Relative speed of two oppositely directed light beams

[Ibix]

It's not necessary to use light beams. If you observe two spaceships moving towards each other at +/- 0.6c then closing speed (rate at which the distance between them is decreasing) is 1.2c in your frame of reference. No physical law is violated.

A point of clarification on this: in Newtonian physics closing speed and relative velocity are equal. Not so in relativistic physics, where the closing speed is limited to $2c$ and relative velocity to $c$.

 

[Grasshopper]

And, as has been pointed out repeatedly, you are analysing this completely wrongly. You need to use Minkowski geometry, not Euclidean geometry.
No. In fact, the 4d analog of distance along a light path is zero (hence the alternative name "null worldline" for lightlike worldlines). And you can't define "speed" through 4d spacetime (Brian Greene notwithstanding) because time isn't a separate thing.

Hey real quick hijack question, then I’ll leave:

If you try to do the “Pythagorean theorem” in Minkowski space for light, it would look like this, right?
$s^2 = (ct)^2 - x_{1}^2 - x_{2}^2 - x_{3}^2 = 0$Or did I miss something?

 

[Nugatory]

Or did I miss something?

Nope, that’s pretty much got it. But the implications of subtracting instead of adding as we do with Euclidean geometry are huge - most of special relativity flows from that one difference.

 

[Ibix]

Or did I miss something?

Looks fine. Pedantically, you want some deltas in there ($\Delta s$, $\Delta t$ etc) because those are differences you are working with. The distinction doesn't matter much here, but becomes important in non-inertial frames and GR.