# General Relativity/Coordinate systems and the comma derivative

In General Relativity we write our (4-dimensional) coordinates as ${\displaystyle (x^{0},x^{1},x^{2},x^{3})}$. The flat Minkowski spacetime coordinates ("Local Lorentz frame") are ${\displaystyle x^{0}=ct}$, ${\displaystyle x^{1}=x}$, ${\displaystyle x^{2}=y}$, and ${\displaystyle x^{3}=z}$, where ${\displaystyle c}$ is the speed of light, ${\displaystyle t}$ is time, and ${\displaystyle x}$, ${\displaystyle y}$, and ${\displaystyle z}$ are the usual 3-dimensional Cartesian space coordinates.

A comma derivative is just a convenient notation for a partial derivative with respect to one of the coordinates. Here are some examples:

1. ${\displaystyle T_{\ \beta ,\gamma }^{\alpha }={\frac {\partial T_{\ \beta }^{\alpha }}{\partial x^{\gamma }}}}$

2. ${\displaystyle f_{,\mu }={\frac {\partial f}{\partial x^{\mu }}}}$

3. ${\displaystyle w_{\ ,\nu }^{\mu }={\frac {\partial w^{\mu }}{\partial x^{\nu }}}}$

4. ${\displaystyle \Gamma _{\ \beta \gamma ,\mu }^{\alpha }={\frac {\partial \Gamma _{\ \beta \gamma }^{\alpha }}{\partial x^{\mu }}}}$

If several indices appear after the comma, they are all taken to be part of the differentiation. Here are some examples:

1. ${\displaystyle S_{\alpha \ ,\mu \nu }^{\ \beta }=\left(S_{\alpha \ ,\mu }^{\ \beta }\right)_{,\nu }={\frac {\partial }{\partial x^{\nu }}}\left({\frac {\partial S_{\alpha }^{\ \beta }}{\partial x^{\mu }}}\right)={\frac {\partial ^{2}S_{\alpha }^{\ \beta }}{\partial x^{\nu }\partial x^{\mu }}}}$

2. ${\displaystyle f_{,\alpha \beta \beta }=\left[\left(f_{,\alpha }\right)_{,\beta }\right]_{,\beta }={\frac {\partial ^{3}f}{\partial ^{2}x^{\beta }\partial x^{\alpha }}}}$

Now, we change coordinate systems via the Jacobian ${\displaystyle x_{\ ,\nu }^{\mu }}$. The transformation rule is ${\displaystyle x^{\bar {\mu }}=x^{\mu }x_{\ ,\mu }^{\bar {\mu }}}$.

Finally, we present the following important theorem:

Theorem: ${\displaystyle x_{\ ,\mu }^{\alpha }x_{\ ,\beta }^{\mu }=\delta _{\beta }^{\alpha }}$

Proof: ${\displaystyle x_{\ ,\mu }^{\alpha }x_{\ ,\beta }^{\mu }=\sum _{\mu =0}^{3}{\frac {\partial x^{\alpha }}{\partial x^{\mu }}}{\frac {\partial x^{\mu }}{\partial x^{\beta }}}}$, which by the chain rule is ${\displaystyle {\frac {\partial x^{\alpha }}{\partial x^{\beta }}}}$, which is of course ${\displaystyle \delta _{\beta }^{\alpha }}$. ${\displaystyle \square }$