Measuring the Spacetime Interval with a Ruler

[Trysse]

TL;DR

Is there a way to measure the spacetime interval directly with a ruler in a Minkowski diagram, rather than calculating it algebraically?

I have checked several textbooks but haven't found a method to measure the spacetime interval directly with a ruler in a Minkowski diagram. Various LLMs also couldn’t provide a reference.

All sources I found define the interval algebraically and emphasize that it differs from Euclidean distance, but none describe a direct geometric measurement method.

Does anyone know of a book or paper that presents such a method?"

 

[weirdoguy]

Well, ruler directly measures length in Euclidean space. Spacetime is non-euclidean, so I don't see why one would think it is possible. There has to bo calculation involved at some point.

 

[Ibix]

There's no way to do it directly (except in the special case that the line of interest is parallel to an axis or the trivial null case) because Minkowski space is not Euclidean. A Minkowski diagram is an honest representation of the Minkowski plane, but it is not an exact replica and cannot be.

@robphy has a useful technique using light cone coordinates that I tried to set out in an earlier version of this post, but I went wrong with the maths somewhere! See here and here and other posts by robphy mentioning "clock diamonds" for the correct version.

 

[Ibix]

I worked out what I did wrong with the maths - I just lost a $\sqrt 2$.

Take a piece of graph paper and turn it 45°. Now draw x and t axes in the normal way - they will lie along the diagonals of the graph paper. One unit on these axes should be where they cross the grid corners - i.e. $\sqrt 2$ times the graph paper's pitch.

Add events to the diagram. Say they are separated by $\Delta x$ and $\Delta t$. The Euclidean separation between them along the two directions of the graph paper's grid (in units of its grid squares) will be $|\Delta x\pm c\Delta t|$. The product of those two is $|\Delta x^2-c^2\Delta t^2|$. You need to keep track of the sign separately, but the graph paper grid being null makes that easy.

Note that this construction defines a rectangle along the graph paper grid whose Euclidean area is numerically equal to the interval between its opposite corners. So you can compare intervals by counting grid squares.

 

[Dale]

TL;DR Summary: Is there a way to measure the spacetime interval directly with a ruler in a Minkowski diagram, rather than calculating it algebraically?

I have checked several textbooks but haven't found a method to measure the spacetime interval directly with a ruler in a Minkowski diagram.

You won’t be able to do that with an ordinary ruler. But you could build a ruler to do so.

Take a clear sheet of plexiglass or similar. On it etch lines at 45 degrees, and a series of hyperbolas that asymptote to those lines. Etch a horizontal and vertical line through the intersection of the 45 degree lines and label each hyperbola with the distance from the intersection at which the hyperbola intersects the vertical or horizontal line

 

[Ibix]

Take a clear sheet of plexiglass or similar. On it etch lines at 45 degrees, and a series of hyperbolas that asymptote to those lines. Etch a horizontal and vertical line through the intersection of the 45 degree lines and label each hyperbola with the distance from the intersection at which the hyperbola intersects the vertical or horizontal line

Printing on acetate might be cheaper, as long as you make sure your printer is OK with it. Otherwise it could cost you a new printer.

You may also want a regular square grid to facilitate alignment with the diagram's axes.

 

[robphy]

Here are slides from a presentation in 2018. (That contact info is not current.)
https://www.aapt.org/docdirectory/m...dGraphPaper-CalculatingWithCausalDiamonds.pdf

In 1+1 dimensions, under a boost (Minkowski or Galilean) [or a rotation in the Euclidean plane], the area is invariant.

 

[DrGreg]

You won’t be able to do that with an ordinary ruler. But you could build a ruler to do so.

Take a clear sheet of plexiglass or similar. On it etch lines at 45 degrees, and a series of hyperbolas that asymptote to those lines. Etch a horizontal and vertical line through the intersection of the 45 degree lines and label each hyperbola with the distance from the intersection at which the hyperbola intersects the vertical or horizontal line

Printing on acetate might be cheaper, as long as you make sure your printer is OK with it. Otherwise it could cost you a new printer.

You may also want a regular square grid to facilitate alignment with the diagram's axes.

This is what the acetate would look like.

• The solid coloured lines measure distance. (E.g. in light-seconds.)
• The dashed coloured lines measure time. (E.g. in seconds.)
• The dotted black lines measure an interval of zero.

Due to symmetry, and the fact that the interval between A and B is the same as the interval between B and A, I've drawn lines to cover only half of the diagram. That would be sufficient in practice. To measure the interval between two events, align one event with the origin at the centre of the acetate, align the axes by rotating the acetate if necessary, and then see which curve the other event lies on (or if it's between curves, estimate the value).

For comparison, the equivalent acetate for Euclidean geometry would look like this:

 

[Dale]

align the axes by rotating the acetate if necessary

Specifically, align the 45 degree asymptotes on the diagram. It is not essential that the horizontal and vertical axes be aligned

 

[robphy]

This summarizes the calculation using causal-diamonds.

