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Dave W.
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 Posted - 12/18/2012 : 19:59:24 [Permalink]
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Originally posted by Machi4velli
...but by 2^10^80 which is of the order of 10^240. |
The 3.3 factor discussed earlier suggests that 21080 is on the order of 10(3×1079), yes?| And yet this is still assuming binary state changes, but I don't understand how this binary change is relevant, I assume there are lots of different ways for it to change, like maybe the 10^6 you mention (meaning we go 10^250*(10^6)^10^250 = 10^15250). |
Oh, I see where I went wrong. Which hypothetical states hypothetical particles are in doesn't contribute to the alleged UPB, it's merely the fact that there are umpty-ump potential state changes per unit time that matters. Yes?
Wait, no. If we roll three six-sided dice once per second for ten seconds, any particular event thus generated is far more likely than if we do the same with 20-sided dice. Dembski talked about state changes without specifying how many possible states there were to change to. 1080 isn't the right answer, because that's analogous to the number of dice being rolled. So we've got the number of dice, and how many times they could possibly have been rolled, but no clue as to the number of sides on any of the dice. How is it possible to calculate a probability without that information?
If each particle is analogous to a coin (a minimum of two states is required for a "state change"), then the largest value the UPB could possibly have must really be 2-10150, right? And that's an upper limit given the least number of states.| I don't understand this well, but if quantum mechanics actually implies that a particle can jump anywhere in a Planck time, and assuming space is quantized, we'd need to make it more like every possible quanta of space for each subset, and multiply by any possible value of momentum (if I'm right to think location and momentum can give us everything we can know), which will be much more than 10^6 (and "more" in a much more astronomically high sense than even the 10^150 number since we'll be taking large exponents of it). This last paragraph being very provisional because as I said I don't know physics well. |
Each particle's wave function includes an open-ended probability spectrum for its location at any given time, and this does imply that it is (for example) possible (though extraordinarily unlikely) for an electron flowing through a wire in my home at one moment to find itself deep in a neutron star on the other side of the galaxy the next moment. It's also been estimated that the probability of all the wave functions in all the particles in a beer can sitting on a table "lining up" just right so that the can's center-of-gravity shifts just enough for the can to topple over is about 10-30 (this is also an example of how quantum effects can directly "leak" into the classical world).
On much smaller distance scales, quantum tunneling (small quantum jumps across thin barriers) is found in many natural processes and is being used in many different technologies today. |
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Machi4velli
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854 Posts |
Posted - 12/18/2012 : 20:02:38 [Permalink]
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Originally posted by Dave W.
Originally posted by JerryB
...He seems to grasp this, you seem not to....how do you calculate much of anything considering quantum particles? |
Quantum physicists calculate zillions of things in their field every day. Are you claiming that they're all wrong to even try? |
Calculate probabilistically! Give a range of regions a particle may be with a degree of uncertainty. It results in bigger areas than points for your predictions, but you can hypothesize a probability distribution and see if it is fit by the data. |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
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Dave W.
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26022 Posts |
Posted - 12/18/2012 : 20:33:22 [Permalink]
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Originally posted by JerryB
I don't know about you Dave.....<:0)
This is SOOooooo.... silly.
There are NO odds that a deck of cards will deal a sequence.....the odds are 100% that it will deal a sequence every time because you either deal the danged cards or you don't...LOL |
What's silly is you ignoring the part where I specified that the probability is based on "The order in which the whole deck comes up." The order is important, or you have to throw out the specificity part of the whole CSI concept. What are the odds on 500 coin tosses? For you to be consistent in your criticisms, you now have to claim that it's 100%, and not 10-150.
