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In my general relativity textbook, the following equation is given:

$$ds^2 = g_{\alpha \beta}(x)dx^{\alpha}dx^{\beta}$$

Which describes the line element $ds^2$ for a metric. $g_{\alpha \beta}$ is a matrix that is the metric itself. This equation implies that a given term of the equation for a line element can only have two differential factors. Either two different linear variables (For example: $dtdr$) or one variable squared (For example: $d\theta^2$).

Intuitively, this makes some sense. Our line element is $ds^2$, so that exponent of 2 could explain why, when you multiply things out, you can only get two differential factors per term, just like squaring any polynomial.

But this doesn't really make sense to me from a more formal perspective. I suppose the issue is that I don't really understand what the exponent of 2 is actually doing. So what is it doing? And what is the exact reason that we can only have two differential factors per term?

(Reading back this question, I'm not sure if everything is clear, so let me know if I can clarify anything or otherwise improve this question).

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    $\begingroup$ It’s a generalization of the Pythagorean Theorem of 2D Euclidean space, $c^2=a^2 + b^2$, to 4D curved spacetime on an infinitesimal scale. So the squares shouldn’t be surprising. In flat spacetime it is just $ds^2=-dt^2+dx^2+dy^2+dz^2$. In Euclidean space, you have probably seen that infinitesimal arc length along a curve is just $ds^2=dx^2+dy^2+dz^2$. $\endgroup$
    – G. Smith
    Commented Feb 21, 2020 at 1:11
  • $\begingroup$ I actually had the exact same question! You may want to take a look at my question about generalizations of Riemannian geometry that I asked on reddit. $\endgroup$
    – MaximusIdeal
    Commented Feb 21, 2020 at 1:58
  • $\begingroup$ One thing I realized from my own question is that it is sort of like asking "why does quantum mechanics use complex numbers but not quaternions?" One answer is that it is just empirically true. However, there is an interesting observation to be made: generalizing something a little bit gives you more features (like going from the real numbers to the complex numbers), but then there is a point where generalizing further somehow takes away features (ex: quaternions fail to have commutativity, and octonions fail associativity!). $\endgroup$
    – MaximusIdeal
    Commented Feb 21, 2020 at 2:05
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/295454/2451 and physics.stackexchange.com/q/233539/2451 $\endgroup$
    – Qmechanic ♦
    Commented Feb 21, 2020 at 3:59
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    $\begingroup$ The above expressione for $ds^2$ is trivially the norm of the differential $ds$ seen as an element of a metric vector space. Scalar products are defined as bilinear applications. Therfore one cannot have anything different from exponent 2. Notice however that this is not a question about physics but about math. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Feb 21, 2020 at 6:07

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The expression for the line element is a generalization of the Pythagorean Theorem of 2D Euclidean space,

$$c^2=a^2+b^2,$$

to 4D curved spacetime, on an infinitesimal scale. So the squares shouldn’t be surprising.

In 3D Euclidean space, you have probably seen that infinitesimal arc length along a curve is just

$$ds^2=dx^2+dy^2+dz^2.$$

In the flat (Minkowskian) spacetime of Special Relativity it is just

$$ds^2=-dt^2+dx^2+dy^2+dz^2.$$

For a general 4D curved spacetime, your expression, written out in $txyz$ coordinates, is

$$\begin{align}ds^2&=g_{tt}(t,x,y,z)\,dt^2+g_{xx}(t,x,y,z)\,dx^2+g_{yy}(t,x,y,z)\,dy^2+g_{zz}(t,x,y,z)\,dz^2 \\ &+2g_{tx}(t,x,y,z)\,dt\,dx+2g_{ty}(t,x,y,z)\,dt\,dy+2g_{tz}(t,x,y,z)\,dt\,dz\\ &+2g_{xy}(t,x,y,z)\,dx\,dy+2g_{xz}(t,x,y,z)\,dx\,dz+2g_{yz}(t,x,y,z)\,dy\,dz. \end{align}$$

The essential idea is that curved space(time) should be, over sufficiently small regions, very similar to flat space(time), in the same way a small patch of the Earth’s spherical surface seems like a plane. This means that you can't have something weird with different powers like

$$ds^2=-A dt^3+B dx^4+C dy^5+D dz^6.$$

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  • $\begingroup$ So, essentially, without this restriction, the equivalence principle wouldn't hold? $\endgroup$
    – RothX
    Commented Feb 21, 2020 at 18:18
  • $\begingroup$ @RothX Yes, I think that's right. $\endgroup$
    – G. Smith
    Commented Feb 21, 2020 at 18:19

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