Understanding definitions

  • Thread starter noamriemer
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  • #1
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Hi guys!
There is something I would like to get your help with...

I am looking at the equation:

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

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

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

The next thing I want to do, is understand what happens if I take [itex]m^2<0[/itex]

Why does this condition mean that the momentum vector would be
[itex]p_\mu=(0,0,0,m)[/itex]
?
Thank you
 

Answers and Replies

  • #2
Nugatory
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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 [itex]\sum[/itex]AiBi

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.
 
  • #3
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Thank you!

But why does
[itex]p_{\mu}=(0,0,0,m)[/itex] relate to [itex]m^2<0[/itex]?

And likewise,

[itex]p_{\mu}=(p,0,0,p)[/itex] relate to [itex]m=0[/itex]?

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

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