# How is the spherical coordinate metric tensor derived? [closed]

I know the values of the metric tensor is $$\eta =\begin{bmatrix} 1&0&0\\ 0&r^{2}&0\\ 0&0&r^{2}\sin\left ( \theta \right )^{2} \end{bmatrix},$$ but how is this derived? Also, is the '(Non)Euclidean'-ness of the spacetime geometry of any relevance to this metric tensor value?

## closed as off-topic by sammy gerbil, ZeroTheHero, Kyle Kanos, Yashas, AccidentalFourierTransformMar 29 '17 at 11:13

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• Would Mathematics be a better home for this question? – Qmechanic Mar 27 '17 at 21:06
• I noticed the answers below assume that you know how to make the coordinate transformation from Cartesian coordinates to spherical coordinates? – Rumplestillskin Mar 28 '17 at 0:33
• You can vote it for migration to the mathse. – peterh Mar 29 '17 at 6:08
• I thought it more adequate for physics since it has more relevance to physics. The mathematical derivation is in the context of physics. – Brian Ko Mar 29 '17 at 10:16

That is simply the metric of an euclidean space, not spacetime, expressed in spherical coordinates. It can be the spacial part of the metric in relativity.

We have this coordinate transfromation:

$$x'^1= x= r\, \sin\theta \,\cos\phi =x^1 \sin(x^2)\cos(x^3)$$

$$x'^2= y= r\, \sin\theta \,\sin\phi =x^1 \sin(x^2)\sin(x^3)$$ $$x'^3= z= r\, \cos\theta = x^\ \cos(x^2)$$

With $\, x^1=r, \quad x^2=\theta, \quad x^3=\phi \quad$ and $\quad x'^1=x, \quad x'^2=y, \quad x'^3=z$

Now you start from

$$\eta_{ij} = \frac{\partial {x'^1}}{\partial {x^i}} \frac{\partial {x'^1}}{\partial {x^j}} +\frac{\partial {x'^2}}{\partial {x^i}}\frac{\partial x'^2}{\partial x^j} + \frac{\partial {x'^3}}{\partial {x^i}}\frac{\partial x'^3}{\partial x^j}$$

And doing it for each component you obtain the result you're looking for. I'll illustrate the case for $\eta_{22}$

$$\eta_{22}= \frac{\partial {x'^1}}{\partial {x^2}} \frac{\partial {x'^1}}{\partial {x^2}} +\frac{\partial {x'^2}}{\partial {x^2}}\frac{\partial x'^2}{\partial x^2} + \frac{\partial {x'^3}}{\partial {x^2}}\frac{\partial x'^3}{\partial x^2} = \\ \frac{\partial {x}}{\partial {\theta}} \frac{\partial {x}}{\partial {\theta}} +\frac{\partial {y}}{\partial {x^2}}\frac{\partial y}{\partial \theta} + \frac{\partial {z}}{\partial {\theta}}\frac{\partial z}{\partial \theta} = \\ r^2 \cos^2\theta \, \cos^2\phi + r^2 \cos^2\theta \sin^2\phi + r^2 \sin^2\theta = r^2$$

Where use has been made of the well known relation $\quad$ $\sin^2 \alpha +\cos^2\alpha=1$

• I just fixed a typo at the last line. I didn't delete first sentence. – lucas Mar 27 '17 at 21:03
• @lucas I know, thanks. It probably happened cause you submitted your edit a few seconds after mine, so you edited the post as it was before I put that first line in. So I edited it again to insert the first line. – Run like hell Mar 27 '17 at 21:04

In spherical one can show that the line element $$ds^2=dx^2+dy^2+dz^2= dr^2+r^2d\theta^2+r^2\sin^2\theta\,d\phi^2= g_{ij}d\xi_id\xi_j$$ with $(\xi_1,\xi_2,\xi_3)=(x,y,z)$ or $(r,\theta,\phi)$, and the usual \begin{align} z&=r\cos\theta\, ,\qquad\qquad\qquad x=r\sin\theta\cos\phi\, ,\quad y=r\sin\theta\sin\phi\, ,\\ dz&=\cos\theta\,dr-r\sin\theta d\theta\qquad\hbox{etc.} \end{align} From $ds^2$ one can just read off the entries as the coefficients of $dr^2$, $d\theta^2$ and $d\phi^2$.