Deriving the Lorentz Commutator and Factor of 2

In summary: I have to ask you a question and I beg you to answer it. Do you remember the beginning of the book? What the author was saying about the aim of the book?In summary, the conversation discusses the process of deriving the Lorentz group elements and representing them using a generator. The conversation also mentions a possible mistake and the correct form of the representation. The importance of normalizing the algebra and using a consistent choice is also discussed. The reader is reminded of the author's aim for the book.
  • #1
spookyfish
53
0
I am trying to derive the algebra and I get a factor of 2 wrong...
Consider the Lorentz group elements near the identity
[tex]
\Lambda_1^\mu\,_\nu = \delta^\mu\,_\nu + \omega_1^\mu\,_\nu, \quad \Lambda_2^\mu\,_\nu = \delta^\mu\,_\nu + \omega_2^\mu\,_\nu
[/tex]
and write a representation as
[tex]
U(\Lambda)=U(1 +\omega)=1-\frac{1}{2}i\omega^{\mu \nu} M_{\mu \nu}
[/tex]
where [itex]M[/itex] is a generator. Now the term [itex]\Lambda \equiv (\Lambda_2^{-1}\Lambda_1^{-1}\Lambda_2 \Lambda_1) [/itex] belongs to the group and up to 2nd order is [itex] 1+[\omega_2,\omega_1] [/itex].
So its representation is
[tex]
U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]
I know this is wrong and I am supposed to get
[tex]
1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]
 
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  • #2
spookyfish said:
I am trying to derive the algebra

What algebra? Is it Lorentz algebra?
 
  • #3
samalkhaiat said:
What algebra? Is it Lorentz algebra?

yes.
 
  • #4
spookyfish said:
I am trying to derive the algebra and I get a factor of 2 wrong...
Consider the Lorentz group elements near the identity
[tex]
\Lambda_1^\mu\,_\nu = \delta^\mu\,_\nu + \omega_1^\mu\,_\nu, \quad \Lambda_2^\mu\,_\nu = \delta^\mu\,_\nu + \omega_2^\mu\,_\nu
[/tex]
and write a representation as
[tex]
U(\Lambda)=U(1 +\omega)=1-\frac{1}{2}i\omega^{\mu \nu} M_{\mu \nu}
[/tex]
where [itex]M[/itex] is a generator. Now the term [itex]\Lambda \equiv (\Lambda_2^{-1}\Lambda_1^{-1}\Lambda_2 \Lambda_1) [/itex] belongs to the group and up to 2nd order is [itex] 1+[\omega_2,\omega_1] [/itex].
So its representation is
[tex]
U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]
I know this is wrong and I am supposed to get
[tex]
1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]

I am not planning to make the calculation, but did you use the fact that for [tex] \Lambda_{\mu}^{\nu} = \delta_{\mu}^{\nu} + \omega_{\mu}^{\nu}[/tex]
then its inverse matrix is: [tex] (\Lambda^{-1})_{\mu}^{\nu} = \delta_{\mu}^{\nu} - \omega_{\mu}^{\nu} +O(\omega^2)[/tex]?
 
  • #5
MathematicalPhysicist said:
I am not planning to make the calculation, but did you use the fact that for [tex] \Lambda_{\mu}^{\nu} = \delta_{\mu}^{\nu} + \omega_{\mu}^{\nu}[/tex]
then its inverse matrix is: [tex] (\Lambda^{-1})_{\mu}^{\nu} = \delta_{\mu}^{\nu} - \omega_{\mu}^{\nu} +O(\omega^2)[/tex]?

Yes. I used it, and I am pretty sure it leads to [itex]\Lambda = 1+ [\omega_2,\omega_1] [/itex]. I am not sure then how it would lead to
[tex]
U(1+[\omega_2,\omega_1])=1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]
(there is a missing factor 1/2 according to my convention)
 
  • #6
Yes, you are, your calculation is right.

