- #1
ChrisVer
Gold Member
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The rapidity is defined as:
[itex]y = \frac{1}{2} ln(\frac{E+p_z}{E-p_z})[/itex]
and for a hard event (hard scattering reaction/two partons reaction) we find that:
[itex]y'= \frac{1}{2} ln(\frac{x_A}{x_B})[/itex]
If the invariant mass is zero the rapidity changes into the pseudorapidity [itex]n[/itex] which depends only on the polar angle [itex]\theta[/itex]:
[itex] n= -ln (tan(\frac{\theta}{2})) [/itex] or [itex]cos (\theta) = tanh(n) [/itex].
Since the pseudorapidity is just a (1-1) transformation of the polar angle theta, why is it prefered? I don't know but I'm losing geometrical intuition of the process when I try to think in [itex]n[/itex] terms...
[itex]y = \frac{1}{2} ln(\frac{E+p_z}{E-p_z})[/itex]
and for a hard event (hard scattering reaction/two partons reaction) we find that:
[itex]y'= \frac{1}{2} ln(\frac{x_A}{x_B})[/itex]
If the invariant mass is zero the rapidity changes into the pseudorapidity [itex]n[/itex] which depends only on the polar angle [itex]\theta[/itex]:
[itex] n= -ln (tan(\frac{\theta}{2})) [/itex] or [itex]cos (\theta) = tanh(n) [/itex].
Since the pseudorapidity is just a (1-1) transformation of the polar angle theta, why is it prefered? I don't know but I'm losing geometrical intuition of the process when I try to think in [itex]n[/itex] terms...