View Full Version : From something, comes nothing...?
1=\phi →\phi/ \infty=0
But,
0= -\phi / \infty ←- \phi=1
If the angle has to be perpendicular over infinity, then Φ → Φ*(∞)|(∞) if there are six degrees of freedom, as in x,y,z and –x,-y and of course –z, and including that Φ → Φ*(∞)|(∞) obeys distributive properties…
I have more to say… give me time…
AlphaNumeric
06-27-08, 02:44 PM
Complete and utter nonsense.
Do you have a job? What do you spend your time doing exactly? Do you sit at home and think "Hey, I'll post more BS on SciForums in the hopes I'll fool some people into thinking I'm not a complete retard".
Well you failed.
Not at all. I wizz around and thoughts on physics always plague my mind.
MacGyver1968
06-27-08, 02:50 PM
I think your wizzing on physics. :)
Now consider tha wave function of the first equation:
\psi >=a_{1}|k_{1}> +a_{2}|k_{2}>
Here,a_i is an acting complex number. Bra is very different to ket in that it does not share the same system as ket. But for every ket, it is said there exists a bra notation so where there is |a> function, there is <k|.
Using the bra-ket rules, the equation |\psi>=a_{1}|k_{1}> +a_{2}|k_{2}> has a new arrangement with bra notation as well,
a_{1}|k{1}> +a_{2}|k_{2}>↔a_{1‘}<k_{1}| + a{2‘}<k{2}|
If we replace \phi with a function of the probabilistic distrubution of the field, \psi, and in reflection of distributive products, such as a_{1}|k{1}> +a_{2}|k_{2}>↔a_{1‘}<k_{1}| + a{2‘}<k{2}| we can hint that one probability must determined another.
So if there was a probability as given:
0=({-\phi} /infty)←({-\phi}=1)
as a negative equator of a solution, if | \Phi>→|-\Phi> so
an abstract operation such as 1=0 can be allowed, so long as 0=1.
Vkothii
06-27-08, 08:11 PM
You should switch coffee brands, mate.
You should switch coffee brands, mate.
Acually, it should be clear that when dealing with \sqrt{1} and \sqrt{-1}, they not only represent negative time directions, that can be even considred as two operations <(t,1)a|b(t,2)> that can be in concordance simultaneously.
Not only that, the very functions themselves seem to propose an unlimited value, of N=0, so that there are positive and negative solutions. Dirac notation, at least as far as i believe is excellent at describing these solutions, even when reflecting on the recently controversial Hamiltonian expression of E \pm=Mc^2.
In a recent thread here, QQ made a valid point. There is no aboslute zero 00*=|0|^2, so there might be something very wrong about what we call time-zero, and space-zero, when reflecting on their simultaneous birth in a single chronon from no where... hence... it can't be an absolute zero.
So, if 0=1, then it is a contradiction only solvable by saying that zero actually has a value of one, and so long as the value of one keeps by the associative principles behind them.
Refere:
J. A. Wheeler and R. P. Feynman, "Interaction with the Absorber as the Mechanism of Radiation," Reviews of Modern Physics, 17, 157–161 (1945).
''Two-Time Measurements, (ADD), 1985,''
J. A. Wheeler and R. P. Feynman, "Classical Electrodynamics in Terms of Direct Interparticle Action," 21, 425–433 (1949).
Bergenheim, M., H. Johansson, B. Granlund, and J. Pedersen. 1996. Experimental Evidence for a Synchronization of sensory Information to Conscious Experience. In Toward a Scientific Basis for Consciousness. S. R. Hameroff, A. W. Kaszniak. and A. C. Scott. (Eds.). The MIT Press, Cambridge, Mass., pp. 303-310.
Cramer, J. G. 1983. Generalized absorber theory and the Einstein-Podolsky-Rosen paradox. In Physical Review D 22:362-376.
Cramer, J. G. 1986. Transactional interpretation of quantum mechanics. In Reviews of Modern Physics 58:647-687.
Libet, B., E. W. Wright, B. Feinstein, and D. K. PearlK. 1979. Subjective Referral of the Timing for a Conscious Sensory Experience. In Brain 102:193-224.
Penrose, R. 1994. Shadows of the Mind. Oxford University Press, New York, p. 387.
AlphaNumeric
06-28-08, 02:25 AM
Complete nonsense.
I bet if I asked you something simple on Dirac notation you'd be unable to do it.
