Drell Yan process


by JoePhysicsNut
Tags: drell yan
JoePhysicsNut
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#1
Jan29-13, 10:41 AM
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The Feynman diagram for Drell Yan has a gamma-star or a Z as the mediator.

Does gamma-star simply mean a photon with a sizeable amount of energy?
Is there a well-defined point in the energy of the mediator when it's a Z instead of a gamma-star? The peak of the Z resonance on an invariant mass plot is 91 GeV, but as a wide resonance it could also have less energy than that.
If a W acts as the mediator, is it still classified as Drell Yan?
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Bill_K
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#2
Jan29-13, 11:06 AM
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Is there a well-defined point in the energy of the mediator when it's a Z instead of a gamma-star? The peak of the Z resonance on an invariant mass plot is 91 GeV, but as a wide resonance it could also have less energy than that.
Fig 4 in this paper shows a nice peak at the Z mass.
mfb
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#3
Jan29-13, 11:08 AM
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It highlights the fact that the photon is virtual (and therefore off-shell).

Is there a well-defined point in the energy of the mediator when it's a Z instead of a gamma-star?
Both are possible, and their relative contribution depends on the energy of the process.

With a W, you need ##q \bar{q'}## (one up-type and one down-type quark). I don't know how this process is called.

JoePhysicsNut
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#4
Jan29-13, 04:57 PM
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Drell Yan process


Quote Quote by mfb View Post
It highlights the fact that the photon is virtual (and therefore off-shell).
Thank you for your replies!

For massive particles, a real particle is one which lies on mass shell i.e. E^2-p^2*c^2=m^2*c^4. For a virtual particle the equals sign doesn't apply and one can get different values for m other than the true value of m.

However, a photon is massless, so what does it mean for a photon to be off-shell?
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#5
Jan29-13, 05:00 PM
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##E \neq p## (or ##E\neq pc## if you don't like c=1) - something which is satisfied for real photons.

Real Photons have the same equation like massive particles, they just have m=0 there.
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#6
Jan29-13, 05:47 PM
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Quote Quote by JoePhysicsNut View Post
Thank you for your replies!

For massive particles, a real particle is one which lies on mass shell i.e. E^2-p^2*c^2=m^2*c^4. For a virtual particle the equals sign doesn't apply and one can get different values for m other than the true value of m.

However, a photon is massless, so what does it mean for a photon to be off-shell?
It also means that for a photon with 4-momentum K, K^2 is NOT 0 as it would be if "on shell".
andrien
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#7
Jan30-13, 12:08 AM
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If a W acts as the mediator, is it still classified as Drell Yan?
it can not be,can you see why.
Bill_K
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#8
Jan30-13, 04:56 AM
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If a W acts as the mediator, is it still classified as Drell Yan?
it can not be,can you see why.
This talk on the observation of Drell-Yan processes at the LHC includes in the discussion those in which W is the intermediate particle. Other references use the term "Drell-Yan-like."

EDIT: Especially see the third slide, "The Drell-Yan processes", in which the diagram given has an intermediate W.
andrien
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#9
Jan30-13, 07:56 AM
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So a u quark and d(bar) quark annihilate each other.
Vanadium 50
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#10
Jan30-13, 10:35 AM
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Technically, Drell-Yan is only virtual photon exchange. However, especially at high mass, it is indistinguishable from Z exchange, so that is often included. The LHC experiments, by analogy, have called q-qbar annihilation through a virtual W "Drell-Yan", but it is strictly speaking not correct.
jtbell
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#11
Jan30-13, 10:46 AM
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I see from Wikipedia that Drell and Yan described this process in 1970. Electroweak theory (including the Z and W) was just getting off the ground at that time, so it seems natural that (a) they considered only photons, and (b) people later tended to generalize to similar processes involving the other gauge bosons.


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