General Relativity Homework 2

Chasse_neige

Problem 1

Suppose $T^{\mu }{}_{\rho \sigma }$ is a tensor under the general coordinate transformation. Please show that

$$T^{\mu }{}_{\rho \sigma ;\nu }\equiv T^{\mu }{}_{\rho \sigma ,\nu }+{\mathrm{\Gamma }}^{\mu }{}_{\nu \tau }{T}^{\tau }{}_{\rho \sigma }-{\mathrm{\Gamma }}^{\kappa }{}_{\rho \nu }{T}^{\mu }{}_{\kappa \sigma }-{\mathrm{\Gamma }}^{\kappa }{}_{\sigma \nu }{T}^{\mu }{}_{\rho \kappa }$$

is also a tensor under general coordinate transformation.

Transformation Rule for Christoffel Symbol

Given that the Christoffel Symbol’s transformation rule is

$$\begin{array}{c}{\mathrm{\Gamma }}^{\prime \lambda }{}_{\rho \sigma }=\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {{\xi }^{\alpha }}^{}}\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\prime \rho }\partial x^{\prime \sigma }}=\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {{\xi }^{\alpha }}^{}}\frac{{\partial }^{}}{\partial {x^{\prime \rho }}^{}}\left(\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\nu }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \sigma }}^{}}\right)\\ =\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {{\xi }^{\alpha }}^{}}(\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\nu }}^{}}\frac{{\partial }^2{x}^{\nu }}{\partial x^{\prime \rho }\partial x^{\prime \sigma }}+\frac{{\partial }^{}{x}^{\mu }}{\partial {x^{\prime \rho }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \sigma }}^{}}\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\mu }\partial x^{\nu }})\\ =\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\mu }}{\partial {x^{\prime \rho }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\kappa }{}_{\mu \nu }+\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\kappa }}^{}}\frac{{\partial }^2{x}^{\kappa }}{\partial x^{\prime \rho }\partial x^{\prime \sigma }}\end{array}$$

Tensor after Transformation

After transformation, the tensor will give out 4 parts.

$$T^{\prime \lambda }{}_{\alpha \beta ;\gamma }\equiv \underset{\text{Part 1}}{\underbrace{T^{\prime \lambda }{}_{\alpha \beta ,\gamma }}}+\underset{\text{Part 2}}{\underbrace{{\mathrm{\Gamma }}^{\prime \lambda }{}_{\gamma \kappa }{T}^{\prime \kappa }{}_{\alpha \beta }}}-\underset{\text{Part 3}}{\underbrace{{\mathrm{\Gamma }}^{\prime \eta }{}_{\alpha \gamma }{T}^{\prime \lambda }{}_{\eta \beta }}}-\underset{\text{Part 4}}{\underbrace{{\mathrm{\Gamma }}^{\prime ϵ}{}_{\gamma \beta }{T}^{\prime \lambda }{}_{\alpha ϵ}}}$$

To deal with this complex tensor, we’ll analyse it one part after another.

The first part

$$\begin{array}{c}{T}^{\prime \lambda }{}_{\alpha \beta ,\gamma }=\frac{{\partial }^{}}{\partial {x^{\prime \gamma }}^{}}\left(\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\mu }{}_{\rho \sigma }\right)\\ =\frac{{\partial }^2{x}^{\prime \lambda }}{\partial x^{\eta }\partial x^{\tau }}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\tau }{}_{\rho \sigma }\text{(Term 1)}\\ +\frac{{\partial }^2{x}^{\rho }}{\partial x^{\prime \gamma }\partial x^{\prime \alpha }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\mu }{}_{\rho \sigma }\text{(Term 2)}\\ +\frac{{\partial }^2{x}^{\sigma }}{\partial x^{\prime \lambda }\partial x^{\prime \beta }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}{T}^{\mu }{}_{\rho \sigma }\text{(Term 3)}\\ +\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\mu }{}_{\rho \sigma ,\nu }\text{(Term 4)}\end{array}$$

gives out 4 terms, and I’ll show in detail how the first three cancel with the $\mathrm{\Gamma }$ parts’ extra terms.

