What is the $ds^2$ notation in relativistic physics?

Question

Could someone please explain me intuitively how $ds^2$ represents distance in relativistic physics?

Answer 1

$s^2 = x^2 + y^2 + z^2 - (ct)^2$, where x, y, z are the usual measures of distance, c is the speed of light, and t is the usual measure of time; then s is the space-time interval.

The space-time interval, s, is a relativistic invariant, giving the same measure between two distinct space-time events, without regard to the relative velocities of observers.

The notation $ds^2$ then refers to a small increment in the space-time interval.

Answer 2

If you wanted to find the length of a curve in 3d you could break it into pieces and replace each piece with a straight line segment and add up the lengths of each piece. This would be an under estimate, but it would get closer and closer as you broke it into smaller and smaller pieces.

To find the length of each piece you could find the $x,$ $y,$ and $z$ coordinates of each end of the line segment and then compute the differences and call them $\Delta x=x_2-x_1,$ $\Delta y=y_2-y_1,$ and $\Delta z=z_2-z_1$. Then you could compute $\Delta s^2=\Delta x^2+\Delta y^2+\Delta z^2$. Then the length is $\sqrt{\Delta s^2}$.

So what's the difference between $\mathrm ds^2$ and $\Delta s^2$? The latter depends on two points whereas the former holds at just one point. For each line segment the latter is correct. But if you start breaking it into small segments you could start using $\mathrm ds^2$ instead and you'll get the wrong answer for thebline segment, but they will approach the right answer as the segments get small. It's a calculus thing. But it's really hard to see the difference if you use the $x,$ $y,$ $z$ coordinate system. So you could use altitude $r$ (measured in meters) and angle from the north pole $\theta$ and angle from the prime meridian $\phi$ (both angles measured in radians, a unit of angle designed to make calculus formulas easier).

Then $$\mathrm ds^2=\mathrm dr^2+r^2\left(\mathrm d\theta^2+(\sin^2\theta)\mathrm d\phi^2\right).$$

To evaluate it you need one $r,$ and $\theta$ to evaluate $r^2$ and $\sin^2\theta$ and then you can compute $\Delta r$ to use for $\mathrm dr$ and compute $\Delta \theta$ for $\mathrm d\theta$ and compute $\Delta \phi$ for $\mathrm d\phi$. This doesn't give you the actual length of that line segment. But the limit of the sum of the lengths will be the right answer.

Now we can talk about relativity. You still want to be able to compute lengths. So you still compute a $\mathrm ds^2$ for each small segment and now you either have to take $\sqrt{-\mathrm ds^2}$ or $\sqrt{\mathrm ds^2}$ to get the length of a little piece because some $\mathrm ds^2$ are positive and some $\mathrm ds^2$ are negative. If you know your curve will have a positive or negative $\mathrm ds^2$ then you'll know which to do.

If you don't know. It might be trickier. But usually you do know. Just like you usually know if you have a ruler or a clock that you are measuring with.

So the $\mathrm d$ part tells you to only use it for small segments and only expect the right answer in the limit and smaller and smaller segments. The square in $\mathrm ds^2$ tells you that you need to take a square root to get a length of the little segment. So the notation is supposed to remind you what to do and how to use it.

Answer 3

ds is an infinitesimal distance along the geodesic. In General Relativity all particles and photons travel along a geodesic, calculated by solving a tensor differential equation, and applying initial conditions.