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We know that (see this wikipedia page) in the metric of Minkowski spacetime: $$ds^2=(dt)^2-(dx)^2-(dy)^2-(dz)^2 \tag{1}$$ and we also know that in spherical coordinates this same metric becomes: $$ds^2=(dt)^2-(dr)^2-r^2(d\theta)^2-r^2\sin^2{\theta}(d\phi)^2 \tag{2}$$ Let's prove this last statement:
We have that: $$\begin{cases}t=t \\ x=r\sin{\theta}\cos{\phi} \\ y=r\sin{\theta}\sin{\phi} \\ z=r\cos{\theta}\end{cases}$$ we can think of $x,y,z$ as functions of $r,\theta,\phi$; so we get: $$dx=\sin{\theta}\cos{\phi}dr+r\cos{\theta}\cos{\phi}d\theta-r\sin{\theta}\sin{\phi}d\phi$$ and so on for $dy,dz$; then we can square to get $(dx)^2,(dy)^2,(dz)^2$ written in terms of $(dr)^2,(d\theta)^2,(d\phi)^2$. Now we can put our findings back into equation (1) and if all goes right we should find equation (2).

However I didn't get to the end of this calculation because the algebra gradually becomes unbearable, especially when you get to the squaring part, where terms with mixed differentials start to pop out. However seems to me that this method should work out fine.

My questions are: This method will lead to the correct solution (2)? And even if this method is indeed correct: is there a better method to demonstrate (2) from (1)? Where better here means simply less algebra.

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  • $\begingroup$ You can draw some diagrams in spherical coordinates and argue that the spatial part of (2) is “obvious” on geometrical grounds. Just think about the orthogonal displacements along $\hat r$, $\hat\theta$, and $\hat\phi$. $\endgroup$ – G. Smith Jul 19 '20 at 19:28
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    $\begingroup$ If you are doing GR, you should not consider a hand-calculation involving 19 terms to be unbearable. It should take a few minutes. $\endgroup$ – G. Smith Jul 19 '20 at 19:39
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Yes! There's a way more simple method of converting the metric into spherical co-ordinates. In cartesian co-ordinates, the expression of the metric is of the form

$$\mathrm ds^2=-c^2\mathrm dt^2+(\text{infinitesimal displacement})^2\tag{1}$$

In cartesian co-ordinates,

$$\text{infinitesimal displacement}=\sqrt{\mathrm dx^2+\mathrm dy^2+\mathrm dz^2}$$

So now, our task is to find such an infinitesimal displacementin spherical co-ordinates. This is a purely mathematical task. Let's start of with a figure.

enter image description here

Image source

In the above image, all the three paths are mutually perpendicular/orthogonal, so the net displacement will just be the

$$\text{infinitesimal displacement}=\sqrt{(\text{path 1})^2+(\text{path 2})^2+(\text{path 3})^2}\tag{2}$$

But it's easy to see that

\begin{align} \text{path 1}&=r\mathrm\: d \theta\\ \text{path 2}&=r\sin \theta \: \mathrm d\phi\\ \text{path 3}&=\mathrm dr \end{align}

And voila, substitute the above expressions into equation $(2)$ and subsequently into equation $(1)$ to get the desired result.