So, the clock-effect/twin-paradox diagram looks like

and an elastic-collision looks like

...all from

Here are slides from a presentation in 2018. (That contact info is not current.)
https://www.aapt.org/docdirectory/m...dGraphPaper-CalculatingWithCausalDiamonds.pdf

In 1+1 dimensions, under a boost (Minkowski or Galilean) [or a rotation in the Euclidean plane], the area is invariant.

One can check that (for example)
the square-interval is

• the black causal diamond has width $u=36$ and height $v=4$ [in terms of the diamonds in this grid... "in this frame"] from counting boxes in the grid.
• Since the area is $uv=144$ [the squared-interval],
  the timelike-diagonal of that diamond has size $\sqrt{uv}=\sqrt{144}=12$ [an invariant].
• The rectangular-coordinates in this frame are
  $\Delta t=\frac{(u+v)}{2}=\frac{(36+4)}{2}=20$
  $\Delta x=\frac{(u-v)}{2}=\frac{(36-4)}{2}=16$
• So, the square-interval is $\Delta t^2 - \Delta x^2= (20)^2-(16)^2=(12)^2$.
  [Indeed, $\Delta t^2 - \Delta x^2= \frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}=uv=(\mbox{diamond area})$]
• [for other frames and their grids,
  we would have $u_A v_A=\mbox{constant}$, which is the equation of a hyperbola.]

[in this frame] the velocity and the Doppler-factor for that diagonal worldline is

• $\beta=\frac{\Delta x}{\Delta t}=\frac{16}{20}=\frac{4}{5}$
• $k=\sqrt{\frac{1+\beta}{1-\beta}}=\sqrt{\frac{1+(4/5)}{1-(4/5)}}=3$
  which can all be expressed in terms of $u$ and $v$:
  $\beta=\frac{\Delta x}{\Delta t}=\frac{(u-v)}{(u+v)}=\frac{36-4}{36+4}=\frac{32}{40}=\frac{4}{5}$
  $k=\sqrt{\frac{1+\beta}{1-\beta}}=\sqrt{\frac{1+\frac{(u-v)}{(u+v)}}{1-\frac{(u-v)}{(u+v)}}}=\sqrt{\frac{u}{v}}=\sqrt{\frac{36}{4}}=3$

[Try this method for another highlighted causal diamond above.]

---

So, the bottom line is that
one can use a rotated-graph paper grid to find by counting boxes in the grid
the size ($\Delta s^2=uv$ =area) and shape ($k^2=\frac{u}{v}$ =aspect ratio)
of the causal diamond for any segment in spacetime.

The rotated-graph paper is essentially a ruler (for areas) and a protractor (for aspect-ratios) for Minkowski spacetime geometry.... by counting boxes.

 

[Trysse]

Thank you for your responses.


The method using the hyperbola is also what at least two LLMs suggested.


Given my original question and the discussion so far, this might sound like a naive question, but I’ll ask it anyway—sincerely:


How sure are you that it is impossible to measure the spacetime interval between two events in a Minkowski diagram using only a ruler?


If you are sure, why is that the case? The obvious response might be “because spacetime is non-Euclidean”, but that’s not quite what I’m asking.


What I mean is: Have you ever seriously tried to find a way to do it? Have you considered whether there might be a geometric approach—perhaps one not commonly discussed?

 

[Ibix]

How sure are you that it is impossible to measure the spacetime interval between two events in a Minkowski diagram using only a ruler?

...you mean something like robphy's method? You can even construct the grid you need with a compass and straightedge if you want to really old-school it.

 

[PeterDonis]

How sure are you that it is impossible to measure the spacetime interval between two events in a Minkowski diagram using only a ruler?

Because, as has already been said, the geometry of spacetime is not Euclidean, and a ruler by itself is an instrument that only works for Euclidean geometry.

Have you ever seriously tried to find a way to do it?

There's no point given that it's proven to be impossible by the argument above.

 

[Trysse]

Imagine an event $O$:

Next, construct a future light cone for $O$:

Then construct some plane that intersects the light cone of $O$ in a circle $o$:

Now select two points that lie in the plane. One inside the light cone ($E_T$) and one outside ($E_S$). Both points represent events for which I want to determine the spacetime interval $s^2$ to the event $O$:

Next, remove the cone for better clarity. You get a circle $o$ and two coplanar points $E_S$ and $E_T$:

Rotate the view to look from above and see the circle undistorted. Remove the blue tinted plane:

Jakob Steiner's Power of a Point theorem describes the relation between the points and the circle.
https://en.wikipedia.org/wiki/Power_of_a_point
I leave it as an exercise for the reader to think about the similarities between the spacetime interval and the power of a point.