Here: 500 (virtual) cards, shuffled by randomly swapping two cards 500,000 times. Any particular sequence will have a probability of occurrence of about 8.2×10-1135, but here's one:173, 354, 408, 108, 96, 347, 480, 176, 22, 289, 15, 58, 146, 110, 43, 261, 435, 288, 175, 243, 229, 28, 228, 269, 273, 462, 14, 122, 200, 331, 446, 413, 139, 214, 89, 481, 307, 221, 305, 51, 456, 76, 255, 160, 321, 485, 447, 388, 430, 235, 284, 148, 298, 445, 382, 286, 377, 423, 400, 487, 272, 475, 427, 9, 428, 140, 156, 164, 350, 356, 291, 174, 249, 72, 275, 132, 417, 426, 111, 202, 414, 394, 415, 84, 309, 468, 245, 368, 222, 340, 215, 115, 407, 126, 90, 86, 88, 341, 118, 98, 403, 66, 145, 137, 61, 171, 493, 339, 234, 306, 53, 65, 45, 17, 158, 329, 3, 381, 246, 439, 315, 322, 181, 26, 128, 209, 198, 116, 353, 117, 138, 450, 252, 47, 489, 490, 241, 64, 163, 325, 100, 357, 71, 367, 282, 260, 453, 49, 242, 195, 134, 224, 97, 424, 124, 438, 482, 410, 56, 270, 220, 23, 133, 278, 218, 167, 337, 352, 449, 361, 230, 465, 130, 476, 401, 2, 312, 12, 238, 345, 459, 432, 311, 402, 39, 48, 232, 248, 169, 189, 179, 121, 379, 42, 206, 183, 497, 85, 217, 384, 191, 280, 267, 219, 212, 36, 333, 374, 204, 496, 383, 297, 127, 392, 187, 93, 494, 83, 185, 455, 363, 123, 239, 109, 330, 101, 207, 370, 302, 386, 257, 411, 247, 99, 473, 385, 151, 55, 227, 7, 409, 102, 395, 27, 303, 24, 441, 77, 471, 80, 147, 444, 161, 454, 231, 197, 425, 213, 372, 404, 162, 365, 41, 119, 324, 443, 300, 155, 375, 378, 294, 240, 348, 256, 466, 10, 172, 226, 440, 277, 419, 287, 292, 104, 153, 205, 157, 479, 469, 73, 458, 344, 135, 355, 32, 208, 31, 182, 150, 81, 492, 495, 254, 332, 310, 54, 60, 451, 50, 274, 105, 406, 186, 30, 79, 387, 437, 320, 461, 319, 82, 251, 46, 94, 196, 20, 120, 201, 314, 263, 95, 299, 113, 168, 190, 317, 154, 180, 431, 380, 59, 103, 290, 152, 327, 396, 376, 457, 452, 199, 144, 237, 304, 397, 91, 391, 398, 474, 279, 500, 8, 364, 142, 373, 29, 285, 335, 433, 283, 5, 68, 328, 271, 253, 11, 268, 4, 399, 323, 316, 436, 296, 295, 18, 422, 136, 107, 336, 129, 216, 420, 460, 264, 463, 334, 6, 70, 57, 366, 250, 170, 477, 38, 35, 472, 211, 498, 412, 40, 464, 193, 342, 159, 25, 112, 276, 62, 210, 313, 19, 75, 125, 359, 13, 434, 67, 393, 488, 225, 34, 390, 265, 470, 416, 418, 358, 16, 369, 499, 343, 448, 486, 484, 293, 349, 308, 258, 188, 69, 63, 421, 346, 92, 87, 360, 192, 326, 429, 203, 106, 301, 467, 33, 141, 149, 405, 389, 442, 244, 143, 165, 114, 1, 194, 177, 52, 338, 236, 21, 262, 259, 74, 491, 281, 184, 266, 131, 178, 351, 166, 483, 233, 318, 37, 78, 478, 362, 223, 371, 44 Took my pathetic computer just 233 milliseconds to shuffle and log the results to a file.
