SL(2,C) to Lorentz in Carmeli's Theory of Spinors

[arkajad]

SL(2,C) to Lorentz in Carmeli's "Theory of Spinors"

On page 56 of "Theory of Spinors", Eq. (3.84a), Carmeli gives the formula for the Lorentz matrix in terms of Pauli matrices and an SL(2,C) matrix g:
\Lambda^{\alpha\beta}=(1/2)Tr(\sigma^\alpha g \sigma^\beta g^*)
His sigma matrices are the standard ones. It seems to me that there should be {\Lambda^\alpha}{\phantom{\alpha}}_\beta on the LHS. For instance, when g is the identity Lorentz transformation, that is
{\Lambda^\alpha}{\phantom{\alpha}}_\beta=\delta^\alpha_\beta
While with his formula we will get the Minkowski metric matrix instead.
Carmeli's conventions about rising and lowering indices are the standard ones.

I am asking, as I do not want to make some trivial mistake myself while writing my own lecture notes.

 

[dextercioby]

Hi Arkajad and welcome back to PF.

For the notes you plan to write, use this book:

https://www.amazon.com/dp/9971503557/?tag=pfamazon01-20

where the right equation is stated and proved:

\Lambda^{\mu}_{~~\nu} = \frac{1}{2}\mbox{Tr}\left(\bar{\sigma}^{\mu}g \sigma_{\nu} g^{\dagger}\right)

g\in\mbox{SL(2,C)}.

 

[arkajad]

Hi Arkajad and welcome back to PF.

For the notes you plan to write, use this book:

https://www.amazon.com/dp/9971503557/?tag=pfamazon01-20

where the right equation is stated and proved:

\Lambda^{\mu}_{~~\nu} = \frac{1}{2}\mbox{Tr}\left(\bar{\sigma}^{\mu}g \sigma_{\nu} g^{\dagger}\right)

g\in\mbox{SL(2,C)}.

Thanks. That confirms my suspicions, since, as I have checked, \bar{\sigma}^{\mu}=\sigma_\mu.
P.S. It is, in fact, not so much "lecture notes", but a book to be published: "Quantum fractals".

 

[samalkhaiat]

On page 56 of "Theory of Spinors", Eq. (3.84a), Carmeli gives the formula for the Lorentz matrix in terms of Pauli matrices and an SL(2,C) matrix g:
\Lambda^{\alpha\beta}=(1/2)Tr(\sigma^\alpha g \sigma^\beta g^*)
His sigma matrices are the standard ones. It seems to me that there should be {\Lambda^\alpha}{\phantom{\alpha}}_\beta on the LHS. For instance, when g is the identity Lorentz transformation, that is
{\Lambda^\alpha}{\phantom{\alpha}}_\beta=\delta^\alpha_\beta
While with his formula we will get the Minkowski metric matrix instead.
Carmeli's conventions about rising and lowering indices are the standard ones.

I am asking, as I do not want to make some trivial mistake myself while writing my own lecture notes.

I don’t see any thing wrong with that. The identity in SL( 2 , C ), i.e., g = I_{ 2 \times 2 }, corresponds to the identity in SO( 1 , 3 ). The Minkowski metric is the identity in the Lorentz group:
<br /> \bar{ x }^{ \alpha } = \Lambda^{ \alpha \beta } x_{ \beta } = \eta^{ \alpha \beta } x_{ \beta } = x^{ \alpha } = I_{ SO( 1 , 3 )} x^{ \alpha } .<br />

Sam

 

[arkajad]

I don’t see any thing wrong with that. The identity in SL( 2 , C ), i.e., g = I_{ 2 \times 2 }, corresponds to the identity in SO( 1 , 3 ). The Minkowski metric is the identity in the Lorentz group:
<br /> \bar{ x }^{ \alpha } = \Lambda^{ \alpha \beta } x_{ \beta } = \eta^{ \alpha \beta } x_{ \beta } = x^{ \alpha } = I_{ SO( 1 , 3 )} x^{ \alpha } .<br />

Sam

According to Carmeli and Malin formula, for g=I_{2\times 2}:

\Lambda^{ \alpha \beta }=\delta^{ \alpha \beta } = \mbox{diag}(1,1,1,1)

(i.e. Kronecker's delta)

instead of (correct)

\Lambda^{ \alpha \beta }=\eta^{ \alpha \beta } = \mbox{diag}(1,-1,-1,-1)

 

[dextercioby]

Yes, the reasoning in 3.83 is wrong, because \frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right) = \delta^{\alpha\beta}, but
x&#039;^{\alpha} \neq \delta^{\alpha\beta}x&#039;_{\beta}

 

[arkajad]

Yes, the reasoning in 3.83 is wrong ...

Thank you. That is also where I have detected the mistake in the derivation.
On the other hand I have checked one by one all 16 expression for {\Lambda^\mu}_{\phantom{\mu}\nu} in Exercise 3.5 on p. 62, and they are all OK.

 

[samalkhaiat]

According to Carmeli and Malin formula, for g=I_{2\times 2}:

\Lambda^{ \alpha \beta }=\delta^{ \alpha \beta } = \mbox{diag}(1,1,1,1)

There is no such object in Minkowski space. When it comes to the Lorentz group, I very much trust Carameli. Also, I know the book in question and I do not remember any such blunder. I guess you are unfamiliar with their conventions or/and calculating the TRACE in a naïve way. When we write the relation
\sigma^{ \mu } \ \sigma^{ \nu } = - \eta^{ \mu \nu } + 2 i \sigma^{ \mu \nu } ,
what we really mean is the following SL( 2 , C ) realation
<br /> \left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } \epsilon_{ A C } + 2 i \left( \sigma^{ \mu \nu } \right)_{ A C } , \ \ \ (1)<br />
where (and this is important)
\left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } , \ \ \ (2)
\left( \sigma^{ \mu \nu } \right)_{ AC } = \left( \sigma^{ \mu \nu } \right)_{ C A } , \ \ \ (3)
and
\epsilon_{ A B } = \epsilon^{ A B } = \left( \begin{array} {cc} 0 &amp; -1 \\ 1 &amp; 0 \end{array} \right) . \ \ \ (4)
Now, to calculate the trace of \sigma^{ \mu } \sigma^{ \nu }, we contract Eq(1) with \epsilon^{ A C } and use Eq(3) and Eq(4)
<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \epsilon^{ A C } \left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } ( - 2 ) + 0 = 2 \eta^{ \mu \nu } . \ \ (5)<br />
Now, if you naively take \mu = \nu = 1 in the above equation, you find 2 = - 2. This is because you are not taking Eq(2) in the consideration. What you should do is the following
<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } ,<br />
where \bar{ \sigma }^{ \nu } = ( I , - \sigma^{ i } ) .
You said the book writes \Lambda^{ \mu \nu} \in SO( 1 , 3 ) as
\Lambda^{ \mu \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } g \sigma^{ \nu } \bar{ g } \right) .
Therefore, for g = \bar{ g } = I \in SL( 2 , C ), you get
\Lambda^{ \mu \nu } ( I ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) .
Now, if you compare this with Eq(5), you find that
\Lambda^{ \mu \nu } ( I_{ 2 \times 2 } ) = \eta^{ \mu \nu } = I \in SO( 1 , 3 ) .
So there is no such object as \delta^{ \mu \nu }.

