Distance b/w 2 Points on Spacetime Diagrams: Meaning?

[Kairos]

With one spatial dimension $x$, the spacetime interval between two events $A$ and $B$ is

$\Delta s= \sqrt{(ct_{B}-ct_{A})^{2}-(x_{B}-x_{A})^{2}}$

I have a technical question: on the plane of the graph (x,y=ct), the ordinary distance between 2 points $A$ and $B$ is

$L= \sqrt{(y_{B}-y_{A})^{2}+(x_{B}-x_{A})^{2}}$

Does this distance have any particular meaning or not?

 

[PeroK]

$L= \sqrt{(y_{B}-y_{A})^{2}+(x_{B}-x_{A})^{2}}$

Does this distance have any particular meaning or not?

$L = \sqrt{(x_{B}-x_{A})^2+(y_{B}-y_{A})^2+(z_{B}-z_{A})^2 }$ has the usual meaning of spatial distance in flat Euclidean space.

 

[Orodruin]

With one spatial dimension $x$, the spacetime interval between two events $A$ and $B$ is

$\Delta s= \sqrt{(ct_{B}-ct_{A})^{2}-(x_{B}-x_{A})^{2}}$

I have a technical question: on the plane of the graph (x,y=ct), the ordinary distance between 2 points $A$ and $B$ is

$L= \sqrt{(y_{B}-y_{A})^{2}+(x_{B}-x_{A})^{2}}$

Does this distance have any particular meaning or not?

No. At least not if, as you did, you assume y = ct.

 

[PeroK]

Precisely a spacetime diagram is drawn on a flat Euclidean sheet,

Minkowski paper is hard to find.

 

[Kairos]

I understand that this measure has no interest in relativity, thank you

 

[Ibix]

Does this distance have any particular meaning or not?

Just to be clear, you are taking two events in a (1+1)d Minkowski spacetime, plotting them on a 2d Euclidean plane, and asking if the Euclidean distance between the two points on the diagram has any significance in the Minkowski plane? No, except in the special cases that $\Delta t$ or $\Delta x$ is zero, in which case $L^2=\pm\Delta s^2$.

(Edit: wow, a lot of posts appeared when I clicked post on this one!)

 

[robphy]

Minkowski paper is hard to find.

I claim that it's actually not hard to find.

However, one does need to know how to use it.
https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

 

[vanhees71]

With one spatial dimension $x$, the spacetime interval between two events $A$ and $B$ is

$\Delta s= \sqrt{(ct_{B}-ct_{A})^{2}-(x_{B}-x_{A})^{2}}$

I have a technical question: on the plane of the graph (x,y=ct), the ordinary distance between 2 points $A$ and $B$ is

$L= \sqrt{(y_{B}-y_{A})^{2}+(x_{B}-x_{A})^{2}}$

Does this distance have any particular meaning or not?

Not within Minkowski geometry, which is the geometry to describe the special-relativistic spacetime.

 

[vanhees71]

Minkowski paper is hard to find.

True, but today we can use a computer to make it easily ourselves. You have to draw the hyperbolae $(ct)^2-x^2=k$ with $k \in \Z$. Then you can draw arbitrary pairs of lines through $ct=x=0$ symmetric to the light-cone lines $x=\pm ct$ ($k=0$) to define arbitrary inertial reference frames (with the correct unit tics marked by the intersections of these axes with the hyperbolae).

 

[Kairos]

Interesting, thank you. And staying on the ordinary paper representation, wouldn't the mathematical tool of imaginary time or space allow to visualize the spacetime intervals as simple Euclidean intervals? (by converting $\Delta s$ to $L$ in my first post)

 

[topsquark]

Interesting, thank you. And staying on the ordinary paper representation, wouldn't the mathematical tool of imaginary time or space allow to visualize the spacetime intervals as simple Euclidean intervals? (by converting $\Delta s$ to $L$ in my first post)

How are you going to represent the effects of an imaginary time axis on an ordinary set of axes? Better to use the hyperbolic graph paper.

-Dan

 

[Vanadium 50]

allow to visualize the spacetime intervals as simple Euclidean intervals

When was the last time you saw a clock read 11:30i? Or a ruler read 12i inches?

