Using invariant hyperbolas to calibrate between reference frames

Question

We can now calibrate the axes of $\tilde{O}$. In fig 1.11.

The axes of $O$ and $\tilde{O}$ are drawn, along with an invariant hyperbola of time like interval $-1$ from the origin. Event $A$ is on the $t-$axis so has $x=0$. Since the hyperbolas the equation $$ -t^2 + x^2 = 1$$, It follows that event $A$ has $t=1$. Similary event $B$ lies on the $\tilde{t}$ axis, so has $\tilde{x}=0$. Since the hyperbola has the equation $$-\tilde{t}^2 + \tilde{x}^2=-1$$ It follows that event $B$ has $\tilde{t}=1$. We have, therefore, used the hyperbolae to calibrate the $\tilde{t}$ axis.

I have the following questions from the above:

1. Precisely what does it mean to calibrate axes? I think it means to establish simultaneity between two different reference frames, but, I'd like a second confirmation as well.

2. Secondly how were events $A$ and $B$ chosen? No mention of that was given in the book to my knowledge.

3. If all else fails, could you recommend another book talking about this calibration procedure between two inertial frames with coordinates?

Answer 1

In units where $c=1$, the Lorentz transformations are \begin{align} \bar{t}&=\frac{t-vx}{\sqrt{1-v^2}},\quad\bar{x}=\frac{x-vt}{\sqrt{1-v^2}}. \end{align} By following a similar process to what I described in this answer, I leave it to you to verify the following facts: in the $(t,x)$ coordinate system,

• The $\bar{t}$-axis is drawn at a slope of $\frac{1}{v}$, while the $\bar{x}$-axis is drawn at a slope of $v$
• level sets of $t$ are horizontal lines, while level sets of $\bar{t}$ are straight lines of slope $v$
• level sets of $x$ are vertical lines while level sets of $\bar{x}$ are straight lines of slope $\frac{1}{v}$
• $-t^2+x^2=-\bar{t}^2+\bar{x}^2$.

Now some basic facts about hyperbolas: the set of points for which $-t^2+x^2=k$ (equivalently $-\bar{t}^2+\bar{x}^2=k$), where $k\in\Bbb{R}$ is a constant fall into three categories:

• $k>0$, in which case the hyperbola opens left-right
• $k=0$, which are straight lines $t=\pm x$ (degenerate hyperbolas)
• $k<0$, in which case the hyperbola opens top-bottom,

and obviously through each point in the plane, there is a unique such hyperbola passing through that point.

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I'm not sure about the purpose of mentioning this calibration procedure using hyperbolas. Generally, when we calibrate something, it means we have some fixed scale, and we're using that to set the scale of something else. But here, we already know the full relationship between $(t,x)$ and $(\bar{t},\bar{x})$, so I'm not sure about the purpose of this. In any case, here's what we can say geometrically: if we consider the point $\mathscr{A}$ with $t(\mathscr{A})=1,x(\mathscr{A})=0$, then you can draw the unique hyperbola through the point $\mathscr{A}$ (i.e the hyperbola with $k=-(1)^2+0^2=-1$). This hyperbola will intersect the $\bar{t}$-axis at a unique point $\mathscr{B}$ which has $\bar{t}(\mathscr{B})=1,\bar{x}(\mathscr{B})=0$. So, by starting from $(t,x)=(1,0)$ we managed to geometrically figure out where the point with $(\bar{t},\bar{x})=(1,0)$ lies. We can keep going: start with the point having $(t,x)=(2,0)$ we can geometrically figure out the point with $(\bar{t},\bar{x})=(2,0)$, next we can start with the point with $(t,x)=(3,0)$ we can figure out where $(\bar{t},\bar{x})=(3,0)$ is and so on.

Similarly, starting with the point such that $(t,x)=(0,1)$ we can draw the hyperbola and figure out where $(\bar{t},\bar{x})=(0,1)$ is, and then so on with $(0,2),(0,3)$ etc. So, by having the scale markings on the $t$ and $x$ axes, we can geometrically figure out the scale markings on the $\bar{t}$ and $\bar{x}$ axes. Having said this, and while this geometry is nice on its own, it doesn't tell you anything which the Lorentz transformations do not already tell you. Hopefully this answers your questions 1 and 2.

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Finally, here's a quibble with your comment following your question 1:

... I think it means to establish simultaneity between two different reference frames, but, I'd like a second confirmation as well.