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    $\begingroup$ It should be stressed that, in contrast to the computation the OP complains about, this is just a heuristic that provides intuition, not a proof. $\endgroup$ – ACuriousMind Jul 20 '20 at 15:45
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    $\begingroup$ @ACuriousMind My answer does provide a proof (I can't see why it doesn't). However, it definitely doesn't prove it the way OP intended to prove it, and that's because the OP is explicitly asking for better methods. $\endgroup$ – user258881 Jul 20 '20 at 16:09
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    $\begingroup$ This is not a proof because the formal mathematical definition of things like $\mathrm{d}s^2$ has no formal relation to the picture you draw. The picture provides intuition, but it doesn't prove anything. $\endgroup$ – ACuriousMind Jul 20 '20 at 16:17
  • $\begingroup$ @ACuriousMind Thanks for your response! But AFAIC, that diagram has only been used to show the "infinitesimal displacement" and it is no way related to the general relativistic aspects of the metric. After all, the problem at hand was just a calculus problem, it was irrelevant to address any well-established physics in this answer, IMO. And in that view, I have used that diagram for better visualization of the three orthogonal displacements constituting the overall displacement. $\endgroup$ – user258881 Jul 20 '20 at 16:24
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$\newcommand{\vect}[3]{\left[ \begin{array}{c} #1 \\ #2 \\ #3 \end{array}\right]} \newcommand{\mat}[9]{\left[ \begin{array}{ccc} #1 & #2 & #3 \\ #4 & #5 & #6 \\ #7 & #8 & #9 \end{array}\right]} \def\st{\sin\th} \def\ct{\cos\th} \def\sf{\sin\f} \def\cf{\cos\f} \def\f{\varphi} \def\th{\theta} \def\VX{{\bf X}} \def\VY{{\bf Y}} \def\MM{{\bf M}} \def\MD{{\bf D}} \def\id{\mathbb{I}}$The other answers provide good geometric intuition. Here we give one way to do the brute force work in an organized fashion. Algebra is good for the soul!

We have \begin{align*} d\VX &= \vect{dx}{dy}{dz} \\ &= \mat{\st\cf}{r\ct\cf}{-r\st\sf} {\st\sf}{r\ct\sf}{r\st\cf} {\ct}{-r\st}{0} \vect{dr}{d\th}{d\f} \\ &= \underbrace{\mat{\st\cf}{\ct\cf}{-\sf} {\st\sf}{\ct\sf}{\cf} {\ct}{-\st}{0}}_\MM \underbrace{\mat{1}{0}{0} {0}{r}{0} {0}{0}{r\st}}_\MD \underbrace{\vect{dr}{d\th}{d\f}}_{d\VY}. \end{align*} Thus, \begin{align*} dx^2+dy^2+dz^2 &= d\VX^T d\VX \\ &= d\VY^T \MD \MM^T \MM \MD d\VY. \end{align*} The krux of the calculation is then in finding $\MM^T \MM$. By inspection, the columns of $\MM$ are orthogonal. By further inspection, the columns are orthonormal. Thus, \begin{align*} dx^2+dy^2+dz^2 &= d\VY^T \MD \id \MD d\VY \\ &= d\VY^T \MD^2 d\VY \\ &= dr^2+r^2d\th^2+r^2\st^2d\f^2. \end{align*}

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There is no time transformation and you can read the metric directly from a diagram. Small coordinate changes $dr$, $d\theta$, $d\phi $ correspond to displacement vectors with magnitudes $dr$, $rd\theta$, $r\sin\theta d\phi $. This is an orthogonal triad, so you can write down your eq $(2)$ immediately

enter image description here

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I understand $(2)$ literally geometric. At any point of a spherical surface of radius r like the earth surface, it is possible to get 3 perpendicular small vectors. One that is vertical to the local surface: $\Delta r$. One locally parallel to North-South direction: $r\Delta \theta$. And finally one parallel to the West-East direction: $r sin(\theta)\Delta \phi$.

Any other direction can be obtained by Pythagoras from that $3$ orthogonal base vectors.

But the algebric brute force method certainly works.

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you can obtain it like this with:

$$\vec{R}=\begin{bmatrix} t \\ x \\ y \\ z \\ \end{bmatrix}=\left[ \begin {array}{c} t\\ r\sin \left( \theta \right) \cos \left( \phi \right) \\ r\sin \left( \theta \right) \sin \left( \phi \right) \\ r\cos \left( \theta \right) \end {array} \right] $$

and

$$\vec{q}=\left[ \begin {array}{c} t\\ r\\ \theta\\ \phi\end {array} \right] \quad, \vec{dq}=\left[ \begin {array}{c} dt\\ dr\\ d\theta\\ d\phi\end {array} \right] $$

the metric $G$ is:

$$G=J^T\,\eta\,J$$

where $$\eta=\left[ \begin {array}{cccc} 1&0&0&0\\ 0&-1&0&0 \\ 0&0&-1&0\\ 0&0&0&-1\end {array} \right] $$ the signature Matrix

