# Definition of rapidity

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1. Mar 22, 2015

### JD96

Hello,

I got interested in the concept of rapidity and would like to know a bit more about it. Unfortunately hyperbolic trigonometric functions are not taught in school, at least not where I'm living, so despite the fact that they preserve many characteristics of ordinary trigonometric functions I don't know why they are defined the way they are.

I know that rapidity can be defined as the arclength of the unit hyperbola or as an hyperbolic angle (and probably many more ways). The latter is expressed as Φ= tanh-1(v/c). I searched for definitions of hyperbolic angles that look like this, but couldn't find one. So I tried to make sense of it using an analogy of "normal" angles: In a cartesian coordinate system S the angle between the y-axis and the y-axis of another coordinate system S', that has been rotated in negative x-direction, is given by α= tan-1(1/m) , where m is the slope of y'. Am I correct that the same holds true for the hyperbolic angle between the ct-, and ct'-axis, namely that Φ= tanh-1(1/m) and since 1/m = v/c one gets the definition of rapidity I wrote above?

Last edited: Mar 22, 2015
2. Mar 22, 2015

### Mentz114

3. Mar 22, 2015

### stevendaryl

Staff Emeritus
Yes, there is a close analogy between ordinary Euclidean geometry and trigonometry and Special Relativity and hyperbolic geometry.

In 2 dimensional space, you can describe points in space by a pair of coordinates $x$ and $y$. If you rotate the x-axis and y-axis through an angle $\theta$, then the new coordinate system is related to the old coordinate system by:

$x' = x cos(\theta) + y sin(\theta)$
$y' = - x sin(theta) + y cos(\theta)$

The slope of the $x'$ axis relative to the $x$ axis is given by a number $m = tan(\theta)$

In SR, with just one spatial dimension, for simplicity, you can describe points in spacetime by a pair of numbers $x$ and $t$. If you change to another reference frame, then the new coordinate system is related to the old coordinate system by:

$x' = x cosh(\theta) - c t sinh(\theta)$
$c t' = -x sinh(\theta) + c t cosh(\theta)$

where $\theta$ is the rapidity. The velocity of the second frame of reference relative to the first is given by:

$v = c tanh(\theta)$

The geometry is very similar, except for a few sign changes, a factor of $c$ here and there, and the replacement of trigonometric functions by hyperbolic trigonometric functions.

4. Mar 22, 2015

### JD96

@Mentz114 : I read the wikipedia article and altough it was a bit intimidating at the beginning I think I am grasping a bit of it. If I understand it correctly I should be able to prove that Φ is given by tanh-1(1/m) by computing (twice) the integral of y=(1+x2)0,5 (the hyperbola y2-x2=1 as a function of x) between x=0 and the place of intersection of the ray and the hyperbola, minus the integral of the ray in that interval. Is that correct? If the answer is yes I will try to do that to convince myself and get practice using hyperbolic functions. Also sorry, if my explanations are confusing to read, since I'm not a native english speaker :)

@stevendaryl: Thanks for your great response, that's a very elegant way of writing the Lorentz transformation!

Edit: I worked out the integral and indeed got the logarithmic expression of tanh-1(1/m). Now I feel a bit more comfortable about using rapidity, again thanks for your responses!

Last edited: Mar 22, 2015
5. Mar 22, 2015

### Mentz114

Your English is fine. The integral of the ray will be zero because $dt^2-dx^2=0$.

6. Mar 22, 2015

### pervect

Staff Emeritus
wiki probably already explained this but, as a consequence of the Eulier identity

$e^{i \theta} = \cos \theta + i \sin \theta$

we can write

cos $\theta$ = $\frac{1}{2} \left( e^{i\ \theta }+ e^{-i \theta} \right)$
sin $\theta$ = $\frac{1}{2i} \left( e^{i\ \theta } - e^{-i \theta} \right)$

and we have as an identity $\cos^2 \theta + \sin^2 \theta = 1$

The hyperbolic trig functions just repliace $i \theta$ with $\theta$

cosh $\theta$ = $\frac{1}{2} \left( e^{ \theta } + e^{ -\theta} \right)$
sinh $\theta$ = $\frac{1}{2} \left( e^{\theta } - e^{-\theta} \right)$

and we have as an identity $\cosh^2 \theta - \sinh^2 \theta = 1$
(You can show this with a bit of algebra if you like, just expand out the terms).

Note that cos and sin are periodic functions with a period of 2 pi and have an absolute value less than 1, while the hyperbolic counterparts are not periodic and go to infinity. As a consequence, if you add Euclidian angles togheter , after $2 \pi$ radians or 360 degrees, you form a complete circle - which you can view as a consequence of the periodicity of the sin and cos functions - or vica versa. However, the hyperbolic functions are not periodic, so you don't have the same concept of adding together angles and forming a complete circle. You can add rapidities together indefinitely, you'll approach infinite rapidity (which is the speed of light) if you do so.

Last edited: Mar 22, 2015
7. Mar 22, 2015

### JD96

I guess I explained it poorly: With the ray I mean a line with arbitrary slope and the integral of the function of that line between 0 and xs (intersection point of the unit hyperbola and the line) is 0,5mxs2. Please correct me if I am misinterpreting your post.

@pervect: I don't know much about complex numbers, however it is interesting to see where those identities come from. Also the last paragraph of your post was helpful and makes a lot of sense to me, since rapidity can be defined as the arclength of the unit hyperbola and the arclength also approaches infinity as the slope of the ct'-axis approaches 1.

8. Mar 22, 2015

### Mentz114

Sorry, you have to draw a picture for me to understand this.

9. Mar 22, 2015

### JD96

Hope this little piece of art clarifies my writing.

