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- Thread starter indigojoker
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- #2

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[tex]

\gamma^5 \equiv i \gamma^0 \gamma^1 \gamma^2 \gamma^3

[/tex]

and

[tex]

\gamma_{\mu} = \eta_{\mu \nu} \gamma^{\nu}

[/tex]

where [itex]\eta_{\mu \nu}[/itex] is the Minkowski metric.

- #3

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I'm not sure why they could be interchanged.

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You will not be able to find all answers to all questions in books. Try to do the calculation by yourself as indicated earlier, it is much more rewarding.

I'm not sure why they could be interchanged.

- #5

Vanadium 50

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Picking pieces out of different books - which may use different conventions - is a recipe for making errors.

- #6

nrqed

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I have no idea what the other posters have in mind...

As far as I know, [tex] \gamma^5 [/tex] and [tex] \gamma_5 [/tex] are exactly the same thing. The 5 here is not a Lorentz index so there is no meaning to having it upstairs or downstairs.

for example, nachtmann (Elementary particle physics) defines

[tex] \gamma_5 = i \gamma^0 \gamma^1 \gamma^2 \gamma^3 [/tex]

Peskin defines [tex] \gamma^5[/tex] exactly the same way.

(But Donoghue et al have a minus sign in the definition)

An important point is that one may write gamma_5 as

[tex] \gamma_5 = \frac{i}{4!} ~\epsilon_{\mu \nu \rho \sigma} \gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma [/tex]

which shows clearly that gamma_5 is a scalar. (well, a pseudoscalar to be more precise since it reverses sign under a reflection in space).

- #7

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It is really important definition if you use dimensional regularization (dimensionality of space-time is [tex]d[/tex]) whereAn important point is that one may write gamma_5 as

[tex] \gamma_5 = \frac{i}{4!} ~\epsilon_{\mu \nu \rho \sigma} \gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma [/tex]

[tex]\eta_{\mu \nu} \gamma^{ \mu } \gamma^{ \nu } = d[/tex]

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