The second part

$$\begin{array}{c}{\mathrm{\Gamma }}^{\prime \lambda }{}_{\gamma \kappa }{T}^{\prime \kappa }{}_{\alpha \beta }=(\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\tau }}{\partial {x^{\prime \kappa }}^{}}{\mathrm{\Gamma }}^{\mu }{}_{\nu \tau }+\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \kappa }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\zeta }}^{}})\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\tau }{}_{\rho \sigma }\\ =\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\tau }}{\partial {x^{\prime \kappa }}^{}}\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{\mathrm{\Gamma }}^{\mu }{}_{\nu \tau }{T}^{\tau }{}_{\rho \sigma }+\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \kappa }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\tau }{}_{\rho \sigma }\end{array}$$

To let the extra term of the second part cancel with the Term 1 of the first part, we’ll have to prove that

$$\begin{array}{}\text{(1)}& \frac{{\partial }^2{x}^{\prime \lambda }}{\partial x^{\eta }\partial x^{\tau }}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \gamma }}^{}}+\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \kappa }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}=0\end{array}$$

Consider

$$\frac{{\partial }^{}}{\partial {x^{\tau }}^{}}{\delta }^{\prime \lambda }{}_{\gamma }=\frac{{\partial }^{}}{\partial {x^{\tau }}^{}}\left(\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\eta }}^{}}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \gamma }}^{}}\right)=0$$$$\begin{array}{c}\frac{{\partial }^{}}{\partial {x^{\tau }}^{}}\left(\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\eta }}^{}}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \gamma }}^{}}\right)\\ =\frac{{\partial }^2{x}^{\prime \lambda }}{\partial x^{\tau }\partial x^{\eta }}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \gamma }}^{}}+\frac{{\partial }^2{x}^{\eta }}{\partial x^{\prime \gamma }\partial x^{\prime \kappa }}\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\eta }}^{}}=0\end{array}$$

So equation (1) is right. We can see that the second part’s extra term cancels with the first part’s Term 1.

The third part

$${\mathrm{\Gamma }}^{\prime \eta }{}_{\alpha \gamma }{T}^{\prime \lambda }{}_{\eta \beta }=\frac{{\partial }^{}{x}^{\prime \eta }}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \beta }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {x^{\prime \eta }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{\mathrm{\Gamma }}^{\kappa }{}_{\rho \nu }{T}^{\mu }{}_{\kappa \sigma }+\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \alpha }\partial x^{\prime \gamma }}\frac{{\partial }^{}{x}^{\prime \eta }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \eta }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\mu }{}_{\rho \sigma }$$

To let the extra term of the third part cancel with the Term 2 of the first part, we’ll have to prove that

$$\begin{array}{}\text{(2)}& \frac{{\partial }^2{x}^{\rho }}{\partial x^{\prime \gamma }\partial x^{\prime \alpha }}=\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \alpha }\partial x^{\prime \gamma }}\frac{{\partial }^{}{x}^{\prime \eta }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \eta }}^{}}\end{array}$$

Consider

$$\frac{{\partial }^{}{x}^{\prime \eta }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \eta }}^{}}={\delta }^{\rho }{}_{\zeta }$$

Therefore

$$\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \alpha }\partial x^{\prime \gamma }}\frac{{\partial }^{}{x}^{\prime \eta }}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \eta }}^{}}=\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \alpha }\partial x^{\prime \gamma }}{\delta }^{\rho }{}_{\zeta }=\frac{{\partial }^2{x}^{\rho }}{\partial x^{\prime \gamma }\partial x^{\prime \alpha }}$$

So this part’s extra term cancels with the first part’s Term 2.

The fourth part

$${\mathrm{\Gamma }}^{\prime ϵ}{}_{\gamma \beta }{T}^{\prime \lambda }{}_{\alpha ϵ}=\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\prime \beta }}{\partial {x^{\nu }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {x^{\prime ϵ}}^{}}{\mathrm{\Gamma }}^{\kappa }{}_{\sigma \nu }{T}^{\mu }{}_{\rho \kappa }+\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \beta }}\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime ϵ}}^{}}{T}^{\mu }{}_{\rho \sigma }$$

To let the extra term of the fourth part cancel with the Term 3 of the first part, we’ll have to prove that

$$\begin{array}{}\text{(3)}& \frac{{\partial }^2{x}^{\sigma }}{\partial x^{\prime \lambda }\partial x^{\prime \beta }}=\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \beta }}\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime ϵ}}^{}}\end{array}$$

Consider

$$\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime ϵ}}^{}}={\delta }^{\sigma }{}_{\zeta }$$

So

$$\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \beta }}\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\zeta }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime ϵ}}^{}}=\frac{{\partial }^2{x}^{\zeta }}{\partial x^{\prime \gamma }\partial x^{\prime \beta }}{\delta }^{\sigma }{}_{\zeta }=\frac{{\partial }^2{x}^{\sigma }}{\partial x^{\prime \lambda }\partial x^{\prime \beta }}$$

So this part’s extra term cancels with the first part’s Term 3.