#### Attached Files:

• ###### hyperbolic angle.png
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10. Mar 22, 2015

### Mentz114

11. Mar 22, 2015

### JD96

I know about spacetime diagrams, but I intended to write about the mathematical definiton of hyperbolic angles. I wanted to prove that tanh-1(1/m), with m being the slope of a line, (in the physical context the ct'-axis) equals the hyperbolic angle and did that by computing the area (shown in the wikipedia article of hyperbolic functions in the picture on the righthand side) using an integral. It worked out, so you don't need to wrap your head about my last posts :)

Last edited: Mar 22, 2015
12. Mar 22, 2015

### Mentz114

OK, but that diagram is a space-time diagram. I was put off by your labelling of the axes as Y, X instead of t, x.

Anyhow, congratulations.

13. Mar 22, 2015

### JD96

Sorry for the confusion I raised, I should have been more precise when formulating my questions...

14. Mar 22, 2015

### HallsofIvy

If you read that article, it does NOT say that hyperbolic functions "operate on angles like the usual sin cos etc." It says they operate on "hyperbolic angles" that are "area of a hyperbolic sector". Trig, or "circular functions" operate on, really, the subtended arclength on the unit circle.

15. Mar 22, 2015

### wle

To add to what stevendaryl wrote: a result of defining the rapidity this way is that it's additive under successive Lorentz boosts in some given spatial direction.

One of the properties of the hyperbolic tangent function is that $$\tanh(\phi_{1} + \phi_{2}) = \frac{\tanh(\phi_{1}) + \tanh(\phi_{2})}{1 + \tanh(\phi_{1}) \tanh(\phi_{2})} \,.$$ If you identify $\frac{u_{1}}{c} = \tanh(\phi_{1})$, $\frac{u_{2}}{c} = \tanh(\phi_{2})$, and $\frac{v}{c} = \tanh(\phi_{1} + \phi_{2})$, you'll recognise this as a different way of writing the Lorentz velocity addition formula: $$v = \frac{u_{1} + u_{2}}{1 + \frac{u_{1} u_{2}}{c^{2}}} \,.$$ It's in this sense that the rapidity is analogous to rotation angle.

16. Mar 22, 2015

### Mentz114

Ouch. That stings but I guess I earned it. I think the OP realized I am an idiot early on so I doubt any harm was done.

17. Mar 22, 2015

### robphy

I think pervect meant $\cosh^2 \theta −\sinh^2 \theta=1$.

Concerning the "hyperbolic angle" in the Wikipedia article,
If I'm not mistaken, just as "the hyperbolic angle is the area of its [unit]-hyperbolic sector",
"the euclidean angle is the area of its unit-circular sector".
The respective angles are equal to the arc-length [according to their respective geometries] on their respective circles.
That is, the rapidity is the Minkowski arc-length along the unit hyperbola,
and the angle is the Euclidean-arc-length along the unit circle.
So, although they are analgous, the rapidity and angle are different measures.
They are related by the slope: slope=tan(angle)=tanh(rapidity).

(One could actually include the Galilean spacetime in this scheme... using Galilean trigonometry suggested by I.M. Yaglom.)

18. Mar 22, 2015

### HallsofIvy

That wasn't quite what I intended! :)

19. Mar 22, 2015

### pervect

Staff Emeritus
Ooos, yes - fixed

20. Mar 23, 2015

### JD96

@wle Very interesting and also practical to know, thanks! I guess one can also reason that this must hold true by considering the arclength definition of rapidity.
@robphy Is the hyperbolic angle between the x'-axis and the hyperbola -(ct)2+x2=1 also called rapidity, or only the area between the hyperbola -(ct)2+x2=-1 and the ct'-axis? I'm asking that because the arclength along the first hyperbola is imaginary.

Also a general question: If cosh(Φ) is equal to the ct-coordinate of the point of intersection of the unit hyperbola (now I'm talking about the timelike one) with the ct'-axis and cosh(Φ)=γ, shouldn't I be able to easily read off the lorentz factor in a Minkowski diagram?

Last edited: Mar 23, 2015
21. Mar 23, 2015

### Staff: Mentor

You can, but because you had to know $\gamma$ to draw the t' axis in the first place it doesn't tell you anything you didn't already know.

22. Mar 23, 2015

### robphy

Recall that analogous to
we have
"Rapidity"="Minkowski angle between future timelike vectors"="intercepted Minkowski spacelike arc length on a Minkowski-circle "/"timelike radius" (so chosen so that rapidity is real-valued).

With the above timelike vectors,
You can define a real-valued Minkowski-angle between their associated spacelike vectors (that is, the spacelike vectors orthogonal to these timelike vectors) analogously (as long as their tips are on the same branch). [Added in edit: in this construction, these spacelike vectors must be coplanar with the timelike vectors.]

Last edited: Mar 23, 2015
23. Mar 23, 2015

### robphy

The gamma factor (Cosh(theta)) is the projection of the unit timelike vector to the unit hyperbola onto your t-axis. That is, construct a Minkowski right triangle with that unit vector as your hypotenuse and with legs parallel to your time-axis and its corresponding space-axis.
The timelike component is the gamma factor.
You can compare the lengths of that projected component and the unit radius vector along the time-axis.

24. Mar 23, 2015

### JD96

@Nugatory
Not if I draw the unit hyperbola and use the fact, that the intersection of the hyperbola and the ct'-axis marks the event, where one unit of proper time has elapsed for the S'-observer, whose x'-coordinate is zero. In that case I only have to know the slope of the ct'-axis. But I agree that this is more complicated than your way.

@robphy Thanks for the great explanation!