# Difference between pseudorapidity and rapidity

1. Jun 30, 2007

### indigojoker

I was wondering what the different between pseudorapidity and rapidity was (pertaining to a high energy accelerator). I know know pesudorapidity is but not sure what rapidity is.

2. Jun 30, 2007

### robphy

3. Jun 30, 2007

### indigojoker

rapidity is a type of "velocity" while pseudorapidity is a type of "angle" that tells me where I am in the detector. That doesnt make any sense at all to me because I dont think one can get m/s from the rapidity formula $$y = \frac{1}{2} ln \left( \frac{E+p_{z}}{E-p_{z}} \right)$$

4. Jul 1, 2007

### robphy

Rapidity $$\theta$$ is a [Minkowski-]angle in spacetime [not space], whose hyperbolic-tangent is related to the spatial-velocity [in space] by:
$$\tanh \theta = \displaystyle\frac{v}{c} = \beta$$

Have you seen
http://pdg.lbl.gov/2007/reviews/kinemarpp.pdf
38. Kinematics, p. 7 , eqns. 38.35-38.37 [where rapidity is called $y$ instead]?
Note (as noted on p.1 of that section) "units are used in which $$\hbar=c=1$$", which not the best way to say it... but it works.

5. Jul 1, 2007

### indigojoker

so where in the detector does rapidity physically describe?

pseudorapidity is described by:
$$\eta=-ln \left( tan \left( \frac{\theta}{2} \right) \right)$$

is the theta in this formula rapidity?

6. Jul 1, 2007

### Hans de Vries

Both are rapidities, where the rapidity $\theta$ is just a measurement of speed:

$$v/c\ =\ \tanh(\theta)$$

The rapidity is zero when v=0, while the rapidity becomes infinite if v=c.
Rapidity is often used to calculate the energy and momentum of a particle:

$$E\ =\ \gamma m c^2\ =\ \frac{mc^2}{\sqrt{1-v^2/c^2}}\ =\ \cosh(\theta) mc^2$$

$$p\ =\ \gamma m v\ =\ \frac{mv}{\sqrt{1-v^2/c^2}}\ = \sinh(\theta) m$$

Pseudorapidity is a rapidity also: If a particle has a small deviation from
the center path of a beam expressed by an angle $\alpha$, and the beam is
assumed to approach the speed of light, then the particle will have a speed
component in the forward direction of the beam which is lower than c.

This speed depends on the angle of deviation and is given by the formula
you quoted. This is a rapidity, To get the speed of the particle you can
simply apply the first formula given above.

So, this is then the speed-component of the particle in the forward
direction of the beam. The total speed of the particle approaches c
a well, like the other particles of the beam.

No, it's the angle of deviation of the center of the beam.

Regards, Hans