# My book on physics... (the second chapter)

Discussion in 'Free Thoughts' started by Reiku, Dec 28, 2011.

1. ### ReikuBannedBanned

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wtf am I talking about, energy is not a vector, it'sa scalar. Reading this back, It is clear I got a bit confused.

This is what happens when I speak to AN.

3. ### AlphaNumericFully ionizedRegistered Senior Member

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You said $E = k+M$. It's wrong for both the reasons I just went into. You explicltly state that expression, which does not follow from the physics, it's something you threw in and it's wrong. There's no two ways about it.

Would you like me to quote you? Here you go,

Since you obviously don't get it, in natural units you set $\hbar =1$ and $c=1$. That reduces the first expression to $\omega = \sqrt{k^{2}+m^{2}}$. What the $k^{2}$ actually is is the square vector norm of the 3 component vector k represents, as I went into in my last post. So even if you could just drop the squares and square roots you'd end up with $||k||$, not k itself. At least $||k||+M$ is a meaningful expression, as it's scalar plus scalar and when you put in the units properly they match. But that's irrelevant seeing as you can't go from $\omega = \sqrt{k^{2}+m^{2}}$ to $\omega = k + M$.

You said it, it's right there for all to see. How you possibly think you can lie your way out of that I have no idea.

Utterly wrong again. Adding in units doesn't change the vector/scalar/matrix structure of an expression, it just rescales things. Rather than saying "1 unit of time" you say "10 seconds" or "1 unit of momentum in the z direction" you say "67 kg.m/s in the z direction".

Now you're showing you don't even know what using natural units means!

This is precisely the problem I and other have with you. You've made a mistake, one which is very basic and which you should just face up to. But rather than doing that you whine about it's someone else's fault and you then compound your errors by trying to explain them away by stating more mistakes!

Properly using units isn't even university level stuff, it's high school.

I point out errors in your posts all the time because you post lengthy posts full of errors all the time. You know full well you don't grasp this stuff properly yet you come here and try to BS people who do know it. Why are you surprised some of us don't stand for it and point out your mistakes?

I'll quote you again and highlight the relevant bit,

You equated them. The first expression is fine. The second is wrong because k is a vector and M is a scalar. Even if we take your k to mean $||k||$ your statement the two expressions are the same is wrong. I'm not misrepresenting you, I'm addressing exactly what you said.

So 'the same' doesn't mean 'the same'? Give me a break!

So now you're saying it isn't that your maths was wrong, you just can't communicate properly?

Thanks, you've just shown you don't understand the Dirac equation either, despite believing you do.

Firstly you don't change the left hand side, so you're mixing units if you change units of the right hand side in the second equation. Secondly you fail to realise that since $\alpha$ and $\beta$ are in the you can't just extract out the scalar coefficients individually in that form, because the spinor is being acts on by matrices. And finally, and most importantly, you cannot ignore that matrices when you talk about taking the square root of the momentum operator in the manner you have.

This final point is the fundamental core of the Dirac equation and its use of matrices and something anyone whose properly covered the Dirac equation will know because it motivates everything Dirac did. Dirac wanted a first order operator $D$ which would square to the wave operator, $-\partial_{t}^{2} + \nabla\cdot \nabla = \eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$. This operator has the same Lorentzian form as the 4-momentum $p_{\mu}p^{\mu} = -E^{2} + \mathbf{p} \cdot \mathbf{p}$, in fact it is the Fourier transform of it so it's entirely equivalent, so we're talking about the same thing.

You can't just use something like $D = \partial_{t} + \sum_{i}\partial_{i}$, which would be akin to your (if you did it properly) $\omega = k+M$ expression. In fact there is no such scalar differential operator. Instead you have to use matrices. Suppose we use matrices such that $D = \gamma^{\mu}\partial_{\mu}$, then $D^{2} = \gamma^{\mu}\gamma^{\nu}\partial_{\mu}\,\partial_{\nu}$, which we need to be equal to $\eta^{\mu\nu}\partial_{\mu}\partial_{\nu}$. Thus, following a few technical things, you arrive at the conclusion that the $\gamma^{\mu}$ matrices satisfy $\gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu} = 2\eta^{\mu\nu} \mathbb{I}_{4}$, this is an example of a Clifford algebra.

Thus we have $D^{2}$ being the wave operator, thus D is the square root. In 1+1 dimensional stuff you'd have $D = \gamma^{0}\partial_{t} + \gamma^{1}\partial_{x}$, which in the notation you use is $D = \beta \partial_{t} + \alpha \partial_{x}$. This is a matrix expression, every single component of $\beta$ has a factor of $\partial_{t}$ in it and likewise for the other term.