Sam

 

[samalkhaiat]

Yes, the reasoning in 3.83 is wrong, because \frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right) = \delta^{\alpha\beta}, but
x&#039;^{\alpha} \neq \delta^{\alpha\beta}x&#039;_{\beta}

Is he doing something along the following lines
<br /> \bar{ x }^{ \mu } = \delta^{ \mu }_{ \rho } \ \bar{ x }^{ \rho } = \delta^{ \mu }_{ \rho } \ \eta^{ \rho \nu } \ \bar{ x }_{ \nu } = \delta^{ \mu }_{ \rho } ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \rho } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } = ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \mu } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } .<br />
If so, why do you think there is an \delta^{ \mu \nu } involved?

Sam

 

[arkajad]

<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = ... = 2 \eta^{ \mu \nu } . \ \ (5)<br />
Sam

Well, their sigma matrices are given explicitly in (3.78):

As you can see - if I am not mistaken:

<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \delta^{ \mu \nu } . \ \ (5)<br />

and NOT

<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \eta^{ \mu \nu } . \ \ (5)<br />

Right?

 

[dextercioby]

There is no such object in Minkowski space. When it comes to the Lorentz group, I very much trust Carameli. Also, I know the book in question and I do not remember any such blunder. I guess you are unfamiliar with their conventions or/and calculating the TRACE in a naïve way. When we write the relation
\sigma^{ \mu } \ \sigma^{ \nu } = - \eta^{ \mu \nu } + 2 i \sigma^{ \mu \nu } ,
what we really mean is the following SL( 2 , C ) realation
<br /> \left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } \epsilon_{ A C } + 2 i \left( \sigma^{ \mu \nu } \right)_{ A C } , \ \ \ (1)<br />
where (and this is important)
\left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } , \ \ \ (2)
\left( \sigma^{ \mu \nu } \right)_{ AC } = \left( \sigma^{ \mu \nu } \right)_{ C A } , \ \ \ (3)
and
\epsilon_{ A B } = \epsilon^{ A B } = \left( \begin{array} {cc} 0 &amp; -1 \\ 1 &amp; 0 \end{array} \right) . \ \ \ (4)
Now, to calculate the trace of \sigma^{ \mu } \sigma^{ \nu }, we contract Eq(1) with \epsilon^{ A C } and use Eq(3) and Eq(4)
<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \epsilon^{ A C } \left( \sigma^{ \mu } \right)_{ A } {}^{ \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = - \eta^{ \mu \nu } ( - 2 ) + 0 = 2 \eta^{ \mu \nu } . \ \ (5)<br />
Now, if you naively take \mu = \nu = 1 in the above equation, you find 2 = - 2. This is because you are not taking Eq(2) in the consideration. What you should do is the following
<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \sigma^{ \nu } \right)_{ C \dot{ B } } = \left( \sigma^{ \mu } \right)^{ C \dot{ B } } \ \left( \bar{ \sigma }^{ \nu } \right)_{ \dot{ B } C } ,<br />
where \bar{ \sigma }^{ \nu } = ( I , - \sigma^{ i } ) .
You said the book writes \Lambda^{ \mu \nu} \in SO( 1 , 3 ) as
\Lambda^{ \mu \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } g \sigma^{ \nu } \bar{ g } \right) .
Therefore, for g = \bar{ g } = I \in SL( 2 , C ), you get
\Lambda^{ \mu \nu } ( I ) = \frac{ 1 }{ 2 } \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) .
Now, if you compare this with Eq(5), you find that
\Lambda^{ \mu \nu } ( I_{ 2 \times 2 } ) = \eta^{ \mu \nu } = I \in SO( 1 , 3 ) .
So there is no such object as \delta^{ \mu \nu }.

Sam

Is he doing something along the following lines
<br /> \bar{ x }^{ \mu } = \delta^{ \mu }_{ \rho } \ \bar{ x }^{ \rho } = \delta^{ \mu }_{ \rho } \ \eta^{ \rho \nu } \ \bar{ x }_{ \nu } = \delta^{ \mu }_{ \rho } ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \rho } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } = ( 1 / 2 ) \mbox{ Tr } ( \sigma^{ \mu } \sigma^{ \nu } ) \ \bar{ x }_{ \nu } .<br />
If so, why do you think there is an \delta^{ \mu \nu } involved?

Sam

Hi Sam,

as usual your argument is impeccable. Indeed, I've overlooked something:

x&#039;^{\alpha} = \delta^{\alpha}_{\beta}x&#039;^{\beta} = \delta^{\alpha}_{\beta}\eta^{\beta\lambda}x&#039;_{\lambda} = \eta^{\alpha\lambda}x&#039;_{\lambda}

therefore

x&#039;^{\alpha} =\eta^{\alpha\beta}x&#039;_{\beta} (as expected, no delta).

and the proof that

\eta^{\alpha\beta} = \frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right)

is exactly how you wrote it, because of the \epsilon metric tensor being antisymmetric for SL(2,C) as opposed to symmetric for SU(2).

 

[arkajad]

and the proof that

\eta^{\alpha\beta} = \frac{1}{2}\mbox{Tr}\left(\sigma^{\alpha}\sigma^{\beta}\right)

is exactly how you wrote it, because of the \epsilon metric tensor being antisymmetric for SL(2,C) as opposed to symmetric for SU(2).

Yet

\eta^{11}=-1

while

\frac{1}{2}\mbox{tr}(\sigma^1\sigma^1)=1

Therefore

1=-1

So, we are getting a contradiction.

 

[dextercioby]

Yes, but actually the <trace> is a little exotic and comes from how spinorial (dotted and undotted) indices act in that product. Their order matters, switching one dotted for one undotted is done with an epsilon matrix. The -1 which 'solves' the contradiction comes from the epsilon above, see the argument by Sam in post #8.

