The Probability of the Universe Being the Exact Way It Is Within the Bounds of the Observable Universe at One Instant (Proposed Formula)

[NowYouKnowFACS]

Hello everyone,

 

For the past few months I have been working on a script for a YouTube channel I am currently working on and writing scripts for.

 

Every day I learn new things that I apply to this formula, and it is something that I feel that a lot of people might find interesting, even if they understand nothing about it.

 

That is beyond the point, the point is the formula. Now, keep in mind that I am a High School student, an in no way commit this to being fact, or in any way/shape/form correct. I am not an astrophysicist or have a degree in theoretical physics.

 

With that said, my formula (And explanation as of this time) is as followed-

 

READ FIRST- UPDATED FORMULA BELOW!

 

IMPORTANT NOTE!- Numbers seen are rounded to the nearest 10^x for the sake of simplification.

 

1:(((10¹⁰⁶)!)+((10¹⁰⁶)(((10¹⁰⁶)-1)!))+((10¹⁰⁶)²(((10¹⁰⁶)-2)!)+...+((10¹⁰⁶)^(10¹⁰⁶)((1)!)+1))

 

Without the numbers, with only the formula itself, this is what it looks like.

 

1:((T!)+((T)⁽ⁿ⁾((T-n)!))+...+1)

1---T!-------n=1----------n--+1

Explanation- (Constants)

T is the total possible particles in the observable universe

(10^106= (10^82)/(10^-24))

n is the number of particles subtracted from total possible

+1 is the possibility of a universe without any particles

! is a factorial sign

 

Explanation- (Actual Explanation)

Well, as I said, I am no scientist, yet, however, I do have an explanation for this.

Most formulas proposed to me for the answer to my question is 10^80!. Now, me being me, I was very unhappy with this. It doesn't take into account for many issues. The universe is not a house of cards. A simple factorial is not a good answer to me. 1/(10^80)! is basically just saying particles are switched, not moved around. The V1.2 of the formula was (roughly) 1/(10^106)!. Now, although better, the universe isn't filled, which is the issue I had. Now, V2.0 was 1/((10^106)!+((10^106)-1)!+...+(1!). V2.1.1 is what you see above.

 

ISSUES WITH THE FORMULA I HAVE FOUND

Once you hit over half of the particles subtracted, the T^n becomes inaccurate.

 

Again, I am not schooled in anything that would make me a valid source on info.

 

--NEW FORMULA REMADE TODAY--

 

Ok, so I believe I have made the corrections necessary to fix the equation to be as fixed as possible with my knowledge.

 

1:((T!)+((T)^n((T-n)!))+...+((T)^(n₁=1/2T(==Max))((T-(n₂=1/2T(Continuing))!)+...+((T)^1(1))+1)

 

Corrections made-

Addressed issue of depreciating possibilities beyond n=1/2T.

n split into n₁ and n₂. n₁ Depreciates by 1 each portion past the halfway mark, while n₂ is continuous.

 

Future things to be addressed-

Find probability based on time, at this point, it only takes into effect one instant in time.

 

Thanks for reading.

 

P.S. Sorry for any grammatical errors or missing info, I was tired when making this.

Any and all forms of criticism and comments are welcome!

 

This is a very fluid and changing formula in my experience, and will most likely continue to be refined and improved with the input and criticisms of others.

Edited October 2, 2014 by NowYouKnowFACS

[ajb]

10^80 is an estimate of the total number of baryons in the observable Universe. This is not the total number of particles, which is not fixed. For example, the number of photons is not a conserved quantity. But I notice you have a factorial sign here? How did you estimate what you call T?

[andrewcellini]

what observations led you to this formula?

[NowYouKnowFACS]

10^80 is an estimate of the total number of baryons in the observable Universe. This is not the total number of particles, which is not fixed. For example, the number of photons is not a conserved quantity. But I notice you have a factorial sign here? How did you estimate what you call T?

First off, I would like to thank you for your reply, and thanks for sparking some research into baryons.

T is honestly a variable for anything that takes up a certain amount of space and can only have a specific number of said objects in a specifically sized container, but the size does not matter in this formula, just the number of things that can fit. I just so happened to use the "particles in the universe" example for this.

That being said, I used (In order to find "T", the number 10^82, which is the top of the range of estimated particles in the universe, according to http://www.universetoday.com/36302/atoms-in-the-universe/ . Now, I used that number, and multiplied that by percent of the universe that is filled, which is 0.0000000000000000000042%, or 4.2x10^-24, if expressed as a decimal, shortened to 10^24 for sake of simplification. Those numbers were taken from http://www.universetoday.com/36302/atoms-in-the-universe/ . Should these 2 be correct, take 10^80, put it over 10^-24, and you get 10^106.

 

I tried to have this be more focused on the equasion, not the actual numbers, as the actual numbers will change depending on how specific you want to be, but the formula *should* stay the same, unless I am proved wrong.