Do you use a specific reference where this equation is given there?

I must say that I myself find QFT quite hard as you can follow my posts in the forum in the last month or so.
 
  • #7
MathematicalPhysicist said:
Yes, you are, your calculation is right.

Do you use a specific reference where this equation is given there?

I must say that I myself find QFT quite hard as you can follow my posts in the forum in the last month or so.

Yes. Sorry, I am using the following reference:
http://www.damtp.cam.ac.uk/user/ho/GNotes.pdf
and trying to fill in the gaps. This is at the end of page 41, and on page 42. My problem is the top of eq. (4.30) given the definition (4.29).
 
  • #8
spookyfish said:
So its representation is
[tex]
U(\Lambda)=U(1+[\omega_2,\omega_1])=1-\frac{1}{2}i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]
I know this is wrong

No, in fact that is the correct form.

and I am supposed to get
[tex]
1-i[\omega_2,\omega_1]^{\mu \nu} M_{\mu \nu}
[/tex]

This is wrong. If you use it, you will have a factor (2) multiplying the RHS of the Lorentz algebra.
 
  • #9
Thank you. you are right, it works out! I stopped when I didn't understand this line, which is a typo, and should have continued in the first place
 
Last edited:
  • #10
spookyfish said:
Thank you. you are right, it works out!

The point of having (1/2) in the [itex]U( \Lambda )[/itex] is to get "mormalized" algebra. There is nothing wrong in choosing [itex]U = 1 + i \omega^{ \mu \nu } M_{ \mu \nu }[/itex], but your algebra, in this case becomes [itex][M , M] = (1/2) ( g M + \cdots )[/itex]. So, when you mormalize [itex]M \rightarrow (1/2) M[/itex] you get the "nice" algebra [itex][M , M] = g M + \cdots [/itex].
The point to remember is this, once you made a choice, you have to use it for all group elements: [itex] ( \Lambda ), ( \Lambda_{1} \Lambda_{2} ) , \cdots , ( \Lambda_{1} \cdots \Lambda_{n} )[/itex].

I stopped when I didn't understand this line, which is a typo, and should have continued in the first place

I think it is lazyness not a "typo", because he repeats the same mistake when he "derives" the Poincare' algebra.

Sam
 

1. What is the Lorentz commutator and factor of 2?

The Lorentz commutator is a mathematical expression used in the theory of special relativity to describe the relationship between space and time. It is represented by the symbol [x, t] and is equal to -i, where i is the imaginary unit. The factor of 2 refers to the fact that the commutator is multiplied by 2 when calculating certain physical quantities in relativity.

2. Why is it important to derive the Lorentz commutator and factor of 2?

Deriving the Lorentz commutator and factor of 2 allows us to understand the fundamental principles of special relativity and how space and time are related. It also helps us make accurate predictions and calculations in various areas of physics, such as electromagnetism and quantum mechanics.

3. How is the Lorentz commutator and factor of 2 derived?

The Lorentz commutator and factor of 2 can be derived using mathematical techniques such as matrix algebra and tensor calculus. It involves manipulating the equations of special relativity, specifically the Lorentz transformations, to find the commutator and factor of 2.

4. What are some real-world applications of the Lorentz commutator and factor of 2?

The Lorentz commutator and factor of 2 have various applications in physics, such as in the study of particle accelerators, where the calculations of particle trajectories and energies rely on these concepts. They are also used in the development of technologies such as GPS systems and nuclear reactors.

5. Are there any controversies surrounding the Lorentz commutator and factor of 2?

There are no major controversies surrounding the Lorentz commutator and factor of 2. However, there have been some debates and alternative theories proposed in the past, such as the theory of Lorentz invariance, which suggests that the factor of 2 may not be necessary in all cases. However, the majority of physicists accept the standard derivation of the Lorentz commutator and factor of 2 as an accurate representation of the relationship between space and time in special relativity.

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