Creation Ex Nihilo

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20 years 9 months ago #7879 by Mac
Replied by Mac on topic Reply from Dan McCoin
North,

1 - Did you not read Prof Tryon's work?

2 - Or did you just not understand it?

3 - Or is it that it simply disagrees with your view and therefore you choose to ignore it or pretend it is false because you say it is.?


"Imagination is more important than Knowledge" -- Albert Einstien

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20 years 9 months ago #7880 by jrich
Replied by jrich on topic Reply from
Mac,
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mac</i>
<br />JRich,

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><b>x' + y' = x + y</b><hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">

ANS: 0 = 0 and you fail to carry signs by masking the operation by deltas * y, etc.

Invalid.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
The relationships between all the variables are consistent with the theory. There are no signs on the variables, BECAUSE THEY ARE VARIABLES, the signs are in the values that they are assigned, just as in your posts. This is simple algebra. If I have made an error, you should have no trouble pointing out the exact step where the error was made. I do not claim infallibility, but you are going to have to be specific and convincing.

I would also welcome others to critique my proof.


JR

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20 years 9 months ago #8250 by jrich
Replied by jrich on topic Reply from
To All,

While doing some research on Tryon I ran across this enjoyable article on this topic.

www.anselm.edu/homepage/dbanach/holt.htm


JR

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20 years 9 months ago #4107 by Mac
Replied by Mac on topic Reply from Dan McCoin
JRich,

It is a wellknown process to use "0"'s in algebraic conversions to cause 2 = 1 for example.

YOur proof does use delta's which are i.e. 0.5 values of the original numbers but my point is you fail to carry the fact that it is 0.5 of a (-) value.

"Imagination is more important than Knowledge" -- Albert Einstien

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20 years 9 months ago #7881 by jrich
Replied by jrich on topic Reply from
Mac,
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Mac</i>
<br />JRich,

It is a wellknown process to use "0"'s in algebraic conversions to cause 2 = 1 for example.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Can you provide an example. I'm not sure I understand what you mean.
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">YOur proof does use delta's which are i.e. 0.5 values of the original numbers but my point is you fail to carry the fact that it is 0.5 of a (-) value.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
I picked the distance d where the object had lost half its gravitational potential energy, thus y' = .5y. The correctness of this is shown by letting y = -1, then y' = -.5, which is half of -1. I am not taking the delta, y - y', where I could have mistakenly changed signs. In fact, I specifically avoided deltas for this reason.


JR

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20 years 9 months ago #7882 by tvanflandern
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by jrich</i>
<br />I would also welcome others to critique my proof.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">I'll accept your invitation, but don't wish to detract from the momentum of the on-going debate. However, there is a flaw in your proof.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">Tryon's theory states that when matter accelerates due to gravity the "negative" gravitational potential energy lost is converted to an equal amount of "positive" rest energy. Let y be the initial gravitational potential energy of an object and x be its initial rest energy, where x + y = 0 (as you insist).<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">This setup is okay even though it uses a non-standard term: "rest energy". Perhaps Tryon explains why he chooses to call it that? In standard physics, we call this "kinetic energy". Then potential energy and kinetic energy are being continually exchanged back-and-forth by the force of gravity, with total energy remaining constant.

<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote">After undergoing gravitational acceleration over a certain distance d the gravitational potential energy is y', where y' = .5y. After the distance d, let x' be the rest energy, where x' &gt; x.<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">Your first sentence here does not specify whether the object is rising or falling in the gravitational field. But your second sentence requires that it be falling. This makes the first criterion impossible because y' must be more negative than y as the object falls. For example, you might have specified y' = 1.5y.

But in that case, everything works out correctly in the end.

Hint: Mac has merged the concept of forms with the concept of properties of forms. Forms cannot be infinite, but their properties certainly can be, as in the examples of slope and countability. In Mac's reasoning, which he applies to objects and not just their properties, if +5 apples are combined with -5 oranges, the result is zero. But that is absurd on its face. The net count (a property) goes to zero, but ten physical entities (forms) remain in existence, whether they are positive or negative. -|Tom|-

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