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# The concept of infinite.

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### #31 ShaunZero

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Posted 01 November 2009 - 05:07 AM

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Here is a "new" one : "Nothing" does not exist. Try to imagine nothing. Try to imagine nothing without it actually being "something". I guarantee that you cannot do it.

You cannot create nothing from something because nothing does not exist.
You cannot create something from nothing because nothing does not exist.
You can create something from something because everything does exist.

This is why there is no beginning.

Nothing is the absence of all things. That is not logically impossible. Just subtract all existence untill there is nothing which exists. What you're saying is almost like saying if you look in a box which contains no objects at all, you can't say "no objects are in this box" because "no objects" does not exist. No objects is technically "nothing", but it's just a way to describe the absence of something.

Just because something has a definition and concept behind it does not mean that it is actually "something". What IS something is the CONCEPT of nothingness.

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The amount of natural numbers is infinite, that is, the set of non-negative integers. Starting from 0 and extending to infinity. 0, 1, 2, 3 ...

I think I would personally label that as a conceptual infinity. I'm talking in terms of something that exists in reality, not just in concept. I'm not saying that infinite has a specific value, but a true infinite "thing" is something that already has an infinite amount of something in existence. That which has a beginning can not be infinite, because there is not an infinite(Still not positing a particular value) amount of time it exists, and we know this due to there being an age of it because of the PRESENT. Something that had no beginning, has an infinite amount of past, and is thus an actual infinite object. A part of it already exists as infinite.

I'm sleepy and just getting off of a really good buzz, so forgive me if I had trouble explaining myself.

Edited by ShaunZero, 01 November 2009 - 05:12 AM.

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### #32 sepulchrave

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Posted 01 November 2009 - 07:52 AM

Raptor, on 31 October 2009 - 08:31 PM, said:

How so?
(I am using a statistical mechanics viewpoint for this proof)

Consider a system of N components. If there is a maximum energy Emax, then that energy occurs only when each fundamental component i has its individual maximum energy (proof by contradiction: if, at energy Emax component i has less than its individual maximum, then the total energy when i assumes its maximum energy is greater than Emax and therefore Emax is not the maximum). Therefore there is only one arrangement of components corresponding to maximum energy. Similarly, there is only one arrangement of components corresponding to minimum energy (usually taken as 0).

There are, however, multiple arrangements corresponding to intermediate energies (proof: for example, component i could have an energy somewhere between minimum and maximum, while all other components have a maximum [or minimum] energy. Alternatively, component i could have a maximum [or minimum] energy and component j could have an energy of (maximum [minimum] energy of j) + (intermediate energy of i in previous case) - (maximum [minimum] energy of i), resulting in the same overall energy).

Now temperature T is defined as: 1/T = dS/dE (where the number of components N is held constant).

Here E is the energy, and S is the entropy, where S = kB ln W, where W is the number of arrangments corresponding to the same energy (usually referred to ``the number of microstates corresponding to the same macrostate'').

Therefore, 1/T = kB d(ln W) /dE = (kB/W) dW/dE

Now, as shown above, W starts at 1 at E = 0 (there is only 1 way of having a minimum energy), ends at 1 for E = Emax (there is only 1 way of having maximum energy), and W > 1 for intermediate E. Therefore, when approaching Emax, W decreases in value - and therefore dW/dE is negative.

Of course if dW/dE is negative, then 1/T is negative and therefore T is negative.

### #33 ships-cat

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Posted 01 November 2009 - 11:33 PM

sepulchrave, on 31 October 2009 - 07:01 PM, said:

A black hole is an object whose mass is such that its gravitational strength exceeds the repulsive force keeping its constituent atoms separate from each other. Classically these atoms then collapse to a singular point.
Ummmm.... noooo... I think you may be mistaken there Sep'

A black hole is simply a planet - or sun - that is very "big", and hence has a strong gravitationl field. So strong, in fact, that light itself cannot escape the 'event horizon'.

