# "Longitudinal Rapidity"

## Main Question or Discussion Point

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 $\omega_x = \cosh \varphi \, \tanh \varphi = \sinh \varphi$:

$f_x = \dfrac{mc^2}{\cosh \varphi} \, \dfrac{d}{d (c \tau)} \sinh \varphi = (mc^2) \, \mathring{\varphi}$.

## Answers and Replies

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