How Do You Derive the Distance Formula for a Hyperbola?

[Adwit]

TL;DR

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?

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.

 

[Gaussian97]

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.

 

[robphy]

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.

 

[Adwit]

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]

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.

 

[SiennaTheGr8]

To be specific, it's the definition of a rectangular hyperbola (in Cartesian x/y coordinates).

 

[robphy]

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]

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, $a=b=1$.

 

[Herman Trivilino]

Where do I find the derivation of distance formula of hyperbola?

Insofar as it's a matter of definition, it can't be derived.

 

[Adwit]

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?

 

[Adwit]

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 ?

 

[Gaussian97]

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 $x^2-y^2=\text{ct.}$

 

[Adwit]

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

 

[Adwit]

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.

 

[vanhees71]

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]

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 $\sqrt{2}$!

 

[Gaussian97]

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
$$\rho = \sqrt{x^2-y^2}$$

 

[Adwit]

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]

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.

 

[Gaussian97]

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 $y=0$.

 

[Adwit]

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 .

 

[Gaussian97]

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]

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]

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]

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.

 

[robphy]

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.

Note that we use position-vs-time diagrams in Galilean physics.
Note further that position-vs-time diagrams have a non-Euclidean geometry.
However, we have learned how to interpret position-vs-time diagrams (by not haphazardly applying Euclidean geometry to it).
Likewise, we must learn to interpret spacetime diagrams (by not haphazardly applying Euclidean geometry to it).

In my opinion, any "doubt" about whether it is such a good tool to learn SRT
is more about how effectively (or not) the diagram is used and interpreted
rather than the diagram itself.

 

[vanhees71]

Well, my students like Minkowski diagrams very much. So obviously it helps them to understand the Lorentz transform and the kinematical consequences (relativity of simultaneity, time translation, length contraction) better, but it's my experience that you have to emphasize the "non-Euclidicity" of the Minkowski plane very much to avoid confusion.

I know you like the Bondi approach and "rotated graph paper" too. I've not tried this approach in teaching though.

Recently by chance I stumbled over another possibility, socalled Loedel diagrams, where you make the $ct$ and $x'$ axes as well as the $ct'$ and $x$ axes perpendicular (in the Euclidean sense). I'm a bit unsure whether this helps a lot didactics wise, but the funny idea is to write the Lorentz transformation in a kind of mixed notation, i.e., as $c t'=f(ct,x')$ and $x=g(ct,x')$:
$$ct'=\sqrt{1-\beta^2} ct-\beta x', \quad x=\sqrt{1-\beta^2} x' + \beta c t.$$
This means $(ct',x)$ is given by a rotation of $(ct,x')$. As I said, I'm not so sure whether this helps a lot, but some people seem to think so:

https://doi.org/10.1088/0143-0807/34/1/67

 

[robphy]

Well, my students like Minkowski diagrams very much. So obviously it helps them to understand the Lorentz transform and the kinematical consequences (relativity of simultaneity, time translation, length contraction) better, but it's my experience that you have to emphasize the "non-Euclidicity" of the Minkowski plane very much to avoid confusion.

Yes, I agree about the necessity to emphasize the "non-Euclidicity".
And that's why I, as a first step,
often emphasize the "non-Euclidicity" for the usual position-vs-time graph [a Galilean spacetime diagram]
since "non-Euclidicity" is not exclusive to special relativity.

Recently by chance I stumbled over another possibility, socalled Loedel diagrams, where you make the $ct$ and $x'$ axes as well as the $ct'$ and $x$ axes perpendicular (in the Euclidean sense). I'm a bit unsure whether this helps a lot didactics wise,
...[snip]

Loedel diagrams and their use of Euclidean trigonometry are quite limited (and clearly wrong if misused)...
but useful only if the goal is to just illuminate or calculate elementary features of special relativity... and not expect to use this as a stepping stone to more advanced treatments.
(My student and I have been working on an article showing how limited Loedel diagrams are...
and are no match for Minkowski diagrams.)

 

[vanhees71]

I agree. I think the unusual mixing of the time coordinate of one frame and space coordinate of the other frame is confusing more than it helps. It spoils the good practice of writing physical laws in a way covariant under Lorentz (or Poincare) transformations.

Concerning your remark about non-relativistic space-time diagrams you are of course right, but there is not so much danger of students getting confused about this non-Euclidicity, because in Newtonian physics you don't get to the idea to introduce "distances" in this time-space plane. That's due to the fact that in Newtonian mechanics you have absolute space and absolute time (a fiber bundle rather than an affine space to describe spacetime).

 

[Dale]

My student and I have been working on an article showing how limited Loedel diagrams are...
and are no match for Minkowski diagrams

Please post that once it is accepted! I would be quite interested in it.