I Trying to understand space-time diagrams

[vanhees71]

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.

 

[robphy]

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.

The minus-sign is needed here if your signature-convention is $(+---)$.

If the spacelike curve lies in the plane of simultaneity of an inertial observer,
the quantity could be called the "spatial distance" along a spatial path,
a Euclidean arc-length.

A special case of this would be the proper-length of an object at rest with respect to this inertial observer.

 

[Freixas]

I've just begun learning about space time diagrams.

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.

There is another approach I came up with to calculate the value of the $ct'$ coordinate, one that doesn't require much math. The experts here may be familiar with it, but I've never seen it anywhere. It relies on Einstein's first postulate: The laws of physics have the same form in all inertial reference frames.

Here is the basic problem. We have an event whose rest coordinates are known; this implies we know the units of the $x$ and $ct$ axes. We want to know the event's $ct'$ value.

I've drawn the $x'$ and $ct'$ axes in red, but not added labeled tick marks. We don't even know the relative velocity of the moving observer (although it could be calculated from the angle of the axes).

The first step is to draw a line parallel to the $x'$ axis through the event. All points on this line have the same $ct'$ value (the same time value as calculated by the moving observer). We don't know this value, but we do know the value where the line crosses the $ct$ axis.

We want to know the length from the origin to A (as has been pointed out, this is not a Euclidean length). Draw another line parallel to the $x$ axis. Again, we know where this line crosses the $ct$ axis. All points on this line have this same time value as calculated by the rest observer.

Using Einstein's first postulate, we know that the ratio 33.33:A must be the same as the ratio of A:25. Each represents how each observer views the other's clock rate. Therefore, we just need to solve for A in the equation $33.33/A = A/25$. The answer is $A = \sqrt{33.33/25} = 28.87$ (approximately).

At one point, I wanted to see if I could explain how to generate a basic Minkowski spacetime diagram (essentially, starting with a rest observer and an observer moving at some given relative velocity, draw the axes, their tick marks, and the worldlines of the two observers) starting from Einstein's S.R. postulates, and without requiring any math. 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]

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.

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).

Hint: You essentially need to find a way to draw hyperbolae.

 

[Freixas]

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).

Intriguing. 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 there is a connection is certain; that the connection is straight-forward is not. It is not enough to come up with a method of finding the tick marks on the moving observer's axes--the reason the method works has to be given, and the explanation can't involve a complex mathematical derivation.

In my system, the explanation directly ties to Einstein's postulate; labeling the point on the $ct'$ axis requires some simple math (you need to know about ratios, square roots, and division). If you have a method that avoids the math altogether, that would be great.

 

[vanhees71]

Einstein's postulates can for sure be fulfilled by simply assuming that the quadratic form on four-vectors, $\eta_{\mu \nu} x^{\mu} x^{\nu}$ with $(\eta_{\mu \nu}=\mathrm{diag}(1,-1,-1,-1)$ is invariant under boosts (i.e., transforming from one inertial frame of reference to another). That's so, because then the front of a spherical em. wave (light signal) obeys $c^2 t^2-\vec{x}^2=0$ in all inertial frames of reference, and indeed with this assumption you can construct Minkowski space with the Lorentz transformations as the symmetry group (or more precisely the Poincare group, because this space is also homogeneous, i.e., translation invariant).

This implies that the temporal/spatial units on the axes of inertial frames depicted in your Minkowski diagram are determined by the corresponding unit hyperbolae $c^2 t^2-x^2=\pm 1$, i.e., the intersection of these hyperbolae determine the unit-tic marks on the time and space axes of all inertial reference frames.

These unit hyperbolae are analogous to the unit circle, with which you construct the units on the axes of Cartesian coordinate systems in the Euclidean plane, and the symmetry in this plane are rotations rather than Lorentz transformations of course.

 

[Freixas]

Einstein's postulates can for sure be fulfilled by simply assuming that the quadratic form on four-vectors

My response to @Kashmir was triggered by the fact that he was starting to learn about spacetime diagrams. I offered my solution to finding the $ct'$ coordinate because I think it's an interesting alternative that applies Einstein's postulate directly.

I do have a personal challenge to explain the basic elements of Minkowsky diagrams to someone with only basic math knowledge (mentioning the "quadratic form of four vectors" would exceed the limits I had in mind) . This is just for my entertainment, and pursuing how one might do that (and what level of knowledge would be allowed) seems like hijacking @Kashmir's thread. If I decide it's worth discussing, I will post my own thread.

 

[Dale]

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.

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.

 

[robphy]

Hint: You essentially need to find a way to draw hyperbolae.

That's what the light-clock diamonds do... they are proxies for hyperbolas.

Constructed using the light-signals in a light-clock,
and motivated by the Bondi k-calculus (using the speed of light postulate and the relativity-postulate).
Their edges are lorentz-invariant directions (in (1+1)-Minkowski spacetime), since they are along eigenvectors of a boost. Since the determinant is one, [since the eigenvalues (stretch factors for the eigenvectors) are reciprocals] the area of the diamond is lorentz-invariant.

Here's my least technical introduction
https://www.physicsforums.com/insights/relativity-on-rotated-graph-paper-a-graphical-motivation/

 

[vanhees71]

They are not proxies but exact. In light-cone coordinates the hyperbolae are parametrized as
$$x^+=\frac{A}{x^-},$$
with $A$ a constant area.

In light-cone coordinates the Lorentz boost reads
$$x^{\prime +}=\kappa x^{+}, \quad x^{\prime -}=\frac{1}{\kappa} x^{-}, \quad \kappa=\exp(-\eta),$$
where $\eta$ is the rapidity. In the usual Lorentzian components it reads
$$x^{\prime 0}=\cosh \eta x^0 - \sinh \eta x^1, \quad x^{\prime 1} =-\sinh \eta x^0 + \cosh \eta x^1.$$
That simple algebra is what's behind your "rotated-graph paper" geometric constructions.

 

[apostolosdt]

Thank you all […]

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]

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?

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.

Then the discussion went on between those participants which was beyond me.