Why Do Square and Rhombus Grids in Spacetime Diagrams Have Equal Areas?

[yinfudan]

I have an interesting finding in spacetime diagram - The area of the square grid for the rest frame O, is the same as the area of the rhombus grid for the inertial frame O'.

More specifically, the area of the square for frame O

(0, 0), (0, 1), (1, 1) and (1, 0) is 1

The area for the rhombus for frame O'

(0, 0), (vγ, γ), (γ+vγ, γ+vγ), and (γ, vγ) is also 1,

no matter what γ or v is.

Is this just a coincidence or does it have mathematics or physics basis?

Thanks!

 

[robphy]

A boost has determinant 1.
With the two lightlike eigenvectors, this can be exploited to develop the Bondi k-calculus.

One can relate the invariant intervals to these areas.

You might like to look through the links on
https://www.physicsforums.com/showthread.php?p=807564#post807564
(the last is my paper).

 

[yinfudan]

A boost has determinant 1.
With the two lightlike eigenvectors, this can be exploited to develop the Bondi k-calculus.

Yes, the determinant of lorentz boost is 1. However, my question here is why it has to be 1. Is it a coincidence or does it have certain meaning in physics? What is the consequence if it is not 1, yet still a linear transform?

 

[George Jones]

Yes, the determinant of lorentz boost is 1. However, my question here is why it has to be 1. Is it a coincidence or does it have certain meaning in physics? What is the consequence if it is not 1, yet still a linear transform?

In order that a linear transformation leave the spacetime interval invariant, it must have determinant \pm1. For space and time orientations to be preserved, the determinant must be plus 1.

 

[yinfudan]

I think I have an easier way to understand why both the area of the square grid for the rest frame O must be equal to the area of the rhombus grid for the moving frame O'.

If the area of the rhombus grid is not the same, say, bigger. Now if I normalize the rhombus grid, that is to make the frame O' axis perpendicular and to make the rhombus grid square with the area equal to one, then, the original square grid for frame O will become rhombus towards the other direction (top left to bottom right). Now let us examine the area of this rhombus for frame O, the area will be smaller than one, and smaller than the square grid for frame O', which is one.

Since frame O and frame O' are both inertial and only have relative speed. They are equally valid and symmetric. So the hypothesis that the area of rhombus grid is bigger must be false, because if it is true, by normalizing the rhombus grid to square, and warping the square grid to rhombus, the newly warped grid will be smaller, which breaks the symmetry.

 

[robphy]

I think I have an easier way to understand why both the area of the square grid for the rest frame O must be equal to the area of the rhombus grid for the moving frame O'.

If the area of the rhombus grid is not the same, say, bigger. Now if I normalize the rhombus grid, that is to make the frame O' axis perpendicular and to make the rhombus grid square with the area equal to one, then, the original square grid for frame O will become rhombus towards the other direction (top left to bottom right). Now let us examine the area of this rhombus for frame O, the area will be smaller than one, and smaller than the square grid for frame O', which is one.

Since frame O and frame O' are both inertial and only have relative speed. They are equally valid and symmetric. So the hypothesis that the area of rhombus grid is bigger must be false, because if it is true, by normalizing the rhombus grid to square, and warping the square grid to rhombus, the newly warped grid will be smaller, which breaks the symmetry.

Would your argument work if you focus on another feature of possible interest... say, the length (drawn on a spacetime diagram) of an inertial timelike-interval of 1-sec (analogous to the 4-velocity of an observer)... or even the length of the diagonal of that square?

I suspect that you are implicitly using additional properties which should really be addressed in your explanation.
In other words, why the area of the rhombus and not something else?

 

[yinfudan]

I suppose it works for length as well:)

The length of one side of the rhombus must be sqrt((1+v²)/(1-v²)).

If longer (or shorter) than this value, when the rhombus is normalized to a unit square, and the original square is warped to a new rhombus, the length of one side of the new rhombus would be shorter (or longer) than this value, which breaks the symmetry.