In conclusion, the transformation for this tensor could be written as

$$\begin{array}{c}{T}^{\prime \lambda }{}_{\alpha \beta ;\gamma }\equiv \underset{\text{Part 1}}{\underbrace{T^{\prime \lambda }{}_{\alpha \beta ,\gamma }}}+\underset{\text{Part 2}}{\underbrace{{\mathrm{\Gamma }}^{\prime \lambda }{}_{\gamma \kappa }{T}^{\prime \kappa }{}_{\alpha \beta }}}-\underset{\text{Part 3}}{\underbrace{{\mathrm{\Gamma }}^{\prime \eta }{}_{\alpha \gamma }{T}^{\prime \lambda }{}_{\eta \beta }}}-\underset{\text{Part 4}}{\underbrace{{\mathrm{\Gamma }}^{\prime ϵ}{}_{\gamma \beta }{T}^{\prime \lambda }{}_{\alpha ϵ}}}\\ =\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{T}^{\mu }{}_{\rho \sigma ,\nu }+\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\nu }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\tau }}{\partial {x^{\prime \kappa }}^{}}\frac{{\partial }^{}{x}^{\prime \kappa }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \beta }}^{}}{\mathrm{\Gamma }}^{\mu }{}_{\nu \tau }{T}^{\tau }{}_{\rho \sigma }\\ -\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\prime \beta }}{\partial {x^{\nu }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {x^{\prime ϵ}}^{}}{\mathrm{\Gamma }}^{\kappa }{}_{\sigma \nu }{T}^{\mu }{}_{\rho \kappa }-\frac{{\partial }^{}{x}^{\prime ϵ}}{\partial {x^{\kappa }}^{}}\frac{{\partial }^{}{x}^{\sigma }}{\partial {x^{\prime \gamma }}^{}}\frac{{\partial }^{}{x}^{\prime \beta }}{\partial {x^{\nu }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\mu }}^{}}\frac{{\partial }^{}{x}^{\rho }}{\partial {x^{\prime \alpha }}^{}}\frac{{\partial }^{}{x}^{\kappa }}{\partial {x^{\prime ϵ}}^{}}{\mathrm{\Gamma }}^{\kappa }{}_{\sigma \nu }{T}^{\mu }{}_{\rho \kappa }\end{array}$$

is a tensor under general coordinate transformation.

Problem 2

Please show that

$$R^{\rho }{}_{\sigma \mu \nu }={\partial }_{\mu }{\mathrm{\Gamma }}^{\rho }{}_{\nu \sigma }-{\partial }_{\nu }{\mathrm{\Gamma }}^{\rho }{}_{\mu \sigma }+{\mathrm{\Gamma }}^{\rho }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\nu \sigma }-{\mathrm{\Gamma }}^{\rho }{}_{\nu \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\mu \sigma }$$

is a tensor under general coordinate transformation.

Computing the Derivative Term

Define $A_{\mu \nu }={\partial }_{\mu }^{\prime }{\mathrm{\Gamma }}^{\prime \rho }{}_{\nu \sigma }+{\mathrm{\Gamma }}^{\prime \rho }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\prime \lambda }{}_{\nu \sigma }$. We can see that $R^{\prime \rho }{}_{\sigma \mu \nu }=A_{\mu \nu }-A_{\nu \mu }$. Therefore, any term in $A_{\mu \nu }$ that is symmetric under the exchange of $\mu \leftrightarrow \nu$ will automatically cancel out. Using the product rule, the chain rule ${\partial }_{\mu }^{\prime }=\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}{\partial }_{\delta }$, and ${\partial }_{\mu }^{\prime }\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}=\frac{{\partial }^2{x}^{\beta }}{\partial x^{\prime \mu }\partial x^{\prime \nu }}$, we differentiate ${\mathrm{\Gamma }}^{\prime \rho }{}_{\nu \sigma }$

$${\partial }_{\mu }^{\prime }{\mathrm{\Gamma }}^{\prime \rho }{}_{\nu \sigma }={\partial }_{\mu }^{\prime }\left(\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\right)+{\partial }_{\mu }^{\prime }\left(\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}\right)$$

