Differential Geometry¶
Metric¶
Definitions¶
Denote the basis in use as \(\hat e_\mu\), then the metric can be written as
if the basis satisfies
Inversed metric
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
Then we can construct the expression of metric of this point under this coordinate system,
For example, in spherical coordinate system,
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
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.
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.,
In this way, the metric tensor for spherical coordinates is
Connection¶
First class connection can be calculated
Second class connection isfootnote{Kevin E. Cahill}
Gradient, Curl, Divergence, etc¶
Gradient¶
Curl¶
For an anti-symmetric tensor, \(a_{\mu\nu}=-a_{\nu\mu}\)
Divergence¶
For an anti-symmetric tensor
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.
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, \(\Delta\equiv \nabla\cdot\nabla\equiv \nabla^2\)