 

[arkajad]

Somewhat desperate checked Carmeli's "Group Theory and Relativity". There it is the way I think it should be:

 

[arkajad]

Yes, but actually the <trace> is a little exotic .

The trace of a matrix is the trace of a matrix. And sigmas are explicitly given matrices. The only "exotic" thing about them is that Carmeli's sign of \sigma^2 ("sigma two") is opposite to the "conventional", most popular sign. But it does not matter for our purpose.

 

[dextercioby]

It does look different, doesn't it ? Yet it's the same author copy-pasting from Ruehl's book...

 

[samalkhaiat]

Well, their sigma matrices are given explicitly in (3.78):

As you can see - if I am not mistaken:

<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \delta^{ \mu \nu } . \ \ (5)<br />

and NOT

<br /> \mbox{ Tr } \left( \sigma^{ \mu } \sigma^{ \nu } \right) = 2 \eta^{ \mu \nu } . \ \ (5)<br />

Right?

You are mistaken. From your first post, I knew that you were calculating the trace in a naïve way. The whole story of post #8 was to point out to you how one should calculate the trace in SL( 2 , C ). You go wrong if you treat ( \sigma^{ \mu } \sigma^{ \nu } )_{ ab } as an SU( 2 ) tensor and calculate the trace using the SU( 2 ) symmetric metric \delta^{ a b }. In SU( 2 ), the trace of the tensor T is
\mbox{ Tr }_{ su( 2 )} ( T ) = \delta^{ a b } T_{ a b } ,
while in SL( 2 , C), you calculate the trace using the antisymmetric metric \epsilon^{ A B }:
\mbox{ Tr }_{ sl( 2 , c ) } ( T ) = \epsilon^{ A B } T_{ A B } .
So, you should not expect the two expressions to give the same number if you don't apply the correct rule for the given group, because the trace is an INVARIANT GROUP OPERATION. Let me give you relevant examples for the above two equations: In SU(2), we have the following (tensor) relation
<br /> \left( \sigma^{ i } \sigma^{ j } \right)_{ a b } = \delta^{ i j } \delta_{ a b } + i \epsilon^{ i j k } \left( \sigma^{ k } \right)_{ a b } .<br />
Therefore the trace is
\mbox{ Tr } \left( \sigma^{ i } \sigma^{ j } \right) = \delta^{ a b } \left( \sigma^{ i } \sigma^{ j } \right)_{ a b } = 2 \delta^{ i j } .
In SL(2,C), the corresponding relations reads
<br /> \left( \sigma^{ i } \sigma^{ j } \right)_{ A B } = - \eta^{ i j } \epsilon_{ A B } + 2 i \left( \sigma^{ i j } \right)_{ A B } ,<br />
and
\mbox{ Tr } ( \sigma^{ i } \sigma^{ j } ) = \epsilon^{ A B } \left( \sigma^{ i } \sigma^{ j } \right)_{ A B } = 2 \eta^{ i j } .

Sam

 

[samalkhaiat]

Yet

\eta^{11}=-1

while

\frac{1}{2}\mbox{tr}(\sigma^1\sigma^1)=1

Therefore

1=-1

So, we are getting a contradiction.

Did you read the paragraph and equation after Eq(5) in post#8? There is no contradiction:
\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) = \frac{ 1 }{ 2 } \left( \sigma^{ 1 } \bar{ \sigma }^{ 1 } \right)^{ A }{}_{ A } = - \frac{ 1 }{ 2 } \left( \sigma^{ 1 } \sigma^{ 1 } \right)_{ A A } = - \frac{ 1 }{ 2 } ( 2 ) = - 1.

Sam

 

[arkajad]

You are mistaken.
Sam

Or Carmeli is mistaken. His formula from one book is different than his formula from the other book.

Do you agree with that?

The traces are the same.

Perhaps you should consider the fact that the trace is the trace - the same in both books.

If Carmeli would have some other trace in mind, he would certainly DEFINE it BEFORE using it, don't you think so?

 

[arkajad]

Did you read the paragraph and equation after Eq(5) in post#8? There is no contradiction:
\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) =\frac{1}{2}\left(\sigma^1\bar{\sigma}^1\right)^A_{\phantom{A}A}

Sam

You see, under the trace there is a product of two matrices: product of \sigma^1 with \sigma^1, Not a product of \sigma^1 with \bar{\sigma}^1.

Therefore your LHS is not equal to your RHS:

\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) \neq\frac{1}{2}\left(\sigma^1\bar{\sigma}^1\right)^A_{\phantom{A}A}

 

[samalkhaiat]

Or Carmeli is mistaken. His formula from one book is different than his formula from the other book.

Do you agree with that?

I don’t have any of Carmeli’s books on me right now. Therefore, I cannot help you with that.
However, you said that Carmeli writes, for Lorentz transformation, the expression
\Lambda^{ \mu \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ \mu } g \sigma^{ \nu } g^{ \dagger } ) . \ \ \ (1)
This expression is correct and unique for any \pm g \in SL( 2 , C ). Even undergraduates should have no problem deriving Eq(1).

The traces are the same.

Perhaps you should consider the fact that the trace is the trace -

Really? So which one of the following would you pick for the trace of the TENSOR T_{ a b }?
Is it T_{ a a }, T^{ a a }, T^{ a }{}_{ a } or T_{ a }{}^{ a }?
Since T is a TENSOR, the answer will depend on the GROUP under which T_{ a b } transforms as a covariant tensor. For some groups, for example SO( n ) and SU( 2 ) upper and lower indices are equivalent (representations), therefore any of the above expression gives the correct answer for the trace. But, if the groups in question were SL( 2 , C ), SO( m , n - m ) or GL( n ) then the answer will be (depending on your convention) either T^{ a }{}_{ a } or T_{ a }{}^{ a }.

In SL( 2 , C ), the \sigma’s are mixed spinor-tensor with (ingeneral)
\left( \sigma^{ \mu } \right)_{ A \dot{ B } } \neq \left( \sigma^{ \mu } \right)^{ A \dot{ B } } .

