it a "plausibility argument". The purpose of a plausibility argument is to render something else that follows it more intuitive. It is *not* a proof. This is why axiom I had to be stated as an axiom, as opposed to, say a theorem.
You state that an axiom is, among other things, "self-evident". While this is part of the general definition of the word axiom, this is not the standard way in which the term is used in physics. The standard use of "axiom" in physics is simply as a formal assumption which requires no further justification. Although a plausibility argument may be given, a plausibility argument is too weak to qualify as a formal justification.
In fact, I challenge you to find anyone who is willing to claim that the axioms of standard quantum mechanics are self-evident. In case you need a refresher, you can find them stated here (NB. Postulates and axioms are used synonymously):
http://vergil.chemistry.gatech.edu/notes/quantrev/node20.html
You won't be able to find anyone who is willing to claim that the axioms or postulates of quantum mechanics are self-evident because they simply aren't. This is really a major part of the problem of the interpretation of quantum mechanics.
Your notion of an axiom was popular up until the beginning of last century when it was thought Euclid's derivation of his geometry proceeded from self-evident axioms and was considered a model for all other sciences to emulate. Alas, David Hilbert showed that Euclid's 'self-evident' axioms actually contained hidden assumptions, and given that it would not be difficult to find people who'd dispute that, say, the speed of light postulate or the principle of equivalence are self-evident, one can make a strong case that none of our fundamental physical theories at present are based on self-evident assumptions.
It may be possible in the future, if we ever obtain a complete understanding of fundamental processes and what they mean, to derive them from self-evident assumptions, but requiring that axioms be self-evident right now is making any approach at a more fundamental understanding unworkable. As mentioned, by your notion of an axiom, we would have to throw out standard quantum mechanics because it is not based on self-evident assumptions.
I did not really need to include the A/V argument in my paper because Axiom I, as stated before, by definition requires no further justification. Nevertheless, I decided to include it in the hope that Axiom I does not appear entirely unmotivated.
Finally, I think it is worthwhile to mention how the evaluation of a scientific argument works in the ideal case. If you are presented with a novel scientific idea, you should ideally evaluate it as follows:
1. you keep your skepticism about the assumptions but try to understand them the best you can
2. You examine the chain of reasoning that leads from the basic assumption to the conclusion.
3. You check whether there are any contradictions, non-sequiturs or other erroneous steps in this chain of reasoning, including the assumptions.
4. You check whether the conclusion fits what is already known, whether it contradicts any known experimental tests, whether it "seems right" etc.
***Only after you have gone through these steps*** do you decide whether to accept the assumptions or not.
You write " I feel free not to accept these assumptions based on the arguments above ...(the rest of the sentence I don't comprehend)" when the arguments you mentioned, to the best I could make out, were themselves based on a misunderstanding of the A/V argument and a misunderstanding of the notion of an axiom in physics, but not on the steps outlined above.
You are certainly welcome not to accept the assumptions, but then why did you bother to continue reading the paper? If you reject the assumptions without having gone through the process above, you are really done right there. We can at that point only agree to disagree. The only problem is that this sort of approach is a lot closer to dogma than to science.
As for your inability to download the paper, please try this url:
http://141.213.232.243/handle/2027.42/83865
go to the bottom of the page and click on "view/open". The paper is supposed to be publicly available. If it is not, I will check with the systems administrator.
Cordially,
Armin
