# Thread: Mathematically Determining a Dimension for the Mind

1. ## Mathematically Determining a Dimension for the Mind

It is possible that consciousness is in fact under perfect conditions to be integrated against a mathematical model from the legendary Pythagorean theorem,

a^2+b^2=c^2

and in any standard course of geometry, you solve it for c, so that c^2√a^2+b^2. Of course, this equation led the way for vector calculus, and describing space as such a vector with three dimensional coordinates.

The human beings mind is an extraordinary instrument. It can quite literally ‘’recreate’’ a three-dimensional world, but somehow, doesn’t exist in it. In fact, no matter how much we deny the latter part, it is well known that what we ‘’observe’’ is not of the outside world, but a representation of it.

There is not many mathematical models of consciousness existing in physics today because of two reasons. One being that string theory has stolen so much time out of the academic world, and two that its not really known how. How do you make a mathematical model of consciousness?

One start, is by describing the ‘’recreated’’ three dimensional phenomenon with similar vector calculus used to describe space. When we do, we can even treat both types of vectors are as totally separate systems… why? Because the mind doesn’t exist in any unique time or external space. What we see, or sense, is purely imaginal.

||V||=√x+y+z,

For a normal three-dimensional space vector calculation, but I am wanting to describe space as well as something ‘’separate’’ to the external dimensions of the human mind, so it would be best to associate imaginary values as well real.

i=x+y+z

And under normal calculative vector products, we can state that the coordinates be described as a row vector with increasing values (a_1, a_2…)

||V|| =i√a_1+a_2+a_3

Giving each increasing value with an imaginary product. But as we introduce a four-vector condition, with one time dimension also acting as an imaginary space dimension, it’s already unique with the human perception, because already, time is an imaginary product of relativity. In other words, time is an imaginary dimension, so we aren’t really going to need to change any condition of the time dimension as seen from a Minkowski spacetime.

Some scientists have gone as far as to claim that perhaps time and mind are two different sides of the same coin, so by making a spacetime model for the mind, under the impression it is ‘’separate’’ from the external world, may not be entirely correct or truthful. But can instead become an excellent tool in bringing consciousness and spacetime together.

η = (0,0,0,1) is the row vector describing a four dimensional vector space under a Poincare Group, with one time dimension, and as many of us will have already seen, is put into a Minkowski Matrix:

….1000
….0100
η =0010
….000-1

Remember, even though this is used to describe spacetime at large, they are so very similar to how can describe the imaginary vectors of what we come to observe and sense. What is great about doing this, is that we can integrate similar concepts of timelike and spacelike qualities to consciousness, when we undergo some chemical or neural change, so that the operators don’t run normally.

What do I mean? Simply that we experience also a spacelike η(v,v,)<0 and timelike η(v,v)>0 condition, so normally, in everyday life [[we usually]] experience the correct flow of time and space (1)… we sense time moving forward without recourse, and we sense our movement in three dimensions through space (2). When does space and time not move accordingly? The answer turns out to be when the mind is not fully aware. When we dream, it is possible that the spacelike conditions and timelike conditions do not operate normally when we are awake (3).

Moving on, there are other ways to describe consciousness. We can totally ignore the fact we observe a three-dimensional vector space, and only focus on the one time dimension, and describe ‘’normal’’ spacetime with an extra time dimension uniquely attached to spacetime. In other words, using the normal Poincare Group and adding on the vector we experience I will describe as Tdi, simply, so that a row vector is given as η =(0,0,0,1,Tdi), so that I can state the following:

a^2 + b^2 + c^2 + tdi^2 - Tdi^2

With the normal time vector described as tdi. I solved the real part of the equation by allowing i^2 = i *k^2 so that the result is

a^2 + b^2 + c^2 - i^2*k^2^2 = 0
a^2 + b^2 +c^2 + k2^2 =0

Solving it normally through algebra, giving a final solution of: a=b=c=k^2=0. There is one last way to integrate this, and that is simply by treating the normal time vector of space as [the] dimension of the mind:

a^2 + b^2 + c^2 + tdi^2

So there is no need to add any extra time dimension at all, so we can allow the one time dimension to have two functions, and would fit in nicely with the postulation that time is somehow mind, made by many scientists over the years.

(1) – I say flow, because flow is a psychological creation of time. It is well understood now that time does not have a flow external of the mind.
(2) – And for those who have heard of ‘’spacetime’’ switching coordinates, so that time is spacelike and space is timelike, so that you begin to move through space in a timelike movement without recourse, and through time as easily as you had moved through space. Unless you have some kind of special mental disability, these specific conditions are purely relativistic in nature.
(3) - The only time I speculate we can describe the internal vectors of perception with zero qualities (0,0,0,0) is when we don’t experience anything at all, such as a total blank condition in our memory.

2. Since the impending source of discussion here is the subject of dimension and vector analysis, i want to point out that the only reality at large is the dimensions the mind creates. If any dimension or vector spacetime is real, in any sense to us, it is the one i have been trying to mathematically describe.