Indeed,
( \sigma^{ 0 } )_{ A \dot{ B } } = ( \sigma^{ 0 } )^{ A \dot{ B } } , ( \sigma^{ 1 } )_{ A \dot{ B } } = - ( \sigma^{ 1 } )^{ A \dot{ B } }, ( \sigma^{ 2 } )_{ A \dot{ B } } = ( \sigma^{ 2 } )^{ A \dot{ B } }, \mbox{ and } \ ( \sigma^{ 3 } )_{ A \dot{ B } } = - ( \sigma^{ 3 } )^{ A \dot{ B } } .<br />
And the SL(2,C) INVARIANT trace is given by
<br /> \mbox{ Tr }_{ sl(2,c) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \sigma^{ \mu } )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ A \dot{ B } } \equiv ( \sigma^{ \mu } )_{ A \dot{ B } } ( \bar{ \sigma }^{ \nu } )^{ \dot{ B } A } \equiv ( \sigma^{ \mu } \sigma^{ \nu } )_{ A }{}^{ A } = 2 \eta^{ \mu \nu }<br />
Notice the POSITIONS of the SL(2, C) indices, in particular, when the BARED sigma appears.


Sam

 

[samalkhaiat]

You see, under the trace there is a product of two matrices: product of \sigma^1 with \sigma^1, Not a product of \sigma^1 with \bar{\sigma}^1.

Therefore your LHS is not equal to your RHS:

\frac{ 1 }{ 2 } \mbox{ Tr } ( \sigma^{ 1 } \sigma^{ 1 } ) \neq\frac{1}{2}\left(\sigma^1\bar{\sigma}^1\right)^A_{\phantom{A}A}

CLEARLY, you do not know much (if any thing) about the spinor-tensor notations of SL( 2 , C ). This is why you are REPEATING and INSISTING on REPEATING your MISTAKES. So, I have to repeat what I have said to you before “READ THE PARAGRAPH AND EQUATION AFTER Eq(5) IN POST #8” or go and read about the tensor notations in SL( 2 , C ). Until then, you have to take, what I have given you about the trace, as a definition.
<br /> \mbox{ Tr }_{ sl(2,c) } ( \sigma^{ i } \sigma^{ j } ) = \left( \sigma^{ i } \bar{ \sigma }^{ j } \right)_{ A }{}^{ A } = - \left( \sigma^{ i } \sigma^{ j } \right)_{ c c } = - \mbox{ Tr }_{ su(2) } ( \sigma^{ i } \sigma^{ j } ) .<br />
Notice, that the LHS gives you \eta^{ i j }, while the trace on the RHS gives you \delta^{ i j }. Thus, we arrive at the identity
\eta^{ i j } = - \delta^{ i j } .
So, there is no monkey business, such as 1 = - 1 in here.
I think, I have given you enough information in this thread. If you want me to continue and do proper business, then you should put some efforts and learn more about SL( 2 , C ).

Sam

 

[arkajad]

Really?

Sam

Really. Indices alpha,beta take values 0,1,2,3. There are no spinor indices in this Carmeli's equation.

So, please, do not mix spinor indices. In Carmeli's book they are introduced only in Chapter 5. The equation I am discussing is in chapter 3.

Indices alpha,beta are space-time indices. You rise and lower them with the 4x4 flat Minkowski metric.

Sigma matrices are explicitly given. They are 2x2 matrices, and under the trace is their product.
Square of each of these matrices is the identity matrix. Its trace is 2.

\Lambda is 4x4 Lorentz transformation. Normally it maps vectors into vectors, thus it has one upper and one lower index. The identity transformation has coefficients \delta^\alpha_\beta

We can rise the lower index to get [\itex]\eta{\lpha\beta}[/itex]

Let me repeat: This is chapter 3 in Crameli's book. No spinor indices, no bars, no epsilons, no dots over spinor indices. All elementary. Rising, lowering of spinor indices comes only in Ch. 5.

Moreover, almost the same is repeated in the other Carmeli's book. Except that this time his formula is correct. In one of his book there is an evident error. Now that I know THAT, I know that the formula with which I have started this thread is erroneous.

Your confusion probably comes from the fact that you do not have Carmeli's book, and I was not clear enough by introducing all the assumptions, notation, index ranges.

Yea, I know, spinor indices MAY BE tricky owing to the antisymmetry of epsilon metric, somewhat strange notation etc. But HERE we do not need all that. We are dealing only with Lorentz 0,1,2,3
indices and a normal trace of a 2x2 matrix.

Anyway: thanks a lot for trying to help me!

 

[dextercioby]

Hi Arkajad, this has developed into an argument, which shouldn't have been the case. So earlier on the first page I understood that the <Tr (...)> is not so straightforward, because the sigma's have both SL(2,C) and Lor(1,3) indices and Carmeli's writing in eqn. 3.8a hides something important, namely that the equation has no spinorial indices, which is wrong. His equation only makes sense when all dotted and undotted indices are placed and a clear meaning to what Tr(...) is given. Also quite 'disturbing' is that the \Lambda has the second index raised, which means that his series of equalities should have been written

\Lambda^{\alpha}_{\ ~ \sigma} \eta^{\sigma\beta} = \Lambda^{\alpha\beta} = ...

Oh, and yes, his treatment in one book is not the same as in the other book. :D Both lack use of SL(2,C) indices. That's why I suggested the book by Mueller-Kirsten and Wiedemann. There's no (from my perspective) better treatment for Lorentz spinors than in the that book.

 

[arkajad]

Both lack use of SL(2,C) indices.

They are needed for the formulation of this particular formula, as well as for its derivation.

I was not aware of the fact that samalkhaiat did not have access to Carmeli's book. If he did,
probably the whole discussion would go smoother. Also if I would have explained all the details
and all the context. So, I have learned somethings. Next time I will think twice before asking for help.
Thanks.

 

[arkajad]

My question to all partcipants of this discussion is this:

In two Carmeli's books we have two different formulas

In "Theory of Spinors" , 2000, (with Malin) , Ch. 3.4.2, p. 56 (3.84a):
\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr }(\sigma^\alpha g\sigma^\beta g\dagger).

In "Group Theory and General Relativity", (1997), Ch. 3-1, p. 36 (3-9a)
\Lambda^{\alpha}_{\phantom{\alpha}\beta}=\frac{1}{2}\mbox{Tr }(\sigma^\alpha g\sigma^\beta g\dagger).

The same context, the same definitions, the same notation: \alpha,\beta=0,1,2,3, g is SL(2,C) matrix, \Lambda - the corresponding Lorentz matrix, \sigma - Pauli matrices, {}^\dagger - Hermitian conjugate of a matrix, Tr - trace of the matrix.

The difference is the eta matrix (Minkowski metric)

\Lambda^{\alpha}_{\phantom{\alpha}\beta}=\Lambda^{\alpha\rho}\eta_{\rho\beta}.

