Spacetime diagram grid area

In summary, the conversation discusses the finding that the area of the square grid for rest frame O is equal to the area of the rhombus grid for inertial frame O', regardless of the values of gamma (γ) and velocity (v). This is due to the determinant of a Lorentz boost always being 1, preserving the spacetime interval and orientation. The conversation also touches on the implications of this finding and whether it applies to other features such as length.
  • #1
yinfudan
26
0
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!
 
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  • #2
Last edited:
  • #3
robphy said:
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?
 
  • #4
yinfudan said:
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 [itex]\pm[/itex]1. For space and time orientations to be preserved, the determinant must be plus 1.
 
  • #5
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.
 
  • #6
yinfudan said:
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?
 
  • #7
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.
 

1. What is a spacetime diagram grid?

A spacetime diagram grid is a visual representation of the relationship between space and time in a particular event or scenario. It is commonly used in physics to illustrate the concept of spacetime, where the vertical axis represents time and the horizontal axis represents space.

2. How is the grid area determined in a spacetime diagram?

The grid area in a spacetime diagram is determined by the units used for time and space. For example, if time is measured in seconds and space is measured in meters, then the grid area would be in square meters-seconds.

3. What is the significance of the grid area in a spacetime diagram?

The grid area in a spacetime diagram represents the interval between two events in spacetime. This interval can be interpreted as the distance traveled by an object in a specific amount of time or the time elapsed between two events at a particular location in space.

4. How does the grid area change in different reference frames?

The grid area in a spacetime diagram can change depending on the reference frame. This is due to the concept of time dilation and length contraction, where the perception of time and space can differ for observers in different reference frames.

5. Can the grid area in a spacetime diagram be negative?

No, the grid area in a spacetime diagram cannot be negative. This is because both time and space are positive quantities, and the grid area is a measure of the interval between two events in spacetime. However, it is possible for the grid area to be zero, which would indicate that the events are occurring at the same time and location.

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