Or, here's 1,000 coin tosses. Any particular sequence will have a probability of occurrence of about 10-301, but here's one:THHTTTHTTHTTTHHTTHHHHHHHTHHHHHTHHTHHHTTTTHTTTHHTTTTTHTHTHTTHTTHHTTHHH
THTTHHTHHTTHTTTTHHTTHTTHHTHHTTHHTTHTTTHHHTHHTHHTTHTHHTHHHHTTHTTHHHTTT
HHHTTTHHTTHTTTTTTTHTHHTHHHTTTTTHTHHHHHTHHHHHHHHTTHHTHHTHTHHHTTTHHHHTT
HTHTHHHTTTTHHTHHHHHTHTHHTTTHHTHTHTHHTTTHTHHTTTHHTHHHTTHHHHHHHHTTTTHHH
TTTTTHHHHTTHHHHTTTTTTTHHTHHHTHTHTHHHTHTTHTTTHHHTTHHTTHHTTHTHTHTHHHTTT
HHHHHHTHHHTHTHTHTHHHHTHHTHTHHHHHHTHHTHTHHHTTTHHTHTTTHHTHTHTHHHHHHTHHH
HHTTHHHHTTHTTTTTHHTTHTHTHHTHTHTHTTTTTTTHHTTHHHHTTTHHTTTHTTHHTHHHTTHHH
HTTHTHTTHTHHTHHHTHTTHHTTTTHHTTHTTTTHTHHTTTHTTTTTHTHHHTTTTTHHTTTHTHHTH
THHTHTHHHTHTTHTHHTHTTTHHHTTHTHTTTTHTTTHTHHHTTHTTHHTHTHHHHHTTHTHTTHHTT
THTTTTTTHTHHHTTHTHTHTTTTHTTHHHHHTHTHTTTTTTTHTTTHTTHHTHTTTHHHTHTTHHTHH
TTHTTTHTHHHHTHHHTHTHTTHTHHHTTHTTTTTHTTTHTTTHTTHTTTTTTHTTTTHHTTTHHHTHT
HHTTHTTHTTHTTHHHHTTHHHTHTTTHHTTTTHTTHTTTTHTHTTTTTHHTTHTTTHHTTTTTHTHTT
THHTTHTTHHHTTHHTTTHHTTHTTHTHHTTHHTTHTHHHHHTTTHTTTTTTHTTHHTTTHHHHTHHHT
TTHHHTHTHTTTHTTTHTTTHTHTTHHTHTTTHTTHHHHHHHHHHHTHTHHHHTTTHTTHHHTHHHTHT
TTHHHHHTTTHTHHHHTTTHHTHHTHHTHHTTHH
Time taken: 12 ms.
Jerry, perhaps you'd like to take my ID Challenge. |
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Dave W.
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Posted - 12/18/2012 : 20:36:14 [Permalink]
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Originally posted by Machi4velli
Calculate probabilistically! |
And they're good at it! Wildly successful, in contrast to Jerry's "how do you calculate much of anything considering quantum particles?" and wishful thinking that Heisenberg somehow made quantum interactions incalculable. |
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Machi4velli
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Posted - 12/18/2012 : 20:51:48 [Permalink]
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Originally posted by Dave W.
Originally posted by Machi4velli
...but by 2^10^80 which is of the order of 10^240. |
The 3.3 factor discussed earlier suggests that 21080 is on the order of 10(3×1079), yes? |
2^10^80 corresponds to the number of subsets of the 10^80 particles in the universe (as far as I can tell Dembski did not consider this, it's just a mathematical fact that there are 2^x distinct subsets of x elements).
But 2^10^80 = 6.668 * 10^240, at least by Google's calculator.
10^(3.3 * 10^79) is much much larger (3.3*10^79 digits), so I'm not sure what this one is.
I suspect any subset of them could undergo some change during a Planck time. So if a change is simply yes-no, (0-1, binary) for each particle, we have 2^10^80 possible changes during a Planck time.
| And yet this is still assuming binary state changes, but I don't understand how this binary change is relevant, I assume there are lots of different ways for it to change, like maybe the 10^6 you mention (meaning we go 10^250*(10^6)^10^250 = 10^15250). |
Oh, I see where I went wrong. Which hypothetical states hypothetical particles are in doesn't contribute to the alleged UPB, it's merely the fact that there are umpty-ump potential state changes per unit time that matters. Yes? |
This is what Dembski must be doing, but I don't understand why binary (change/no change) is supposed to be relevant to reality.