At least one of the formulas must be wrong. My contention is that the second formula is right, the first one is wrong. Probably other readers will be also puzzled, will google, and will come to this thread. So, we should reach a clear consensus - which one is right and which one is wrong.

P.S. I wrote to Carmeli and my email got returned. I checked: Carmeli died in 2007. I wrote to Malin, did not receive any reply.

 

[dextercioby]

Apparently there's no consensus among the ones that do not use SL(2,C) indices when writing the equality, even though they are obviously not the same. Check out Barut's book on electrodynamics and field theory vs. Bogoliubov's et al. text on axiomatical field theory (1975, English).

 

[arkajad]

Check out Barut's book on electrodynamics and field theory vs. Bogoliubov's et al. text on axiomatical field theory (1975, English).

Checked both. Both give the second formula. (Their sigmas with lower index are exactly the same as Carmeli's sigmas with upper index - up to an irrelevant here sign of \sigma^2).

 

[dextercioby]

The other sources I've checked use sigma bar. They also reduce to the second formula.

 

[arkajad]

In fact the derivation is straighforward.

1) Define \Lambda in terms of an SL(2,C) matrix A by
A\sigma_\mu A^\dagger=\sigma_\nu\,\Lambda^\nu_{\phantom{\nu}\mu}
2) Check that
\mbox{Tr }(\sigma_\mu \sigma_\nu )= 2\delta_{\mu\nu}
3) Multiply the first equation by \sigma_\rho and take the trace of both sides. Use the second equation. Rename the indices.

 

[dextercioby]

Yes, but you lack covariance. The indices summed over must be one up, one down, so that 1 should be

A \sigma_{\mu} A^{\dagger} = \sigma^{\nu} \eta_{\nu\lambda} \Lambda^{\lambda}_{\ ~ \mu}

and \delta_{\mu\nu} \equiv \eta_{\mu\nu}.

 

[arkajad]

But we do not demand covariance from sigma matrices. They are just a convenient fixed basis in the space of all Hermitian matrices. Once this is remembered, the rest follows. Carmeli defines his basis having upper indices, others define their basis with lower indices. Does not matter. It is just a basis that is numbered. Perhaps it should be numbered neutrally as \sigma(\mu). Then there would be less reasons for a confusion.

 

[arkajad]

Perhaps I should add a clarification. We have Minkowski space M and its dual M*. Usually a basis in a vector space is written using lower indices for numbering, basis in the duac space with upper indices. Usually we identify M with Hermitian 2x2 matrices. Then we choose a basis in the space of Hermitian matrices - the Pauli sigma matrices - with lower indices for numbering of basis elements. So, we have \sigma_\mu matrices defined. If we want to introduce some \sigma^\nu - we have to define them first. It is up to us how we define them, but first they have to be defined. Then consequences from our definion can be derived.

But, on the other hand, we could also decide to identify the dual space with Hermitian matrices. Again, we need to introduce a basis, this time it can be convenient to use upper indices for numbering the elements of the basis.

Of course M and M* can be identified using eta, If we decide to do this, we must take care of consistency of all our definitions.

In some books the indices numbering a basis (a frame, or "moving frame" in differential geometry context, or "tetrad") the indices numbering the basis element are written differently that indices of components of tensors. This is done for a good reason - to avoid confusion. Physicists tend to think that any object carrying an index is necessarily a tensor. But this not true. The idex may simply number the four vectors of a fixed tetrad. Textbooks written by mathematicians are, I think, in this respect, more reliable than textbooks written by physicists. These later more often rely on the fact that reader will be able to figure out the meaning of the symbols used from the context.

 

[Bill_K]

In some books the indices numbering a basis (a frame, or "moving frame" in differential geometry context, or "tetrad") the indices numbering the basis element are written differently that indices of components of tensors. This is done for a good reason - to avoid confusion. Physicists tend to think that any object carrying an index is necessarily a tensor. But this not true. The idex may simply number the four vectors of a fixed tetrad.

As you say, we often encounter quantities having two different types of indices, such as a tetrad basis e_a^μ. However, rather than regarding the tetrad index as "fixed", a better viewpoint is that both types of index have a transformation group associated with them. In fact there is a local group that transforms one set of basis vectors into another, which may be taken to be the Lorentz group or something more general. The tetrad index is transformed under this group, and may also be raised and lowered using an appropriate metric.

Another common situation in which two types of indices are encountered is the embedding of surfaces in a space of higher dimension, with some indices following the transformations of the spatial coordinates and some following the surface coordinates.

 

[arkajad]

A However, rather than regarding the tetrad index as "fixed", a better viewpoint is that both types of index have a transformation group associated with them.

Indeed, you are right. I would just add to what you just wrote that this point of view becomes especially important in gauge theories, for instnce when we want to consider gravitation as a gauge theory of the Lorentz group (or, even better, of SL(2.C)) .

 

[dextercioby]

Arkajad, do share with us if you get a reply from the co-author of <Theory of Spinors>. Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

 

[arkajad]

I checked. He is 76 years old professor emeritus, probably busy writing another book after his "Nature Loves to Hide: Quantum Physics and Reality, a Western Perspective". So, I am not going to disturb him any more with such a small detail that we can figure out ourselves. It is a good exercise anyway. Forces you to think.

 

[arkajad]

Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

Oh, yes. These authors are funny! They introduce the sigmas tilded, with an upper index. They are identical to those ones with lower index. And there is an explicit warning the reader that this is not the same as using eta to change the index position.

 

[Bill_K]

Arkajad, do share with us if you get a reply from the co-author of <Theory of Spinors>. Another solid reference on SL(2,C) spinors is the book by Sexl and Urbantke: <Relativity, Groups, Particles>, where the section relevant to this thread is on page 236.

Is there still a question remaining on this? My impression is that the OP had been answered completely, several times, beaten to death in fact! But for what it's worth, I'll summarize it all over again.

The Pauli matrices σ^μ_AB' are a conversion between the Lorentz vector and the (1/2, 1/2) representation of SL(2,C). While they can be represented as a fixed set of numbers, they are not limited to these values, and under a change of basis transform appropriately according to all three of their indices. Yes, covariance must be maintained.


A vector V^μ corresponds to a spinor V^AB' by V^μ = σ^μ_AB' V^AB', or conversely V^AB' = V^μ σ^μ_AB'. (I'm ignoring any leading 1/2's, which in the present context serve only as a distraction. Fill them in if you like.)

The norm of the vector can be written either V_μ V^μ or V_AB' V^AB', requiring the identity η_μν = σ_{μ AB'} σ_ν ^AB'.

