# Is this geometric argument enough to show that special relativity assumes flat spacetime?

I am preparing myself to teach a class about special relativity a few weeks from now. To make sure they'll understand that spacetime must be flat for special relativity to work, I came up with the following argument, which is based on the famous derivation of time dilation where a moving reference frame ($$S'$$) sees light take a longer path because of the motion between the reference frames. Basically, this derivation.

To make some context and define notation, here is the picture I am using: So there is a reference frame $$S$$, stationary relative to a light source. Another frame $$S'$$ moves away from the source with velocity $$v$$. Light is emitted and takes a time $$t$$ in the frame $$S$$ to reach the 'ceiling'. If we enter the reference frame $$S'$$ we see light take a different path due to the motion of frame $$S$$ (this causes the frame $$S'$$ to measure a time $$t'$$ until light reaches the 'ceiling'). The famous picture for this effect is: Now this is where my argument begins:

When we draw the motion of light in both frames, we can create a right triangle as shown. But to derive the time dilation formula we end up having to assume that the Pythagorean Theorem holds! Since the Lorentz transformations for special relativity depend on this step (and so all of Special Relativity does), it appears that all of Special Relativity depends on assuming that the Pythagorean Theorem will work.

But the Pythagorean theorem only holds if the geometry is euclidean! Therefore, Special Relativity will only work if space has euclidean geometry.

From here I would present the idea that to generalize Relativity to any geometry we'd need to use General Relativity (GR). But GR itself says that for small enough regions, spacetime can always be approximately flat, which means Special Relativity does hold on laboratory experiments, particle interactions etc. on Earth due to space being approximately euclidean for those cases.

And this is the argument I want to present. I am afraid, however, that this argument might not be correct for some reason (perhaps there is some mistake on the relationship between flat spacetime and euclidean space, but I'm not sure). I have an acceptable Special Relativity background so complex explanations and other (correct) arguments for this are welcome. My problem is that I am not so familiar with the math of GR, and this is could become a big problem for any arguments I come up here.

So I ask: is there a mistake in this argument? Or does it hold? Is it sufficient? If not, please correct me and/or link me to an actually good argument which I could present. My students have also never seen a tensor in their lives, so if possible avoid them. But anyway, all help is welcome!

• Consider that the equation you presented here might be confusing to your students, since you're mixing t and t' from the different reference frames on the right hand side. The drawing might also make it seem like the light originates from two different points in space time. – Halbeard Nov 30 '18 at 20:37
• A very rigorous and tightly reasoned treatment that does a good job on this sort of thing is Bertel Laurent, Introduction to spacetime: a first course on relativity. Another approach that may be of interest is in my SR book, lightandmatter.com/sr (sections 2.2 and 2.5). It's relatively easy to take a typical exposition of SR and pick out one or more places where flatness has been implicitly assumed but never explicitly invoked. It's more difficult to build up all the scaffolding without noticing that you've made such implicit assumptions. – Ben Crowell Nov 30 '18 at 20:50

No, flat space doesn't guarantee flat spacetime. For example, consider a spacetime with a weak gravitational field, corresponding to gravitational potential $$\phi$$. It can be described by the metric $$g_{00} = - (1 + 2 \phi), \quad g_{ij} = \delta_{ij}$$ which has perfectly flat space but non-flat spacetime.