Investigations into the infinitesimal Lorentz transformation

No, the rotations are incorrect. For a rotation about the z-axis by an angle ##\theta##, the matrix form should be$$\Lambda_{\nu}^{\mu} =\left( \begin{array}{cccc}\text{cos}\ \theta & -\text{sin}\ \theta & 0 & 0 \\\text{sin}\ \theta & \text{cos}\ \theta & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \end{array} \right).$$For an infinitesimal rotation, the matrix form should be$$\Lambda_{\nu}^{\mu
  • #1
spaghetti3451
1,344
33

Homework Statement


[/B]
A Lorentz transformation ##x^{\mu} \rightarrow x'^{\mu} = {\Lambda^{\mu}}_{\nu}x^{\nu}## is such that it preserves the Minkowski metric ##\eta_{\mu\nu}##, meaning that ##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}## for all ##x##. Show that this implies that ##\eta_{\mu\nu} = \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##.

Use this result to show that an infinitesimal transformation of the form ##{\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}## is a Lorentz transformation when ##\omega^{\mu\nu}## is antisymmetric: i.e. ##\omega^{\mu\nu}=-\omega^{\nu\mu}##.

Write down the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a rotation through an infinitesimal angle ##\theta## about the ##x^{3}##-axis.

Do the same for a boost along the ##x^{1}##-axis by an infinitesimal velocity ##v##.

Homework Equations



3. The Attempt at a Solution [/B]

##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\rho}x^{\rho})({\Lambda^{\mu}}_{\sigma}x^{\sigma})##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}({\Lambda^{\rho}}_{\mu}x^{\mu})({\Lambda^{\sigma}}_{\nu}x^{\nu})##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}##

##\implies \eta_{\mu\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}##Am I correct so far?
 
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  • #2
failexam said:
##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}##

##\implies \eta_{\mu\nu}=\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}##

Am I correct so far?
Yes, I think so. But someone might want to see explicitly how you get the last line from the next to last line.
 
  • #3
Well, the next to last line consists of a sum of terms ##\eta_{\mu\nu}x^{\mu}x^{\nu}## and ##\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}x^{\mu}x^{\nu}## with all possible combinations of values ##0,1,2,3## for the indices ##\mu## and ##\nu##.

However, the last line only contains the terms ##\eta_{\mu\nu}## and ##\eta_{\rho\sigma}{\Lambda^{\rho}}_{\mu}{\Lambda^{\sigma}}_{\nu}## for a specific value of ##\mu## and ##\nu##.

I thought that this was obvious, so I decided to skip the explanation. Isn't my reasoning sound, though?
 
  • #4
If I have two matrices ##A_{\mu \nu}## and ##B_{\mu \nu}## that satisfy ##A_{\mu \nu}x^{\mu}x^{\nu} = B_{\mu \nu}x^{\mu}x^{\nu}## for all possible ##x^{\mu}##, can I conclude that ##A_{\alpha \beta} = B_{\alpha \beta}## for all ##\alpha## and ##\beta##?
 
  • #5
I think so, yes.
 
  • #6
Try to show explicitly that ##A_{12} = B_{12}##.
 
  • #7
Oh wait. The correct relation is ##A_{12}+A_{21}=B_{12}+B_{21}##.
 
  • #8
failexam said:
Oh wait. The correct relation is ##A_{12}+A_{21}=B_{12}+B_{21}##.
Right.
 
  • #9
Well! In that case, I need to rewrite my solution:

##\eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}x'^{\mu}x'^{\nu}##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\mu\nu}({\Lambda^{\mu}}_{\sigma}x^{\sigma})({\Lambda^{\nu}}_{\tau}x^{\tau})##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}({\Lambda^{\sigma}}_{\mu}x^{\mu})({\Lambda^{\tau}}_{\nu}x^{\nu})##

##\implies \eta_{\mu\nu}x^{\mu}x^{\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}x^{\mu}x^{\nu}##

##\implies \eta_{\mu\nu}+\eta_{\nu\mu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\nu}{\Lambda^{\tau}}_{\mu}##, since ##x^{\mu}x^{\nu}=x^{\nu}x^{\mu}##

##\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\tau\sigma}{\Lambda^{\tau}}_{\mu}{\Lambda^{\sigma}}_{\nu}##, since the metric tensor is symmetric, i.e. ##\eta^{\mu\nu}=\eta^{\nu\mu}##

##\implies \eta_{\mu\nu}+\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}+\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

##\implies 2 \eta_{\mu\nu}= 2 \eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

Is my solution correct?
 
Last edited:
  • #10
Yes. The symmetry of the metric tensor is used here.
 
  • #11
Next, I need to use the result ##\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}## to show that an infinitesimal transformation of the form ##{\Lambda^{\mu}}_{\nu}={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}## is a Lorentz transformation when ##\omega^{\mu\nu}## is antisymmetric: i.e. ##\omega^{\mu\nu}=-\omega^{\nu\mu}##:

##\eta_{\mu\nu}=\eta_{\sigma\tau}{\Lambda^{\sigma}}_{\mu}{\Lambda^{\tau}}_{\nu}##

##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}+{\omega^{\sigma}}_{\mu})({\delta^{\tau}}_{\nu}+{\omega^{\tau}}_{\nu})##

##\implies \eta_{\mu\nu}=\eta_{\sigma\tau}({\delta^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\delta^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\delta^{\tau}}_{\nu}+{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu})##

##\implies \eta_{\mu\nu}=\eta_{\mu\nu}+\eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu}+\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}##

##\implies \eta_{\mu\tau}{\omega^{\tau}}_{\nu}+\eta_{\nu\sigma}{\omega^{\sigma}}_{\mu} = 0##, where we neglect the term ##\eta_{\sigma\tau}{\omega^{\sigma}}_{\mu}{\omega^{\tau}}_{\nu}## because it is of second order in the infinitesimal ##{\omega^{\mu}}_{\nu}##.