Wait, no. If we roll three six-sided dice once per second for ten seconds, any particular event thus generated is far more likely than if we do the same with 20-sided dice. Dembski talked about state changes without specifying how many possible states there were to change to. 1080 isn't the right answer, because that's analogous to the number of dice being rolled. So we've got the number of dice, and how many times they could possibly have been rolled, but no clue as to the number of sides on any of the dice. How is it possible to calculate a probability without that information?
If each particle is analogous to a coin (a minimum of two states is required for a "state change"), then the largest value the UPB could possibly have must really be 2-10150, right? And that's an upper limit given the least number of states. |
Yes, he's telling us exactly how many die there are (and at the end multiplying by 10^8 for some reason), which is why to me it's so perplexing as to why it's supposed to be useful.
If it's 10^6 for each particle, my number is actually wrong, but what I said is an upper bound. Really, for a single Planck time, the maximum number of possible state changes with 10^6 possible states to change to for each particle would be
(10^6)^(a_1) + (10^6)^(a_2) + ... + (10^6)^(a_(2^10^80))
where
a_i is the number of elements in the ith possible combination of the 10^80 particles in the universe that change state
i goes from 1 to 2^10^80 to handle each possible combination of particles in the universe.
so each term is this:
(number of states that each particle that changes may change to)^(number of particles that change)
So if we had 2 dice, one white and one black, and we rolled some random number of them (maybe none), in a time increment, we would have
2^2 distinct rolls
- we could roll no dice (1 subset)
- we could roll the white one or the black one only (2 subsets)
- we could roll both (1 subset)
1+2+1 = 4 die roll choices (this would be the number of possible choices of particles to change)
And now for the possible results (using my formula)
no die: 1 possible result
white die only: 6^1 = 6 possible rolls
black die only: 6^1 = 6 possible rolls
both dice: 6^2 possible rolls (we're not summing them, just counting the possible orderings of rolls)
1+6+6+36 = 49 total possible outcomes
So this would be -- in a universe of 2 particles, any subset of them undergo a change and each particle that changes has 6 possible states to change to, resulting 49 is the maximum number of changes that could occur in 1 Planck time.
If what we've said about Dembski's argument is correct (at least prior to multiplying by 10^8), he would have this be 2, not 49.
Let me calculate it up and I can give us a good compact formula for this. |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
| Edited by - Machi4velli on 12/18/2012 20:54:05 |
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Machi4velli
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854 Posts |
Posted - 12/18/2012 : 21:08:52 [Permalink]
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Oh, if you simplify it using binomial theorem, we can find something very simple:
For a universe of k particles and each particle has s possible states to change to,
(# subsets of size 0)*s^0 + (# subsets of size 1)*s^1 + ... + (# subsets of size k)*s^k
But # subsets of size i is just the number of combinations of i elements out of k = kCi = k!/(i!*(k-i)!), a binomial coefficient, so we have
sum: kCi * s^i for i = 0 to k
But by the binomial theorem, this is valued at (1 + s)^k
With s = 10^6, k = 10^80
(1 + 10^6)^(10^80)
which has around 6*10^80 digits.... corresponding to the number of possible state changes in a single Planck time. (If it's binary, (1 + 1)^(10^80), same as the power set's cardinality)
Then take this to the power 10^45*10^17 = 10^62 to get the number of possible state changes since the beginning of time. (2^10^80^10^62 for binary.) |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
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Machi4velli
SFN Regular
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854 Posts |
Posted - 12/18/2012 : 21:18:35 [Permalink]
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| The assumption that each particle has the same number of states to change into is not really true, but we can scale that result to the a minimum of 1 gives us that 2^10^80^10^62 = 7.5*10^149310, so that must be a lower bound of number of possible state changes since the beginning of the universe. |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
| Edited by - Machi4velli on 12/18/2012 21:20:48 |
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Dave W.
