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indigojoker

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- Thread starter indigojoker
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indigojoker

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humanino

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If Halzen & Martin's introductory textbook does not go deep enough, you probably need any textbook entitled "introduction to quantum field theory". Or "Dynamics of the Standard Model" (Donoghue, Golowich & Holstein @ Cambridge University Press) might do if you want phenomenology. If you are enclined towards theory, try Pokorsky's "Gauge Field Theories" @ same editor.

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indigojoker

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Are there examples where they do not make this assumption?

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humanino

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Later on when they talk about interferences in electron-positron annihilation for instance. But as you reach the end of this book, you probably need a thicker oneAre there examples where they do not make this assumption?

Therefore, congrats to you, because Halzen & Martin is a nice summary of basic particle physics.

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indigojoker

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Say the cross section for some process is: [tex]\sigma = \frac{G^2 s}{\pi} [/tex] (H+M 12.60)

Then using IVB theory, we take out the [tex]\frac{G^2}{4}[/tex] constant when calculating the amplitude and replace it with [tex]\left(\frac{g^2}{M_W^2+q^2}\right)^2[/tex] so when all is said and done, we are left with the cross section as: [tex]\sigma=\left(\frac{g^2}{M_W^2+q^2}\right)^2 \frac{4s}{\pi}[/tex]

I think this makes sense because the first cross section allows [tex]\sigma[/tex] to go to infinity as s becomes large while using the IMV theory, the cross section is corrected at large s by the q^2 on the denominator, thus giving a finite total cross section.

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nrqed

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Say the cross section for some process is: [tex]\sigma = \frac{G^2 s}{\pi} [/tex] (H+M 12.60)

Then using IVB theory, we take out the [tex]\frac{G^2}{4}[/tex] constant when calculating the amplitude and replace it with [tex]\left(\frac{g^2}{M_W^2+q^2}\right)^2[/tex] so when all is said and done, we are left with the cross section as: [tex]\sigma=\left(\frac{g^2}{M_W^2+q^2}\right)^2 \frac{4s}{\pi}[/tex]

I think this makes sense because the first cross section allows [tex]\sigma[/tex] to go to infinity as s becomes large while using the IMV theory, the cross section is corrected at large s by the q^2 on the denominator, thus giving a finite total cross section.

Hi indigojoker,

It's definitely true that the IVB model improves greatly the high energy behavior.

It turns out, though, that it is nonrenormalizable if I recall correctly. I am not sure about this but I think that essentially, the difference between the IVB and the Standard Model weak interaction is that in the IVB model there is no relation between between the couplings to the charged and neutral carriers. In the weak interaction, there is of course a definite relation between them. Another difference (I think but am not sure) is that I think the original IVB model had only vector coupling ([tex] \gamma^\mu [/tex] ).

The book by Aitchison and Hey discusses a little bit the IVB model (Gauge theories in Particle Physics)

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