So technically we cannot measure the spacetime interval with a ruler, in the same way, we cannot measure the power $\Pi$ of $E_S$ or $E_T$. However, we can measure the squareroot i.e. $\sqrt{\Pi(E_S)}$ respectively $\sqrt{\Pi(E_T)}$.

For $E_T$ this is $1/2$ of the shortest cord through $E_T$:

For $E_S$ this is the tangent distance to $o$.

or in the spacetime diagram:

So if you want to be pedantic you must draw a square to represent $s^2$, which you technically cannot measure with a ruler. But I think this is not the point of this exercise.

I apologize for the trick question. I only wanted to make sure I do not tell you something you already knew.

 

[Ibix]

So, to be clear, you're going to get $\Delta x$ and $\Delta t$ between two events. Then you're going to draw a circle of radius $\Delta t$ and add a point a distance $\Delta x$ away from the circle center. Then you're going to draw a chord or a tangent through that point and measure its length, which will be $\sqrt{|\Delta s^2|}$.

Ok. That's not using a Minkowski diagram, which was part of your specification in the OP, but it will work. It seems overcomplicated to me - you need to draw a diagram for each pair of events of interest and you need to know $\Delta x$ and $\Delta t$, at which point I'd have thought numerical computation (even if you have to use Newton-Rapheson on paper for a square root) would be quicker and more precise.

 

[Trysse]

I don't draw the circle. The circle is constructed from intersecting the cone and the plane.
All you see above is the same GeoGebra model.

In some pictures I have only removed some elemnts. The 2d pictures are from the 2d view of GeoGebra. So it is always the same circle.

Overall it is not about practicability. It is about seeing the geometric relations.

And as the last picture shows, you can see and measure the distance in the Minkowski diagram.

It is rather the other way around: The minkowski diagram is not necessary. I could measure the same distance on a sphere and a point.

 

[Ibix]

It is about seeing the geometric relations.

Ok. I suspect adding a third dimension to a two dimensional problem is unlikely to generate insight, but it's your time.

 

[robphy]

I don't draw the circle. The circle is constructed from intersecting the cone and the plane.
All you see above is the same GeoGebra model.

In some pictures I have only removed some elemnts. The 2d pictures are from the 2d view of GeoGebra. So it is always the same circle.

Overall it is not about practicability. It is about seeing the geometric relations.

And as the last picture shows, you can see and measure the distance in the Minkowski diagram.

It is rather the other way around: The minkowski diagram is not necessary. I could measure the same distance on a sphere and a point.

GeoGebra is a good tool to test your construction and calculation.

• Given a pair of events on a Minkowski spacetime diagram,
  do your construction and report the relevant "ruler" measurements.
  Then, calculate the square-interval from those "ruler" measurements.
• Perform a boost and do the construction on the boosted pair of events.
  Hopefully you get the same square-interval.
  In the process, you might be able to see how your "ruler" measurements transform.
  

(I used GeoGebra to study my clock-diamond constructions
and create interactive visualizations to give physical and mathematical intuition about what is going on.
https://www.geogebra.org/u/robphy
You might like Circular Light Clocks (https://www.geogebra.org/m/pr63mk3j),
Calibrating Bob's Light Clock (https://www.geogebra.org/m/kvfsq664), and
Relativity on Rotated Graph Paper (https://www.geogebra.org/m/HYD7hB9v).)

 

[Renato Iraldi]

TL;DR Summary: Is there a way to measure the spacetime interval directly with a ruler in a Minkowski diagram, rather than calculating it algebraically?

I have checked several textbooks but haven't found a method to measure the spacetime interval directly with a ruler in a Minkowski diagram. Various LLMs also couldn’t provide a reference.

All sources I found define the interval algebraically and emphasize that it differs from Euclidean distance, but none describe a direct geometric measurement method.

Does anyone know of a book or paper that presents such a method?"

there is a way to measure the results of a lorentz transformation with a ruler but in Epstein diagrams.
see https://qr.ae/pAhiVq

 

[Renato Iraldi]

Because, as has already been said, the geometry of spacetime is not Euclidean, and a ruler by itself is an instrument that only works for Euclidean geometry.


There's no point given that it's proven to be impossible by the argument above.

The geometry of spacetime is coherent with a Euclidean geometry, this is proven by the fact that a lorentz transformation may be made using euclidean geometry
and measuring the result with a ruler. see https://qr.ae/pAhiVq

 

[Ibix]

there is a way to measure the results of a lorentz transformation with a ruler but in Epstein diagrams.
see https://qr.ae/pAhiVq

You have to be very careful measuring distances on a space-proper-time diagram, because apart from along the worldlines you don't actually have distances. For example, if two of your worldlines cross this does not mean that the objects ever had zero distance between them.

I must say I've never seen the utility of space-proper-time diagrams.

 

[PeterDonis]

Thread closed for moderation.