 

Now, you asked about the factorial. It is the only way I can [myself] come up with this, or any similar equation, with my limited knowledge. let me explain why. Using the simplest portion of the equation, T!, this is my reasoning. Since the universe in this portion is completely filled, each MUST replace one another. Assuming this, one can use a factorial to figure out how many possibilities there are, given that number. I can explain the others if wanted.

 

 

what observations led you to this formula?

Well, in addition to said above, let me explain it like how I did to my friend today.

 

Let's take the first part, this assumes the entire universe is filled with particles, junk, whatever, just the max that you can possibly fit into the universe, only given space constraints. So, the first part is just all the combinations of all the possibilities of where they can go, kinda like a deck of cards. Assuming all these, you get T!.

Let's go to the next part. Now you need to know what n₁ and n₂ are. n is the portion number to the right of T!, and represents the number of particles/things/whatever is removed from the space. Once you hit the halfway mark, this is where the n₁ and n₂ splits come into place. n₁ is difficult to explain, but it is the highest number of possibilities you can possibly have. once you take more than this, you are basically repeating the opposite of what you just did, so you go back down to 1. n₂ is where the number just continues, as you are just removing particles/things/whatever. You continue down until you get to (T^1)(T-(T-1)!), or really just T. After that, you just add 1, for the universe with no particles.

 

Thanks for reading, and keeps the questions and comments rolling.

 

Please also keep in mind to focus on the formula itself, as T is really just, as explained above, the most things that can possibly fit into a box.

For figuring out the question proposed, what would be the best thing to use as T? Any input?

Edited October 3, 2014 by NowYouKnowFACS

[NowYouKnowFACS]

Whoever gave me a -1, would you care to explain? Is there something I missed? Something I did wrong? Something you don't understand? I would like to fix whatever you find error in.

[John Cuthber]

"The Probability of the Universe Being the Exact Way It Is"

= 1 exactly, by observation.

[andrewcellini]

nowyouknowFACS, how is that an answer to "what observation led you to this conclusion?"

[NowYouKnowFACS]

"The Probability of the Universe Being the Exact Way It Is"

= 1 exactly, by observation.

I was explaining probability as 1 to Total possible outcomes. You are correct in saying we observe 1 outcome, but I am speaking about theoretical possible universes in the same size as ours.

 

nowyouknowFACS, how is that an answer to "what observation led you to this conclusion?"

Ok, sorry, let me redo this, I have no idea why I misunderstood the question.

 

Observations-

1)There are X number of particles in our universe, and only X percent (decimal) is filled, so the possible total (Expressed as T) would be Particles/percent(decimal).

2)Very similarly to a deck of cards, you have a specific set of 52 cards, which takes up the same amount of space, which to find out the number of possibilities, you do 52!, which is what I based T! off of.

3)Not every possibility of the universe is completely filled, so to account for this, ((T)^n((T-n)!)) was made. It is essentially the same as T, except particles are removed. Now, the left portion is to account for which particles are removed. Each particle is assigned a number, and cannot be replaced, so the possibility (If I am not mistaken) can be expressed as T^n, to show the possibilities of what arrangement of particles can be removed. This is then multiplied by (T-n)!.

4)If only 1 particle remains, there are not (T)^(T-1) possibilities of arrangements, so that is why I explained where n₁=1/2T is the max for that portion of the equation.

5)You must add 1 for the possibility of a universe absent of particles.

 

If there are any other questions, feel free to ask.

 

I would like to reiterate I am a high school student. I am not a college graduate. This is all speculation, and I am in no way/shape/form any way an expert on the subject.

 

If I am wrong, please explain why, so I can either correct the misinterpretation, or fix or even invalidate my equation.

[Moontanman]

I was explaining probability as 1 to Total possible outcomes. You are correct in saying we observe 1 outcome, but I am speaking about theoretical possible universes in the same size as ours.

 

Ok, sorry, let me redo this, I have no idea why I misunderstood the question.

 

Observations-

1)There are X number of particles in our universe, and only X percent (decimal) is filled, so the possible total (Expressed as T) would be Particles/percent(decimal).

2)Very similarly to a deck of cards, you have a specific set of 52 cards, which takes up the same amount of space, which to find out the number of possibilities, you do 52!, which is what I based T! off of.

3)Not every possibility of the universe is completely filled, so to account for this, ((T)^n((T-n)!)) was made. It is essentially the same as T, except particles are removed. Now, the left portion is to account for which particles are removed. Each particle is assigned a number, and cannot be replaced, so the possibility (If I am not mistaken) can be expressed as T^n, to show the possibilities of what arrangement of particles can be removed. This is then multiplied by (T-n)!.

4)If only 1 particle remains, there are not (T)^(T-1) possibilities of arrangements, so that is why I explained where n₁=1/2T is the max for that portion of the equation.

5)You must add 1 for the possibility of a universe absent of particles.

 

If there are any other questions, feel free to ask.

 

I would like to reiterate I am a high school student. I am not a college graduate. This is all speculation, and I am in no way/shape/form any way an expert on the subject.

