Differential Geometry



Denote the basis in use as \(\hat e_\mu\), then the metric can be written as

\[g_{\mu\nu}=\hat e_\mu \hat \cdot e_\nu\]

if the basis satisfies

Inversed metric

\[g_{\mu\lambda}g^{\lambda\nu}=\delta_\mu^\nu = g_\mu^\nu\]

How to calculate the metric

Let’s check the definition of metric again.

If we choose a basis \(\hat e_\mu\), then a vector (at one certain point) in this coordinate system is

\[x^a=x^\mu \hat e_\mu\]

Then we can construct the expression of metric of this point under this coordinate system,

\[g_{\mu\nu}=\hat e_\mu\cdot \hat e_\nu\]

For example, in spherical coordinate system,

(6)\[\vec x=r\sin \theta\cos\phi \hat e_x+r\sin\theta\sin\phi \hat e_y+r\cos\theta \hat e_z\]

Now we have to find the basis under spherical coordinate system. Assume the basis is \(\hat e_r, \hat e_\theta, \hat e_\phi\). Choose some scale factors \(h_r=1, h_\theta=r, h_\phi=r\sin\theta\). Then the basis is

\[\hat e_r=\frac{\partial \vec x}{h_r\partial r}=\hat e_x \sin\theta\cos\phi+\hat e_y \sin\theta\sin\phi+\hat e_z \cos\theta,\]

etc. Then collect the terms in formula (6) is we get \(\vec x=r\hat e_r\), this is incomplete. So we check the derivative.

\[ \begin{align}\begin{aligned}\mathrm d\vec x = \hat e_x (\mathrm dr \sin\theta\cos\phi+r\cos\theta\cos\phi\mathrm d\theta-r\sin\theta\sin\phi\mathrm d\phi)\\\hat e_y (\mathrm dr\sin\theta\sin\phi+r\cos\theta\sin\phi\mathrm d\theta+r\sin\theta\cos\phi\mathrm d\phi)\\\hat e_z (\mathrm dr\cos\theta-r\sin\theta\mathrm d\theta)\\ = \mathrm dr(\hat e_x\sin\theta\cos\phi +\hat e_y \sin\theta\sin\phi -\hat e_z \cos\theta)\\\mathrm d\theta (\hat e_x\cos\theta\cos\phi +\hat e_y \cos\theta\sin\phi - \hat e_z \sin\theta)r\\\mathrm d\phi (-\hat e_x\sin\phi +\hat e_y \cos\phi)r\sin\theta\\=\hat e_r\mathrm dr+\hat e_\theta r\mathrm d\theta +\hat e_\phi r\sin\theta\mathrm d \phi\end{aligned}\end{align} \]

Once we reach here, the component (\(e_r ,e_\theta, e_\phi\)) of the point under the spherical coordinates system basis (\(\hat e_r, \hat e_\theta, \hat e_\phi\)) at this point are clear, i.e.,

\[\begin{split}\mathrm d\vec x = \hat e_r\mathrm d r+\hat e_\theta r\mathrm d \theta+\hat e_\phi r\sin\theta \mathrm d\phi \\ = e_r\mathrm d r+e_\theta \mathrm d\theta+e_\phi \mathrm d\phi\end{split}\]

In this way, the metric tensor for spherical coordinates is

\[\begin{split}g_{\mu\nu}=(e_\mu\cdot e_\nu) = \begin{pmatrix} 1 & 0 & 0 \\ 0 & r^2 & 0 \\ 0 & 0 & r^2 \sin^2\theta \end{pmatrix}\end{split}\]


First class connection can be calculated

\[\Gamma^\mu_{\phantom{\mu}\nu\lambda}=\hat e^\mu\cdot \hat e_{\mu,\lambda}\]

Second class connection isfootnote{Kevin E. Cahill}


Gradient, Curl, Divergence, etc


\[T^b_{\phantom bc;a}= \nabla_aT^b_{\phantom bc}=T^b_{\phantom bc,a}+\Gamma^b_{ad}T^d_{\phantom dc}-\Gamma^d_{ac}T^b_{\phantom bd}\]


For an anti-symmetric tensor, \(a_{\mu\nu}=-a_{\nu\mu}\)

\[\begin{split}\mathrm{Curl}_{\mu\nu\tau}(a_{\mu\nu}) \equiv a_{\mu\nu;\tau}+a_{\nu\tau;\mu}+a_{\tau\mu;\nu} \\ = a_{\mu\nu,\tau}+a_{\nu\tau,\mu}+a_{\tau\mu,\nu}\end{split}\]


\[\begin{split}\mathrm{div}_\nu(a^{\mu\nu})&\equiv a^{\mu\nu}_{\phantom{\mu\nu};\nu} \\ & = \frac{\partial a^{\mu\nu}}{\partial x^\nu}+\Gamma^\mu_{\nu\tau}a^{\tau\nu}+\Gamma^\nu_{\nu\tau}a^{\mu\tau} \\ & = \frac1{\sqrt{-g}}\frac{\partial}{\partial x^\nu}(\sqrt{-g}a^{\mu\nu})+\Gamma^\mu_{\nu\lambda}a^{\nu\lambda}\end{split}\]

For an anti-symmetric tensor

\[\mathrm {div}(a^{\mu\nu})=\frac1{\sqrt{-g}}\frac{\partial}{\partial x^\nu}(\sqrt{-g}a^{\mu\nu})\]

Annotation Using the relation \(g=g_{\mu\nu}A_{\mu\nu}\), \(A_{\mu\nu}\) is the algebraic complement, we can prove the following two equalities.

\[V^\mu_{\phantom\mu;\mu}=\frac1{\sqrt{-g}}\frac{\partial}{\partial x^\mu}(\sqrt{-g}V^\mu)\]

In some simple case, all the three kind of operation can be demonstrated by different applications of the del operator, which \(\nabla\equiv \hat x\partial_x+\hat y\partial_y+\hat z \partial_z\).

  • Gradient, \(\nabla f\), in which \(f\) is a scalar.
  • Divergence, \(\nabla\cdot \vec v\)
  • Curl, \(\nabla \times \vec v\)
  • Laplacian, \(\Delta\equiv \nabla\cdot\nabla\equiv \nabla^2\)