This ability to take square roots is done in terms of differential operators acting on spinors. When you Fourier transform the spinor equations you end up with spinor equations which still have matrices in them but now the derivatives are become scalar multiples. But the scalar multiplies still appear in every component of the relevant matrices, you cannot just drop the matrices and declare the scalar components satisfy the equation on their own, else you end up with the ridiculous result you had.

You've dug yourself a hole, showing you don't understand the context and meaning of the things you're reading which pertain to square roots of such expressions. I know what is being referred to because I spent time and effort learning it years ago and I can see why someone who doesn't understand the whole role of matrices and all the suppressed vector indices, matrices, spinor indices etc which float around would make such a mistake.

You stated previously you believed your understanding of the Dirac equation pretty good. It is demonstrably not. What you're showing you don't understand is the founding principle from which all of the properties and structures of the Dirac equation follow, the motivation behind what Dirac did and why its different from 'simpler' quantum field theory.

No doubt you'll make some excuse, try to dig yourself out and perhaps even complain about biased mods and how I always smack you down but that fact remains you are time and again showing how you don't understand what you're talking about. You've picked up snippets from here and there and you're trying to interpret things like "Square root of the Dirac equation" in terms you understand, which are so far short of the truth you don't even realise you completely fail to grasp the concepts. This completely undermines threads like this, where you try to argue you are in a position to explain things to people. You're not and even worse, you don't realise you're not and thus you have no problems spreading mistakes because you don't even realise you're making them.

I'm more than happy to have another physicist who has studied the Dirac equation come into this thread and give you a second opinion, so you can't play the "Oh it's just because you don't like me!" card.

Last edited: Jan 8, 2012

5. ### ReikuBannedBanned

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11,238
Yes, I know. Post 41 admits that when I read that back, I was completely off with the energy is a vector part... my eye's screwed and I almost could not believe I wrote that. Anyway... will get to the rest of your post soon.

7. ### ReikuBannedBanned

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I am equation omega as energy... why am I doing this, is there a reason? If you state omega is a vector, then omega can't be energy...

but it is because E= \hbar \omega, which is were the omega comes from... I seriously think you are the one who has made the mistake AN. You say

''M is a scalar, k and omega are vectors.''

But omega cannot be a vector because it is energy... or atleast, I am under the impression this is were omega comes from... in fact I am 99% positive. k is momentum, and M is mass, so they are very relevant when understanding the wave equations deriving the dirac equation.

8. ### ReikuBannedBanned

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In fact, I knew something was positively bugging me, and it was you all along, you sent me off on the wrong track. Omega is energy. You've equated it as a vector, which is wrong.

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10. ### AlphaNumericFully ionizedRegistered Senior Member

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So rather than adding 2 vectors and a scalar you're adding 2 scalars and a vector. Oh that's much better. Yes, I have no problem saying I was mistaken in referring to $\omega$ as a vector, I was thinking about too many things. But that doesn't negate my complete demolishing of your nonsense. The large part of my post which you quoted in post 44 is still completely applicable.

11. ### ReikuBannedBanned

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well, I shouldn't have jumped from $omega = k+M$ from $\omega = \sqrt{k^2 +M^2}$ I admit my mistake there. I will fix it for my presentation.

Now, as for your mistake, I was dignified when I pointed it out to you. Look at the way you reacted when you thought I had not noticed what I said about energy being a vector? That's a fine line you tread being condescending in your replies when you are subject to simple mistakes as well.

(Of course I realized this before AN posted and thought wtf... )

12. ### AlphaNumericFully ionizedRegistered Senior Member

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In fact it gets worse, because the matrices you've given for your $\alpha,\beta$ are not a representation of the right Clifford algebra! Since you'll no doubt not get it, let's use your representations of those matrices to demonstrate.

The Dirac matrices need to satisfy $\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}$ where $\{a,b\} = ab+ba$. Thus in 1+1 dimensions one of the two matrices squares to the identity, the other minus the identity! Your matrices don't do that! If you want a proper representation of the Clifford algebra $Cl_{1,1}$ you can pick the following, for the signature (-+),

$\gamma^{1} = \alpha = \left( \begin{array} 0 & 1 \\ 1 & 0 \end{array} \right)$ and $\gamma^{0} = \beta = \left( \begin{array} i & 0 \\ 0 & -i \end{array} \right)$.