and $$J=\frac{\partial \vec R}{\partial \vec q}=\left[ \begin {array}{cccc} 1&0&0&0\\ 0&\sin \left( \theta \right) \cos \left( \phi \right) &r\cos \left( \theta \right) \cos \left( \phi \right) &-r\sin \left( \theta \right) \sin \left( \phi \right) \\ 0&\sin \left( \theta \right) \sin \left( \phi \right) &r\cos \left( \theta \right) \sin \left( \phi \right) &r\sin \left( \theta \right) \cos \left( \phi \right) \\ 0&\cos \left( \theta \right) &-r\sin \left( \theta \right) &0\end {array} \right] $$

thus:

$$G=\left[ \begin {array}{cccc} 1&0&0&0\\ 0&-1&0&0 \\ 0&0&-{r}^{2}&0\\ 0&0&0&-{r}^{2} \left( \sin \left( \theta \right) \right) ^{2}\end {array} \right] $$

and the line element

$$ds^2=\vec{dq}^T\,G\,\vec{dq}={{\it dt}}^{2}-{{\it dr}}^{2}-{d\theta }^{2}{r}^{2}-{d\phi }^{2}{r}^{2 } \left( \sin \left( \theta \right) \right) ^{2} $$

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When I do these calculations, I do not expand the squares but do book-keeping and simplifications in my head.

We get $$ \begin{cases} dx = \color{red}{dr \sin\theta \cos\phi} \color{green}{+ r \cos\theta\,d\theta \cos\phi} \color{blue}{- r \sin\theta \sin\phi\, d\phi} \\ dy = \color{red}{dr \sin\theta \sin\phi} \color{green}{+ r \cos\theta\,d\theta \sin\phi} \color{blue}{+ r \sin\theta \cos\phi \, d\phi} \\ dz = \color{red}{dr \cos\theta} \color{green}{- r \sin\theta \, d\theta} \end{cases} $$

When you calculate $dx^2+dy^2+dz^2$ the $\color{red}{\text{red}}$ parts squared will sum to $\color{red}{dr^2},$ the $\color{green}{\text{green}}$ parts squared to $\color{green}{r^2 \, d\theta^2},$ and the $\color{blue}{\text{blue}}$ parts squared to $\color{blue}{r^2 \sin^2\theta \, d\phi^2}.$ Then, checking the cross-terms, e.g. $\color{red}{\text{red}}$-$\color{green}{\text{green}}$ one can see that they all cancel.

Thus, $dx^2+dy^2+dz^2 = dr^2 + r^2 \, d\theta^2 + r^2 \sin^2\theta \, d\phi^2.$

If you can not handle the calculations in your head, take at least one kind of terms at a time:

$\color{red}{\text{red squared}}$: $$ (dr \sin\theta \cos\phi)^2 + (dr \sin\theta \sin\phi)^2 + (dr \cos\theta)^2 \\ = dr^2 \sin^2\theta \cos^2\phi + dr^2 \sin^2\theta \sin^2\phi + dr^2 \cos^2\theta \\ = dr^2 \sin^2\theta + dr^2 \cos^2\theta \\ = dr^2 $$

$\color{red}{\text{red}}\text{-}\color{green}{\text{green}}\text{ cross-term}$: $$ 2 \, dr \sin\theta \cos\phi \, r \cos\theta \, d\theta \cos\phi + 2 \, dr \sin\theta \sin\phi \, r \cos\theta \, d\theta \sin\phi - 2 \, dr \cos\theta \, r \sin\theta \, d\theta \\ = 2 \, r \, dr \, d\theta \sin\theta \cos\theta \cos^2\phi + 2 \, r \, dr \, d\theta \sin\theta \cos\theta \sin^2\phi - 2 \, r \, dr \, d\theta \sin\theta \cos\theta \\ = 2 \, r \, dr \, d\theta \sin\theta \cos\theta - 2 \, r \, dr \, d\theta \sin\theta \cos\theta \\ = 0. $$ and so on.

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