Gravigyro-Magnetic Equations with the Angle Between Spin States and a Force Equation

Discussion in 'Pseudoscience Archive' started by Reiku, Feb 28, 2012.

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1. ReikuBannedBanned

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Did you follow any links as well? Do you have motz' paper handy? You would see more relevance as well if you followed the papers linked and went out your way to find motz paper.

3. ReikuBannedBanned

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And the ij notation on the matrix is in fact saying you can decompose the equation to suit for particle i or particle j. You think you're right and you are so far from it. Tensor calculus on the head AN!

5. ReikuBannedBanned

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I'll tell you what AN, how about if I give some demonstrations of this ''decomposing the equation for particle i and j?''

From now on, I am going to drop the ij notation on all the other values because it seems to have confused the hell out of you.

7. ReikuBannedBanned

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I said I would drop the notation on other values, but... some of them need to be expressed this way to get an idea of what is happening

$V = \sum^{N-1}_{i=1} \sum^{N}_{i+1} g(r_{ij})$

This is a potential.

The force then between two particles can be given as

$F_{ij} = - \frac{\partial V (r_{ij})}{\partial r_{ij}} \hat{n}_{ij}$

Notice, $r_{ij}$ is simply the distance between $i$ and $j$, then it becomes noticable that I am using this subscript on values as markers.

Let's decompose some equations for the particles $i$ and $j$ respectively - let's begin with a simple example for the force equation

$M_ia_i = \sum^{N}_{j=1, j \ne i} F_{ij}$

which describes the motion of particle $i$ and for particle $j$ we have

$M_ja_j = \sum^{N}_{i=1, i \ne \j} F_{ij}$

As you can see, the $ij$-notation is doing nothing of the sort of calculations alphanumeric was insinuating.

Let's do it again, this time for the density relationship for the spin along a certain axis:

$\frac{\nabla^2 \phi}{G}((\hat{n} \cdot \vec{\sigma}_{ij}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi}{G} \vec{\theta}_{ij} \rightarrow T_{\mu \nu} \delta^{\mu \nu}$

Can be docomposed for $i$ and $j$ as:

$\frac{\nabla^2 \phi}{G}((\hat{n} \cdot \vec{\sigma}_{i}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi}{G} \vec{\theta}_{i} \rightarrow T_{\mu \nu} \delta^{\mu \nu}$

and

$\frac{\nabla^2 \phi}{G}((\hat{n} \cdot \vec{\sigma}_{j}) \begin{bmatrix} \alpha \\ \beta \end{bmatrix}) = \frac{\nabla^2 \phi}{G} \vec{\theta}_j \rightarrow T_{\mu \nu} \delta^{\mu \nu}$

Come to think about it, did you not even ask yourself why in the first part of the OP why I was making the reader accustomed to knowing this notation for particle spins for particles $i$ or $j$?

8. ReikuBannedBanned

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Hmmm.... Now my memory tells me that James not long ago posted a quetsion to me involving matrix multiplication....

So I call you a flat-out liar in this respect.

9. ReikuBannedBanned

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Ok, so why did I create the force along an axis equation?

I'm going to explain some of the mathematics of the matrices. It's pretty much standard in QM to calculate $(\sigma \cdot \hat{n})$ gives you a four by four matrix, it is also a dot product of the form $n_1\sigma + n_2 \sigma_2 + n_3\sigma_3$. The matrix is

$\begin{bmatrix}\ (n_3) & (n_{-}) \\ (n_{+}) & (-n_3) \end{bmatrix}$

Some definitions of these entries are

$n_1 - in_2 = n_{-}$

$n_1 + in_2 = n_{+}$

and

$(\sigma + \hat{n})^2 =1$

we can have

$n_{+}n_{-} = n_{1}^{2} + n_{2}^{2}$

and

$+n_{3}^{2} - n_{3}^{2} = 1-n_{3}^{2}$

This is all standard stuff. Now draw two lines perpendicular to each other in unit length, one might as what is the probability that $\hat{m}$ is $+$ and $\hat{n}$ is $-$, well, knowing that $|\sigma \cdot n = 1>$ then the probability is

$|<\sigma \cdot m = 1 | \sigma \cdot n = 1>|^2$

Now

$\begin{bmatrix}\ (n_3) & (n_{-}) \\ (n_{+}) & (-n_3) \end{bmatrix} \begin{bmatrix}\ \alpha \\ \beta \end{bmatrix} = \begin{bmatrix}\ \alpha \\ \beta \end{bmatrix}$

this is just

$\begin{bmatrix}\ 1 \\ z \end{bmatrix}$

and

$n_3 + n_z = 1$, thus $n_z = 1-n_3$ then $z= \frac{1-n_3}{n}$ thus

$\begin{bmatrix}\ 1 \\ z \end{bmatrix} = \begin{bmatrix}\ 1 \\ \frac{1-n_z}{n} \end{bmatrix}$

We are actually on our way to calculate the angle between the two unit vectors. Well, actually... there is quite a bit of rough and tumble with some algebra first. I am going to skip about 5 operations to get

$= \begin{bmatrix}\ \sqrt{\frac{1+n_3}{2}} \\ \frac{1-n_3}{n_{-}} \sqrt{\frac{1 + n_3}{2} \end{bmatrix}$

which gives

$= \frac{2}{1+n_{3}}$

I skip it because it is very tedious with a lot of matrix latex involved. Yuk.

