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I'm having trouble understanding just what the hyperbolas on a space-time diagram actually signify. From what I understand, these hyperbolas trace out the equation $S^2=x^2-(ct)^2$, which is the equation for a hyperbola. Fine so far. Furthermore, the plot is supposed to represent the value of $S$, which is the invariant interval.

When I think of a spacetime event, though, I think of a line that begins at the origin and that moves out the upper right of the diagram, in between the $ct$-axis and the 45% Null line of the speed of light. I can see how different frames of reference could represent different coordinate shifts of the Spacetime diagram which would yield different values of $x$ and $ct$, keeping the value of $S$ invariant. What I am struggling with is understanding what the hyperbolic shape of the plot of $S$ on the spacetime means. What is it saying? How do I read it? Assuming we are only dealing with time-like events, what do different values of $S$ signify? Is there only that single hyperbola that the value of $S$ falls along for a given spacetime event, or is that curve part of a larger family of curves?

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  • $\begingroup$ Although special relativity involves the geometry and the trigonometry of the hyperbola (as the flat noneuclidean Minkowskian geometry), it does not directly involve “hyperbolic geometry” (the curved lobachevskian geometry). $\endgroup$
    – robphy
    Commented Sep 27, 2020 at 4:40

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If you draw a spacetime diagram, and mark some point $P = (t, x)$ on it, then the hyperbola that passes through $P$ is the set of all possible points that $P$ can be Lorentz transformed to. In other words, for any other inertial frame, $S'$, with any relative velocity $v$ the Lorentz transformation of $P$ will lie on the hyperbola through $P$.

Spacetime diagram

For example take the point $P$, which has a timelike interval from the origin. Any Lorentz transformation can only move $P$ along the hyperbola through $P$. So we can immediately see that $t'_P \ge 0$ i.e. we can't have a failure of causality. By contrast take the point $Q$ that has a spacelike interval from the origin. A quick glance is enough to tell us that $t'_Q$ can be greater than or less than zero.

The point of diagrams like this is to aid understanding in circumstances where our intuition will frequently let us down. Because a Lorentz transformation can only move points along the hyperbola through them it gives us a quick visual guide as to what the effect of the transformation will be.

Response to comment:

If $\Delta s$ is the interval from $O$ to $P$, then $\Delta s$ is also the interval from $O$ to $P'$ and indeed from $O$ to any point on the hyperbola. This is because we've defined the hyperbola as the points satifying:

$$ s^2 = x^2 - (ct)^2 $$

for constant $s$. The point $O$, $(0, 0)$, doesn't move under a Lorentz transformation i.e. $O' = O = (0, 0)$, so that means the interval $O \rightarrow P$ is the same as $O' \rightarrow P'$ for all possible Lorentz transformations.

The physical significance of this is that the interval $\Delta s$ is an invarient under Lorentz transformations, which of course we already knew. This is indeed a fundamental statement about the geometry of spacetime. Euclidean space is defined by Pythagoras' theorem:

$$ ds^2 = dx^2 + dy^2 + dz^2 $$

i.e. the metric is $(+++)$. Minkowski spacetime is defined by:

$$ ds^2 = -c^2dt^2 + dx^2 + dy^2 + dz^2 $$

i.e. the metric is $(-+++)$ (or $(+---)$ if you prefer). This is telling us that the geometry of Minkowski spacetime is fundamentally different to Euclidean space.

All the effects seen in SR can be derived simply from this statement of the metric, so it is the single most fundamental thing you need to understand about SR.

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  • $\begingroup$ Great, thanks for the visual aid, John. I think I get it now. Each of the Pprimes that lie along the hyperbola represent different frames of reference traveling at different velocities relative to the spatial locale of the event (proper time frame). The P values asymptote at the Null line and can travel no faster. Fascinating Spock. How interesting that these Lorentz transformations would trace out exactly the shape of a hyperbola, complete with the asypmtote at c. That must mean something deep. Does that reflect the intrinsic geometry of space? Or perhaps "space-time" rather? $\endgroup$
    – DiracSea
    Commented Jun 5, 2014 at 9:22
  • $\begingroup$ @DiracSea: I've edited my answer to respond to your comment $\endgroup$
    – John Rennie
    Commented Jun 5, 2014 at 9:36
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    $\begingroup$ @DiracSea: Extra 2D geometric analogy: Euclidean rotation is $e^{i\theta}(x+iy)$ and Lorentz boost is $e^{j\eta}(t+jx)$, where $i^2 = -1$ and $j^2 = +1$. In particular: $$\begin{bmatrix}t'\\x'\end{bmatrix}= \begin{bmatrix}\cosh\eta &-\sinh\eta \\-\sinh\eta & \cosh\eta\\\end{bmatrix} \begin{bmatrix}t\\ x\end{bmatrix}$$ is an obvious analogue of Euclidean rotation, with velocity $v = dx/dt = \tanh\eta$ taking the role of slope $m = dy/dx = \tan\theta$. For comparison, Galilean boost would be $e^{\epsilon v}(t+\epsilon x)$, $\epsilon^2 = 0$. $\endgroup$
    – Stan Liou
    Commented Jun 5, 2014 at 10:10
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    $\begingroup$ @DiracSea as an aside, Stan Liou is referring to the Algebra of Split Complex Numbers $\{x+j\,y|\,x,\,y\in\mathbb{R};\,j^2=1\}$, which is not a field like $\mathbb{R}$ or $\mathbb{C}$ because there are nonzero divisors of nought. These latter correspond to the null (lightlike) vectors in STR. $\endgroup$
    – Selene Routley
    Commented Jun 5, 2014 at 10:25
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@Ittiandro , one can actually see the invariant interval in the hyperbola.

