ATANH

Application:

An application of the hyperbolic arctangent function, atanh(x), can be found in the field of special relativity, specifically when dealing with the addition of velocities. The function is also known as arctanh(x) or artanh(x).

Let's consider two objects moving along the same line. According to classical physics, if a train moves at velocity v1​ relative to the ground, and a person walks inside the train at a velocity v2​ relative to the train, the person's velocity relative to the ground would simply be v=v1​+v2​.

However, special relativity dictates that this simple addition is incorrect, especially as velocities approach the speed of light, c. The correct formula for the addition of velocities is:

$v=\frac{v_1+v_2}{1+\frac{v_1{v}_2}{c^2}}$

where v is the resulting velocity relative to the ground.

The hyperbolic arctangent function emerges when we use a different way to represent velocities, called "rapidity." Rapidity, denoted by ϕ, is defined such that the velocity v is given by: v=ctanh(ϕ) where tanh(ϕ) is the hyperbolic tangent function. This relationship allows us to rewrite the velocity addition formula in a very elegant way.

If we have two rapidities ϕ1​ and ϕ2​ corresponding to velocities v1​ and v2​, the rapidity of the combined velocity, ϕ, is simply the sum of the individual rapidities: ϕ=ϕ1​+ϕ2​

This is a much simpler addition rule. To find the final velocity v, we simply take the hyperbolic tangent of the combined rapidity: v=ctanh(ϕ1​+ϕ2​). The atanh(x) function is the inverse of the tanh(x) function. We use it to convert a velocity ratio, β=v/c, back into its corresponding rapidity ϕ: ϕ=atanh(v/c​)

Example: A Spaceship and an Astronaut

Imagine a spaceship traveling at v1​=0.5c relative to a space station. An astronaut inside the spaceship launches a probe at v2​=0.8c relative to the spaceship, in the same direction. We want to find the probe's velocity, v, relative to the space station.

Using Classical Velocity Addition

v=v1​+v2​=0.5c+0.8c=1.3c This is impossible since nothing can travel faster than the speed of light, c.

Using Relativistic Velocity Addition

$v=\frac{0.5c+0.8c}{1+\frac{(0.5c)(0.8c)}{c^2}}=\frac{1.3c}{1+\frac{0.4c^2}{c^2}}=\frac{1.3c}{1.4}\approx 0.9286c$

Using Rapidity and atanh(x)

First, we calculate the rapidities for the two velocities using the atanh(x) function, where x=v/c.

• Rapidity of the spaceship:

${\varphi }_1=\text{atanh}\left(\frac{0.5c}{c}\right)=\text{atanh}(0.5)\approx 0.549$

• Rapidity of the probe relative to the spaceship:

${\varphi }_2=\text{atanh}\left(\frac{0.8c}{c}\right)=\text{atanh}(0.8)\approx 1.099$

Next, we add the rapidities to find the total rapidity:

$\varphi ={\varphi }_1+{\varphi }_2\approx 0.549+1.099=1.648$

Finally, we convert this total rapidity back to a velocity:

$v=c\tanh (\varphi )=c\tanh (1.648)\approx c(0.9286)=0.9286c$

As you can see, both relativistic methods give the same result. The use of atanh(x) and rapidity simplifies the addition of velocities to a straightforward sum, which is a powerful and elegant way to handle relativistic transformations.

Here is a table showing the relationship between velocity ratio (v/c), rapidity (ϕ), and the resulting combined velocity.