# Distance between two events

I would like to start from the very basic such as Pythagoras Theorem. If I want to calculate distance between two points in a right-angled Triangle We use $$c^{2}= a^{2}+b^{2}$$; Using this We are quite capable to calculate the distance between two points in a 2D space. For 3D this formula is slightly modified and We use $$ds^{2}=dx^{2}+dy^{2}+dz^{2}$$. But for a 4D we used to calculate distance (rather interval between two events) through this equation $$\boxed{ds^{2}=dx^{2}+dy^{2}+dz^{2}-c^{2}dt^{2}}$$

Can anybody explain this modification?(looking for mathematical motivation)

• This is defined to be the "metric interval", and equals zero for light. physics.stackexchange.com/questions/304799/… – ProfRob Jul 6 '20 at 6:47
• It encapsulates the fact that the speed of light is the same for all observers. Different observers ascribe, x, y, z and t coordinates to events differently, but in way that preserves this "distance" (just as non-moving observers rotated from one another may ascribe different x,y and z coordinates, but will agree on distances). So all observers agree that lightrays connect points at "distance" zero. – Steve Linton Jul 6 '20 at 8:38
• I think the important thing for any answer to cover is: exactly what dt represents there. For example, in a static system everyone sees the same spatial values because they are on the "same" clock. – Carl Witthoft Jul 6 '20 at 15:17
• This is the interval that all observers in inertial reference frames will agree is the same, since they also measure the same speed of light as @SteveLinton notes. There’s a mathematical motivation for it here: en.wikipedia.org/wiki/…. – ELNJ Jul 7 '20 at 12:08

We could define a distance $$ds'$$ in four-dimensional space-time as

$$(ds')^2 = (dx)^2 + (dy)^2 + (dz)^2 + (cdt)^2$$

This would be a Euclidean distance. The problem is that experimental observation shows that observers in different inertial reference frames disagree when they measure this $$ds'$$ distance between the same two points (events) in spacetime. This suggests it is not a physically meaningful value, since it depends on how the observer is moving.

However, if we use the Minkowski distance $$ds$$ defined by

$$(ds)^2 = (dx)^2 + (dy)^2 + (dz)^2 - (cdt)^2$$

we find that observers in different inertial frames agree on the value of $$ds$$ (at least if we restrict ourselves to flat spacetime), so this is an objective value (independent of how the observer is moving) and so more useful.

An everyday analogy is that when talking about locations in England (say) it is more useful to use the objective measure "distance from the centre of London" than the subjective measure "distance from where I am right now".