 

If I am wrong, please explain why, so I can either correct the misinterpretation, or fix or even invalidate my equation.

 

 

There is no way you can calculate what the probability of the universe is from one data point. As John said, the probability of the universe being the way it is = 1

[andrewcellini]

Observations-

1)There are X number of particles in our universe, and only X percent (decimal) is filled, so the possible total (Expressed as T) would be Particles/percent(decimal).

2)Very similarly to a deck of cards, you have a specific set of 52 cards, which takes up the same amount of space, which to find out the number of possibilities, you do 52!, which is what I based T! off of.

3)Not every possibility of the universe is completely filled, so to account for this, ((T)^n((T-n)!)) was made. It is essentially the same as T, except particles are removed. Now, the left portion is to account for which particles are removed. Each particle is assigned a number, and cannot be replaced, so the possibility (If I am not mistaken) can be expressed as T^n, to show the possibilities of what arrangement of particles can be removed. This is then multiplied by (T-n)!.

4)If only 1 particle remains, there are not (T)^(T-1) possibilities of arrangements, so that is why I explained where n₁=1/2T is the max for that portion of the equation.

5)You must add 1 for the possibility of a universe absent of particles.

this seems more like a train of thought rather than observations.

[ajb]

1)There are X number of particles in our universe, and only X percent (decimal) is filled, so the possible total (Expressed as T) would be Particles/percent(decimal).

The number of particles in the observable universe is not constant, for example there is no conservation of photon number. Anyway you can make a guess at the baryon number, which we did as something like 10^{80}. Okay...

 

Then what do you mean by "X percent (decimal) is filled"? Do you mean something more like the average density of matter in the Universe? This is something like a few hydrogen atoms per cubic meter.

[NowYouKnowFACS]

The number of particles in the observable universe is not constant, for example there is no conservation of photon number. Anyway you can make a guess at the baryon number, which we did as something like 10^{80}. Okay...

 

Then what do you mean by "X percent (decimal) is filled"? Do you mean something more like the average density of matter in the Universe? This is something like a few hydrogen atoms per cubic meter.

Let me start off by stating what I have tried to say many times. The numbers used are irrelevant, and as you said, cannot be calculated. To find more of what I was saying, look here -http://www.universet...n-the-universe/- This is where I got that number, and an explanation of it.

 

HOWEVER- this is not the point of all this. I have decided to ignore the questions based on the numbers themselves, as they are irrelevant to the equation itself.

 

In fact, let's rename this to "The possibilities of objects in a box", as that is more what this equation is.

 

You have T total objects that can fit. There are T! number of possibilities should it be filled. This is like the deck of cards example.

You remove one object, so there are T^1 possibilities of which one could be removed, and there are then (T-1)! possibilities of objects in the box.

You remove objects until there are half the objects left(n=1/2T). At this point, the highest total possibilities for T^n has been reached, and will then return to 1 (or 0).

You go all the way until you have all but 1 object removed. You then have T possibilities, as it can be any of the objects.

There is also a possibility of no objects being in the box, and therefor, must add 1.

 

This is a simple, non-dealing-with-the-universe explanation of the equation. No questions dealing with numbers themselves will be answered, as it is considered irrelevant to the equation itself, unless it is used to disprove the equation, which, as of now, is:

1:((T!)+((T)^n((T-n)!))+...+((T)^(n₁=1/2T(==Max))((T-(n₂=1/2T(Continuing))!)+...+((T)^1(1))+1)

 

I am sorry for my frustration, however, almost all of the questions have not been regarding the actual equation.

Sorry to andrewcellini for my ineptitude to answer his question satisfactorily. (This is a sincere statement)

Edited October 7, 2014 by NowYouKnowFACS

[andrewcellini]

how would you know the probability of a given configuration of objects in the box unless you opened it many many times?

Edited October 7, 2014 by andrewcellini

[Bignose]

In fact, let's rename this to "The possibilities of objects in a box", as that is more what this equation is.

 

You have T total objects that can fit. There are T! number of possibilities should it be filled. This is like the deck of cards example.

I think this analogy fails. Because an object in a box doesn't have to have a fixed position. I.e. an object could be located at x=1.0 just as well as at x=1.000000001 just as well at x = 1.4 and so on. An infinite number of choices actually.

 

However, a card in a deck has to be in an ordinal position from 1 to 52. Not an infinite number of choices.

 

Usually things like "probability of finding a particle in a certain state" uses some kind of Liouville/Hamiltonian description. http://en.wikipedia.org/wiki/Liouville's_theorem_(Hamiltonian)

 

You define terms in probabilities of finding particle 1 within the element of state space [math]d\mathbf{x}_1[/math], finding particle 2 within the element of state space [math]d\mathbf{x}_2[/math], and so on.

Edited October 7, 2014 by Bignose

[ajb]

It is not only that we have an an infinite number of "places to fill", we don't have to restrict ourself to putting just one particle at each location. Bosons are allowed to "share the same state"...

 

In short we think the equation is of no cosmological relevance.