# QFT Lorentz transformation question

1. Sep 22, 2005

### 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.

2. Sep 22, 2005

### vanesch

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

3. Sep 22, 2005

### emob2p

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

4. Sep 23, 2005

### 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.

5. Sep 23, 2005

### 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?

6. Sep 23, 2005

### George Jones

Staff Emeritus
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

7. Sep 23, 2005

### George Jones

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

$$\left[ \begin{array}{ccc} 0 & w \\ w & 0 \\ \end{array} \right]$$

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

Regards,
George

8. Sep 23, 2005

### Hans de Vries

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

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

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

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

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

Last edited: Sep 23, 2005
9. Sep 23, 2005

### 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?

10. Sep 23, 2005

### Hans de Vries

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

Regard, Hans

11. Sep 30, 2009

### tayyaba aftab

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

12. Oct 1, 2009

### Fredrik

Staff Emeritus
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.

Last edited: Oct 1, 2009
13. Oct 1, 2009

### Fredrik

Staff Emeritus
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:

$$\eta=\Lambda^T\eta\Lambda=(I+\omega+\mathcal O(\theta^2))^T\eta(I+\omega+\mathcal O(\theta^2))=\eta+\omega^T\eta+\eta\omega+\mathcal O(\theta^2)[/itex] Since this holds for all $\theta$, the nth order terms on the right must be equal to the nth order terms on the left for all n. In particular, this means that [tex](\eta\omega)^T=-\eta\omega$$

Note that this is an exact equality, even though it came from a Taylor expansion.

Also, keep this in mind:
This means that $\omega_{\mu\nu}$ are actually the components of $\eta\omega$, not $\omega$. That's why it's anti-symmetric.

Last edited: Oct 1, 2009
14. Oct 1, 2009

### Fredrik

Staff Emeritus
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

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

15. Oct 1, 2009

### George Jones

Staff Emeritus
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,

http://www.amazon.com/Relativity-Gr...=sr_1_1?ie=UTF8&s=books&qid=1254394610&sr=1-1.

16. Oct 3, 2009

### 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,

http://www.amazon.com/Relativity-Gro...4394610&sr=1-1. "

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