That relative minus sign is a real thing. You can't define/wish it away.

 

[PeterDonis]

wouldn't the mathematical tool of imaginary time or space allow to visualize the spacetime intervals as simple Euclidean intervals?

First, this is a mathematical trick that doesn't change the physics.

Second, this trick only works in flat spacetime. It doesn't work in curved spacetime, so it can't be generalized to GR. That makes its usefulness limited even as a mathematical trick.

 

[robphy]

Interesting, thank you. And staying on the ordinary paper representation, wouldn't the mathematical tool of imaginary time or space allow to visualize the spacetime intervals as simple Euclidean intervals? (by converting $\Delta s$ to $L$ in my first post)

It's not clear what you mean.
Given two points with coordinates (x_A,ct_A) and (x_B,ct_B),
you can plot them on graph paper, regardless of any distance function.
For each pair of points, you can define

• a function \Delta s (as you defined above) which assigns a number to a pair
• another function L (as you defined above) which assigns generally a different number to the same pair
• (you can define yet more functions)

The definitions of such functions doesn't change the appearance on the graph paper.
What such functions do is (for example) select
the set of events with (say) \Delta s =1,
which are different the set of events such that L=1.

Maybe you seek a different representation... which (to me) hasn't been clearly defined.

 

[Kairos]

It's not clear what you mean.
Given two points with coordinates (x_A,ct_A) and (x_B,ct_B),
you can plot them on graph paper, regardless of any distance function.
For each pair of points, you can define

• a function \Delta s (as you defined above) which assigns a number to a pair
• another function L (as you defined above) which assigns generally a different number to the same pair
• (you can define yet more functions)

The definitions of such functions doesn't change the appearance on the graph paper.
What such functions do is (for example) select
the set of events with (say) \Delta s =1,
which are different the set of events such that L=1.

Maybe you seek a different representation... which (to me) hasn't been clearly defined.

Yes, that was my question: if we could use imaginary time as a simple tool to make another representation (not to change the physics!) where $L=1$ directly visible on the new representation, would be obtained with the same pair as $\Delta s =1$ (not directly visible on Minkowski diagram). According to the different answers, I forget this vague and probably wrong idea! Thank you.

 

[Ibix]

According to the different answers, I forget this vague and probably wrong idea!

The problem is that you would need to be able to draw a line on a piece of paper with imaginary length, so that its contribution to the squared Euclidean length is negative. That would be a neat trick if you could do it...

 

[robphy]

$L=1$ directly visible on the new representation...

How is $L=1$ directly visible? by using a ruler you can rotate, or by imagining circles of radius 1?

$\Delta s =1$ (not directly visible on Minkowski diagram)

With a little guidance, imagination, and some practice counting,
one can visualize the squared-interval $\Delta s^2$ on a Minkowski spacetime diagram.

Rather than visualize hyperbolas, one can visualize causal diamonds (parallelograms with lightlike sides, formed from the intersections of lightcones). Thus, these are Lorentz invariant.

• Below is a triplet of timelike segments from event O with $\Delta s^2=\Delta (ct)^2 - \Delta x^2=4$.
  The diamonds have area 4 and these are segments are their timelike diagonals. So, $\Delta s^2=+4$. (The corners opposite to event O do lie on a hyperbola of "radius" 2 "centered" at O.)
  (Instead of a ruler you can rotate,
  imagine a deformable constant-area parallelogram with sides parallel to the gridlines.)
• There is also a triplet of spacelike (dotted) segments with $\Delta s^2= -4$.
  These diamonds have area 4 and these are segments are their spacelike diagonals.
• $\Delta s_{AZ}^2=+36$, $\Delta s_{MN}^2=-36$, $\Delta s_{AM}^2=0$, $\Delta s_{AN}^2=0$, $\Delta s_{MZ}^2=0$, $\Delta s_{NZ}^2=0$
• $\Delta s_{OZ}^2=132$ (count the grid diamonds)
  $\Delta s_{OZ}^2=11.5^2-0.5^2$ (count along the grid diamond diagonals)

details: https://www.physicsforums.com/insights/relativity-rotated-graph-paper/