Drawing spacetime diagrams assuming absolute time (page-6,Section 1.5, Schutz)

Question

Suppose an observer $O$ uses the coordinates $t,x$ as above, and that another observer $\tilde{O}$ with coordinates $\tilde{t}, \tilde{x}$ is moving with velocity $v$ in the $x$ direction relative to $O$. Where do the coordinate axes for $\tilde{t}$ and $\tilde{x}$ go in the spacetime diagram of $O$?

$\tilde{t}$ axis: This is the locus of events at constant $\tilde{x}=0$ , which is the locus of the origins of $\tilde{O}$'s spatial coordinates. This is $\tilde{O}$'s world as shown in fig $1.2$

Note: Slightly paraphrased.

In this above picture I am a bit confused when I try to compare it with how things would like in Newtonian case. In the Newtonian case, we assume that there is an absolute time function, so that suggests that even when we implant the coordinate grid of an observer whose coordinate grids are $(\tilde{t},\tilde{x})$ onto the $(t,x)$ plane, we should have the $\tilde{t}$ axis coinciding with the $t$ axis.

However, according to Schutz's arguement, it suggests that axis of second observer should be bent relative to first observer even Newtonian case. Since the points of $\tilde{x}=0$ would still be a sloped line against the time axis.

What am I missing here?

Related

Answer 1

On a Newtonian or Galilean Spacetime diagram (the PHY 101 position vs time graph), the circle is a horizontal line in your diagram. (In a Minkowski diagram, the circle is a hyperbola.)

Lines of simultaneity (which correspond to the definition of “perpendicular” to inertial worldlines) are defined by tangents to circles.

In the Galilean case, all tangents to the unit Galilean circle conincide. This implies absolute time.

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UPDATE

I added a set of links to your other question on hyperbolas in special relativity: https://physics.stackexchange.com/a/709878/148184 .

The question here about absolute time is about the E=0 case in my:
robphy's spacetime diagrammer for relativity v.8e-2021 (time upward) https://www.desmos.com/calculator/emqe6uyzha (play with the E-slider)

Answer 2

If you're considering a Newtonian/Galilean change of coordinates, the formulas are related as \begin{align} \bar{t}=t,\quad\bar{x}=x-vt. \end{align} This means take any point $p$ in the spacetime. It has coordinate representations $(t(p),x(p))$ and $(\bar{t}(p),\bar{x}(p))$. These tuples of numbers are then related by $\bar{t}(p)=t(p)$ and $\bar{x}(p)=x(p)-v\cdot t(p)$.

The fact that $\bar{t}=t$ implies that in the picture it is the $\bar{x}$ and $x$ axes which coincide. Why? It's a simple matter of unwinding definitions \begin{align} \text{$p$ lies on $x$-axis} &\iff t(p)=0\\ &\iff \bar{t}(p)=0\tag{since $t=\bar{t}$}\\ &\iff \text{$p$ lies on $\bar{x}$-axis} \end{align}

So what you're confusing is that the functions $t,\bar{t}$ being equal does not mean their axes are aligned (because the axis defined by a coordinate function describes where all the remaining coordinates vanish). The functions $t,\bar{t}$ being equal implies that their level-sets are the same. So, if on that piece of paper if you draw a horizontal line, then that tells you all the points of constant $t$, and hence equivalently, all the points of constant $\bar{t}$. So, on the $(t,x)$ plane:

• Level sets of $t$ are straight horizontal lines
• Level sets of $x$ are straight vertical lines
• Level sets of $\bar{t}$ are straight horizontal lines
• Level sets of $\bar{x}$ are straight lines of slope $\frac{1}{v}$ (we have the fraction because of the way the axes are drawn).

So, if I give you an arbitrary point $p$ in the $(t,x)$ plane, you can immediately tell me what its various coordinates $(t(p),x(p))$ and $(\bar{t}(p),\bar{x}(p))$ are (see figure below). Given the point $p$, draw a horizontal line passing through $p$; this intersects the $t$-axis at a point $q$. Now, draw a vertical line through $p$; this will intersect the $x$-axis at some point $s$. Lastly, draw a line of slope $\frac{1}{v}$ passing through $p$; this will intersect the $x$-axis at a point $r$. So, putting together everything I've said:

• $t(p)=t(q)$ and $x(p)=x(s)$
• $\bar{t}(p)=t(p)=t(q)$ (first equality is because $\bar{t}=t$ by hypothesis) and $\bar{x}(p)=\bar{x}(r)=x(r)$ (the final equality is because $t(r)=0$).

So, the coordinates of $p$ are: $(t(p),x(p))=(t(q),x(s))$ and $(\bar{t}(p),\bar{x}(p))=(t(q),x(r))$, and the tuples on the RHS of these equations are immediately read-off from the picture. So, if you have explicit scales on your picture, you can immediately tell what the numerical values are.