Expanding this gives 6 terms

$$\begin{array}{c}{\partial }_{\mu }^{\prime }{\mathrm{\Gamma }}^{\prime \rho }{}_{\nu \sigma }=\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\prime \rho }}{\partial x^{\delta }\partial x^{\alpha }}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term 1)}\\ +\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^2{x}^{\beta }}{\partial x^{\prime \mu }\partial x^{\prime \nu }}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term 2)}\\ +\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^2{x}^{\gamma }}{\partial x^{\prime \mu }\partial x^{\prime \sigma }}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term 3)}\\ +\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}{\partial }_{\delta }{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term 4)}\\ +\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\prime \rho }}{\partial x^{\delta }\partial x^{\alpha }}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}\text{(Term 5)}\\ +\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^3{x}^{\alpha }}{\partial x^{\prime \mu }\partial x^{\prime \nu }\partial x^{\prime \sigma }}\text{(Term 6)}\end{array}$$

Computing the Product Term

Next, we compute ${\mathrm{\Gamma }}^{\prime \rho }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\prime \lambda }{}_{\nu \sigma }$

$${\mathrm{\Gamma }}^{\prime \rho }{}_{\mu \lambda }=\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \lambda }}^{}}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \eta }+\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^2{x}^{ϵ}}{\partial x^{\prime \mu }\partial x^{\prime \lambda }}$$$${\mathrm{\Gamma }}^{\prime \lambda }{}_{\nu \sigma }=\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }+\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}$$

Multiplying these gives 4 terms. Note that we can collapse $\frac{{\partial }^{}{x}^{\eta }}{\partial {x^{\prime \lambda }}^{}}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\alpha }}^{}}={\delta }_{\alpha }^{\eta }$

$$\begin{array}{c}{\mathrm{\Gamma }}^{\prime \rho }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\prime \lambda }{}_{\nu \sigma }=\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \alpha }{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term A)}\\ +\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \alpha }\text{(Term B)}\\ +\left(\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^2{x}^{ϵ}}{\partial x^{\prime \mu }\partial x^{\prime \lambda }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\alpha }}^{}}\right)\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }\text{(Term C)}\\ +\left(\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^2{x}^{ϵ}}{\partial x^{\prime \mu }\partial x^{\prime \lambda }}\frac{{\partial }^{}{x}^{\prime \lambda }}{\partial {x^{\alpha }}^{}}\right)\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}\text{(Term D)}\end{array}$$

Now, substitute the Lemma into Terms C and D. They become

$$\text{Term C}=-\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\prime \rho }}{\partial x^{\delta }\partial x^{\alpha }}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }$$$$\text{Term D}=-\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\prime \rho }}{\partial x^{\delta }\partial x^{\alpha }}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}$$

Cancellation of Extra Terms

When we sum $A_{\mu \nu }={\partial }_{\mu }^{\prime }{\mathrm{\Gamma }}^{\prime \rho }{}_{\nu \sigma }+{\mathrm{\Gamma }}^{\prime \rho }{}_{\mu \lambda }{\mathrm{\Gamma }}^{\prime \lambda }{}_{\nu \sigma }$, Term 1 cancels perfectly with Term C and Term 5 cancels perfectly with Term D.

The remaining components of $A_{\mu \nu }$ are the pure tensor terms (Term 4 and Term A) and the leftover "extra" non-tensor terms (Term 2, Term 6, Term 3 and Term B). Group the non-tensor terms remaining in $A_{\mu \nu }$

$$S_{\mu \nu }=\underset{\text{Term 2}}{\underbrace{\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^2{x}^{\beta }}{\partial x^{\prime \mu }\partial x^{\prime \nu }}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }}}+\underset{\text{Term 6}}{\underbrace{\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^3{x}^{\alpha }}{\partial x^{\prime \mu }\partial x^{\prime \nu }\partial x^{\prime \sigma }}}}+\underset{\text{Term 3}}{\underbrace{\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^2{x}^{\gamma }}{\partial x^{\prime \mu }\partial x^{\prime \sigma }}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }}}+\underset{\text{Term B}}{\underbrace{\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\alpha }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \alpha }}}$$

