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IMHO there is an element of overthinking in this thread. Especially for a first exposure to spacetime diagrams.
Maybe the level of the thread should go to "B".PeroK said:IMHO there is an element of overthinking in this thread. Especially for a first exposure to spacetime diagrams.
And this is supposed to be simpler than teaching (multi-)linear algebra with the Minkowski product as fundamental form of a Minkowski space? For me the deliberation from drawing Minkowski diagrams in high school to tensor algebra and calculus was a revelation.robphy said:These diagrams may help you appreciate that "rapidity" can be thought of as an "angle",
when used properly. As with any tool, one must learn how to use it and how not-touse it.
Note the use of an arc of the "Minkowski circle"...
so we don't fall into the trap of thinking in terms of Euclidean geometry, as @vanhees71 warns us about.
From https://www.physicsforums.com/threads/velocity-through-spacetime.957038/post-6068213
View attachment 314576
From my answer to https://physics.stackexchange.com/questions/494698/a-best-definition-of-proper-acceleration
View attachment 314567
View attachment 314568
From
https://physics.stackexchange.com/questions/667907/lorentz-transform-moving-length-to-moving-length
View attachment 314569View attachment 314570
In my opinion, since we teach students to use trigonometry to solve free-body diagrams,
we can leverage that experience in trying to teach relativity via hyperbolic-trigonometry
(on the way to possibly having them work with vectors and tensors and use invariant thinking).
There has to be something to show the "geometry" encoded by the tensor-algebra.
As an example of the power of trigonometry, look at my answer to this question:
https://physics.stackexchange.com/q...-out-of-knowledge-kinetic-energy-in-lab-frame
The question asks for a derivation of
##|\vec{p}^{\text{com}}_{\pi}|=\frac{M_{p}\sqrt{T^{L}_{\pi}(T^{L}_{\pi}+2M_{\pi})}}{\sqrt{(M_{p}+M_{\pi})^{2}+2T^{L}_{\pi} M_{p}}}##
and wonders about whether it is dependent on the frame of reference.
My answer gives an answer by interpreting the formula given
as the altitude of a triangle CB in connection with the Law of Sines in Minkowski spacetime.
View attachment 314571Here is the muon problem solved trigonometrically (however, no angular arcs shown)
from https://physics.stackexchange.com/q...-distance-traveled-and-the-times-it-takes-why
View attachment 314575
To play around with the analogy,
try https://www.desmos.com/calculator/emqe6uyzha
Set ##E= -1##, then play around.
Move the slider slowly back to ##E= +1##.
See https://www.physicsforums.com/threads/why-is-time-orthogonal-to-space.969731/post-6159973 for a description of an earlier version of this.
Yes.vanhees71 said:And this is supposed to be simpler than teaching (multi-)linear algebra with the Minkowski product as fundamental form of a Minkowski space? For me the deliberation from drawing Minkowski diagrams in high school to tensor algebra and calculus was a revelation.
Yes.Orodruin said:Just a comment regarding angle as the arc length of the unit circle vs area in the case of the Minkowski geometry:
It is of course possible to reformulate the Euclidean case in terms of the area of the circle sector as well rather than the arc length of the circle segment (the two being related by a factor of two). This makes the approaches even more analogous.
(That's a boost in light-cone coordinates, in the Bondi k-calculus.)"If we now apply to the plane the linear transformation
##\alpha'=k\alpha\qquad \beta'=k^{-1}\beta\qquad(12)##
the hyperbola is transformed into itself. The points P and Q are moved to any arbitrary position on the curve, but the ratio ##\alpha_2:\alpha_1## is unaltered. It is easily shown that in this transformation the area of any triangle, and hence the area of any figure in the plane, is unchanged ; so that u, sinh u and cosh u are unchanged. They are therefore all functions of the ratio ##\alpha_2:\alpha_1## alone. Hence sinh u and cosh u are functions of u alone, and the definitions here given are a proper generalization of the usual definitions."
In the spirit of my previous post (mentioning the unified geometrical viewpoint),Kashmir said:Attempt to find the hyperbolic angle (rapidity) of points on the line ##x'##:
...
Hence all points on the line ##x'## have the same hyperbolic angle ##\theta##.Correct?
Yes, I'll try that way.robphy said:In the spirit of my previous post (mentioning the unified geometrical viewpoint),
repeat your strategy for a circular angle [noting any slight variations that are necessary].
Does it work?
Kashmir said:Attempt to find the hyperbolic angle (rapidity) of points on the line ##x'##:
View attachment 314620
(Taking ##ct=y##.)
