What Are Casimir Operators and Rest Reference Conditions in Particle Physics?

  • Context: Graduate 
  • Thread starter Thread starter noamriemer
  • Start date Start date
  • Tags Tags
    Definitions
Click For Summary
SUMMARY

The discussion focuses on the application of Casimir Operators in particle physics, specifically the equation W^{\mu}=-\frac{1}{2} \varepsilon^{\mu\nu\lambda\sigma}M_{\nu\lambda}p_{\sigma}. Participants explore the implications of analyzing a particle in its rest reference frame, represented as p_\mu=(m,0,0,0), and how this leads to the condition W^\mu =\frac {1} {2} m\varepsilon^{\mu\nu\lambda0}M_{\nu\lambda}. The conversation also delves into the significance of the condition m^2<0, which results in the momentum vector p_\mu=(0,0,0,m), and the relationship of p_{\mu}=(p,0,0,p) to massless particles.

PREREQUISITES
  • Understanding of Casimir Operators in quantum field theory
  • Familiarity with the Einstein summation convention
  • Knowledge of particle momentum representations
  • Basic concepts of relativistic mass and energy relations
NEXT STEPS
  • Study the implications of Casimir Operators in quantum field theory
  • Learn about the Einstein summation convention in detail
  • Research the properties of massless particles in particle physics
  • Explore the relationship between energy and momentum in relativistic contexts
USEFUL FOR

Physicists, graduate students in theoretical physics, and researchers interested in quantum field theory and particle dynamics.

noamriemer
Messages
50
Reaction score
0
Hi guys!
There is something I would like to get your help with...

I am looking at the equation:

W^{\mu}=-\frac{1}{2} \varepsilon^{\mu\nu\lambda\sigma}M_{\nu\lambda}p_{\sigma}

Which is, if I understand correctly,a Casimir Operator.
Now, I wish to look at a particle in its rest reference, meaning,
p_\mu=(m,0,0,0)

Why would these conditions yield :
W^\mu =\frac {1} {2} m\varepsilon^{\mu\nu\lambda0}M_{\nu\lambda}
?
I can seem to understand how the indices change...

The next thing I want to do, is understand what happens if I take m^2&lt;0

Why does this condition mean that the momentum vector would be
p_\mu=(0,0,0,m)
?
Thank you
 
Physics news on Phys.org
The expression you've quoted uses the Einstein summation convention, in which repeated indices are summed over: AiBi is a convenient short way of writing \sumAiBi

And because the only non-zero element of p is p0, when you do the summation over σ, all the terms are zero except the one in which σ is zero.
 
Thank you!

But why does
p_{\mu}=(0,0,0,m) relate to m^2&lt;0?

And likewise,

p_{\mu}=(p,0,0,p) relate to m=0?

I understand why
p_{\mu}=(m,0,0,0) relate to massive particle,
My logic here is p_0=E and E\approx m
and \vec{p}=0 (because we are looking at the reference frame)
But same logic does not work for me regarding the two eq. above...
Thank you!
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
4K
  • · Replies 24 ·
Replies
24
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 14 ·
Replies
14
Views
3K