Proving Commutation Relation in Poincaré Transformation

Click For Summary

Homework Help Overview

The discussion revolves around proving the commutation relations in the context of Poincaré transformations, specifically focusing on the generators in the scalar field representation. The original poster has made attempts to derive these relations but is struggling with certain aspects of the algebra involved.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are exploring the relationship between derivatives of coordinates and the implications of index notation. The original poster questions the transition from one form of the derivative to another and seeks clarification on the underlying principles.

Discussion Status

There is an ongoing exploration of the nuances of index notation and the implications of using covariant versus contravariant indices. Some participants have provided insights into the correct interpretations of these mathematical expressions, while others are seeking further clarification on specific points of confusion.

Contextual Notes

Participants note the importance of the Minkowski metric in their discussions and highlight that the assumptions made in this context may not hold in more complex scenarios, such as General Relativity.

Petraa
Messages
20
Reaction score
0

Homework Statement



Given a Poincaré transformation, Lorentz+translation, I have to find the Poincaré generators in the scalar field representation and then prove that the commutation relations.

I've done the first part but I can't prove the commutation relations.

Homework Equations



P_{\mu}=i\partial_{\mu}

M_{\mu\nu}=i\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right)

The Attempt at a Solution



For example for the mixed commutator after doing some straight-forward algebra

\left[M_{\mu\nu},P_{\rho}\right]=i^{2}\left[\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right),\partial_{\rho}\right]=\left[\partial_{\rho},\left(x_{\mu}\partial_{\nu}-x_{\nu}\partial_{\mu}\right)\right]=\partial_{\rho}x_{\mu}\partial_{\nu}-\partial_{\rho}x_{\nu}\partial_{\mu}

Now if we recall the definition of the generator of the translations <br /> P_{\mu}=i\partial_{\mu}\implies\partial_{\mu}=\frac{P_{\mu}}{i}=-iP_{\mu}<br />
\left[M_{\mu\nu},P_{\rho}\right]=\partial_{\rho}x_{\mu}\partial_{\nu}-\partial_{\rho}x_{\nu}\partial_{\mu}=\partial_{\rho}x_{\mu}\left(-iP_{\nu}\right)-\partial_{\rho}x_{\nu}\left(-iP_{\mu}\right)=i\left(\partial_{\rho}x_{\nu}P_{\mu}-\partial_{\rho}x_{\mu}P_{\nu}\right)

I know the results of the commutators from the Poincaré algebra so <br /> \partial_{\rho}x_{\mu}=g_{\rho\mu} but I don't understand it. I thought that
<br /> \partial_{\rho}x_{\mu}=\delta_{\rho\mu}

Any help in order to prove the penultimate relation ? Because I don't know how to go from
<br /> \partial_{\rho}x_{\mu}=\delta_{\rho\mu} to
<br /> \partial_{\rho}x_{\mu}=g_{\rho\mu}

Thanks
 
Physics news on Phys.org
Petraa said:
I thought that
<br /> \partial_{\rho}x_{\mu}=\delta_{\rho\mu}

This is not true. What is true is ##\partial_\rho x^\mu = \delta_\rho^\mu##, which is not the same thing.
 
@Orodruin Can you develop it a little more pls ? Because I have problems with this little quibbling of index notation and such
 
The coordinate with covariant index does not fulfil the relation you quoted. It only holds for a contravariant index and when you lower it using the metric you get the relation you were looking for. Placing it on a more appropriate level is impossible if you do not specify exactly which part you are having problems with and what your current level of understanding is.
 
So basically \partial_{\rho}x_{\nu}=\partial_{\rho}g_{v\alpha}x^{\alpha}=g_{v\alpha}\partial_{\rho}x^{\alpha}=g_{v\alpha}\delta_{\rho}^{\alpha}=g_{\nu\rho}=g_{\rho\nu}

Any mistake ?
 
Just beware that things will not work out like this once you start looking at GR and the metric becomes coordinate dependent ... The assumption here is that you are working with the Minkowski metric.
 
Orodruin said:
Just beware that things will not work out like this once you start looking at GR and the metric becomes coordinate dependent ... The assumption here is that you are working with the Minkowski metric.

Yes, indeed I'm using the Minkowski metric. The part where I have problems, as I just realized right now, is the difference between \partial_{\alpha}x_{\nu}
\partial_{\rho}x^{\alpha}
\partial^{\rho}x_{\alpha}
\partial^{\rho}x^{\alpha}

If you know any book for dummies like me where this topic is covered I would appreciate that. Thank you for your time
 
The only thing to realize is essentially that ##x^\mu## is different from ##x_\mu## and that
$$
\partial_\mu x^\nu \equiv \frac{\partial x^\nu}{\partial x^\mu}$$
and there really is not much more to it. You can work it all out from there.
 

Similar threads

Replies
2
Views
2K
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
8K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K