3. ||V||=√x+y+z,
You mean $||V|| = \sqrt{x^{2}+y^{2}+z^{2}}$
For a normal three-dimensional space vector calculation, but I am wanting to describe space as well as something ‘’separate’’ to the external dimensions of the human mind, so it would be best to associate imaginary values as well real.

i=x+y+z
You mean something more like $x^{2}+y^{2}+z^{2} = -1$. After all, $\sqrt{x^{2}+y^{2}+z^{2}} \not= x+y+z$.
And under normal calculative vector products, we can state that the coordinates be described as a row vector with increasing values (a_1, a_2…)
I'm forced to ask you just wnat vector calculus you've done. Inner products are not something taught in high school. So which book or lecture course did you learn it from?
η = (0,0,0,1) is the row vector describing a four dimensional vector space under a Poincare Group,
No, it's 'a' row vector and you cannot tell anything about how a vector transforms under a particular group. For instance, x = (0,0,0,1) can be in any 4 dimensional vector space. If it's Euclidean then it's not going to transform under the Poincare group, it would be the Galilean group instead.

You define the transformations by giving the metric.
and as many of us will have already seen, is put into a Minkowski Matrix:

….1000
….0100
η =0010
….000-1
Don't use the same symbol for a vector and a metric in the same space.
We can totally ignore the fact we observe a three-dimensional vector space
4 dimensional. The vector and metric you just gave are 4 dimensional.
With the normal time vector described as tdi. I solved the real part of the equation by allowing i^2 = i *k^2 so that the result is

a^2 + b^2 + c^2 - i^2*k^2^2 = 0
a^2 + b^2 +c^2 + k2^2 =0

Solving it normally through algebra, giving a final solution of: a=b=c=k^2=0.
Completely inconsistent. For a start, i is a constant, you cannot consider di. It's like saying d5. You mean idT. If you have $x^{2}-t^{2}$, you cannot show that it's got no real solutions by saying "Let t = iT, so $x^{2}+T^{2}=0$ so x=T=0. If the original x and t are real then doing a Wick rotation gives you a complex T. So you cannot assume that $x^{2}+T^{2}=0$ gives x=T=0. Clearly x=t is a real solution to the original equation so real solutions exist.

What vector calculus do you actually know?
(2) – And for those who have heard of ‘’spacetime’’ switching coordinates, so that time is spacelike and space is timelike, so that you begin to move through space in a timelike movement without recourse, and through time as easily as you had moved through space. Unless you have some kind of special mental disability, these specific conditions are purely relativistic in nature.
It's a choice of signature. You can have any of the following :

$\eta = \textrm{diagonal}(1,1,1,-1)$
$\eta = \textrm{diagonal}(-1,-1,-1,1)$
$\eta = \textrm{diagonal}(-1,1,1,1)$
$\eta = \textrm{diagonal}(1,-1,-1,1)$

provided you are consistent with your conventions. It's nothing to do with moving through time as easily as you move through space, it's just deciding wether you want to work with minus signs in particular places in your equations.

But then you wouldn't know that, you've never done any relativity.

*cue Reiku having a cry because I point out he doesn't know the things he talks about*

4. Well, this is better. At least your talking about the subject. The first set of equations where meant to have the squares. Yes.

I was taught vector calculus when i first started college. In fact, i am sure it was one of the first things we were taught about.

What do you mean by using the same symbol?

And you can consider Tdi, because T and d are not important like i, or atleast, they are not subject to any particular value. They are only describing the function of the constant, in these equations. They can be solved this way, because i had a scientist look at them, so i am quite happy.

And no i won't cry. I am more than overjoyed you are tackling the subject, and not directing all your attention to me for a change.

5. And the latter, i will not answer correctly. You just can't help having a stab, but i will say relativity is my strongest area.

6. I also want to note that Td and td are only descriptions, and shouldn't be treated as important other than (short notation) -- did you pick up what td and Td even meant? I doubt you did, because you are treating them as real variables by giving solving them as x=t=0. In the original sum, a^2+b^2+c^2+tdi-Tdi, it is almost better to give it a value of: a^2+b^2+c^2+i_1+i_2, however, it doesn't specify what i_2 and i_1 have in common, if they have anything in common at all. Its like pulling two variables out of the aether and hoping they are understood through mere calculus alone.

7. Originally Posted by Reiku
I was taught vector calculus when i first started college. In fact, i am sure it was one of the first things we were taught about.
Mechanics taught in A Level maths is not what I call 'vector calculus'. It's like saying "I have done A Level English" when you've just learnt the alphabet. This is vector calculus.

And you certainly aren't taught it in physics at high school level!
Originally Posted by Reiku
What do you mean by using the same symbol?
You called both a vector and the metric $\eta$. Didn't you notice?
Originally Posted by Reiku
And you can consider Tdi, because T and d are not important like i, or atleast, they are not subject to any particular value. They are only describing the function of the constant, in these equations. They can be solved this way, because i had a scientist look at them, so i am quite happy.
No, you didn't have a scientist look at them. At least not a competant one. 'd' is not a variable, it doesn't take a value. It means that dx or dy or d[something] are infinitesimal variations of x, y or [something]. You don't rearrange things to give d = .