It can be interpreted as the invariant area of a causal diamond, with one vertex on the origin and the other vertex on the hyperbola, with sides parallel to the light-cone (the asymptotes of the hyperbola).

Here's screenshot from a GeoGebra visualization I used at conference presentation https://www.geogebra.org/m/HYD7hB9v#material/qAt9EQgd

Invariant area under hyperbola

Since the Lorentz boost has unit determinant, the area of the diamond is unchanged by a boost. The edges of the diamond are formed by the eigenvectors of the boost, and the eigenvalues are the Doppler factor and its reciprocal (whose product is equal to the determinant).

From an article I contributed
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
By counting the boxes, you can verify that all of the "diamonds" have the same area. robphy-hyperbola-area

I have exploited this fact to develop a method of doing graphical calculations in special relativity:

Relativity on rotated graph paper
American Journal of Physics 84, 344 (2016); https://doi.org/10.1119/1.4943251

We demonstrate a method for constructing spacetime diagrams for special relativity on graph paper that has been rotated by 45°. The diagonal grid lines represent light-flash worldlines in Minkowski spacetime, and the boxes in the grid (called “clock diamonds”) represent units of measurement corresponding to the ticks of an inertial observer's light clock. We show that many quantitative results can be read off a spacetime diagram simply by counting boxes, with very little algebra. In particular, we show that the squared interval between two events is equal to the signed area of the parallelogram on the grid (called the “causal diamond”) with opposite vertices corresponding to those events. We use the Doppler effect—without explicit use of the Doppler formula—to motivate the method.

Here's a screenshot from another GeoGebra visualization in that collection:
https://www.geogebra.org/m/HYD7hB9v#material/VrQgQq9R robphy-RRGP-examples

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Thanks for referring me to Math.Jax. Unfortunately, I don t have a clue how to use it, but perhaps it is not necessary, as my question was very simple and, as I later found, A.Wheeler already answered it with his usual clarity in his textbook ( An introduction to Spacetime physics).

THe reason why the Hyperbola diagram does not allow to see the invariant interval , but shows a number of varying lengths from the vertex O to the hyperbola curve, is simply because the hyperbola diagram is based on the Euclidean geometry, which is inadequate to represent the the four-dimensions non-euclidean spacetime curved geometry. He says that the Hyperbola diagram is , in his words, a LIE !

Thanks anyway

Franco

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  • $\begingroup$ "Because the paper picture of spacetime is a lie! The length of an arrow on a piece of paper is Euclidean, related to the sum of squares of the space separations of the endpoints in two perpendicular directions. Euclidean geometry works fine if what is being represented is flat space, for example the map of a township. But Euclidean geometry is the wrong geometry and betrays us when we try to lay out time along one direction on the page. Instead we need to use Lorentz geometry of spacetime." (Taylor and Wheeler, Spacetime Physics 2ed, p 143.) $\endgroup$
    – robphy
    Commented Sep 29, 2020 at 3:11
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I'd just like to justify John Rennie's claim that the hyperbola is the set of points where the point can boosted to:

In an $x-y$ plane, a Lorentz boost is defined as:

$$ x' = \gamma(x-vy)$$

$$ y' = \gamma (y-v \frac{x}{c^2})$$

With the inverse boosts as:

$$ x = \gamma(x'+vy')$$

$$ y= \gamma(y'+v \frac{x'}{c^2} )$$

With $\gamma = \frac{1}{\sqrt{1- \frac{v^2}{c^2}}}$, and $c$ is a constant (speed of light in relativity).

Dividing the two:

$$ \frac{x'}{x} = \frac{x - v y}{x' + v y'}$$

$$ \frac{y'}{y} = \frac{y - v \frac{x}{c^2}}{ y' + v \frac{x'}{c^2}}$$

Isolate for $v$ in each individual equation above and equate, and we have:

$$ y'^2 - \frac{x'^2}{c^2} = y^2 - \frac{x^2}{c^2}$$

So, suppose we set an initial point $(x,y)$ , then the $(y',x')$ pairs solving would give a hyperbola.

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  • $\begingroup$ It would be more illuminating by rewriting as $(x'/c) = \gamma((x/c)-(v/c)y)$ and $y = \gamma(y-(v/c)(x/c))$. Add the two and subtract the two to obtain a set of decoupled equations: $((x'/c)+y')=\gamma(1-(v/c))((x/c)+y)$ and $((x'/c)-y')=\gamma(1+(v/c))((x/c)+y)$. The product of these two equations yields the equation of the hyperbola. The scaling factors in the decoupled equations are the eigenvalues of the boost, whose product equals 1 since the determinant equals 1. The larger scaling factor (larger eigenvalue) is the Doppler factor. $\endgroup$
    – robphy
    Commented Jun 8, 2022 at 18:19

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