What is the role of the longitudinal rapidity in relativistic contexts?

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In summary, longitudinal rapidity is a measure of the rapidity of a particle's motion in the direction of its motion and is defined as the logarithm of the ratio of the particle's energy to its longitudinal momentum. It differs from regular rapidity in that it is only measured in the direction of motion and is invariant under Lorentz transformations. The formula for calculating longitudinal rapidity is y = 1/2 * ln((E + pz)/(E - pz)), where E is the particle's energy and pz is its longitudinal momentum. It is important in particle physics for understanding high energy collisions and describing the distribution and kinematics of particles. Longitudinal rapidity can be negative, indicating motion in the opposite direction of momentum, or
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SiennaTheGr8
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In relativity contexts, "rapidity" is usually defined as ##\phi = \tanh^{-1} \beta##, always positive like ##\beta##. One sometimes also encounters the rapidity vector, ##\vec \phi = \phi \hat \phi = \phi \hat \beta##, with Cartesian components like ##\phi_x = \phi \cos{\theta} = \tanh^{-1} (\beta) \cos{\theta} ## (where ##\theta## is the angle between ##\vec \phi## and the positive ##x##-direction).

That's all well and good, but I'd argue it's of limited use. For more flexibility, we want a signed rapidity-like parameter, and the components of the rapidity vector don't quite fit the bill (explained below).

The quantity we're looking for is what particle physicists (unfortunately) call "rapidity." To avoid confusion, I'll call it longitudinal rapidity and label it ##\varphi## (curly phi). I want to make the case that this longitudinal rapidity deserves more recognition in relativistic contexts.

The longitudinal rapidity is defined as the inverse hyperbolic tangent of the Cartesian velocity-component parallel to a given boost axis (normalized velocity, I should say). So if the boost is along the ##x##-axis, the longitudinal rapidity is ##\varphi = \tanh^{-1} \beta_x = \tanh^{-1} (\beta \cos \theta)##, where ##\theta## has the same meaning it had a few paragraphs up. Naturally ##\varphi## takes the sign of ##\beta_x## (hyperbolic tangent and its inverse are odd functions).

We see that in general, ##\varphi \neq \phi_x##. The exception is when ##\phi_x## is the only non-zero component of ##\vec \phi## (motion in the ##\pm \, x##-direction).

What's nice about the longitudinal rapidity (as opposed to ##\phi_x##) is that it transforms additively under the boost in question, and—consequently—that changes/differences in it are invariant under the boost. The additivity of the transformation follows simply from the transformation rule for ##\beta_x## under a boost along the ##x##-axis:

##\beta^\prime_x = \dfrac{\beta_x - \beta_{\textrm{boost}}}{1 - \beta_x \, \beta_{\textrm{boost}}}##

##\tanh \varphi^\prime = \dfrac{\tanh \varphi - \tanh \phi_{\textrm{boost}}}{1 - \tanh \varphi \, \tanh \phi_{\textrm{boost}}} = \tanh (\varphi - \phi_{\textrm{boost}})##,

so that ##\varphi^\prime = \varphi - \phi_{\textrm{boost}}##. The invariance of changes/differences in longitudinal rapidity under said boost then follows by transforming ##\varphi_2 - \varphi_1##:

##\varphi^\prime_2 - \varphi^\prime_1 = (\varphi_2 - \phi_{\textrm{boost}}) - (\varphi_1 - \phi_{\textrm{boost}}) = \varphi_2 - \varphi_1##.Here is a neat application, an elegant demonstration that the Cartesian component of three-force ##\vec f## that's parallel to an object's three-velocity ##\vec \beta## is invariant under a boost in the ##\pm \, \hat \beta## direction (for constant ##m##). An overdot is a ##ct##-derivative (coordinate time), and an over-circle is a ##c \tau##-derivative (proper time). The celerity is ##\vec \omega = \gamma \vec \beta##.

##\vec f = \dot{\vec p} c = (mc^2) \, \dot{\vec \omega}##

##f_x = (mc^2) \, \dot{\omega}_x = \dfrac{(mc^2) \, \mathring{\omega}_x}{\gamma} ##.

With the ##x##-axis as boost axis, the longitudinal rapidity is ##\varphi = \tanh^{-1} \beta_x##, and so the ##x##-component of the celerity is ##\omega_x = \gamma \beta_x = \cosh \phi \, \tanh \varphi##. Now impose the condition that the three-velocity is in the ##\pm \, x##-direction. In this special case (as we covered above), ##\varphi = \phi_x = \pm \, \phi##, and that gives ##\cosh \varphi = \cosh \phi = \gamma## (even function). So ##\omega_x = \cosh \varphi \, \tanh \varphi = \sinh \varphi##:

##f_x = \dfrac{mc^2}{\cosh \varphi} \, \dfrac{d}{d (c \tau)} \sinh \varphi = (mc^2) \, \mathring{\varphi}##.
 
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  • #2
Can you provide some references for the rapidity vector?
 

1. What is longitudinal rapidity?

Longitudinal rapidity is a measure of the rapidity of a particle's motion in the direction of its motion. It is defined as the logarithm of the ratio of the particle's energy to its longitudinal momentum.

2. How is longitudinal rapidity different from regular rapidity?

Longitudinal rapidity is a measure of rapidity in the direction of motion, while regular rapidity is a measure of rapidity in any direction. Longitudinal rapidity is also invariant under Lorentz transformations, while regular rapidity is not.

3. What is the formula for calculating longitudinal rapidity?

The formula for calculating longitudinal rapidity is: y = 1/2 * ln((E + pz)/(E - pz)), where E is the particle's energy and pz is its longitudinal momentum.

4. Why is longitudinal rapidity important in particle physics?

Longitudinal rapidity is important in particle physics because it helps us understand the behavior of particles in high energy collisions. It is a useful tool for describing the distribution and kinematics of particles in these collisions.

5. Can longitudinal rapidity be negative?

Yes, longitudinal rapidity can be negative. This indicates that the particle is moving in the opposite direction of its momentum. A positive longitudinal rapidity indicates that the particle is moving in the same direction as its momentum.

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