For instance, if $\frac{dy}{dx} = 5$ this doesn't mean that its $\frac{d \times y}{d \times x} = 5$ so $\frac{y}{x} = 5$.

It means that an infinitesimal change in the y variable is 5 times larger than the same infinitesimal step in the x direction (this is make rigorous using limits). If you go further and start developing differential p-form notation, as used in differential geometry, then you end up with 'd' being a differential operator. You don't solve for differential operators.

This demonstrates you don't know basic calculus or vectors and that either your 'scientist' friend isn't a very good scientist or you're lying in order to try to justify your nonsense.

I read differential geometry books for fun. You aren't going to BS your way past me.
Originally Posted by Reiku
And the latter, i will not answer correctly. You just can't help having a stab, but i will say relativity is my strongest area.
Relativity taught in college isn't going to allow you to say "I can do relativity". Otherwise I'm a qualified chemist. I got 100% in several of my A Level chemistry exams.

You are on the first step in a long road to becoming able to say "I can do [subject]". You aren't there yet.

8. You're a total arsehole alphanumeric.

9. Then feel free to spend a short time answering 2 or 3 of the questions in that vector calc. link I posted. Prove I'm wrong about you.

10. The mind occupies no dimension because it doesn't exist. This illusion is generated by a physical process. It consists of information.

11. The mind occupies some kind of imaginary dimension. All i ask is that you consider is that the mind has its own degrees of freedom, which is also the definition of a dimension.

12. You're mistaking the usual use of the word 'imaginary' with the mathematical use of the word 'imaginary'.

And even then, you cannot do the maths properly.

How long do you need to do 2 or 3 of those questions? Lunchtime? Not much effort to prove me wrong.

If I am wrong...

13. Actually, i explained that the use of imaginary could be decieving, because we have real and imaginary concepts. I've already spoke about this alphanumeric, so i asked it be treated differently.

14. And you took the latter equations out of context, giving T and t and d actual values when they are totally valueless.

15. Errr... you're the one claiming that you can solve and equation for 'd'. I specifically explained why that's nonsense. Or don't you remember?

You're the one saying you can solve T and t so they must be 'something'.

Why are you avoiding answering any of those questions? Come on, it takes only a few minutes to answer say question 3. The answer is just 1 line. Is 3 minutes too much to ask for you to put your physics where your mouth is? Every time I ask these questions I keep worrying "God, they are so easy, what if he can do them!" but then Euler reminds me over MSN that while we think they are easy, you don't. I've yet to see any crank answer any of them.

16. Your ego is astounding alphanumeric. What are you getting from this? Why are you consulting your lackies who equally don't know anything about my personal life, or what i do daily at college mmm?

As i have explained, i will not entertain pet homework theories. I have 16 equations to do in four days, plus five physics homework problems, and you want me to take time to solve the problems you throw at me... i wouldn't entertain it AN.

And, you will also recall me saying that Tdi an tdi are nothing but i_1 and i_2, so they are calculatable.

17. Originally Posted by Reiku
What are you getting from this?
That warm fuzzy feeling associated with endorphins.

Ah yeah, there it is...
Originally Posted by Reiku
Why are you consulting your lackies who equally don't know anything about my personal life, or what i do daily at college mmm?
Because we all come across cranks in our time online. Cranks are everywhere.

And we know what you don't know, you don't do vector calculus.
Originally Posted by Reiku
As i have explained, i will not entertain pet homework theories. I have 16 equations to do in four days, plus five physics homework problems, and you want me to take time to solve the problems you throw at me... i wouldn't entertain it AN.
I just did question 3 on that sheet. It took me 4 minutes, while watching 'House' and eating breakfast. Here's the solution :

$|\sigma - \lambda I_{3}| = 0$ gives eigenvalues for $\sigma$. Solving this gives $\lambda$ = 0, 3, 3.

Hence, the eigenvector with eigenvalue 0 will be the direction with no current flow. Solving $\sigma \mathbf{v} = 0$ for $\mathbf{v}$ gives v = (1,1,1).

The two directions with $\lambda = 3$ are v = (1,0,-1) and (-1,1,0). Since any linear combination of these is also a solution, be get there's an entire plane, defined by the equation $v_{1}+v_{2}+v_{3}=0$, which has maximal current.

Done. See, easy to do when you know how. And I've known how for 5 years. I'm way beyond that now.

But you couldn't even do that. Do you have so little spare time you couldn't manage that? Yet you have time to write pointless essays about BS like the original post in this thread.

18. An overload of endorphines can cause high hallucigenic properties. I wonder if this is what you seem to suffer from.

19. Still, I'd rather be high on endorphines and truthful about my mathematical ability than normal and a liar whose only way of getting endorphines is to pretend to be well read on quantum mechanics and vector calculus but cannot do either of them and resorts to writing BS essay after BS essay.

Endorphines doesn't explain my maths and physics knowledge. You don't see smack addicts spontaneously becoming the next Einstein. What explains your continued lies, avoidance and ignorance?

20. You watch who you're calling a liar, without any conclusive evidence to suggest otherwise. You base the wild claims on overlaods of self-ego and unsupported conclusions. If you recieve ignorance, its because you are inticing the ignorance to begin with.

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