How do I make spacetime diagram consistent with length contraction?

Question

Say the Earth and a distant galaxy is a distance $l$ away from each other. There is also a spaceship starting its journey at the Earth with a speed $u$ towards the galaxy.

I tried to draw a spacetime diagram of the situation (see below), with the $x$ and $ct$ axes representing a stationary frame of an outside observer, and the $x'$ and $ct'$ axes represents the rest frame of the ship (i.e. the ship's worldline is the $ct'$ line). I have also drawn the galaxy world line.

I also showed the length $l$ and $l'$ on the diagram, where $l'$ represents the distance someone on the ship would think they need to travel to reach the galaxy. From length contraction, I know that $l'$ is supposed to be shorter than $l$, but my spacetime diagram seems to suggest otherwise.

What is wrong with my logic or diagram?

Answer 1

In general, be cautious about diagrams. In these cases, beware of

1. A change of topology, e.g. in flat projections of the Earth.

2. A change in metrics, e.g. the implicit metric on certain atlas projections vs the natural metric on a sphere.

In the diagram above, the topology is not a problem. Minkowski space, denoted $\mathbb{R}^{1,d}$, is topologically the same as $\mathbb{R}^{d+1}$, so drawing the diagram on a flat piece of paper is topologically OK.

The problem you have comes from point 2: when one draws a diagram on flat paper, one implicitly wants to use the standard Euclidean metric $g = dt^2 + dx^2$. But in special relativity, we must use the Minkowski metric $g = -dt^2 + dx^2$.

For example, a line segment from $(t_0,x_0)$ to $(t_1,x_1)$ in Minkowski space has length $-\Delta t^2 + \Delta x^2$. As we rotate a line that originally points in the $x$ direction to the $t$ direction decreases, its length - as measured by the Minkowski metric - decreases (even though the length is preserved by the Euclidean metric). As the line segment becomes vertical, its length even becomes negative. (Of course, in such cases this length cannot correspond with a physical length of an object.) Note that the Minkowski metric does not even preserve ordering of lengths relative to the Euclidean metric.

In your case, you will find that the physical length does contract if you account for the Minkowski metric, even though the Euclidean length increases.

Finally, note that we can't use standard trigonometry for triangle in the $(t,x)$-plane: the length of the hypotenuse of a right angle triangle with sides $\Delta x$ and $\Delta t$ is $\sqrt{-\Delta t^2 + \Delta x^2}$. However, if you use the hyperbolic $\sinh$ and $\cosh$ instead, you will find that you can mimic standard trigonometry in the $(t,x)$. plane (exercise). For example, $l = l' \cosh \alpha$ in your diagram.

I seem to remember a wonderful little book by Wald called Spacetime and Gravity (or something like that), which explains this well.

Answer 2

Your diagram is correct. To see how length contraction emerges, first write down the expression for $L'$ in terms of $x$ and $t$: $$L'=\int_{x=0}^{x=L}\sqrt{ds^2}=\int_{x=0}^{x=L}\sqrt{dx^2-dt^2}$$ Drawn in this view, the $x'$ axis is $t=vx$, so $dt=vdx$. Hence: $$L'=\int_{x=0}^{x=L}\sqrt{dx^2-v^2dx^2}=\sqrt{1-v^2}\int_{x=0}^{x=L}dx=\gamma^{-1}L.$$ I think what you were confused about is the apparent "paradox" that $L'$ appears longer than $L$. This is certainly true if distances are measured like $ds^2=dx^2+dt^2$. But, distances in Minkowski spacetime are actually measured like $ds^2=dx^2-dt^2$, so it makes perfect sense that $L'<L$. A way to see this is to sketch out $$x^2-t^2=L^2,$$ the locus of spacelike points equidistant from the origin with length equal to $L$. Not surprisingly, $L'$ lies short of this hyperbola.

Answer 3

The tick marks on an axis scale as

$$ \sqrt{\frac{1+\beta^2}{1-\beta^2}}$$