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  Why is pseudorapidity defined as $-\log \tan \theta/2$

+ 6 like - 0 dislike

Why the log? Is it there to make the growth of the function slower?

As this is a common experimental observable, it doesn't seem reasonable to take the range from $[0,\infty)$ to $(-\infty,\infty)$ (For a particle emitted along the beam axis after collision $\theta = 0$ wouldn't be better to have a number that says how close it is to zero rather than one that says how large a number it is. I hope that makes the question clear.)

This post imported from StackExchange Physics at 2014-07-18 05:00 (UCT), posted by SE-user yayu
asked Jul 14, 2011 in Theoretical Physics by yayu (100 points) [ no revision ]

2 Answers

+ 21 like - 0 dislike

The advantage of this particular definition is that differences in pseudorapidity are invariant under boosts along the $z$ axis. Specifically, consider a Lorentz transformation corresponding to a boost by velocity $\beta\hat{z}$. Since the tangent is basically a transverse distance over a longitudinal distance, it transforms under a Lorentz boost along the longitudinal axis with a factor $\gamma$:

$$\tan\phi \sim\frac{\Delta x_T}{\Delta x_L} \qquad\to\qquad \tan\phi' = \frac{\Delta x_T}{\Delta x_L/\gamma} = \gamma\tan\phi$$

and if you put that in the pseudorapidity formula, you find

$$\eta = -\log\biggl(\tan\frac{\theta}{2}\biggr) \qquad\to\qquad \eta' = -\log\biggl(\tan\frac{\theta}{2}\biggr) - \log\gamma$$

The first term is a function of the particle's trajectory, but the second term is a function of the boost parameter only - it doesn't depend on the particle at all. So if you have two particles coming out of a collision with pseudorapidities $\eta_1$ and $\eta_2$, the $\log\gamma$ term is the same for both, and thus when you take the difference it cancels out:

$$\eta_1 - \eta_2 = \eta_1' - \eta_2'$$

The reason why it's so important to keep the pseudorapidity difference invariant is that in particle physics, people like to make plots called "lego plots" which show the directional distribution of the particles detected after a collision. When you do this, you could plot the particle detections vs. the polar and azimuthal angles $\theta$ and $\phi$. But if you use pseudorapidity instead of the angle $\theta$, the fact that $\Delta\eta$ is invariant means that you can perform a Lorentz boost on your data by just translating the whole graph along the $\eta$ axis.

This is useful because in hadron collisions, it's often the case that one of the individual quarks or gluons involved in the collision may have a lot more momentum than the other, so all the particles produced come out near one end of the detector. But by translating the graph appropriately, you can effectively shift to the center-of-mass frame of the colliding quarks or gluons, where the particles come out symmetrically distributed, and it's a lot easier to analyze.

By the way, the reason it's $\tan\frac{\theta}{2}$ rather than just $\tan\theta$ is that we'd like a jet which comes out of the collision at $\theta \approx \frac{\pi}{2}$, where the resolution is best, to have the same shape in the lego plot as it does in physical space. In particular, a circular jet should appear circular on the graph. This requires that the two coordinates be scaled the same way. If we were using $(\theta,\phi)$ as the directional coordinates, this wouldn't be a problem, since both are measured in radians, so we just need to choose a scaling for $\eta$ such that a small increment in $\theta$ near $\frac{\pi}{2}$ corresponds to the same numerical increment in $\eta$:

$$\biggl|\frac{\mathrm{d}\eta}{\mathrm{d}\theta}\biggr|\biggl(\theta=\frac{\pi}{2}\biggr) = 1$$

The argument in the first part of the post (along with some common-sense conditions) basically requires that psuedorapidity be defined as

$$\eta = -\log(\tan a\theta)$$

for some constant $a$. Plugging into the derivative gives

$$\biggl|\frac{\mathrm{d}\eta}{\mathrm{d}\theta}\biggr|_{\frac{\pi}{2}} = \biggl|\frac{a\,\sec^2(a\theta)}{\tan(a\theta)}\biggr|_{\frac{\pi}{2}} = \biggl|\frac{a}{\sin(a\theta)\cos(a\theta)}\biggr|_{\frac{\pi}{2}} = \frac{2a}{\sin(2a\pi/2)} = 1$$

(I'm abusing the notation a bit by using vertical bars for both absolute value and substitution, but hopefully the meaning is clear.)

This is a transcendental equation, so you can't solve it analytically, but with a little mathematical reasoning it's not hard to convince yourself that $a = \pm\frac{1}{2}$ are the only nonzero solutions. Having a negative argument to a logarithm brings in an extra imaginary term, though it would cancel out anyway, so we might as well choose the positive one.

This post imported from StackExchange Physics at 2014-07-18 05:00 (UCT), posted by SE-user David Z
answered Jul 14, 2011 by David Z (660 points) [ no revision ]
Thanks a lot, David. This is a great answer. I had been looking for an involved experimental perspective as well and this is better than most stuff I read.

This post imported from StackExchange Physics at 2014-07-18 05:00 (UCT), posted by SE-user yayu
+ 2 like - 0 dislike

Oh god.. I remember stressing over this one when I started in particle physics. :)

Regarding the use of the log: This is just a hyperbolic trig thing. First write down the pseudorapidity using momenta perpendicular and parallel to the beam line like they do in the second expression here:


Then use the inverse hyperbolic identities you find here:


and you can get back to the angle.

Regarding it's use as an experimental observable: There are two reasons high energy guys prefer to use it (from wikipedia):

  • Pseudorapidity depends only on the polar angle of its trajectory, and not on the energy of the particle. (nice sometimes when you're comparing data from one detector to the next)

  • In hadron collider physics, the rapidity (or pseudorapidity) is preferred over the polar angle because, loosely speaking, particle production is constant as a function of rapidity. One speaks of the "forward" direction in a hadron collider experiment, which refers to regions of the detector that are close to the beam axis, at high | | .

This post imported from StackExchange Physics at 2014-07-18 05:00 (UCT), posted by SE-user unclejamil
answered Jul 14, 2011 by unclejamil (140 points) [ no revision ]

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