Equation of ##x'## is given as ##y Cosh\theta=x Sinh\theta ## so
On the line ##x'## we've a point ##y/x=Tanh \theta## ----(1)
This point lies on a hyperbola ##x^2 -y^2=R^2## with ##R^2=x^2(1-Tanh\theta)##
The area of sector is ##
A = \frac{1}{2}x\sqrt{x^2 -R^2} - \int _R ^x \sqrt{x^2 -R^2}dx##
View attachment 314619
##A=\frac{R^2}{2} \ln \left(\frac{x+\sqrt{x^2-R^2}}{R}\right)##
We've defined ##2A/R^2##( Twice area divided by Minkowski length) as the hyperbolic angle ##\beta##, so:##\beta=\ln \left(\frac{x+y}{R}\right)## -----(2)
For all points on the ##x'## axis we can write ##x=R cosh u, y=R sinh u##-----(3)
we then have from (2) and (3) :
##\beta=u## ----(4)
But also ##y/x = tanh u## and ##y/x=Tanh\theta##( from (1))
So ##β=u=\theta##
##β=\theta##
Hence all points on the line ##x'## have the same hyperbolic angle ##\theta##.Correct?
In addition: It is also possible to define the hyperbolic angle as the Lorentzian arc length of the "unit" hyperbola segment.Orodruin said:Just a comment regarding angle as the arc length of the unit circle vs area in the case of the Minkowski geometry:
It is of course possible to reformulate the Euclidean case in terms of the area of the circle sector as well rather than the arc length of the circle segment (the two being related by a factor of two). This makes the approaches even more analogous.
Indeed. This is actually my usual approach.Sagittarius A-Star said:In addition: It is also possible to define the hyperbolic angle as the Lorentzian arc length of the "unit" hyperbola segment.
Source:
https://en.wikipedia.org/wiki/Hyperbolic_angle#Relation_To_The_Minkowski_Line_Element
You can do the corresponding thing for the spacelike unit hyperbolae and then it is of course not proper time.vanhees71 said:I'd however stress once more that it's not a usual arclength, and that's why I'd call it "proper time" and only "proper time", which is defined for time-like world lines.
The minus-sign is needed here if your signature-convention is ##(+---)##.vanhees71 said:For a spacelike curve you must define it of course with an additional sign under the square root. I can't remember that I've ever needed such a quantity.
I'm going all the way back to the original OP. The question was about whether a particular formula generated the value of the ##ct'## coordinate. That question has been answered in spades by all the experts.Kashmir said:I've just begun learning about space time diagrams.
The way to think about this is to ask yourself how you would do it in Euclidean space with a straightedge and a compass (no graded ruler, which effectively doubles as both).Freixas said:Labeling the units for the axes of the moving observer eluded me for some time; the method shown here fails to avoid all math--I still need a division and a square root--but it's as simple as I could get.
Orodruin said:The way to think about this is to ask yourself how you would do it in Euclidean space with a straightedge and a compass (no graded ruler, which effectively doubles as both).
vanhees71 said:Einstein's postulates can for sure be fulfilled by simply assuming that the quadratic form on four-vectors
It comes almost directly from the 2nd postulate. Start with the spacetime interval: $$\Delta s^2 = -c^2 \Delta t^2 +\Delta x^2 + \Delta y^2 + \Delta z^2$$ Note that this is the equation for a hyperboloid. If ##\Delta s^2 > 0 ## it is a hyperboloid of one sheet, and if ##\Delta s^2 <0 ## it is a hyperboloid of two sheets. Now, for the special case of ##\Delta s^2 = 0## we have a degenerate hyperboloid, a cone. Up to this point this is just geometry. If we rearrange the special case cone then we can get: $$c^2 \Delta t^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$$ which we recognize as a sphere with radius ##c \Delta t##. In other words, this is the equation of a flash of light (which we call a light cone). The light expands outward in a sphere at a speed of ##c##. The first postulate is embodied in the fact that ##\Delta s^2## is invariant, meaning that if something travels at the speed of light in any direction in one reference frame then it travels at the speed of light in all reference frames.Freixas said:But before I go off to drawing hyperbolas by geometric construction, I would have to figure out how to relate that to Einstein's postulates.
That's what the light-clock diamonds do... they are proxies for hyperbolas.Orodruin said:Hint: You essentially need to find a way to draw hyperbolae.
You asked for help. Now, in turn, you can help this Forum. Honestly, how much did this thread manage to illuminate the topic you started?Kashmir said:Thank you all […]
It helped clear that the diagram was to be interpreted not read off directly. And the angles were not circular but hyperbolic and how we calculate it.apostolosdt said:You asked for help. Now, in turn, you can help this Forum. Honestly, how much did this thread manage to illuminate the topic you started?