I'm sure Carmeli covers this adequately in Chapter 5, but in Chapter 3 he suppresses the spinor indices and tries to write it as a matrix equation. Well it can't be written as a matrix equation without pulling some tricks! The indices simply do not agree, and they must. We need to understand that in Chapter 3 the matrix operations (product and trace) are not the usual ones. The best you can do perhaps is

σ_{μ AB'} σ_ν ^AB' = σ_{μ AB'} ε^B'D' σ_{ν CD'} ε^AC = - σ_{μ AB'} ε^B'D' σ^T_{ν D'C} ε^CA = - Tr(σ_μ ε σ^T_ν ε)

(Where T is the transpose. Or since σ is Hermitian the transpose could be replaced by the complex conjugate.)

 

[arkajad]

The indices simply do not agree, and they must

In fact, they don't. Example: Carmeli's 1977 book and other textbooks mentioned. But we must remember their meaning. For instance, if I introduce a fixed tetrad e_\mu of four tangent vectors, this index does not transform under change of coordinates. It is not a tensorial index. Once you remember the meaning - you are safe.

Physicists quite often instead of doing geometry do what I call "indexology". Which is the art of operating on indices without understanding what is geometrical meaning of these operations. Sometimes this leads to results that are hard to get using geometrical pictures. Most of the time the results of such calculations are correct. But other times it may lead to confusion and errors.

 

[samalkhaiat]

This thread shows how easily one gets into troubles if one confuses TENSOR CONTRACTION, i.e., A_{ a r } B^{ c r }, with the corresponding MATRIX MULTIPLICATION, i.e., A_{ a r } B^{ r c }. The two expressions become identical, if and only if the followings are true:
1) the first and the second index both belong to identical or equivalent representations of the group in the question, and
2) B^{ c r } = B^{ r c }.
We will see that none of these conditions is satisfied in SL( 2 , C ). But, before doing the tensor notations of SL( 2 , C ), I would like to derive the relation between the Lorentz transformation,
x^{ \mu } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ x^{ \nu } , \ \ \Lambda \in SO( 1 , 3 ) , \ \ (1)
and the corresponding SL( 2 , C ) transformation of the Hermitian matrix x_{ \mu } \sigma^{ \mu } = \eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau }:
<br /> \eta_{ \mu \tau } \ x^{ \mu } \ \sigma^{ \tau } \rightarrow \eta_{ \nu \tau } \ x^{ \nu } \ g \ \sigma^{ \tau } \ g^{ \dagger } , \ \ ( g , g^{ \dagger } ) \in SL( 2 , C ) \ (2)<br />
I choose to define
<br /> \eta^{ \tau \mu } \ \sigma_{ \mu } = \sigma^{ \tau } \equiv ( I , \vec{ \sigma } ) ,<br />
I also define
<br /> \eta^{ \tau \mu } \ \bar{ \sigma }_{ \mu } = \bar{ \sigma }^{ \tau } \equiv ( I , - \vec{ \sigma } ) .<br />
Notice that \sigma_{ \mu } and \bar{ \sigma }^{ \nu } are the same matrices. However, writing \sigma_{ \mu } = \bar{ \sigma }^{ \mu } would mess up your covariant expressions (some thing that you would not want to do). So, to stay out of troubles we will treat \sigma^{ \mu } and \bar{ \sigma }^{ \nu } as two independent objects.
Now, by direct matrix multiplication, we can easily verify that
\mbox{ Tr } ( \sigma^{ \tau } \bar{ \sigma }^{ \rho } ) = 2 \ \eta^{ \tau \rho } . \ \ (3)
Contracting both side of Eq(1) with \eta_{ \mu \tau } \sigma^{ \tau } gives
\eta_{ \mu \tau } \ x^{ \mu }\ \sigma^{ \tau } \rightarrow \Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ x^{ \nu } \ \sigma^{ \tau } . \ \ \ (4)
Comparing the RHS of Eq(4) with that of Eq(2) leads to
\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } \ \sigma^{ \tau } = \eta_{ \nu \tau } \ g \ \sigma^{ \tau } \ g^{ \dagger } .
Multiplying both sides by \bar{ \sigma }^{ \rho } and taking the trace, we find (using Eq(3)) that
\Lambda^{ \mu }{}_{ \nu } \ \eta_{ \mu \tau } ( 2 \ \eta^{ \tau \rho }<br /> ) = \mbox{ Tr } (g \ \eta_{ \nu \tau } \ \sigma^{ \tau } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) ,
or
2 \Lambda^{ \mu }{}_{ \nu } \ \delta^{ \rho }_{ \mu } = \mbox{ Tr } ( g \ \sigma_{ \nu } \ g^{ \dagger } \ \bar{ \sigma }^{ \rho } ) .
Thus
\Lambda^{ \rho }{}_{ \nu } ( g ) = \frac{ 1 }{ 2 } \mbox{ Tr } ( \bar{ \sigma }^{ \rho } \ g \ \sigma_{ \nu } \ g^{ \dagger } ).

***********

Tensor Contraction & Trace in SL( 2 , C)

