# Can the roots of spacetime interval be linearly added up?

It seems ridiculous to add up roots of two intervals to get the total interval. Considering 3 events, $\Delta s_{13}$ shouldn't be equal to $\Delta s_{12}$+$\Delta s_{23}$. But if we consider the events in a moving particle's worldline, it seems reasonable to add up the proper times. What's wrong with the logic?

The square root form for $\Delta s$ in terms of $v$ is correct for infinitesimal intervals, and really you should integrate it to find a finite spacetime interval: $$\Delta s_{12}=\int_1^2 \sqrt{ds^2}.$$ So indeed if you consider three events, $$\Delta s_{13}=\int_1^3 \sqrt{ds^2}=\int_1^2 \sqrt{ds^2}+\int_2^3 \sqrt{ds^2}=\Delta s_{12}+\Delta s_{23}.$$

Of course if your trajectory has constant velocity the integral simplifies and it looks like you're summing the two square roots.

• To put it in a slightly more wordy but common-sense way: usually in 3D space when we add two vectors we have $|\vec v_1+\vec v_2|<|\vec v_1|+|\vec v_2|$ but when the vectors point in the same direction the $<$ turns into $=$, and in full generality we say $\le$. Spacetime has an analogous sense in which two 4-vectors can point in the "same direction," in which case their addition formula is direct. As long as events happen at the same place in a coordinate system, those spacetime intervals are precisely parallel and can be added just like $|\vec v_1+\vec v_2|=|\vec v_1|+|\vec v_2|$ can in 3D. Commented Nov 28, 2016 at 22:10
• @CRDrost That's a good exposition - put it as a separate answer - I'd upvote it - even if it does seem trivial to you. There are loads of highschool and freshmen physicists on this site and one shouldn't underestimate the worth of good technical writing even for very simple things. Commented Nov 28, 2016 at 22:35
• I think it's important to translate the two separate comments: what I was saying is that $\Delta s$ is not the norm of any vector, so you need to calculate its length in spacetime through the integral above. What you're saying is that if $\Delta s$ can be identified with the norm of a vector, the trajectory is a straight line, and then the sum of the norms is the norm of the sum if you break the trajectory into two separate parts. Commented Nov 28, 2016 at 22:52
• @user121664 Thanks for your answer first. But shouldn't $\Delta s_{12}=((\Delta x)^2-(c\Delta t)^2)^{1/2}$? Why is it stated as the integral of infinitesimal roots of intervals? Commented Nov 28, 2016 at 23:04
• That formula is valid to calculate the spacetime distance between two events along a straight line. If the events 1,2,3 are along a straight line, then you can "sum the square roots", as you say, which is what CR was getting at. If the particle is not moving in a straight line, that formula is not valid, and you need to use the integral form in my answer. Since that is the general rule, proper times do sum. See also the answer by @Electrodynamist. Commented Nov 28, 2016 at 23:43

For timelike intervals (the intervals that can be travelled by a particle),

$d\tau^2 = g_{\alpha \beta} dx^{\alpha}dx^{\beta}$

Hence, $d\tau = \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}$

Thus, the total interval $\tau_{13} = \int_1^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}$

Therefore,

$\tau_{13} = \int_1^2 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}} + \int_2^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}} = \tau_{12} + \tau_{23}$ . Always.

The point is $\tau_{13}^2 \neq \tau_{12}^2+\tau_{23}^2$.

Because, $d\tau^2 = {g_{\alpha \beta} dx^{\alpha}dx^{\beta}} = g_{\alpha\beta}U^{\alpha}U^{\beta} d\lambda^2$

($\lambda$ is the affine parameter used to paramatrize the curve (path) along which the particle travels and $U^{\mu} := \dfrac{dx^{\mu}}{d\lambda}$)

Thus, in order to integrate, we first take the square root and then only perform the integration. There is no meaning to directly saying $\int_1^3 d\tau^2$.

So, $\tau_{13}^2 = \big(\int_1^3 \sqrt{g_{\alpha \beta} dx^{\alpha}dx^{\beta}}\big)^2$ NOT that $\tau_{13} = \sqrt{\int_1^3 {g_{\alpha \beta} dx^{\alpha}dx^{\beta}}}$.

'Reduced Mathematics' Version: Your intuition is right. $\tau_{13} = \tau_{12} + \tau_{23}$. Reason is simply that we first take square root of $d\tau^2$ and then integrate - not the other way around. Thus, $\tau_{13}^2 \neq \tau_{12}^2+\tau_{23}^2$ but $\tau_{13} = \tau_{12} + \tau_{23}$.