Why is the time axis in a space-time diagram a distance

[studiot]

3 hours ago, joigus said:

Probably just a rhetorical question, but interesting nonetheless.

The answer, very much having to do with the excellent points made, is all about symmetry groups --from a mathematical physics perspective.

In economics, eg., you have graphs representing marginal utility --dimensionless, if I remember correctly-- plotted against number of units of goods --again, dimensionless. In electrical engineering tho, you have graphs plotting intensity against voltage --both dimensionful. But the subtle mathematical point is that in neither of those cases is there a group of symmetry relating different, but equaly valid, observational stances --classes of valid observers. In the physics of space-time, there is such class of valid observations, and the measurements of time and space of one observer are mixed up with the measurements of time and space of another, so it is natural to define the parameters of the transformation as dimensionless numbers.

You could insist that (I,V) (intensity and voltage) be a 'vector', but it doesn't make much sense unless you can define transformations that take one stance to another and relate them in a linear way:

I'=aI+bV

V'=cI+dV

with a,b,c,d being the dimensionless parameters that mix them up.

That's the reason why in mathematical physics you tend to be more careful and say: Such and such quantities are a vector under SO(3) (rotations); or a vector under O(3) (rotations and inversions-reflections), etc.

In the case of Minkowski vectors, it's all about inertial observers. It's what we call the 'Lorentz group'.

I had hesitated about introducing group theory, but now you have gone and done it. +1

I doubt there is much group theory in Engineering, but the following development of your linear transformations may be accessible to Caruthers.

3 hours ago, joigus said:

I'=aI+bV

V'=cI+dV

Note that linear transformations (in the sense of Linear Analysis or Linear Algebra) are the simplest but not the only possible transformations.

Linear transformations work so some texts say we need a transformation so we will start with the simplest a linear transformation and leave it at that.

However @Caruthers you may be familiar with  (linear) rotations in mech eng theory so I will link to them as follows.

If we change the coordinates from x, y, z, τ to x', y' z', θ τ' by rotating an angle the coordinate axes in the plane (x, τ) through an angle θ from τ towards x, where

τ = ict    -   it is common and convenient to introduce the new variable tau as ict here

(Again I am moving the axes this time, not the position vector so moving the axes from tau towards x is equivalent to moving the vector from x towards tau)

then the transformation becomes

x' = xcosθ - τsinθ
τ' = τcosθ + xsinθ
y' = y
z' = z

Using the familiar circular trig functions but in 4D.

Comparing these with the Lorenz version in xyzt and x'y'z't'  (ie ordinary time) 

[math]x' = \frac{{x - vt}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}[/math]

[math]t' = \frac{{x - \frac{{vx}}{{{c^2}}}}}{{\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}[/math]

[math]y' = y[/math]

[math]z' = z[/math]

we see that

[math]\cos \theta  = \sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} [/math]

and

[math]\sin \theta  = \frac{{ - iv}}{{c\sqrt {1 - \frac{{{v^2}}}{{{c^2}}}} }}[/math]

so theta is an imaginary angle since cosθ is greater than 1 and sinθ recovers the imaginary symbol i.

If you are familiar with hyperbolic trig functions you will immediately recognise that this can also be done with hyperbolic sine and cosine (sinh and cosh) and real angles.

 

These are all the different approaches you will find in the texts.

Good luck with your bedtime reading with your candle in your boat cabin.

 

 

Edited December 2, 2021 by studiot

[joigus]

19 hours ago, studiot said:

Using the familiar circular trig functions but in 4D.

Comparing these with the Lorenz version in xyzt and x'y'z't'  (ie ordinary time) 

[...]

Wonderful complement to what I was saying

[johnsankey]

Space-time, Minkowski vectors, is mathematics not physics.

To be mathematics, all you have to be is logically consistent.

To be physics, it has to be observable.