The matrices \sigma^{ \mu } behave like mixed spinor-tensors indexed by A = 1,2 and \dot{B} = \dot{1}, \dot{2} which label the fundamental and anti-fundamental (conjugate) representations respectively,
( \sigma^{ \mu } )_{ A \dot{B} } = ( I , \vec{ \sigma } )_{ A \dot{B} } .
Complex conjugation turns undotted indices into dotted ones and vice versa. For example, the complex conjugate of the spinor-tensor \Psi_{ A B \dot{C}} is given by
\overline{ \Psi_{ A B \dot{C} } } = \bar{ \Psi }_{ \dot{A} \dot{B} C } .
Therefore, the hermiticity of \sigma^{ \mu } means, in the SL( 2 , C ) index notations, that
\overline{ ( \sigma^{ \mu } )_{ A \dot{B} } } \equiv ( \bar{ \sigma }^{ \mu } )_{ \dot{A} B } = ( \sigma^{ \mu } )_{ B \dot{A} } . \ \ (5)
Raising and lowering of the indices is done here by the invariant (metric) spinors
\epsilon_{ A B } = \epsilon^{ A B } = \epsilon_{ \dot{A} \dot{B} } = <br /> \epsilon^{ \dot{A} \dot{B} } , \ \ (6a)
which are the 2-dimensional Levi-Civita symbol, i.e. ,
<br /> \epsilon_{ A B } = - \epsilon_{ B A } = \left( \begin{array} {rr} 0 &amp; 1 \\ -1 &amp; 0 \end{array} \right) . \ \ (6b)<br />
They satisfy the relations
\epsilon_{ A B } \epsilon^{ C B } = \epsilon^{ C }{}_{ A } = - \epsilon_{ A }{}^{ C } = \delta^{ C }_{ A } . \ \ (6c)
The invariance of the metric spinor \epsilon_{ A B } under SL( 2 , C), which is a necessary condition for the consistency of the relations (6), follows from the definition of a determinant
\det | g | \epsilon_{ A B } = g_{ A }{}^{ C } \ g_{ B }{}^{ D } \ \epsilon_{ C D } ,
and the fact that, for all g \in SL( 2 , C ), \det | g | = 1.
The invariant trace of tensors is, therefore, obtained by contracting all indices with the appropriate spinor metric. For example, the trace of the tensor T^{ A B } is given by
\mbox{Tr}_{ sl(2,C) } ( T ) = \epsilon_{ A B } T^{ A B } . \ \ \ (7)
It is clear from this that a traceless matrix, if interpreted as tensor, needs not be traceless tensor. For instance, as an anti-symmetric matrix, the Levi-Civita symbol is traceless, i.e. ,
\mbox{Tr}_{ M } ( \epsilon ) = 0 ,
while the tensor \epsilon^{ A B }, as seen from Eq(7) and Eq(6c), has a non-vanishing trace,
\mbox{Tr}_{ sl(2,C) } ( \epsilon ) = \epsilon^{ A }{}_{ A } = - \epsilon_{ A }{}^{ A } = 2 .
Another example is the Pauli matrices \sigma^{ i } themselves. While \mbox{Tr}_{ M } ( \sigma^{ i } ) = 0, \mbox{Tr}_{ sl(2,C) } ( \sigma^{ i } ) does not exist (not even defined). However, the tensor ( \sigma^{ \mu } \sigma^{ \nu }<br /> )_{ A B } has a trace given by
<br /> \mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \sigma^{ \mu } )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ A \dot{ B } } = 2 \eta^{ \mu \nu } . \ \ \ (8)<br />
This follows directly from the Dirac algebra (upon contracting the indices A and C):
<br /> \{ \sigma^{ \mu } , \sigma^{ \nu } \}^{ C }_{ A } \equiv ( \sigma^{ \mu} )_{ A \dot{ B } } ( \sigma^{ \nu } )^{ C \dot{ B } } + ( \sigma^{ \nu } )_{ A \dot{ B } } ( \sigma^{ \mu } )^{ C \dot{ B } } = 2 \ \delta^{ C }_{ A } \eta^{ \mu \nu } .<br />
The hermiticity condition Eq(5) allows us to rewrite the trace in Eq(8) as that of matrix multiplication
<br /> \mbox{Tr}_{ sl(2,C) } ( \sigma^{ \mu } \sigma^{ \nu } ) = ( \bar{ \sigma }^{ \mu } )_{ \dot{ B } A } ( \sigma^{ \nu } )^{ A \dot{ B } } = \mbox{Tr}_{ M } ( \bar{ \sigma }^{ \mu } \sigma^{ \nu } ) .<br />
Let us now show that the complex conjugation defined in Eq(5) is consistent with our (index free) definition of \bar{ \sigma }^{ \mu } = ( I , - \vec{ \sigma } ). Consider
( \sigma^{ \mu } )^{ A \dot{ B } } = \epsilon^{ A C } \ \epsilon^{ \dot{ B } \dot{ D } } \ ( \sigma^{ \mu } )_{ C \dot{ D } } ,
now take the complex conjugate of both sides,
<br /> \overline{ ( \sigma^{ \mu } )^{ A \dot{ B } } } = \overline{ \epsilon^{ A C } } \ \overline{ \epsilon^{ \dot{ B } \dot{ D } } } \ \overline{ ( \sigma^{ \mu } )_{ C \dot{ D } } } .<br />
Using the rule of complex conjugation and Eq(5), we find
<br /> ( \bar{ \sigma }^{ \mu } )^{ \dot{ A } B } = - \epsilon^{ B D } \ ( \sigma^{ \mu } )_{ D \dot{ C } } \ \epsilon^{ \dot{ C } \dot{ A } } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ B \dot{ A } } .<br />
Therefore
\bar{ \sigma }^{ \mu } = - ( \epsilon \ \sigma^{ \mu } \ \epsilon )^{ t } .
Since \epsilon^{ t } = \epsilon^{ -1 } = - \epsilon, then
\bar{ \sigma }^{ \mu } = - \epsilon \ ( \sigma^{ \mu } )^{ t } \ \epsilon = ( I , - \vec{ \sigma } ) .
I finish this post by writing the completeness relations with respect to the trace
\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )^{ C \dot{ D } } = 2 \ \delta^{ C }_{ A } \ \delta^{ \dot{ D } }_{ \dot{ B } } ,
which follow from
\eta_{ \mu \nu } \ ( \sigma^{ \mu } )_{ A \dot{ B } } \ ( \sigma^{ \nu } )_{ C \dot{ D } } = 2 \ \epsilon_{ A C } \ \epsilon_{ \dot{ B } \dot{ D } } .

Sam

 

[arkajad]

This thread shows how easily one gets into troubles if one confuses TENSOR CONTRACTION, i.e., A_{ a r } B^{ c r }, with the corresponding MATRIX MULTIPLICATION
...
Sam

What is not clear from your post is: who has confused these two things and exactly where?
I think being absolutely clear about this point will be of value for those who will read this thread.
Otherwise they can be confused.

I also understand that you agree that there is an error in Carmeli-Malin book, while there is no error in the book by Carmeli alone. Even if you did not spell it clearly. Yes?

 

[samalkhaiat]

What is not clear from your post is: who has confused these two things and exactly where?
I think being absolutely clear about this point will be of value for those who will read this thread.
Otherwise they can be confused.

I leave this to the readers. After all these posts, it shouldn’t be hard question to answer: who got “1 = -1” and where, and who said “trace is a trace” and where.

I also understand that you agree that there is an error in Carmeli-Malin book, while there is no error in the book by Carmeli alone. Even if you did not spell it clearly. Yes?