Now we indeed get that $\gamma^{0}\gamma^{0} = \eta^{00}\mathbb{I} = -\mathbb{I}$ and $\gamma^{1}\gamma^{1} = \eta^{11}\mathbb{I} = +\mathbb{I}$

But then we come to another problem with your matrices, you fail to realise that $\alpha \beta + \beta \alpha$ isn't what it should be, which is zero. Squaring to the identity is only part of the properties needed for the Dirac equation. In order to get the D operator to square to something resembling the wave operator the cross terms have to vanish, which is done by matrix algebra. It's easy to check that my version has $\alpha\beta + \beta\alpha = 0\mathbb{I}$, as needed.

So what is the Dirac operator? $D = \beta \partial_{t} + \alpha \partial_{x} = \left( \begin{array} i\partial_{t} & \partial_{x} \\ \partial_{x} & -i\partial_{t} \end{array} \right)$. Now what happens if we square this operator? $D^{2} = \left( \begin{array} i\partial_{t} & \partial_{x} \\ \partial_{x} & -i\partial_{t} \end{array} \right) \left( \begin{array} i\partial_{t} & \partial_{x} \\ \partial_{x} & -i\partial_{t} \end{array} \right) = \left( \begin{array} -\partial_{t}^{2}+\partial_{x}^{2} & 0 \\ 0 & -\partial_{t}^{2}+\partial_{x}^{2} \end{array} \right) = \mathbb{I}(-\partial_{t}^{2}+\partial_{x}^{2})$

So we don't end up with a scalar differential operator, we end up with a matrix whose non-zero components are all the same differential operator, the wave equation. Thus if something satisfies the Dirac equation it will satisfy the wave equation component by component, because the matrix in question is the identity matrix and it doesn't mix components.

So there's another fundamental mistake you made, which only further shows that you don't have a working understanding of the mathematical expressions you're posting nor do you realise the role the square root concept plays in all of this.

I'd say that's enough for one day, no doubt you'll provide another load of miscopied nonsense to refute while I sleep.

13. ### ReikuBannedBanned

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Right hold on... your going way to fast for me... I am not as good at maths as you and you know this... I need to chew on this!

As for your comment about the square root sign, I do understand it's importance and I have noted above saying:

well, I shouldn't have jumped from $omega = k+M$ from $\omega = \sqrt{k^2 +M^2}$ I admit my mistake there. I will fix it for my presentation.

Now, as for your mistake, I was dignified when I pointed it out to you. Look at the way you reacted when you thought I had not noticed what I said about energy being a vector? That's a fine line you tread being condescending in your replies when you are subject to simple mistakes as well.

(Of course I realized this before AN posted and thought wtf... )

14. ### AlphaNumericFully ionizedRegistered Senior Member

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Do you admit that the large section of my post which you quoted is entirely accurate? It's entirely unrelated to the index structure of $\omega$, it pertains to the over arching conceptual motivation for the Dirac equation itself, something you've shown you don't understand.

I ask again because I get the feeling you're trying to weasal out of addressing it by saying "Well you admitted your criticism wasn't correct so case closed", when that minor slip up on my part has absolutely no impact on the entirety of post 42. Added to that what I just posted about your Dirac matrices your complete lack of understanding is pretty clear. In fact....

...that gives me the impression either you're utterly oblivious to just how wrong you are and how my post lays it out or you're being deliberately obtuse to avoid facing up to those problems I've highlighted. There's a lot more problems in your posts than just the scalar/vector thing and considering you quote me explaining such a problem you really have no excuse to ignore them.

Your equation is still wrong, so it's neither here nor there. And the vast majority of my post, including the large chunk you quoted, explained how you were wrong in a different but far more fundamental manner.

I can only assume you're being deliberately obtuse because I explained at length your large mistakes and now you are trying to brush them as because I can make little mistakes like thinking about several things at once (you do make a lot of errors that I was thinking about how to cover succinctly) and explaining how an incorrect equation was incorrect in not precisely the right manner.

So this has nothing to do with me slamming you for little mistakes, I'm going into great detail about HUGE mistakes you're making. You've said you believe you're pretty knowledgeable about the Dirac equation and now you're showing you don't understand the reason for the form it takes, you don't understand why the Dirac matrices are what they are, you don't understand the equations they satisfy and you have a fundamental lack of grasp of the origin of the Dirac equation, both conceptually and mathematically. That is why I'm posting rather curt posts. You put yourself up as knowledgeable when everyone, including you, knows full well you aren't and I'm now proving you aren't.