Relabel dummy indices in Term B ($ϵ\to \alpha ,\zeta \to \beta ,\alpha \to \gamma$) to make it directly comparable to Term 3. The sum of Term 3 and Term B is

$$\text{Term 3}+\text{Term B}=\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }(\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^2{x}^{\gamma }}{\partial x^{\prime \mu }\partial x^{\prime \sigma }}+\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^2{x}^{\gamma }}{\partial x^{\prime \nu }\partial x^{\prime \sigma }})$$

Look at the expression for $S_{\mu \nu }$. Because mixed partial derivatives commute, $\frac{{\partial }^2{x}^{\beta }}{\partial x^{\prime \mu }\partial x^{\prime \nu }}$ and $\frac{{\partial }^3{x}^{\alpha }}{\partial x^{\prime \mu }\partial x^{\prime \nu }\partial x^{\prime \sigma }}$ are symmetric under $\mu \leftrightarrow \nu$. In our combined (Term 3 + Term B), swapping $\mu \leftrightarrow \nu$ simply swaps the two pieces inside the parenthesis. Therefore, $S_{\mu \nu }$ is completely symmetric.

Antisymmetrizing to obtain the Tensor

Recall that the Riemann tensor is $R^{\prime \rho }{}_{\sigma \mu \nu }=A_{\mu \nu }-A_{\nu \mu }$. When we subtract the $\mu \leftrightarrow \nu$ terms, $S_{\mu \nu }-S_{\nu \mu }=0$. All extra terms vanish and the only survivors are the pure tensor terms (Term 4 and Term A)

$$R^{\prime \rho }{}_{\sigma \mu \nu }=(\text{Term 4}(\mu ,\nu )-\text{Term 4}(\nu ,\mu ))+(\text{Term A}(\mu ,\nu )-\text{Term A}(\nu ,\mu ))$$

Plugging in Term 4

$$\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\partial }_{\delta }{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }-\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\partial }_{\delta }{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }$$

Swap dummy indices $\delta \leftrightarrow \beta$ in the subtracted term

$$\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\alpha }}^{}}\frac{{\partial }^{}{x}^{\delta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}({\partial }_{\delta }{\mathrm{\Gamma }}^{\alpha }{}_{\beta \gamma }-{\partial }_{\beta }{\mathrm{\Gamma }}^{\alpha }{}_{\delta \gamma })$$

Plugging in Term A

$$\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\beta \gamma }-\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}{\mathrm{\Gamma }}^{ϵ}{}_{\zeta \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\beta \gamma }$$

Swap dummy indices $\zeta \leftrightarrow \beta$ in the subtracted term

$$\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{ϵ}}^{}}\frac{{\partial }^{}{x}^{\zeta }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\frac{{\partial }^{}{x}^{\gamma }}{\partial {x^{\prime \sigma }}^{}}({\mathrm{\Gamma }}^{ϵ}{}_{\zeta \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\beta \gamma }-{\mathrm{\Gamma }}^{ϵ}{}_{\beta \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\zeta \gamma })$$

Finally, renaming the global dummy indices to $(ϵ\to \tau ,\zeta \to \alpha ,\beta \to \beta ,\gamma \to \omega )$ to match the right side conventions

$$R^{\prime \rho }{}_{\sigma \mu \nu }=\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\omega }}{\partial {x^{\prime \sigma }}^{}}\frac{{\partial }^{}{x}^{\alpha }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}\underset{R^{\tau }{}_{\omega \alpha \beta }}{\underbrace{({\partial }_{\alpha }{\mathrm{\Gamma }}^{\tau }{}_{\beta \omega }-{\partial }_{\beta }{\mathrm{\Gamma }}^{\tau }{}_{\alpha \omega }+{\mathrm{\Gamma }}^{\tau }{}_{\alpha \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\beta \omega }-{\mathrm{\Gamma }}^{\tau }{}_{\beta \lambda }{\mathrm{\Gamma }}^{\lambda }{}_{\alpha \omega })}}$$

Thus, we are left strictly with the required transformation law

$$R^{\prime \rho }{}_{\sigma \mu \nu }=\frac{{\partial }^{}{x}^{\prime \rho }}{\partial {x^{\tau }}^{}}\frac{{\partial }^{}{x}^{\omega }}{\partial {x^{\prime \sigma }}^{}}\frac{{\partial }^{}{x}^{\alpha }}{\partial {x^{\prime \mu }}^{}}\frac{{\partial }^{}{x}^{\beta }}{\partial {x^{\prime \nu }}^{}}{R}^{\tau }{}_{\omega \alpha \beta }$$

which shows that the Riemann tensor is a tensor.