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26022 Posts |
Posted - 12/18/2012 : 21:58:32 [Permalink]
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Originally posted by Machi4velli
Originally posted by Dave W.
Originally posted by Machi4velli
...but by 2^10^80 which is of the order of 10^240. |
The 3.3 factor discussed earlier suggests that 21080 is on the order of 10(3×1079), yes? |
2^10^80 corresponds to the number of subsets of the 10^80 particles in the universe (as far as I can tell Dembski did not consider this, it's just a mathematical fact that there are 2^x distinct subsets of x elements).
But 2^10^80 = 6.668 * 10^240, at least by Google's calculator.
10^(3.3 * 10^79) is much much larger (3.3*10^79 digits), so I'm not sure what this one is. |
2(1080) = 10(3×1079).
(210)80 = 102480 = 6.67×10240.
Order-of-operations difficulty. Use parentheses. Your 2^10^80 should evaluate to the much larger number (parentheses around 1080).| I suspect any subset of them could undergo some change during a Planck time. So if a change is simply yes-no, (0-1, binary) for each particle, we have 2^10^80 possible changes during a Planck time. |
Which is a BFN with about ten thousand quadrillion quadrillion quadrillion quadrillion quadrillion digits in it, not the paltry 150 of Demski's calculation.This is what Dembski must be doing, but I don't understand why binary (change/no change) is supposed to be relevant to reality.
...
Yes, he's telling us exactly how many die there are (and at the end multiplying by 10^8 for some reason), which is why to me it's so perplexing as to why it's supposed to be useful. |
Good luck getting such information out of him (or out of Jerry).| If it's 10^6 for each particle... |
Nonono. That's a number I grabbed out of thin air, asking Jerry, "why not a million states instead of just two?" He had no answer, of course.
I don't know how many "states" any particular kind of particle could attain. If we understand that the number of Baryons approximates the number of atoms in the universe (since the vast majority of Baryons are free protons, otherwise known as hydrogen ions), we could (potentially) simplify things and talk about "states" as electron orbitals, and energy transfers (via photons) would change the state of an atom. But it doesn't simplify things much, because for just the first 4 energy levels of hydrogen's single electron, there are 21 distinct orbitals, depending upon the angular momentum and magnetic quantum numbers. I can't even find a complete table of hydrogen electron orbitals, from ground state to ionization, so I imagine it's huge.
If we were to use momentum as part of a particle's state, then that's an essentially continuous variable. As is the polarization of photons. For all practical purposes, if these values were to be included in trying to calculate the number of states of a particle, the answer would be infinite.
But Demski wasn't even content with two. He essentially asked us to consider particles to be one-sided dice (by leaving the number of states out of the calculation altogether), and then somehow built a probability calculation on that premise, and now Jerry is trumpeting its utility without understanding even one of our criticisms. |
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Dave W.
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Posted - 12/18/2012 : 22:10:59 [Permalink]
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Originally posted by Machi4velli
The assumption that each particle has the same number of states to change into is not really true, but we can scale that result to the a minimum of 1 gives us that 2^10^80^10^62 = 7.5*10^149310, so that must be a lower bound of number of possible state changes since the beginning of the universe. |
I think (2(1080))(1062) is going to be much larger than 7.5×10149310. |
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Machi4velli
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Posted - 12/18/2012 : 22:43:19 [Permalink]
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Uh yep, I was sticking in unwarranted parentheses lol,
2^(10^80) = 10^(3.01 * 10^79)
Then we have that to the power 10^62 and we can multiply the (3.01*10^79) and 10^62 to get the final result
10^(3.01*10^141) I believe, for the number of possible state changes assuming they're binary since the beginning of the universe.