Your "Oh you're being condescending, you make little mistakes too!" is such a laughable attempt to avoid facing up to your huge mistakes you must be aware of your mistakes and you're desperately trying to avoid addressing them. Else the only alternative is you're really stupid enough to think such an excuse would work.

15. ### ReikuBannedBanned

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Yes, the rest of your post was accurate. I am not being obtuse, just pointing out that to err alphanumeric, is practically human. That even those with a PhD are open to error, even simple one's, the kind you like to make condescending remarks about to other people. Just nip your attitude in the bud, seriously, that is all I am saying. As for your last post, there are a few queeries... I am not sure yet if I understand the whole thing. I wasn't aware that the clifford algebra would be wrong... did you mean the matrices, or the algebra itself?

16. ### ReikuBannedBanned

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Well, see... I understand immediately from your post that the dirac matrices need to satisfy

$\{\gamma^{\mu},\gamma^{\nu}\} = 2\eta^{\mu\nu}\mathbb{I}$

That is simply

$\gamma^{\mu}\gamma^{\nu} + \gamma^{\nu}\gamma^{\mu}$

That just stands as the algebra satisfying $C\ell_{1,3}(\mathbf{C})$. What is unclear to me is why my algbra is wrong... I satisfied the conditions required for the algebra, so this is why I am asking whether it is the matrices that have been written wrong, or whether the entire algebra is wrong.

17. ### ReikuBannedBanned

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Perhaps that is were we differ... Are you sure it is a C_11 algebra. See which algebra I thought it was above.

18. ### AlphaNumericFully ionizedRegistered Senior Member

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But you've shown you don't even understand the motivation for the Dirac equation or what makes it special compared to other things like the wave equation, how Dirac constructed it or how to apply it. Your matrices don't satisfy the correct equations which are central to the Dirac equation. You don't understand how the square rooting is done and why the matrices are essential.

These are fundamental things, yet you claim you grasp the Dirac equation well, well enough to be telling other people about it. You've gotten much of the fundamental things wrong.

Your $\alpha,\beta$ don't. Work it out for yourself. You're using $\beta = \mathbb{I}$ so $\alpha \beta + \beta \alpha = 2\alpha$, which is obviously not the required 0. Even when I do the calculations explicitly for you you don't get it!

As for $Cl_{1,n}$, it means the algebra defined by the binary product $\{ \gamma^{\mu} , \gamma^{\nu} \}$ outputs either the 0 matrix or $\pm 2 \mathbb{I}$. The coefficient is defined by the 1+n dimensional Lorentzian metric $\eta^{\mu\nu}$. You don't even recognise the notation, despite it being common place in the literature pertaining to the Dirac equation.

Your matrices should have satisfied $\{ \gamma^{a} , \gamma^{b} \} = 2\eta^{ab}\mathbb{I}$. They don't. They satisfy $\{ \gamma^{a} , \gamma^{a} \} = 2\mathbb{I}$ and the cross combination doesn't equal the required zero.

You really should be able to do all of this yourself if you understood the Dirac equation. Even if you could do all of this it wouldn't be called a 'working understanding', that's how far you are from grasping this.

Understand? This isn't a little slip up on your part, it's a huge gap in understanding of the fundamental nature of the Dirac equation. I accidentally referred to $\omega$ as a vector because I was talking about how you've screwed up scalars/vectors/matrices in regards to a spinor equation. I obviously know that frequency and periods are scalars, you have demonstrated you don't know about the properties of the Dirac equation you're trying to present yourself as knowledgeable on.

19. ### ReikuBannedBanned

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Could you elucidate on your use of $Cl_{1,1}$ I am not familiar with it. As far as I am aware, the clifford algebra that deals with your conditions are actually $C \ell_{1,3}(\mathbf{C})$?

20. ### ReikuBannedBanned

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Oh sorry, you did.

21. ### ReikuBannedBanned

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I guess I should ask if Cl_11 and Cl_13(C) are synonymous? You call it a binary product but I don't think that is the true representation for the binary product you have mentioned. Note, I have never seen your notation before, I understand it as the one I wrote.

22. ### ReikuBannedBanned

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Right I see the problem concerning the algebra... I have a funny feeling I have latexed it wrong. Be back soon after I try and calculate the right condition.

23. ### ReikuBannedBanned

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Right, I've figured out the mistake.

Beta should be

01
10

That would fix the algebra. After your big presentation of mathematics, that was simple enough. Talk about making a mountain out of a molehill.