No, this makes no sense. Simultaneity is not a notion to be related between two observers. Simultaneity is a concept within one observer's frame. In the $(t,x)$ coordinate system, simultaneity is described by the level sets of $t$. Or more precisely, we say two events $p,q$ in the spacetime are simultaneous with respect to $(t,x)$ if and only if they lie in the same level set of $t$ (i.e if and only if $t(p)=t(q)$). In the case we're considering, these are the horizontal lines, as I mentioned above. We can obviously make an analogous definition for the barred coordinates: two events $p,q$ are said to be simultaneous with respect to $(\bar{t},\bar{x})$ if and only if they lie in the same level set of $\bar{t}$ (i.e $\bar{t}(p)=\bar{t}(q)$), and geometrically this happens if and only if they lie on the same straight line of slope $v$.

Answer 2

Given the tickmarks on an inertial worldline, calibration is the establishment of the corresponding tickmarks along another inertial worldline using a hyperbola (akin to using a circle, as one might do with a compass).

The hyperbola represents an equal separation from the origin in special relativity.

As an alternative to using hyperbolas, one can use a radar measurement procedure. They are equivalent for inertial observers.

(The radar measurement is related to a family of equal-area “causal diamonds” on a spacetime diagram.)

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UPDATE

My visualization might help: robphy's spacetime diagrammer for relativity v.8e-2021 (time upward) https://www.desmos.com/calculator/emqe6uyzha (play with the E-slider)

The idea is that the hyperbola is the "circle (locus of constant separation)" in the position-vs-time graph for special relativity (the Minkowski spacetime diagram).

By appropriate constructions analogous for the circle (E=-1 case), practically all of the textbook formulas of special relativity (E=+1) and Galilean relativity (E=0) arise.

In my approach, the "circle" encodes the "transformation (rotation or boost)" as well as other aspects of the geometry. Similar to how we teach Euclidean geometry by describing circles... before we describe rotation matrices.
Implicitly, the unit "circle" represents the metric at that event.

• One of my favorite quotes:
  Mermin ( "Lapses in Relativistic Pedagogy" http://aapt.scitation.org/doi/10.1119/1.17728 ) "...Lorentz transformation doesn't belong in a first exposure to special relativity. Indispensable as it is later on, its very conciseness and power serve to obscure the subtle interconnnectedness of spatial and temporal measurements that makes the whole business work. Only a loonie would start with real orthogonal matrices to explain rotations to somebody who had never heard of them before, but that's how we often teach relativity. You learn from the beginning how to operate machinery that gives you the right answer but you acquire little insight into what you're doing with it."

For connections of the hyperbola of special relativity to "radar methods" and "causal diamonds",
I'll direct you to

• Hermann Bondi's "Relativity and Common Sense: A New Approach to Einstein"
• my presentation of Bondi's ideas in my blog contribution https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/
• my article "Relativity on Rotated Graph Paper", Am. J. Phys. 84, 344 (2016); http://dx.doi.org/10.1119/1.4943251 .
• and my blog-contribution about it
  https://www.physicsforums.com/insights/relativity-rotated-graph-paper/

Answer 3

I think I get it a bit more, so what we do is firstly place the $\tilde{t}$ and $\tilde{x}$ axis in the $t-x$ plane. The exact way of placing this is discussed earlier in the book (page-6). Now, consider a point lying one unit along the $t$ axis and none along the x-axis $(0,1)$, we have that this point lies on the hyperbola:

$$ -t^2 + x^2 = -1 \tag{1}$$

This could be thought of as the set of points at negative unit space time interval from the origin. Now, the interesting point is, since the space time interval is invariant, in the $\tilde{t} - \tilde{x}$ plane, again it will be that the set of points having unit negative space time interval to origin is given as:

$$ - \tilde{t}^2 + \tilde{x}^2 = - 1 \tag{2}$$

Let's go back and look at our space time diagram from Schutz:

We see that curly $B$ also lies on the same hyperbola of constant space time interval as $A$, but what are the coordinates of it? Well firstly we know that it must satisfy (2) and further we know that $\tilde{x}=0$ on the line it is on. Hence,

$$ \tilde{t}=1$$

Note that it was very much neccesary that I wrote (1) in the way of (2) , otherwise I couldn't have done the $\tilde{x}=0$ substitution.

Finally, we can derive now the time dilation. Firstly, we see the $\tilde{t}$ is given by the equation $0=\tilde{x} = x-vt$, th

Answer 4

TrystinFreedom,I've read you nonsense (from textbooks):"But the basic point is the same as in the Spacetime Diagram analysis: you have two twins who take different paths through spacetime, and those paths have different lengths, and the lengths of the paths are the amounts that each twin ages during the trip. So the different ages of the twins when they meet again are no more mysterious than the fact that, if two twins take road trips between, say, New York and Los Angeles by different routes with different lengths, their odometers will read different mileages when they meet at the end even if they were the same at the start.

Reference: https://www.physicsforums.com/insights/when-discussing-the-twin-paradox-read-this-first/" You can put your cartoons in a .. toilet!I doubt you'v heard about "wine logische Schwache/Weekness (by A.Einstein,1920,Germ.Nat.Scient.Society)