Theorem of finite matter

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Offline jeffreyH

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Theorem of finite matter
« on: 29/08/2015 19:45:16 »
If we take any quantity and divide it by itself it is obviously unity. As in [tex]\frac{x}{x}\,=1[/tex]. If we consider infinity as a quantity then we can formulate the abstract algebraic formula [tex]g\,=\,\frac{\infty}{\infty}\frac{GM}{r^2}[/tex]. We can then decompose this unity to arrive at [tex]\frac{g}{\infty}\,=\,\frac{GM}{\infty r^2}[/tex]. Here the left hand side is zero and the right hand side has an infinite radial value. These values are imposed upon the equation forcing a particular result but show the relationship of the gravitational field to infinity.

However, we can also formulate the equation thus, [tex]g\infty\,=\,\frac{\infty GM}{r^2}[/tex]. This then relates to an infinite mass energy. Such as that required to escape a black hole. Here radial distance from the source has no effect on the strength of force. This is because by implication there is no point outside the source across an infinite distance. This result indicates that if there were an infinite amount of matter it would have to occupy an infinite amount of space. Thus there cannot be an infinite amount of matter in the universe.

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Offline chiralSPO

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Re: Theorem of finite matter
« Reply #1 on: 29/08/2015 19:54:53 »
∞/∞ is not necessarily 1

for instance:

lim        2n = 2
n→∞     n

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Offline jeffreyH

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Re: Theorem of finite matter
« Reply #2 on: 29/08/2015 19:56:59 »
Which surely equates to multiplying 2 by unity.

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Offline jeffreyH

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Re: Theorem of finite matter
« Reply #3 on: 29/08/2015 20:16:33 »
∞/∞ is not necessarily 1

for instance:

lim        2n = 2
n→∞     n

If you had stated instead:

[tex]\lim_{n \to \infty}\,\,\frac{2n}{-n}\,=\,-2[/tex]

then you would have a difference and be reformulating the equation in terms that could ultimately become complex and therefore amenable to quantum mechanics.

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Offline jeffreyH

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Re: Theorem of finite matter
« Reply #4 on: 30/08/2015 17:29:01 »
As an extension to the above the abstraction could be changed to the following.

[tex]\frac{g}{\infty i}\,=\,\frac{GM}{-\infty i r^2}[/tex]

So that we are using a complex infinity with its complex conjugate. The complex conjugate indicating a negative direction along the radius.
« Last Edit: 30/08/2015 17:31:26 by jeffreyH »

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Offline alancalverd

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Re: Theorem of finite matter
« Reply #5 on: 31/08/2015 00:29:22 »
[tex]\frac{g}{\infty}\,=\,\frac{GM}{\infty r^2}[/tex]. Here the left hand side is zero and the right hand side has an infinite radial value.

No,the RHS = 0.
helping to stem the tide of ignorance

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Offline jeffreyH

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Re: Theorem of finite matter
« Reply #6 on: 31/08/2015 00:48:01 »
[tex]\frac{g}{\infty}\,=\,\frac{GM}{\infty r^2}[/tex]. Here the left hand side is zero and the right hand side has an infinite radial value.

No,the RHS = 0.


Yes. Sorry I didn't make that clear.