A beam of light, shone up from the surface of a black hole, falls back under the influence of gravity and hits the surface again... just like a bullet from a handgun would do, if fired from the surface of the earth.

A beam of light passing near to the "black hole" has it's path twisted by the gravitational field. If it passes close enough, then it's path is twisted into an inward spiral, and it eventually spirals down into the planet/sun.

Imagine being on the space shuttle and firing a bullet towards the earth. If you fire it TOO directly towards the earth, it will be captured by the earths gravitational field, and ultimately get dragged down to the planets surface.

Aim just a little higher, and the bullet will 'graze' the upper atmosphere, but will have enough energy to carry on moving, and continue forwards into space... albeit with a defelcted path. Such is the case with light and a black hole.

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Gravitational energy from a point particle is given by ~ 1/r, if the next particle is at the same spot than r is vanishingly small and the mutual gravitational energy between them is infinite.
Gravity - as a force - is a MINNOW compared to the forces within the nucleus of an atom. Now amount of squeezing would change that.

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Well... technically 1/0 is `undefined'. 1/x -> infinity as x -> 0 though. (Just to be unnecessarily pedantic.)
Actually... no.. it is NOT pedantic, and you are right to highlight it.

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### #34 Raptor

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Posted 01 November 2009 - 11:55 PM

sepulchrave, on 01 November 2009 - 07:52 AM, said:

(I am using a statistical mechanics viewpoint for this proof)

Consider a system of N components. If there is a maximum energy Emax, then that energy occurs only when each fundamental component i has its individual maximum energy (proof by contradiction: if, at energy Emax component i has less than its individual maximum, then the total energy when i assumes its maximum energy is greater than Emax and therefore Emax is not the maximum). Therefore there is only one arrangement of components corresponding to maximum energy. Similarly, there is only one arrangement of components corresponding to minimum energy (usually taken as 0).

There are, however, multiple arrangements corresponding to intermediate energies (proof: for example, component i could have an energy somewhere between minimum and maximum, while all other components have a maximum [or minimum] energy. Alternatively, component i could have a maximum [or minimum] energy and component j could have an energy of (maximum [minimum] energy of j) + (intermediate energy of i in previous case) - (maximum [minimum] energy of i), resulting in the same overall energy).

Now temperature T is defined as: 1/T = dS/dE (where the number of components N is held constant).

Here E is the energy, and S is the entropy, where S = kB ln W, where W is the number of arrangments corresponding to the same energy (usually referred to ``the number of microstates corresponding to the same macrostate'').

Therefore, 1/T = kB d(ln W) /dE = (kB/W) dW/dE

Now, as shown above, W starts at 1 at E = 0 (there is only 1 way of having a minimum energy), ends at 1 for E = Emax (there is only 1 way of having maximum energy), and W > 1 for intermediate E. Therefore, when approaching Emax, W decreases in value - and therefore dW/dE is negative.

Of course if dW/dE is negative, then 1/T is negative and therefore T is negative.

Thanks for the response, I see you put a lot of work in to it. I'll return when I have the time to give it the attention it deserves.

ships-cat, on 01 November 2009 - 11:33 PM, said:

Ummmm.... noooo... I think you may be mistaken there Sep'

A black hole is simply a planet - or sun - that is very "big", and hence has a strong gravitationl field. So strong, in fact, that light itself cannot escape the 'event horizon'.

A beam of light, shone up from the surface of a black hole, falls back under the influence of gravity and hits the surface again... just like a bullet from a handgun would do, if fired from the surface of the earth.

You definitely have the right idea, SC, but I think you're slightly mistaken.

A black hole is an object that has an escape velocity greater than or equal to the speed of light. Your typical planet or star isn't capable of generating such a force. It's only when a very massive star comes to the end of its life and undergoes a gravitational collapse that its mass becomes so dense as to occupy a single zero-volume point in space, that its gravitational force and therefore escape velocity climb so high that light becomes unable to escape from the gravitational well.

