QFT Lorentz transformation question

[emob2p]

Hi,
Today in QFT class, we examined an infinitesimal Lorentz transformation. From this we showed that a LT is represented by an anti-symmetric matrix. But this means that the diagonal elements should all be zeros. My question is how this reduces to the standard LT with gamma, gamma, one, one running along the diagonal and +/- v*gamma in the 12, 21 elements. Where is my thinking confused? Thanks.

 

[vanesch]

Hi,
Today in QFT class, we examined an infinitesimal Lorentz transformation. From this we showed that a LT is represented by an anti-symmetric matrix.

Wasn't it: the unit matrix + an infinitesimal antisymmetric matrix ?

 

[emob2p]

Yes that is correct. But then that would make the diagonal elements of their sum all 1. Where do the gammas come from?

 

[dextercioby]

Beware

\Lambda^{\mu}{}_{\nu}=\delta^{\mu}_{\nu}+\epsilon^{\mu}{}_{\nu}

BUT

\epsilon^{\mu\lambda}=\epsilon^{\mu}{}_{\nu}\eta^{\nu\lambda}

AND

\epsilon^{\mu\lambda}=-\epsilon^{\lambda\mu}

Do you see the difference ?

Daniel.

 

[emob2p]

I guess I see the difference, but how does that help the original question. Isn't eta(nu, lambda) just a the unity matrix with a -1 as the time-time element?

 

[George Jones]

The exponential map maps a Lie algrbra into a neighbourhood of the identity of a corresponding Lie group. An infinitesimal Lorentz transformation is an element of the Lorentz Lie algebra so(1,3), while a Lorentz transformation is an element of the Lorentz Lie group SO(1,3). A lie algebra element that has trace zero exponentiates to a Lie group element that has determinant +1 - in this case a proper Lorentz transformation.

Regards,
George

 

[George Jones]

As an example, consider the infinitesimal Lorentz tranformation that is zero everywhere except in 2x2 block in the upper left, which is

\left[ \begin{array}{ccc}<br /> 0 &amp; w \\<br /> w &amp; 0 \\<br /> \end{array} \right]<br />

Exponentiate this 2x2 matrix (by diagonalizing first), and you'll see a standard boost expressed using hyperbolic trig.

Regards,
George

 

[Hans de Vries]

Yes that is correct. But then that would make the diagonal elements of their sum all 1. Where do the gammas come from?

Repeated multiplication by (I+\delta) leads to the Lorentz Transform when \delta is symmetric.


\mbox{Lorentz Transform}\ \ =\ \ \lim_{\delta \rightarrow 0} <br /> \left(\left| <br /> \begin{array}{cccc} <br /> 1 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 1 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 1 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 1 <br /> \end{array} <br /> \right| +\left| <br /> \begin{array}{cccc} <br /> 0 &amp; \delta &amp; 0 &amp; 0 \\ <br /> \delta &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 <br /> \end{array} <br /> \right|\right)^{\psi/\delta}\ \ =\ \ \left( <br /> \begin{array}{cccc} <br /> \cosh{\psi} &amp; \sinh{\psi} &amp; 0 &amp; 0 \\ <br /> \sinh{\psi} &amp; \cosh{\psi} &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 1 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 1 <br /> \end{array} <br /> \right) <br />


Repeated multiplication by (I+\delta) leads to Spatial Rotation when \delta is anti-symmetric.


\mbox{Spatial Rotation}\ \ \ \ \ \ =\ \ \lim_{\delta \rightarrow 0}<br /> \left(\left| <br /> \begin{array}{cccc} <br /> 1 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 1 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 1 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 1 <br /> \end{array} <br /> \right| +\left| <br /> \begin{array}{cccc} <br /> 0 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; \delta &amp; 0 \\ <br /> 0 &amp; -\delta &amp; 0 &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 0 <br /> \end{array} <br /> \right|\right)^{\phi/\delta}\ \ =\ \ \left( <br /> \begin{array}{cccc} <br /> 1 &amp; 0 &amp; 0 &amp; 0 \\ <br /> 0 &amp; \cos{\phi} &amp; -\sin{\phi} &amp; 0 \\ <br /> 0 &amp; \sin{\phi} &amp; \cos{\phi} &amp; 0 \\ <br /> 0 &amp; 0 &amp; 0 &amp; 1 <br /> \end{array} <br /> \right) <br />



The angle \phi and the boost \psi are proportional to "the number of times you multiply"
The boost \psi is related to the speed v by:

\tanh{(v/c)} = \psi

Substitute this in the Lorentz Transform matrix and you get the usual form with the gammas.



Regards, Hans

 

[emob2p]

Won't those equalities only hold as psi and phi go to infinity? In other words, don't you get the eqaulities only after an infinite number of infinitesimal transformations?

 

[Hans de Vries]

Won't those equalities only hold as psi and phi go to infinity? In other words, don't you get the eqaulities only after an infinite number of infinitesimal transformations?

Correct, it's a limit, so you should read \psi/\delta and \phi/\delta in the exponents with \delta \rightarrow 0

Regard, Hans

 

[tayyaba aftab]

can anybody kindly suggest some link or book for understanding lorentz group and lorentz from scratch.

 

[Fredrik]

can anybody kindly suggest some link or book for understanding lorentz group and lorentz from scratch.

