Ibix said:
It's the square of the elapsed time and the square of the spatial distance.
No, it isn't. It's the squares of
coordinate differentials. That makes a difference. See below.
Paige_Turner said:
"One subtracts elapsed time from the spatial distance."
No. One subtracts the
coordinate time differential from the
coordinate spatial differentials. Coordinate time differential is not the same as elapsed time, and coordinate spatial differentials are not the same as spatial distance.
In other words, if you are interpreting ##dx^2## as "spatial distance in the x direction", and similarly for ##dy^2## and ##dz^2##, and if you are interpreting ##dt^2## as "elapsed time", you are incorrect. Those are just coordinate differentials, not actual measurements.
Let me elaborate on this a bit. Let's leave out the ##y## and ##z## directions for simplicity and just consider motion in the x direction. Suppose we have a straight line in spacetime that goes from ##(x,t)=(0,0)## to ##(x,t)=(1,2)##. We compute the squared interval as ##ds^2=dx^2−dt^2=1^2−2^2=−3##. What is this telling us? Per my previous post, it is telling us (a) that we have an interval of ##\sqrt{3}## (b) which we interpret as an elapsed time (proper time) on a clock that follows the given path through spacetime, i.e., a clock whose worldline is a straight line from ##(0,0)## to ##(1,2)##.
Notice that the above, in itself, tells us
nothing about
any spatial distance. If we want to interpret ##dx## as a spatial distance, we have to add more observers to our scenario: we need to have two observers, one at ##x=0## and one at ##x=1##, who are at rest relative to each other and in the frame in which we are doing our calculations. Then we could say that ##dx=1## is the spatial distance between those two observers, because if we take both of those observers at the same coordinate time, say ##t=0##, then the squared interval between them is ##ds^2=dx^2−dt^2=1^2−0^2=1##, which, per my previous post, we interpret (a) as an interval of ##1## (b) which is a spatial distance, i.e., the length measured on a ruler along the spacelike line from ##(0,0)## to ##(1,0)##. Then, since the two observers are at rest in our frame, we could say that this is also the spatial distance that the clock travels; but note that the clock does
not go from ##(0,0)## to ##(1,0)##, it goes from ##(0,0)## to ##(1,2)##. So interpreting ##dx## as a spatial distance requires us to look at a
different interval (a different line through spacetime) than the one the clock is traveling on.
Similar remarks apply if we want to interpret ##dt##, the coordinate time difference, as an "elapsed time"; it is
not the elapsed time for the clock itself (that is ##\sqrt{3}##, as we have already shown), and interpreting it as "elapsed time" for either of our other two observers (the one at rest at ##x=0## or the one at rest at ##x=1##) requires looking at
different spacetime intervals--different lines through spacetime--than the one the clock is traveling on. (I'll leave you to think about which lines through spacetime those are, and how to calculate their intervals to get an elapsed time of ##2##.)
I agree that, once you have gone through the above reasoning, there is a strong temptation to think about the coordinate differentials ##dx## and ##dt## as "spatial distance" and "elapsed time" anyway, because it seems innocuous--after all, they turn out to be the same, right? But it isn't innocuous. Aside from the fact that doing that seems to be causing you confusion, it also is a bad habit to get into because all of this nice, simple arrangement where coordinate intervals end up being exactly the same as spatial distances or elapsed times in some frame stops working when you add gravity to the mix and have to use curved spacetime (i.e., general relativity) instead of flat spacetime (which is what we're dealing with here). In a curved spacetime, in general, coordinate differentials have no physical meaning, and thinking of them as having physical meaning causes all sorts of problems. (Note that we can even find coordinates in flat spacetime that raise similar issues--for example, null coordinates or Rindler coordinates.)
Also, in curved spacetime, the metric can get much more complex than just putting a minus sign in front of ##dt^2##. So asking "why" the metric takes a particular form gets even more problematic--unless you're willing to accept the only real answer, which is "because that's what is required to correctly describe the spacetime geometry". We don't start with coordinates; we start with a spacetime geometry, and we have to figure out how to describe it using coordinates. The minus sign in the metric of flat Minkowski spacetime is there because we've found that it's necessary to correctly describe the geometry of flat Minkowski spacetime using Cartesian ##x##, ##y##, ##z##, ##t## coordinates. That's really all there is to it.