##\implies \omega_{\mu\nu}+\omega_{\nu\mu}=0##

##\implies \omega_{\mu\nu}=-\omega_{\nu\mu}##

##\implies \omega^{\mu\nu}=-\omega^{\nu\mu}##

so that ##\omega^{\mu\nu}## is antisymmetric.

Is my solution correct?
 
  • #12
Yes.
 
  • #13
Finally, the remaining parts of the problem:

The matrix form for ##{\Lambda^{\mu}}_{\nu}## that corresponds to a rotation through a finite angle ##\theta## about the ##x^{3}##-axis is given by

## {\Lambda^{\mu}}_{\nu} =
\left( \begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & \text{cos}\ \theta & -\text{sin}\ \theta & 0 \\
0 & \text{sin}\ \theta & \text{cos}\ \theta & 0 \\
0 & 0 & 0 & 1 \end{array} \right) =
\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 & -\theta & 0 \\
0 & \theta & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,
##

where we used ##\text{sin}\ \theta = \theta - \frac{\theta^{3}}{3!}+\cdots## and ##\text{cos}\ \theta = 1 + \frac{\theta^{2}}{2!}+\cdots ## and we only kept terms up to linear order in ##\theta## in the expansion of ##{\Lambda^{\mu}}_{\nu}##,

so the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a rotation through an infinitesimal angle ##\theta## about the ##x^{3}##-axis is given by

##{\omega^{\mu}}_{\nu}=
\left( \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & -\theta & 0 \\
0 & \theta & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right).##
The matrix form for ##{\Lambda^{\mu}}_{\nu}## that corresponds to a boost along the ##x^{1}##-axis by a finite velocity ##v## is given by

## {\Lambda^{\mu}}_{\nu} =
\left( \begin{array}{cccc}
\gamma & -\gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right) =
\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 & -v & 0 & 0 \\
v & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ,
##

where we used ##\gamma =\frac{1}{\sqrt{1-v^{2}}}=(1-v^{2})^{-1/2}=1+\frac{v^{2}}{2}+\dots ## and ##\gamma v =v\frac{1}{\sqrt{1-v^{2}}}=v(1-v^{2})^{-1/2}=v+\frac{v^{3}}{2}+\dots ## and we only kept terms up to linear order in ##v## in the expansion of ##{\Lambda^{\mu}}_{\nu}##,

so the matrix form for ##{\omega^{\mu}}_{\nu}## that corresponds to a boost along the ##x^{1}##-axis by an infinitesimal velocity ##v## is given by

##{\omega^{\mu}}_{\nu}=
\left( \begin{array}{cccc}
0 & -v & 0 & 0 \\
v & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right).##Is this solution correct?
 
  • #14
failexam said:
The matrix form for ##{\Lambda^{\mu}}_{\nu}## that corresponds to a boost along the ##x^{1}##-axis by a finite velocity ##v## is given by

## {\Lambda^{\mu}}_{\nu} =
\left( \begin{array}{cccc}
\gamma & -\gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right) =
\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 & -v & 0 & 0 \\
v & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right) + \cdots
##
Check the sign in the first column of the last matrix on the right side.
 
  • #15
It is a typo. We should instead have the following:

##{\Lambda^{\mu}}_{\nu} =
\left( \begin{array}{cccc}
\gamma & -\gamma v & 0 & 0 \\
-\gamma v & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \end{array} \right) =
\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 & -v & 0 & 0 \\
-v & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right) + \cdots ={\delta^{\mu}}_{\nu}+{\omega^{\mu}}_{\nu}+\cdots ##

so that

##{\omega^{\mu}}_{\nu}=
\left( \begin{array}{cccc}
0 & -v & 0 & 0 \\
-v & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \end{array} \right)##.

I believe everything else is correct, isn't it?
 
  • #16
Looks good.
 

1. What is the infinitesimal Lorentz transformation?

The infinitesimal Lorentz transformation is a mathematical concept used in physics to describe how coordinates and velocities change between two reference frames in special relativity. It is a small, incremental transformation that is used to study changes in space and time at a very small scale.

2. How is the infinitesimal Lorentz transformation derived?

The infinitesimal Lorentz transformation is derived using a combination of mathematics and the principles of special relativity. It involves manipulating equations such as the Lorentz transformation and the Lorentz factor to account for very small changes in space and time.

3. What is the significance of the infinitesimal Lorentz transformation?

The infinitesimal Lorentz transformation is significant because it allows scientists to study the effects of special relativity at a very small scale. This is important for understanding the behavior of particles at high speeds and in strong gravitational fields, and for developing theories such as quantum field theory.

4. How is the infinitesimal Lorentz transformation used in experiments?

The infinitesimal Lorentz transformation is used in experiments in various ways, depending on the specific field of study. For example, it is used in particle physics to study the behavior of particles in accelerators, and in astrophysics to understand the behavior of objects in strong gravitational fields.

5. Are there any limitations to the infinitesimal Lorentz transformation?

Like any mathematical model, the infinitesimal Lorentz transformation has its limitations. It is based on the principles of special relativity, which may not accurately describe phenomena at very small scales or in extreme conditions. Additionally, the infinitesimal Lorentz transformation does not take into account the effects of quantum mechanics, which are necessary for a complete understanding of the universe.

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