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

Right, I will clear this up for future readers since I cannot edit the post:

An algebra method of decomposing this equation can make

\(\omega^2 = (\alpha k + M \beta)(\alpha k + M \beta)\)

Now we are going to expand the right hand side of this equation

\((\alpha k + M \beta)(\alpha k + M \beta) = \alpha^2k^2 + M^2 \beta^2 + \alpha \beta k M + \beta \alpha kM\)

Since we know the matrices require that \(k^2 + M^2\) then we can see the first term \(\alpha^2k^2 +M^2 \beta^2 \) already satisfies the equation, so the rest can be neglected \(\alpha \beta k M + \beta \alpha k M\). Because of this however, we can clearly see that there is some kind of commutation relation happening in the equation, because \(\beta^2=1\) would need to be satisfied and also \(\alpha^2 = 1\) would need to be satisfied, again because \((\alpha \beta + \beta \alpha)km\) becomes invalid.

The alpha matrix is given as

\(\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \alpha\)

so \(\alpha^2\) is given as

\(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)

\(\beta\) is not the same as this because it satisfies the clifford algebra \(\alpha \beta + \beta \alpha = 0\). This means it's matrix has the form of

\(\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} = \beta\)


Right now? I think it is.

 

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Now by multiplying the two matrices \(\alpha\) and \(\beta\) will give

01
-10

 

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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.

I'd like you to still answer this...

You later defined it as Cl_1n which is different to what you said originally because it seems very obvious that you could fill in the ''n'' with 3. But that is not what you wrote, you write Cl_11...

Mistake AN?

 

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I just explained them, I gave their definition! Don't you understand simple matrix expressions?

It is the definition of the algebra.

Funny, it appears in the literature everywhere. I have always suspected you have a particular source you're getting this from because the alpha/beta notation is quite an old one and very rarely used in the literature now.

I see you're still trying to avoid facing up to the fact your problem is not just one of making constant mathematical slip ups but one of not understanding the motivating concepts and reasons behind the Dirac equation.

It isn't a molehill, you don't understand fundamental concepts behind the Dirac equation and you demonstrably can't do the mathematics. It's hard to get much worse.

You're still not getting it. Just because the left side acting on the spinor is the same as the right side acting on the spinor doesn't mean you can equate them in the manner you have.

For example, given a matrix A it'll have an eigenvector v with eigenvalue a. Thus by definition we can write \(Av = av\). This does not imply that A=a. Instead it implies that the action of A on v is the same as the action of a on v.

It's a common mistake people with no working experience of this stuff make. You helped to confuse yourself by working in the momentum representation. If you had stayed in the usual representation you'd start with the spinor wave equation of

\(\partial_{t}^{2}\psi = (\alpha \partial_{x} + M \beta)(\alpha \partial_{x} + M \beta)\psi\)

Clearly now you can't say that this implies \(\partial_{t}^{2} = (\alpha \partial_{x} + M \beta)(\alpha \partial_{x} + M \beta)\), you have a bunch of derivatives just hanging there which rings alarm bells with most people (this is generally a bad thing to have, though some formalisms don't mind it). Clearly you can't just equate partial derivatives, they are operators, not states or functions. The Dirac equation says their action on a particular type of spinor is the same, which is very different.

This is stuff you'd gain from doing actual work with the equation, rather than just copying what you've read someone else write and trying to interpret it using your pre-university level understanding of mathematics.

Firstly \(\alpha \beta + \beta \alpha\) doesn't become 'invalid', its just zero. Secondly \(\alpha \beta + \beta \alpha = 0\) isn't enough to give a Clifford algebra, you have to define the binary product on everything.

Wow, what did you do, wait 40 minutes? And after I'd said I was going to bed. Jesus, learn some patience. And how to use the edit button! 8 posts in a row is completely unnecessary.

And you're also failing to understand what a binary product is, which is hardly surprising since you haven't done the introductory courses in groups and algebras which would be expected of people covering the Dirac equation. How much more of this is it going to take before you realise you don't have a good understanding of this stuff? You only think you do because you knock out a couple of paragraphs of material you're parroting from other sources without understanding. As soon as you try to do anything yourself you fail miserably.

A binary product is something which takes in 2 things and outputs 1 thing. Addition on the integers is a binary product, an integer plus another integer is a third integer. The binary product \(\{ \gamma^{\mu} , \gamma^{\nu} \} = 2\eta^{\mu\nu}\mathbb{I}\) takes in two matrices and spits out a third. All algebras need such a thing. Lie algebras have the Lie bracket. The cross product in \(\mathbb{R}^{3}\) is another example. From that you can do all the necessary work with the Dirac matrices because it tells you everything you need to know, just as defining the Lie bracket tells you everything you need to know about a Lie algebra.

Clearly \(Cl_{1,1}\) and \(Cl_{1,3}\) are not synoymous from the definition, they have different dimensions because they have different numbers of matrices in their definitions. \(Cl_{n,m}\) has n+m single index matrices, ie \(\gamma^{a}\). These can form combinations \(\gamma^{ab} \propto \gamma^{[a}\gamma^{b]}\) and \(\gamma^{abc} = \gamma^{[a}\gamma^{b}\gamma^{c]}\) etc. Consider all possible combinations and include the identity matrix then you have an algebra with \(2^{n+m}\) members.

Can you tell me why? If you're knowledgeable about the Dirac equation and you honestly have enough experience with the matrix operators to be in a position to be writing a book which talks about them then you really should know. You should have seen everything I've just said before. You should know what the notation means. You should know the answer immediately.

You're working with the 1+1 dimensional Dirac equation, while the standard one is the 1+3 dimensional Dirac equation. Notice the link? 1+1 dimensions, \(Cl_{1,1}\)? Or 1+3 dimensions and \(Cl_{1,3}\). It's all just different cases of a general formalism.

In fact, giving an overview of the Dirac equation should be done in a general manner. None of the algebra is made significantly worse. In fact, the motivating structure is made much much clearer. Perhaps that is part of the reason why you don't get it, you don't understand enough simple notation to streamline the whole thing.

See how many times you fall flat on your face? You're writing about stuff Cambridge covers in its 4th year and you're getting it wrong again and again, even the fundamental concepts, never mind the constant stream of basic algebraic screw ups you make. You're clearly lacking in some very basic understanding of prerequisite material, never mind the Dirac equation itself. You're trying to argue it's possible for people to grasp this stuff without doing the boring details and you're making yourself evidence against your claim! And the fact you're aiming this at 14~16 year olds is ridiculous. You have brought up mathematics well beyond your understanding (and riddled with errors) and you expect people with less knowledge of maths and physics than even you to grasp it properly?

You need to spend your time more wisely. Get a book on matrices, vectors and differential equations. There's plenty of A Level (or Highers, since you live in Scotland) books on their basics which you could start with. That and you need to learn to use the edit button. Once you've managed that then you can move onto actual university level stuff. Until then you're just deluding yourself, deceiving others and getting no closer to that physics career you've implied you want. No doubt you'll tell me to mind my own business but I do wonder what you do with yourself day to day at the moment. Do you have a job? You regularly post in the middle of the night and clearly you've got plenty of time on your hands. If only you'd use some of that time more constructively.

 

It's right though now. I know it is.

As for the expression Cl_11, that is not what defines the Clifford Algebra, it is Cl_13(C) which is why I asked you if they were synonymous. I have never seen your expression. Mine works correctly, I have never seen yours.

I have had enough of these discussions now. Thank you for your patience. -ish.