# Hyperbolic Geometry

• I

## Summary:

This problem is from the book "The Geometry Of Special Relativity" written by Tevian Dray. How is the distance function formula come from? Does it have any proof?

## Main Question or Discussion Point

Can anyone derive the distance formula of a hyperbola for me, please? I have not found the derivation on the internet. I can't get any clue from the picture of hyperbola. Last edited:

Related Special and General Relativity News on Phys.org
How do you define a hyperbola?
Because if you define it like the curves with $x^2-y^2=\text{const}$ or by the curves with parametrization $x=\cosh\theta, \quad y=\sinh\theta$, then is very easy to prove.

Last edited:
• Klystron and vanhees71
robphy
Homework Helper
Gold Member
The title should really be "Hyperbolic Trigonometry" ("Hyperbola Geometry" is okay. But "hyperbolic geometry" [a curved space that violates the Parallel postulate] is not a good name) ,
where the hyperbola plays the role of the circle as the curve of constant separation from a point. In principle, the hyperbola can thought of as the experimental result of where "1 tick" occurs for all wristwatches carried by inertial observers that experienced a common event (call it O) on a spacetime diagram.

Under a Lorentz transformation, all events on that hyperbola are transformed along that hyperbola (just as a rotation transforms points on a circle to other points on that circle). A "degenerate hyperbola with radius 0" is along the lightcone, which is associated with the speed of light.

UPDATE: Assuming the future-timelike unit hyperbola (representing "1 tick"), use the spacelike-arclength on that unit hyperbola to define the Minkowski-angle (the rapidity) $\theta$. This Minkowski-angle is not the same as the Euclidean-angle on a given diagram... but they are related by $\tanh\theta=\tan\theta_{euc}$ ... but only for this diagram.

Last edited:
• Orodruin
Don't bring special relativity or any physics ( ct or Lorentz transformation ) in here. I want to see the problem from the mathematical perspective, not physics perspective. Where do I find the derivation of distance formula of hyperbola? Forget physics. How do I derive it mathematically?

Orodruin
Staff Emeritus
Homework Helper
Gold Member
Don't bring special relativity or any physics in here. I want to see the problem from the mathematical perspective, not physics perspective. Where do I find the derivation of distance formula of hyperbola? Forget physics. How do I derive it mathematically?
You have it backwards. Just as $x^2 + y^2 = R^2$ is the definition of a circle, $x^2 -y^2 = \alpha$ is the definition of a hyperbola.

• Motore
• Klystron and Orodruin
robphy
Homework Helper
Gold Member
To be specific, it's the definition of a rectangular hyperbola (in Cartesian x/y coordinates).
True, but I think it's fair to say that this aspect is a convenient "coordinate choice", not a physical one.
There's no physical reason why the lightcone opening angle has to look like 90-degrees on a diagram.
(Note that the lightlike-directions (say future-frontward and future-rearward) are not Minkowski-orthogonal.)

vanhees71
Gold Member
2019 Award
Don't bring special relativity or any physics ( ct or Lorentz transformation ) in here. I want to see the problem from the mathematical perspective, not physics perspective. Where do I find the derivation of distance formula of hyperbola? Forget physics. How do I derive it mathematically?
The question is, from which assumptions you want to derive the equation
$$x^2-y^2=\text{const}$$
from.

One way is to use the geometrical definition of an hyperbola as the set of points, for which the magnitude of the difference from two points, the focal points $F_1$ and $F_2$, is constant, i.e., for any point $P$ on the hyperbola
$$||F_1 P|-|F_2 P||=2a=\text{const}.$$
Now let $(x,y)$ the coordinates of the points $P$ on the hyperbola and make $F_1$ and $F_2$ be located at $(\pm e,0)$. Then the equation reads
$$\sqrt{(x+e)^2+y^2}-\sqrt{(x-e)^2+y^2}=\pm 2 a.$$
After some simple but a bit elaborate algebra you get the equation
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$$
where $b=\sqrt{e^2-a^2}$.

Your example is the special case, $a=b=1$.

• Klystron
Mister T
Gold Member
Where do I find the derivation of distance formula of hyperbola?
Insofar as it's a matter of definition, it can't be derived.

How do you define a hyperbola?
Because if you define it like the curves with $x^2-y^2=\text{const}$ or by the curves with parametrization $x=\cosh\theta, \quad y=\sinh\theta$, then is very easy to prove.
I do the math. Constant becomes 1. But the screenshot of my question shows that constant is some kind of squared distance ρ ^2. Can you show me where is this distance in the hyperbola picture?

vanhees71, I know your math. I just want to know where is ρ in the picture of the hyperbola ? In the case of circle geometry, we see distance r. Where is the distance ρ? Is it from the origin point to point B ?

