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Hyperbolic geometry in SR

Physics Asked by DiracSea on December 1, 2020

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?

3 Answers

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.

Answered by John Rennie on December 1, 2020

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

Answered by Ittiandro on December 1, 2020

@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

Answered by robphy on December 1, 2020

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