Imaginary Pythagoras

[DaveC426913]

TL;DR

Is this geometry valid?

I posted this in the Lame Math thread, but it's got me thinking.

Is there any validity to this? Or is it really just a mathematical trick?

Naively, I see that i² + plus 1² does equal zero².

But does this have a meaning?

I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero?

Ibix offered a rendering of the diagram using what I assume is matrix* notation:

which I assume makes the apparent paradox go away, but does that mean the first diagram is not valid?

* never learned matrices

I suppose it is theoretically possible to have a triangle with zero length hypotenuse if you look at it edge-on in an abstract 3D space - i.e. the two axes are superimposed.

 

[Mark44]

I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero?

A triangle in the complex plane would have the lengths of all of its sides as real numbers; i.e., as the magnitudes of the various quantities. $|i| = 1$.

Ibix offered a rendering of the diagram using what I assume is matrix* notation:

I'm not sure what the equation scrawled in the diagram is saying. Is the left side of the equation $2^{ab}$? If so, I don't understand how that can be equal to a matrix.

I suppose it is theoretically possible to have a triangle with zero length hypotenuse

If both legs are equal in size and one lies on top of the other, the hypotenuse of such a triangle would be zero, but that's not a very interesting triangle.

 

[fresh_42]

The problem here is, that $i$ is not a length. If you write it as real vectors like in the Gaußian plane of complex numbers, then it becomes wrong. If you consider it as a complex equation, then it is correct, but lacks the interpretation in the real world.

 

[kuruman]

You have two vectors in the complex plane:
$Z_1=(0,i)$ and $Z_2=(1,0)$.
Noting that $Z_1-Z_2 = (-1,i)$, the magnitude of the hypotenuse squared is
$|Z_1-Z_2|^2=(-1,i)^*\cdot(-1,i)=(-1,-i)\cdot(-1,i)=1+1=2.$
This is the sum of the magnitudes-squared of the sides, $|Z_1|^2+|Z_2|^2.$

 

[PeroK]

TL;DR Summary: Is this geometry valid?

I posted this in the Lame Math thread, but it's got me thinking.

View attachment 365364

Is there any validity to this? Or is it really just a mathematical trick?

Naively, I see that i² + plus 1² does equal zero².

But does this have a meaning?

That's one representation of Minkowski geometry, where the vertical axis is time and the horizontal axis is spatial. The spacetime distance in $c =1$ units is given by:
$$(\Delta s)^2 = -(\Delta t)^2 +(\Delta x)^2$$This is sometimes represented using "imaginary" time.

PS the paths of zero distance are those followed by light.

 

[DaveC426913]

That's one representation of Minkowski geometry, where the vertical axis is time and the horizontal axis is spatial. The spacetime distance in $c =1$ units is given by:
$$(\Delta s)^2 = -(\Delta t)^2 +(\Delta x)^2$$This is sometimes represented using "imaginary" time.

PS the paths of zero distance are those followed by light.

Fascinating crossover. Can you suggest where I can read up on this more? (I'm not post-secondary edumacated, so the hard maths is over my head, but the physics may not be*)

*yes, the irony of the notion that these can be separated does not escape me.

 

[Ibix]

Can you suggest where I can read up on this more? (

@bcrowell's Relativity for Poets is a decent maths-free resource, and free to download. https://lightandmatter.com/poets/

 

[fresh_42]

This is essentially the difference between $\mathbb{R}^2$ and $\mathbb{C},$ and what distinguishes real from complex analysis: squares are no longer automatically non-negative. This has a significant impact on functions.

If you are interested in Minkowski spaces and a notation with imaginary time, then you should read about the Wick rotation. It translates between the two.

 

[Hornbein]

I recommend
Reflections on Relativity
Book by Kevin Brown

 

[PeroK]

I recommend correcting the thread title! I assumed someone would have done it by now.

 

[A.T.]

I recommend correcting the thread title! I assumed someone would have done it by now.

I assumed a pun attempt. "Pitagyros" is one I remember from my school days during more racist times.

 

[sbrothy]

I'm also rather un-edumacated so thanks for these suggestions.

 

[zerodish]

There are many valid mathematical systems in use. You will need to define the rules where this is valid. Argand chose to make the imaginary axis at right angels to the real axis in the Argand Plane. The imaginary axis is associated with the real axis. The Pythagorean Theorum does not work in the Argand Plane since it is essentially a 1 dimensional object. However if you use the imaginary axis of one Argand Plane with the real axis of a different Argand Plane the example you showed is valid. Two Argand Planes can intersect in a single point at right angels to each other forming a four dimensional coordinate system. Which is what you have done.

 

[robphy]

Note the last diagram from my Insight https://www.physicsforums.com/insights/relativity-rotated-graph-paper/
As others have said, the 0-hypotenuse could suggest a lightlike vector in special relativity.
I don't use complex-numbers in special-relativity.
Below, I use real numbers with a Minkowski spacetime metric.
The vertical-3 is for a timelike-vector and the horizontal-3 is for a spacelike vector.

(The segments count the number of light-clock diamonds along the segment.
The choice of which diagonal distinguishes timelike from spacelike.
All light-clock diamonds have the same area... and are traced out by a standard light-clock carried by the piecewise-inertial observer.
The square-interval is the area of the causal-diamond between two events. It is zero for DF since the diamond is degenerate.)

Try Triangle-4 (You may have to scroll down.) Use the tool in Geogebra toolbar to decorate a segment with light-clock diamonds.
You can drop your own endpoints. Return to the cursor in the toolbar to move the endpoints of your segment.