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I have not studied physics but I was browsing Carroll's relativity book and randomly stumbled upon the following which I would like to understand mathematically. It says

$$ds^{2} = 0 = - \left( 1 - \frac{2GM}{r} \right) dt^{2} + \left( 1 - \frac{2GM}{r} \right)^{-1} dr^{2},$$ from which we see that $$\frac{dt}{dr} = \pm \left( 1 - \frac{2GM}{r} \right)^{-1}.$$

I struggle to understand how this was obtained. For simplicity, call $a = \left( 1 - \frac{2GM}{r} \right)$. It is immediate that $$\frac{dt^2}{dr^2} = a^{-2}$$ however I don't see why this is the same as $(\frac{dt}{dr})^2$. Unless I completely misunderstand the notation, mathematically wouldn't $\frac{dt^2}{dr^2}$ mean that we differentiate the function $t^2$ with respect to the variable $r^2$? For example, in calculus if we have a function $f(x)$ then $\frac{df^2}{dx^2}$ is clearly not the same as $(f'(x))^2$. What am I fundamentally misundersanding?

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    $\begingroup$ In general relativity there is some terrible notation. For example $dt^2 $ is not $d(t^2)$ but rather $dt^2 = (dt)^2 = (dt) \cdot (dt)$. But even then the argument seems to be purely formal. $\endgroup$
    – jd27
    Commented Apr 27 at 9:31
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    $\begingroup$ To add a bit more to the comment above: one should see $dt$ (and $ds$ etc) as functions on the tangent spaces ($dt$ and $dr$ are linear functions there), $dt^2$ means (as mentioned above) $(dt)^2$; the equation $dr/dt = \dots$ is again an equation on the tangent spaces (or tangent bundle, if you know the terminology) and it is the equation for the isotropic vectors. $\endgroup$
    – user8268
    Commented Apr 27 at 9:50
  • $\begingroup$ These operations are as benign as separation of variables when solving certain ODEs. Rigor comes back when we check that $t(r)$ solves Carroll's ODE. Hence the short cut worked, terrible notation or not. $\endgroup$
    – Kurt G.
    Commented Apr 27 at 13:17
  • $\begingroup$ Oh I see, thanks! $\endgroup$
    – Esoog
    Commented Apr 27 at 16:15
  • $\begingroup$ In differential geometry, $(dt)^2$ is the symmetric product of $dt$ and $dt$. So this acts on pairs of vectors, $(dt)^2(u, v) = u_t v_t$. You can probably justify the author's solution this way. $\endgroup$
    – Kakashi
    Commented Apr 27 at 17:49

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This can be made rigorous, but this is one of the instances in which the notation is chosen so that the calculations get intuitive.

$ds^2$ is just a bookkeeping notation to express the metric tensor, which you can think of as a non-positive definite inner product between vectors defined on spacetime. Now the actual calculation Carroll is doing is computing the inner product between the four-velocity (the four-dimensional analog of velocity) with itself. The four-velocity has components $$u^\mu = \left(\frac{dt}{d\lambda},\frac{dr}{d\lambda},\frac{d\theta}{d\lambda},\frac{d\phi}{d\lambda}\right)$$ for some parameter $\lambda$ labeling the curve to which the four-velocity is parallel. From Carroll's expression, I can tell he is assuming that $\frac{d\theta}{d\lambda} = \frac{d\phi}{d\lambda} = 0$. Calculating the inner product of $u^\mu$ with itself then tells you that $$0 = - \left( 1 - \frac{2GM}{r} \right) \left(\frac{dt}{d\lambda}\right)^{2} + \left( 1 - \frac{2GM}{r} \right)^{-1} \left(\frac{dr}{d\lambda}\right)^{2},$$ where he had previously assumed that the inner product of $u^\mu$ with itself was zero. Now, both $t$ and $r$ can be understood as functions of a single variable $\lambda$. You can then use the chain rule in the expression above to get to Carroll's final expression.

Of course, when doing physics we are not interested in all of these intermediate steps: they only slow you down. So we work with this notation that allows you to do all of these procedures way more quickly, so the notation ends up doing the hard work for you.

I want to point out that this is good notation. You make the notation do the hard work for you. If you want to write rigorous proof, this notation might be bad, but when we use it we do not want to write rigorous proofs: we want to do calculations. Hence, we are just using Axiom 7 (Suggestiveness II) of this beautiful answer by Terry Tao. The notation already implements many important results that we use all the time so we can make short cuts and get to the result faster and with less work. This is the same basic idea behind Dirac notation in quantum mechanics.

Each notation has its purpose. If you use it for the wrong purpose, of course it will be bad. But using it for its intended purpose can simplify many calculations.

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