Last edited:
I do the math. Constant becomes 1. But the screenshot of my question shows that constant is some kind of squared distance. Can you show me where is this distance in the hyperbola picture?
Sorry, there was no math to be done, when I wrote the parametrization I should have written $x=a \cosh \theta$ and $y=a\sinh \theta$.
The fundamental points in all this are:
What do we call distance?
In Euclidean geometry, we say that two points that belong to the same circle (centred at the origin) are at the same distance (from the origin). The equation for a circle is simply $x^2+y^2=\text{ct.}$ (here and elsewhere in my post $\text{ct.}$ means "constant" nothing to do with time and speed of light) then we can define the distance as the square root of this constant, so for example, any point that belongs to the circle $x^2+y^2=2$ will be at a distance $\sqrt{2}$.

In Hyperbolic geometry, we say that two points are at the same distance is they belong to the same hyperbola, the mathematical equation for a hyperbola is just $x^2-y^2 = \text{ct.}$, and as in EG we define this constant to be de distance.

What is your definition of a hyperbola?

Of course all this depends on $x^2-y^2=\text{ct.}$ being your definition of hyperbola, so for us to help you further we need to know what is your precise definition of a hyperbola, whatever it is, there is a way to prove that is equivalent to $x^2-y^2=\text{ct.}$

Last edited:
Gaussian97, Of course, there was math to be done. This is the math.

#### Attachments

• 45.4 KB Views: 11
Gaussian97 , In the case of circle geometry, we see distance r. Where is the distance ρ in hyperbola picture? Is it from the origin point to point B ? Now I hope you understand my question.

#### Attachments

• 44.8 KB Views: 9
vanhees71
Gold Member
2019 Award
vanhees71, I know your math. I just want to know where is ρ in the picture of the hyperbola ? In the case of circle geometry, we see distance r. Where is the distance ρ? Is it from the origin point to point B ?
Yes, because in the primed basis the equation for the hyperbola reads the same as for the unprimed (it's form-invariant, which is the very point of Lorentz transformations):
$$x^{\prime 2}-y^{\prime 2}=\rho^2,$$
which implies that $y'=0$ leads to $x'=\rho$ (for this branch of the hyperbola at the right half-plane as drawn in your figure).

This also shows you the important point that, when reading a Minkowski diagram, you must forget about Euclidean length you are used to from elementary school on. That's why I always have been a bit in doubt whether Minkowski diagrams are such a good tool to learn SRT.

vanhees71
Gold Member
2019 Award
Sorry, there was no math to be done, when I wrote the parametrization I should have written $x=a \cosh \theta$ and $y=a\sinh \theta$.
The fundamental points in all this are:
What do we call distance?
In Euclidean geometry, we say that two points that belong to the same circle (centred at the origin) are at the same distance (from the origin). The equation for a circle is simply $x^2+y^2=\text{ct.}$ (here and elsewhere in my post $\text{ct.}$ means "constant" nothing to do with time and speed of light) then we can define the distance as this constant, so for example, any point that belongs to the circle $x^2+y^2=2$ will be at a distance 2.

In Hyperbolic geometry, we say that two points are at the same distance is they belong to the same hyperbola, the mathematical equation for a hyperbola is just $x^2-y^2 = \text{ct.}$, and as in EG we define this constant to be de distance.

What is your definition of a hyperbola?

Of course all this depends on $x^2-y^2=\text{ct.}$ being your definition of hyperbola, so for us to help you further we need to know what is your precise definition of a hyperbola, whatever it is, there is a way to prove that is equivalent to $x^2-y^2=\text{ct.}$
It's of course $x^2-y^2=(c t)^2$ and the circle $x^2+y^2=2$ has radius $\sqrt{2}$!

the circle $x^2+y^2=2$ has radius $\sqrt{2}$!
Sure! I have edited my post, sorry!

@Adwit What do you mean that we "see" distance $r$? Distance is not a thing you can see, what you see are two points (usually connected by a line) and you can write an $r$ to express that they are at a distance $r$.

Then, following the same reasoning, you can join the two points (the origin and the point B) by a line and write a $\rho$ next to it to tell that these two points are at a distance $\rho$ from each other. But in the same way, you can draw a line to the point A and write $\rho'$ next to it. In general, you can draw a line to any point with cartesian coordinates $(x,y)$, and write a $\rho$ next to it to express that this point is at a distance $\rho$ from the origin, and then, you can compute the numerical value of $\rho$ using
$$\rho = \sqrt{x^2-y^2}$$

Sure! I have edited my post, sorry!