No, I did not agree:
1) I did read the book “Theory of Spinors” 12 years ago and I do not remember seeing any errors in it. However, I do remember that half of the book was about SL( 2 , C ) and its representations and the second half was about the spinor formulation of various field theories.
2) You thought that the following equations (from the book) were wrong
\mbox{Tr} ( \sigma^{ \alpha } \sigma^{ \beta } ) = 2 \eta^{ \alpha \beta } , \ \ \ \mbox{Tr} ( \sigma^{ \alpha } g \sigma^{ \beta } g^{ \dagger }) = 2 \Lambda^{ \alpha \beta } .
I have shown you, several times, that the equations are correct and consistent with each other if the trace is an SL( 2 , C ) trace.
3) Moshe Carmeli was a very careful writer and teacher. I have read many of his good published paper and three of his books. I believe that he published more that 30 paper and 5 books. He spent his life working on SO( 1 , 3 ) and SL( 2 ,C ). In the early 1970’s he formulated his SL( 2 ,C ) gauge theory of gravitation and showed that it is equivalent to Einstein’s theory of gravitation. Now days, we speak of the Carameli’s Lagrangian.

Sam

 

[arkajad]

3) Moshe Carmeli was a very careful writer and teacher.

Sam

If so how you explain the fact that in "Theory of Spinors" he has

\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)

While in "Group Theory and relativity" he has

\Lambda^\alpha_{\phantom{\alpha}\beta}=<br /> \frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)
Do you notice the difference? I mean, not on the right hand side, right hand sides are the same. But on the left hand side there is a very small and subtle difference. If you see this difference, how are you going to explain it? Just address this simple point. I will really appreciate it. Other readers of these pages will appreciate it as well, I am sure.

Right hand sides are the same, left hand sides are different. Not only different, but if you for a moment suppose thath they are equal, you will easily get 1=0. So they are not just different, they contradict one to another.

Remember: alpha and beta spacetime indices. Sigmas are Pauli matrices. They are exactly the same in those two books.

 

[samalkhaiat]

If so how you explain the fact that in "Theory of Spinors" he has

\Lambda^{\alpha\beta}=\frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)

This equation is correct, if the trace is that of tensor contraction, i.e., when we treat the sigmas as SL(2,C) mixed tensors. In other words, the trace is what I denoted by \mbox{Tr}_{ sl(2,C) }.

While in "Group Theory and relativity" he has

\Lambda^\alpha_{\phantom{\alpha}\beta}=<br /> \frac{1}{2}\mbox{Tr}(\sigma^{\alpha} g\sigma^\beta g\dagger)
Do you notice the difference?

I have never seen this book. However, by looking at this non-covariant equation, I guess he used the (non-covariant) matrix equality between \sigma^{ \beta } and \bar{ \sigma }_{ \beta } which I warned about in post#41. So, if you plug
\sigma^{ \beta } = \bar{ \sigma }_{ \beta }
back into his equation, you get the correct and covariant equation
2 \ \Lambda^{ \alpha }{}_{ \beta } = \mbox{Tr}_{M} ( \sigma^{ \alpha } \ g \ \bar{ \sigma }_{ \beta } \ g^{ \dagger } )
In this equation, the trace is just that of ordinary matrix multiplication.

I mean, not on the right hand side, right hand sides are the same.

On the contrary, the RHS’s are different. The trace in the first equation is \mbox{Tr}_{sl(2,C)} while the trace in the second equation is \mbox{Tr}_{M}.

Sam

 

[arkajad]

On the contrary, the RHS’s are different.
Sam

On the contrary, they are exactly the same. I did copy and paste. And the traces are exactly the same. Though you would have to check the book in order to understand that this is the case. I strongly suggest you do it. Otherwise you will have to guess. Guessing is never a proof of anything.

"Theory of Spinors":

"Group Theory and General Relativity":

 

[samalkhaiat]

On the contrary, they are exactly the same. I did copy and paste. And the traces are exactly the same.

These are meaningless sentences. Denoting the trace by \mbox{Tr} is the same, how one calculates the trace is the difference.

Though you would have to check the book in order to understand that this is the case. I strongly suggest you do it.

I don’t need to see any book, because I am an expert on the subject.

Otherwise you will have to guess. Guessing is never a proof of anything.

Clearly, now you are playing politics with me, because my posts left you with no meaningful questions. Sir, I am a theoretical physicist not politician.

Good Luck

Sam

 

[arkajad]

Sir, I am a theoretical physicist not politician.

Good Luck

Sam

Even theoretical physicists should use logic. My question was about particular formulas in particular books. You did not address my questions. You have addressed your own questions. I understand you were trying to help, and I thank you for your effort. But you did not help. Perhaps one of the reasons was that you have the wrong idea that experts can't make errors. They can and they do. Studying history of science can help to fix this erroneous idea. Usually, when two experts have contradicting ideas about some subject, a third and a fourth expert is being called. But here we do not have too many experts as it appears. Perhaps because it is summer hollidays time.

As another expert on the subject let me state my hypothesis: Carmeli's error in "Theory of Spinors" was due to Malin. Malin was editing, as it appears from the almost identical content of parts of two books, sometimes by simple copy and paste. He was struck but apparent lack consistency in Carmeli's formula. He had a wrong idea that the equation must be covariant. You also seem to share this wrong idea. But these equations do not have to be covariant for the simple idea that they are not dealing with tensors. On the left hand side are the components of the Lorentz matrix. Lorentz matrix is a numerical object - not a tensor. It does not depend on the frame of reference. It operates on frames, but it is not a frame. So, Malin had this wrong idea, and he "fixed" Carmeli's formula without much thinking. Then he had to "fix" the proof. To prove a wrong formula he had to make an error in his proof. Which he did.

That is my hypothesis. When the third expert will come - it will be interesting to know his/her opinion about my hypothesis.

 

[arkajad]

Here is the evidence of the copy and paste job in the two Carmeli's book. Word by word, with just two "little" differences:

Only one of them can be right. Now I know it is Carmeli alone who is right. Carmeli-Malin is wrong. But the proofs are wrong in both. Probably Malin did not notice the mistake in the proof, and wrongly corrected the right formula incorrectly assuming that the proof was correct.

 

[Bill_K]

I have never before seen such a big deal made out of such a trivial point. You've repeatedly been given full explanations, all the way back to dextercioby's #2. So I can't imagine what good it would do to repeat it again, since you are apparently only interested in publicizing Carmeli's "mistake" rather than understanding it.

Nevertheless, the point, once again, is that Tr(σ^μσ^ν) is not a covariant quantity unless you do one of two things: a) replace one of the σ's, by defining a barred quantity σ^−μ which reverses the sign of one component, and writing Tr(σ^−μσ^ν), or b) redefine the Tr operation itself to implicitly include this replacement. There is nothing especially strange about redefining Tr. Doing this is similar to what we do by redefining the dot product operation u·v to include the minus sign when dealing with 4-vectors.