Proper parentheses: (2^(10^80))^(10^62), which is what I think you've written (the scripts look weird)
Go figure, we go beyond the x and y of (1 + x)^y and a math guy gets confused :) |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
| Edited by - Machi4velli on 12/18/2012 22:44:32 |
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Machi4velli
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854 Posts |
Posted - 12/18/2012 : 22:54:51 [Permalink]
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Math > computation! Easy way:
((1+x)^y)^(w*z)
for
x = # possible states for a particle to change to
y = # particles in the universe
w = # Planck times in a second
z = # seconds since beginning of the universe
(1 + x)^(wyz) = (1 + x)^(10^80 * 10^45 * 10^17) = (1 + x)^(10^142)
For binary, x = 1,
2^(10^142) ~= 10^(3.01 * 10^141) |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
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Dave W.
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26022 Posts |
Posted - 12/18/2012 : 23:00:19 [Permalink]
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Originally posted by Machi4velli
For binary, x = 1,
2^(10^142) ~= 10^(3.01 * 10^141) |
And a googolplex is only 1010100... |
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Dave W.
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Posted - 12/18/2012 : 23:02:17 [Permalink]
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Originally posted by Machi4velli
Proper parentheses: (2^(10^80))^(10^62), which is what I think you've written (the scripts look weird) |
Oh, and yes: that's exactly what I wrote.| Go figure, we go beyond the x and y of (1 + x)^y and a math guy gets confused :) |
Since I'm not a "math guy," I'm pretty proud of myself right now.  |
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Machi4velli
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Posted - 12/18/2012 : 23:05:10 [Permalink]
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Originally posted by Dave W.
If we were to use momentum as part of a particle's state, then that's an essentially continuous variable. As is the polarization of photons. For all practical purposes, if these values were to be included in trying to calculate the number of states of a particle, the answer would be infinite. |
We could bound this! But it doesn't help Dembski much :)
In any case, we showed that a 3.01*10^79 digit number of distinct events could occur in a single Planck time even in a binary world ignoring momentum, which makes the 1 in 10^150 odds seem miniscule (edit: or massive odds I should say!). |
"Truth does not change because it is, or is not, believed by a majority of the people."
-Giordano Bruno
"The greatest enemy of knowledge is not ignorance, but the illusion of knowledge."
-Stephen Hawking
"Seeking what is true is not seeking what is desirable"
-Albert Camus |
| Edited by - Machi4velli on 12/18/2012 23:16:25 |
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Dave W.
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26022 Posts |
Posted - 12/18/2012 : 23:54:55 [Permalink]
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Originally posted by Machi4velli
But a continuous random variable couldn't exist in a quantized universe. |
As I mentioned earlier, there's still some scientific debate over whether space/time is quantized or not. Is there good reason to think we cannot have continuous space/time while simultaneously having energy emitted from excited atoms in discrete, predictable amounts (quanta)?| I realize you realize this, but we could probably still calculate some bound on the number of possible states. |
Assuming a quantized universe, we could note that particle momentum is typically measured in electron Volts (eV), but aren't quantized at that unit, since visible-light photons have energies ranging from 1.6 to 3.4 eV. In that same Wikipedia article, the extreme examples are 1.6 eV on the low end and 3×1020 eV on the high end for individual particles, so the range is huge. And given that the energies of extra-low-frequency radio photons are measured in femto-eV, it suggests that there may be at least 1035 different states for particle momentum (a range from near 0 to 1020 with a resolution of at most 10-15).
Maybe the quantization is logarithmic, and so the number of possibilities is much, much smaller (with possible states at the high end much farther apart than at the low end), but I've seen no evidence of that.| In any case, we showed that a 3.01*10^79 digit number of distinct events could occur in a single Planck time even in a binary world ignoring momentum, which makes the 1 in 10^150 odds seem miniscule (edit: or massive odds I should say!). |
Indeed.
But note that long ago, Jerry tried to convince me that -X=Y and X=-Y are not equivalent equations (those weren't the equations in question, but get the gist across). Given that his familiarity with basic algebra is so poor, I can only conclude that the statistical discussion you and I have been having is way over his head, and none of our calculations will make an iota of difference to his Dembski worship. |
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Our creator was a computer