Some other text I read referenced Ohnuki, but it's out of print, so you'd have to buy it used or look for it in a library. This one by Anadijiban Das looks good too. I haven't read either of these books, so I can only say that they appear to contain what you're looking for (but you should look inside and find out for yourself). I learned most of this stuff by reading chapter 2 of Weinberg's QFT book. You can definitely learn a lot from there too.

 

[Fredrik]

If we view the components of \Lambda as functions of six parameters, representing a velocity difference and a rotation, chosen so that we get the identity transformation when all parameters are 0, we can Taylor expand Lambda like this:

\Lambda(\theta)=\Lambda(0)+\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0 \Lambda(\theta)+\mathcal O(\theta^2)

where \mathcal O(\theta^2) represents all terms of second order and higher. Now let's define

\omega=\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0 \Lambda(\theta)

and use this Taylor expansion in the equation that defines a Lorentz transformation:

\eta=\Lambda^T\eta\Lambda=(I+\omega+\mathcal O(\theta^2))^T\eta(I+\omega+\mathcal O(\theta^2))=\eta+\omega^T\eta+\eta\omega+\mathcal O(\theta^2)[/itex]<br /> <br /> Since this holds for all \theta, the nth order terms on the right must be equal to the nth order terms on the left for all n. In particular, this means that<br /> <br /> (\eta\omega)^T=-\eta\omega<br /> <br /> Note that this is an exact equality, even though it came from a Taylor expansion. <br /> <br /> Also, keep this in mind:<br /> <blockquote data-attributes="" data-quote="Fredrik" data-source="post: 0" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> Fredrik said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Recall that the components on row \mu, column \nu of the matrices<br /> <br /> \Lambda, \Lambda^T, \eta, \eta^{-1}, \omega<br /> <br /> are written as<br /> <br /> \Lambda^\mu{}_\nu, \Lambda^\nu{}_\mu, \eta_{\mu\nu}, \eta^{\mu\nu}, \omega^\mu{}_\nu<br /> <br /> and that \eta^{-1} and \eta and used to raise and lower indices. </div> </div> </blockquote>This means that \omega_{\mu\nu} are actually the components of \eta\omega, not \omega. That&#039;s why it&#039;s anti-symmetric.

 

[Fredrik]

We can also Taylor expand the operator U(\Lambda)) that represents the Lorentz transformation \Lambda:

U(\Lambda(\theta))=U(\Lambda(0))+\theta^a\frac{\partial}{\partial\theta^a}\bigg|_0 U(\Lambda(\theta))+\mathcal O(\theta ^2)

=I+\theta^a\frac{\partial}{\partial\Lambda^\mu{}_\nu}\bigg|_I U(\Lambda) \frac{\partial}{\partial\theta^a}\bigg|_0\Lambda^\mu{}_\nu+\mathcal O(\theta ^2)

=I+\omega^\mu{}_\nu\frac{\partial}{\partial\Lambda^\mu{}_\nu}\bigg|_I U(\Lambda)+\mathcal O(\theta ^2)

Now define

X_\mu{}^\nu=\frac{\partial}{\partial\Lambda^\mu{}_\nu}\bigg|_I U(\Lambda)

and note that

\omega^\mu{}_\nu X_\mu{}^\nu=\eta^{\mu\rho}\omega_{\rho\nu}\eta_{\mu\sigma} X^{\sigma\nu}=\omega_{\rho\nu}\delta^\rho_\sigma X^{\sigma\nu}=\omega_{\sigma\nu}X^{\sigma\nu}

=\omega_{\sigma\nu}\left(\frac{X^{\sigma\nu}+X^{\nu\sigma}}{2}+\frac{X^{\sigma\nu}-X^{\nu\sigma}}{2}\right)=\omega_{\sigma\nu}\left(\frac{X^{\sigma\nu}-X^{\nu\sigma}}{2}\right)

=\frac{i}{2}\omega_{\sigma\nu}(-i (X^{\sigma\nu}-X^{\nu\sigma}))

(Recall that we always have A_{\mu\nu}S^{\mu\nu}=0 when A is anti-symmetric and S is symmetric).

This means that we can write

U(\Lambda(\theta))=I+\frac{i}{2}\omega_{\mu\nu}J^{\mu\nu}+\mathcal O(\theta^2)

if we define

J^{\mu\nu}=-i (X^{\mu\nu}-X^{\nu\mu})

 

[George Jones]

can anybody kindly suggest some link or book for understanding lorentz group and lorentz from scratch.

At what level and with respect to what? Classical relativity? Quantum field theory?

A nice book is Relativity, Groups, and Particles: Special Relativity and Relativistic Symmetry in Field and Particle Physics by Roman U. Sexl and Helmuth K. Urbantke,

https://www.amazon.com/dp/3211834435/?tag=pfamazon01-20.

 

[tayyaba aftab]

orignally posted by George Jones
"At what level and with respect to what? Classical relativity? Quantum field theory?

A nice book is Relativity, Groups, and Particles: Special Relativity and Relativistic Symmetry in Field and Particle Physics by Roman U. Sexl and Helmuth K. Urbantke,

https://www.amazon.com/Relativity-Gro...4394610&sr=1-1&tag=pfamazon01-20. "

i want to start from classical to qft.so that i make my concepts very clear.