If I have a spacetime diagram with $t$ on the vertical axis and $x$ on the horizontal axis, does calculating an area in this diagram have any physical significance?
The reason I ask is because I'm trying to figure out if the $u$ in all the $\cosh(u)$ and others hyperbolic trigonometry functions has a physical meaning.
In a (1+1)-Minkowski spacetime, the area of "causal diamond" of OQ (where Q is in the causal future of O) is equal to the squared-interval from O to Q.
(The causal diamond of OQ is the intersection of the future of O with the past of Q.)
In the diagram above, O is at the origin-event, and event Q is 5 ticks (upward) and 3 sticks (to the right)… count the grid-diamonds along their timelike diagonals, then diamonds along their spacelike diagonals. The squared-interval is $5^2-3^2=16$, which is the number of light-clock diamonds in the causal diamond of OQ (the blue shaded region OPQR).
I have exploited this fact to develop a method of doing graphical calculations in special relativity:
"Relativity on rotated graph paper"
American Journal of Physics 84, 344 (2016);
https://doi.org/10.1119/1.4943251
related:
The reason I ask is because I'm trying to figure out if the in all the cosh() and others hyperbolic trigonometry functions has a physical meaning.
Here's a geometric interpretation of $u$:
If $u$ is the relative-rapidity between two future-pointing timelike rays, then $u$ is proportional to the area of the hyperbolic sector in the unit-hyperbola. $\cosh u$ is related to the area of certain triangles.
See:
https://en.wikipedia.org/wiki/Rapidity#Area_of_a_hyperbolic_sector
See also:
I just found this:
update:
A physical interpretation of $u$ is less clear in this case
because $u$ is the Minkowski arc-length along a spacelike curve (the corresponding intercepted-arc along the unit-hyperbola).
(Note that $u$ is the natural logarithm of the Doppler-factor $k=\exp u$ [featured in the Bondi k-calculus]. Since in (1+1)-Minkowski spacetime the rapidity $u$ is additive [since unit-hyperbolic sector-areas are additive], the Doppler factor $k$ is multiplicative.)
This connects back to your other question about rapidity:
How important is the concept of rapidity in relativity?
Alternatively, one could consider a geometrically similar case of the worldline of uniformly-accelerated observer (a hyperbola with timelike tangents), where $u$ is proportional to the proper-time elapsed along the arc of that timelike-hyperbola.
Suppose a cube is moving with respect to you.
In a diagram of $3$ dimensional space, area has a physical meaning. But people are usually doing classical physics. If everyone measures parallel to the edges of the cube. Everyone would agree on the area of the sides of the cube, no matter how they move. The area is invariant.
On the other hand, you could choose a coordinate system at an angle and calculate how the area of the side appeared to you from that angle. Different frames of reference at different angles would give you different apparent areas. You could say there is some physical meaning to the apparent areas, but any such meaning would depend on the coordinate system you chose.
If you know the angle of the cube's axes in your coordinate system, you could use that information to calculate the true area.
Space time has $4$ dimensions. Only $2$ fit on paper, so usually people only try to draw $2$. Typically they draw $1$ spatial dimension, and that is typically the direction of travel of something. The other is typically time. This is because such a space time diagram shows interesting relationships between time and distance in that direction.
But if you are interested in areas, you could draw a space time diagram with two spatial directions. It isn't as useful, but you could do it.
Suppose you draw a $2$ D space time choosing a frame at rest with respect to the cube. You pick two spatial axes along the edges of the cube. Since the cube is at rest, you get the expected area for the side of the cube.
Suppose you pick a frame at rest with respect to yourself, with one axis in the direction of the cube's motion and the other perpendicular, both aprallel to edges of the cube. You would find a smaller area than you expect because the cube is Lorentz contracted in the direction you see it traveling.
Suppose you picked two spatial dimensions perpendicular to the direction of travel. Neither direction is contracted from your frame of reference, so you get the same area as the rest frame. But another observer moving sideways would not agree with you.
So these area do have the same sort of physical meaning as classical measurement at an angle. They are the areas you measure from your frame of reference. If you know the velocity, you can calculate the rest lengths of the sides and calculate the area in the cube's rest frame.
Area is a frame dependent quantity, a measure of how things appear to you. The rest area is an invariant quantity that you can calculate. Any observer in any frame of reference could find that rest area.
The name itself of "hyperbolic functions" is due to the fact that the functions $cosh(t)$ and $sinh(t)$ play, in the parametric representation of the equilateral hyperbola $x^{2}-y^{2}= a^{2}$ the same role as the functions $cos(t)$ and $sin(t)$ for the circle $x^{2}+y^{2}=a^{2}$
The parametric representation of circle is: $x=a\cos(t) , y= a \sin(t)$
and, for the hyperbola $x=a\cosh(t), y=a\sinh(t)$
as it is easy to see using the relation: $cosh(t)^{2}-sinh(t)^{2}=1$
The geometrical significance of the parameter $t$ in both cases, for the circle and for the hyperbola, is identical.
If we designate $S$ the area of a portion of the circle and by $S_{0}$ the area of the entire circle ($S_{0}=\pi a^{2}$), we have: $t= 2\pi \frac{S}{S_{0}}$
Let us now assume that S denotes the area of an analogous sector of the equilateral hyperbola. We have $S= area\, OMN - area\, AMN$, M a point on the heperbolus and N its projection on the axis (Ox) and A the point of intersection of the hyperbola with the axis (Ox).
$S=\frac{1}{2}xy-\int_{a}^{x} ydx=\frac{1}{2}x\sqrt{x^{2}-a^{2}}-\int_{a}^{x}\sqrt{x^{2}-a^{2}}\,dx$ $S=\frac{1}{2}a^{2}\ln(\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,)$
If now we put, denoting again by $S_{0}$ the area of the circle $t= 2\pi \frac{S}{S_{0}}=\ln(\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,) $
we easily find that:
$e^{t}=\frac{x}{a}+\sqrt{\frac{x^{2}}{a^{2}}-1}\,\,\,\,\;, e^{-t}=\frac{x}{a}-\sqrt{\frac{x^{2}}{a^{2}}-1}$
hence, adding term to term and multiplying by $\frac{a}{2}$:
$x=\frac{a}{2}(e^{t}+e^{-t})=a\cosh(t)$
$y=\sqrt{x^{2}-a^{2}}=\sqrt{a^{2}\cosh^{2}(t)-a^{2}}=a\sinh(t)$
Reference: Higher Mathematics Course,Volume I, V.Smirnov