### #35 danydandan

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Posted 01 November 2009 - 11:57 PM

I feel like im sticking my nose in
You could look at it like this that you can divide infinite in to different states
nearly infinite, truly infinite, infinitely infinite

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### #36 sepulchrave

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Posted 02 November 2009 - 02:09 AM

ships-cat, on 01 November 2009 - 11:33 PM, said:

Ummmm.... noooo... I think you may be mistaken there Sep'

A black hole is simply a planet - or sun - that is very "big", and hence has a strong gravitationl field. So strong, in fact, that light itself cannot escape the 'event horizon'.
Not to my knowledge. In Classical Mechanics, sure, but in General Relativity the gravitational field required to bend light requires mass-energy densities that are within their own Schwarzschild radius. Once this happens all the matter is ultimately doomed to fall into a singularity1.

ships-cat, on 01 November 2009 - 11:33 PM, said:

Gravity - as a force - is a MINNOW compared to the forces within the nucleus of an atom. Now amount of squeezing would change that.
First part, yes. Second part, no. Gravity is always mutually attractive. All other forces are equal-and-opposite. Nuclear force can always be at least partially shielded. Gravity cannot.

The pressures from electrostatic force, nuclear force, and the ``degeneracy'' pressure from the Pauli exclusion principle have been calculated, and any body of mass exceeding the Chandrasekar limit has the potential to create a singularity.

In the above statements I have referenced:
• S. M. Carroll, An Introduction to General Relativity, Spacetime and Geometry (Addison Wesley, San Francisco) 2004, pgs. 229-236
• R. K. Pathria, Statistical Mechanics (Elsevier, Oxford) 2nd Edition, 1996, pgs. 219-223

Now I agree that the idea of an actual singularity is a bit bizarre and hopefully a proper unification of QM and GR will eliminate the possibility of a singularity. However, as it currently stands, GR predicts that singularities exist within every black hole.

1 It has been pointed out that the Schwarzschild radius for the mass of the observable universe exceeds the observable radius of the universe, and therefore perhaps we are all in a black hole. If this is the case then (barring the expansion of the universe) we are all doomed to eventually fall through a black hole. The moral of this story is that while supermassive black holes are relatively `tame' creatures, they still (according to GR, anyway) have, or will eventually have, a singularity somewhere in them.

### #37 maximaldecimal

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Posted 02 November 2009 - 04:42 AM

sepulchrave, on 31 October 2009 - 05:45 AM, said:

What physics is that?

It's a joke a phisics professor friend told me once. We cannot prove the existence of zero since it is the absence of a thing. Not a thing and thus nonexistent except in concept and representation.  So he doesn't believe in zero. Since it doesn't exist exept on paper, I had to agree to disagree, but thats beside the point I think.  If something were less then everything then it would be finite. So infinity is comprised of all the numbers otherwise it would be finite.  It would also be represented in zero or negative numerical connotations and the rest of the values as well.  We can with numbers counting up prove an infinite string can be begun but unless it is measured it is only a finite representation of these infinite numerical values. Like the line with one point analogy.  The line is infinite in vector but not in scope.  These are still finite representations.  This is also complicated by the time factor. If something has a begining it is finite.  So the only thing that would be truely infinite would be something that was not created ergo zero, it would have to be both or else it would have an end and not infinite.  In my opinion it is everything.  So to answer the question it's not really anyones physics more like hypothetical philosophic rhetoric.

Edited by maximaldecimal, 02 November 2009 - 04:59 AM.

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### #38 sepulchrave

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Posted 02 November 2009 - 08:42 PM

maximaldecimal, on 02 November 2009 - 04:42 AM, said:

It's a joke a phisics professor friend told me once. We cannot prove the existence of zero since it is the absence of a thing.
I see. That's a fair point.

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