@Adwit What do you mean that we "see" distance $r$? Distance is not a thing you can see, what you see are two points (usually connected by a line) and you can write an $r$ to express that they are at a distance $r$.
Of course, distance is a thing you can see. When I draw a line which connects two points, that line is the distance either it is in hyperbola or circle. That is what I mean by saying " we see distance". Understand?
That pictured line is the distance. Everybody in the world thinks that the connector line is distance. Why don't you think that way? You can see distance. Distance is not only something to measure, but also something to see, something to feel.

PeterDonis
Mentor
2019 Award
Can you show me where is this distance in the hyperbola picture?
It's the distance along the $x$ axis from the origin to the hyperbola--in other words, the distance of "closest approach" of the hyperbola to the origin.

• Klystron and Dale
Where is the distance ρ in hyperbola picture? Is it from the origin point to point B ? Now I hope you understand my question.
Then I don't understand your question, if $\rho$ is the distance to point B, obviously you see it in your picture as the line connecting the origin with point B. But if $\rho$ is the distance to point A, obviously you see it in your picture as the line connecting the origin with point A.

Maybe you are asking at which point in the parabola does the "hyperbolic distance" equal the "Euclidean distance" then, of course, it's in the only point in your parabola for which $y=0$.

In the picture, ρ is not written. I didn't know either ρ is the connector line from the origin to point B or A or it is any other line ? I didn't know which line is ρ . I wanted to know which line indicates ρ .

In your original post, the image says:
Measure the "squared distance" of the point B with coordinates $(x,y)$ from the origin by using the definition $\rho^2 = x^2 - y^2$

So, in this case, $\rho$ refers to point B and therefore you should draw it as the line connecting the origin with point B.

PeterDonis
Mentor
2019 Award
I wanted to know which line indicates ρ .
See my post #19.

in this case, $\rho$ refers to point B
No, it doesn't, not the way the OP is asking. In terms of actual Euclidean distance on the diagram, $\rho$ is the distance I described in post #19. Point B is not that Euclidean distance from the origin.

PeterDonis
Mentor
2019 Award
In the picture, ρ is not written.
The definition of "distance" for hyperbolic geometry, as the image you posted said, is not Euclidean distance. In terms of hyperbolic "distance", every point on the hyperbola is the same "distance" $\rho$ from the origin. But this notion of "distance" is not Euclidean distance. In terms of Euclidean distance, as I said in post #19, $\rho$ is the "distance of closest approach" of the hyperbola to the origin. As the hyperbola is drawn in the diagram, that means the distance along the $x$ or $y$ axis from the origin to the hyperbola.

vanhees71
Gold Member
2019 Award
Then I don't understand your question, if $\rho$ is the distance to point B, obviously you see it in your picture as the line connecting the origin with point B. But if $\rho$ is the distance to point A, obviously you see it in your picture as the line connecting the origin with point A.

Maybe you are asking at which point in the parabola does the "hyperbolic distance" equal the "Euclidean distance" then, of course, it's in the only point in your parabola for which $y=0$.
That's my point! You have to forget about the Euclidean idea of distances as soon as you draw a minkowski diagram. The hyperbolae drawn in #1 are the locations of points of "constant distance" in the sense of the Minkowski pseudo metric.

The hyperbolae
$$(c t)^2-x^2=A=\text{const}$$
can however be time-like, i.e., $A>0$ and space-like $A<0$. For $A=0$ they degenerate to two straight lines through the origin $x=\pm c t$, the light-cone.

The Lorentz transformations in this Minkowski plane are those linear mappings that keep this bilinear form form-invariant, i.e.,
$$\begin{pmatrix} c t' \\ x' \end{pmatrix} = \begin{pmatrix} \cosh \alpha & -\sinh \alpha \\ -\sinh \alpha & \cosh \alpha \end{pmatrix} \begin{pmatrix} c t \\ x \end{pmatrix}$$
you have for all vectors $(c t,x)^{\text{T}}$
$$(c t')^2-x^{\prime 2}=c^2 t^2-x^2.$$
The hyperbolae have the same form wrt. both coordinates and thus indeed in the Minkowsi diagram drawn in #1 $OB$ has the same "length" (in the sense of the Minkowski plane) as the origin to the intersection of the hyperbola with the $x$ axis, i.e., the Euclidean distances you are "seeing" in the drawing are misleading, because of course if you read the diagram as in the Euclidean plane obviously the two distances are different, but what counts in the Minkowski diagram is the "distance" in the sense of the Minkowski bilinear form and not that of the Euclidean bilinear form you are used to from Euclidean geometry.