Anyway, the dot product of $n$ and $m$ is given with a constant

$1+ n_1m_1 + n_2 m_2 + n_3m_3$

which produces the angle between the vectors

$\frac{1 + n \cdot m}{2} = \frac{1+ cos \theta_{nm}}{2}$

and Viola, that is that part finished. Whilst this is standard, the new relationship of the force and the angle relationship presented itself to me when understanding that

$-\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \mu(\hat{n} \cdot \vec{\sigma}_{ij}) = -\frac{\partial^2 V^2 (r_{ij})^2}{\partial^2 r^{2}_{ij}} \begin{bmatrix}\ \mu(n_3) & \mu(n_{-}) \\ \mu(n_{+}) & \mu(-n_3) \end{bmatrix}$

could in principle be calculated because the force equation:

$F_{ij} = -\frac{\partial V(r_ij)}{\partial r_{ij}}\hat{n}_{ij}$

Had the unit vector in it. It just goes that the relationship was allowed to be given as such.

Last edited: Mar 2, 2012
10. ReikuBannedBanned

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As soon as you know your angle, you can calculate the original distance $r_{ij}$ in your force equation using some simple trigonometric identities so the equation definately has its uses.

11. ReikuBannedBanned

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Oh! And don't forget that you may freely exchange the notations $\theta_{nm}$ and $\theta_{ij}$. Don't mistake it as a tensor calculation like AN did.

12. ReikuBannedBanned

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I've made this statement about the trig a few times I should give a demonstration at least. We may calculate the component:

$comp (\hat{m} \hat{n}) = ||\hat{m}|| cos \theta = \hat{n} \cdot \frac{\hat{m}}{||\hat{m}||}$

which is how the trigonomentry comes in.

13. BalerionBannedBanned

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I think AN pretty much summed this thread up.

Cesspool time.

14. funkstarratsknufValued Senior Member

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Complete and utter bullshit. No pass.

15. ReikuBannedBanned

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CRAP?

Susskind... you know that scientist... top scientist... was his teachings... that reallly good scientist, most of us like?

CRAP... go say it to his face lol

16. ReikuBannedBanned

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The (standard stuff of course) was his teachings. The rest was my realizations.

17. ReikuBannedBanned

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Tell you what Funstar, shall I find the video for you, so you can work out the angle for yourself... you really need to stop being so aggresive. You've clearly demonstrated time and time again you don't even know physics yourself to refute it.

But hold on, I will find it. I have the video link somewhere...

18. ReikuBannedBanned

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Well, can't find it right now. There are so many video's of his which is the problem. i've watched and took notes from every one of them... there is well over 30 videos there. But it's only a matter of time. I will find it.

But I still refute your claim it is crap. It certainly isn't if I have learned it off him, which I have because I have ''susskind lecture'' written at the top of it.

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21. AlphaNumericFully ionizedRegistered Senior Member

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You think $\mu,\nu$ means tensor and i,j doesn't? Clearly you've never done any actual relativity, as the form an index takes is entirely at the authors discretion. Personally I use $\mu,\nu$ to represent 3+1 dimensional coordinates, $n,m$ to mean 6 dimensional coordinates on compact spaces, N,M to mean 9+1 dimensional coordinates and i,j to represent 3 dimensional coordinates. But that's just me, other authors have other preferences.

By saying "A tensor is normally written as $G_{\mu\nu}$ you illustrate how little reading you've done and how completely inexperienced you are. I would guess that you've picked one paper/video to copy stuff from and you think the notation used there is somehow universal. It's a clear sign of how ignorant you are.

As for it being Newton's constant, you would have no need to give it indices. If all terms are multiplied by Newton's constant then you just have an overall factor of G, $G_{ij}$ is meaningless. Again, you're clutching at straws to make excuses for your mistakes. It's as you always do.

This thread, like all your others, is completely disingenuous, pointing that out and pointing out your dishonesty and mistakes isn't trolling. Calling a liar a liar is not something they want to hear but it isn't trolling. The liar being a repeated and deliberate liar is trolling. Why do you think you get banned so often?

I know enough physics that you can't try that argument against me and you ignore me too. And your "I'll go find a video!" thing always backfires. It either shows you're just mindlessly copying from someone, as you copied Susskind in your discussion with James, or you're actually mistakenly trying to paraphrase source material you don't understand.

22. ReikuBannedBanned

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huh?

I said ''normally written''. Not ''is written''.

I know it's the morning AN, but please, open those eyes.

23. ReikuBannedBanned

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AN

''This thread, like all your others, is completely disingenuous, pointing that out and pointing out your dishonesty and mistakes isn't trolling. Calling a liar a liar is not something they want to hear but it isn't trolling. The liar being a repeated and deliberate liar is trolling. Why do you think you get banned so often? ''

You're behaviour towards me recently is more aggressive and progressively odd. I don't read my thread and it says ''liar'', unless I have said something which presents me as such. I haven't.

I have used quantum mechanics as I have been taught, I've taught myself. My force equation (I've demonstrated above) had some good applications, such as being able to calculate the distance between two particles using trigonometric identities. It calculates the force between particles and the local force along a spin axis.