Your expressions aren't even coherent. The ones you don't **** up you just make up. I bet you cannot actually do anything with such notation.
Just because you read something on Wikipedia doesn't mean you even know how to copy and paste the equations here. Euler saw Penrose give a talk recently and Penrose was being interviewed by a TV crew afterwards. The majority of his talk had been based on work involving fibre bundles and the TV crew said "Can you give the viewers at home some example of how to easily grasp this concept" and Penrose, without skipping a beat just looked him in the face and said "No."
And?
Dirac notation is very hard. You shouldn;t degrade one for at least trying, even if one has it right or not. Remember, what i told you about being a modest scientist... perhaps a tad of integrity about them?
In other words, boasting and therefore flaming.
AlphaNumeric
06-28-08, 03:10 AM
Dirac notation is very hard. You shouldn;t degrade one for at least trying, even if one has it right or not.The thing is you're not 'at least trying', you're using something you know you don't understand to make claims about something else you don't understand.
If you started a thread, like QH does, saying "I'm trying to do question 1 (http://www.damtp.cam.ac.uk/user/examples/D18a.pdf) but I'm a bit stuck on how to express the Hamiltonian in the right form" I'd be happy to help. Instead, you post thread after thread where you make claims about conciousness and the universe using Dirac notation and then ignore correction.
The problem is you aren't trying.
Remember, what i told you about being a modest scientistWhy don't you show some modesty, admit you don't know Dirac notation and then we can start a thread where some of us who do know it guide you through learning it. I'm always happy to help. You're just not happy to learn.
Anyway... back to ignore my friend.
...........\\
Now using my differential equations, to solve two paradoxes… guess which ones they are… :) I express:
(\int \psi (\hbar/i)(\phi / \phi x) \psi*(\hbar/i) (\phi / \phi x) \psi(x)d(x))-(<xp>)=\int \psi x(\hbar/i)(\phi \phi x) \psi*dx
Only if \int \psi(\hbar/i)(\phi/\phi x) \psi*x dx I not used in the equations as a substitute for (<xp>)=\int \psi x(\hbar/i)(\phi \phi x) \psi*dx
To make something real in Dirac Bra-Ket notation, these principles must appy:
<xp>*=<px>*
<px>*=<xp>*
>>> where ^* it means acting conjugates…
So since you can only deduct these as:
<xp>=<xp>*+i\hbar + <px>*=<px>-i\hbar
They give real directions, and negative directions for objects through spacetime.
AlphaNumeric
06-28-08, 05:06 AM
What a suprise, I challenge Reiku to put his physics where his mouth is an he runs away (http://www.sciforums.com/showthread.php?t=82501).
To make something real in Dirac Bra-Ket notation, these principles must appy:
<xp>*=<px>*
<px>*=<xp>*
>>> where ^* it means acting conjugates…You only need one of those equations, since they are the same equation with the left and right hand sides swapped.
So since you can only deduct these as:
<xp>=<xp>*+i\hbar + <px>*=<px>-i\hbarNo, you start with [x,p] = i.\hbar and then take the expectation of those, giving you
\langle xp - px \rangle = i\hbar. The real or complex nature of \langle xp\rangle is irrelevent, since it's true irrespective of the value of <xp> or <px>.
And your equations are wrong :
[x,p] = ihbar (which I'll call h from here on in)
<xp-px> = ih
<xp>-<px> = ih
<xp> = ih + <px>
If <xp>=<xp>* and <px>* = <px> (which is a specific case, not general) then taking complex conjugates we get
<xp>* = -ih + <px>*
<xp> = -ih + <px>
But <xp> = ih + <px> too, so adding them gives
<xp> = <px>
so
<[x,p]> = 0
But by definition of quantum commutation relations <[x,p]> = ih
Contradiction.
Therefore <xp> and <px> cannot be real.
Yet again, the simplest application of Dirac notation shows Reiku is not only ignorant but flat out wrong. And since I'm on his ignore list, he'll continue to post nonsense despite his BS plain for all to see.
I'm sorry, but i am going to have to report this to then Supermoderator.
AlphaNumeric
06-28-08, 07:25 PM
Why? I have posted actual quantum mechanics. I've used actual mathematical physics to demonstrate your claims are false.
Why should I not be allowed to post physics in reply to your pseudoscience? Could it be you just don't like being shown to be wrong? Then stop posting stuff you just make up. You give me ammunition against you every time you start one of these threads.
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