General Relativity Homework 1

Chasse_neige

Problem 1

The energy-momentum tensor is defined as

$$T^{\alpha \beta }=\sum _n{p}_n^{\alpha }\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{t}^{}}{\delta }^3(x-x_n(t))$$

where $x_n(t)$ and $p_n^{\alpha }(t)$ are the position and four-momentum of the $n$-th particle at the time $t$. Please show that $T^{\alpha \beta }$ is a tensor under the Lorentz transformation.

$T^{\alpha \beta }$ is a tensor is equivalent to $\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{t}^{}}{\delta }^3(x-x_n(t))$ is a vector, which I’ll prove here:

When transfering from frame $S$ to $S^{\prime }$, this part becomes

$$\frac{{\mathrm{d}}^{}{x}^{\prime \beta }}{\mathrm{d}{t^{\prime }}^{}}{\delta }^3(x^{\prime }-x_n^{\prime }(t^{\prime }))$$

Try to expand the 3-dim delta function into 4-dim

$${\delta }^4(x-x_n(\tau ))=\delta (t-t_n(\tau ))\cdot {\delta }^3(x-x_n(\tau ))$$

where $\tau$ is the proper time for the n’th particle. Using the properties of the delta function

$$\delta (t-t_n(\tau ))=\frac{\delta (\tau )}{\left|\frac{{\mathrm{d}}^{}(t-t_n(\tau ))}{\mathrm{d}{\tau }^{}}\right|}=\frac{\delta (\tau )}{\left|\frac{{\mathrm{d}}^{}{t}_n}{\mathrm{d}{\tau }^{}}\right|}$$

Therefore, we can deduce that

$$\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{t}^{}}{\delta }^3(x-x_n(t))=\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{t}^{}}\int {\mathrm{d}}^{}\tau {\delta }^4(x-x_n(\tau ))\left|\frac{{\mathrm{d}}^{}{t}_n}{\mathrm{d}{\tau }^{}}\right|=\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{\tau }^{}}\int {\mathrm{d}}^{}\tau {\delta }^4(x-x_n(\tau ))$$

So we can see that $\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{\tau }^{}}=u^{\beta }$ is a vector and $\int {\mathrm{d}}^{}\tau {\delta }^4(x-x_n(\tau ))$ is a scalar, which means $\frac{{\mathrm{d}}^{}{x}^{\beta }}{\mathrm{d}{t}^{}}{\delta }^3(x-x_n(t))$ is a vector and $T^{\alpha \beta }$ is a tensor.

Problem 2

We use ${\xi }^{\alpha }$ to denote an inertial coordinate system in which a massive particle moves freely. In these coordinates the equation of motion is

$$\frac{{\mathrm{d}}^2{\xi }^{\alpha }}{\mathrm{d}{\tau }^2}=0$$

We now describe the same motion using a general coordinate system $x^{\mu }$. Then, we have

$$0=\frac{{\mathrm{d}}^2{\xi }^{\alpha }}{\mathrm{d}{\tau }^2}=\frac{{\mathrm{d}}^{}}{\mathrm{d}{\tau }^{}}\left(\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\mu }}^{}}\frac{{\mathrm{d}}^{}{x}^{\mu }}{\mathrm{d}{\tau }^{}}\right)$$

Please finish the calculation and show that in the general coordinate system,

$$\frac{{\mathrm{d}}^2{x}^{\mu }}{\mathrm{d}{\tau }^2}+{\mathrm{\Gamma }}^{\mu }{}_{\rho \sigma }\frac{{\mathrm{d}}^{}{x}^{\rho }}{\mathrm{d}{\tau }^{}}\frac{{\mathrm{d}}^{}{x}^{\sigma }}{\mathrm{d}{\tau }^{}}=0$$

where the connection $\mathrm{\Gamma }$ is defined as

$${\mathrm{\Gamma }}^{\mu }{}_{\rho \sigma }=\frac{\partial x^{\mu }}{\partial {\xi }^{\alpha }}\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\rho }\partial x^{\sigma }}$$

(Please note that in the class the definition of $\mathrm{\Gamma }$ missed a minus sign.)

Expanding the formula

$$0=\frac{{\mathrm{d}}^2{\xi }^{\alpha }}{\mathrm{d}{\tau }^2}=\frac{{\mathrm{d}}^{}}{\mathrm{d}{\tau }^{}}\left(\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\mu }}^{}}\frac{{\mathrm{d}}^{}{x}^{\mu }}{\mathrm{d}{\tau }^{}}\right)=\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\rho }\partial x^{\sigma }}\frac{{\mathrm{d}}^{}{x}^{\rho }}{\mathrm{d}{\tau }^{}}\frac{{\mathrm{d}}^{}{x}^{\sigma }}{\mathrm{d}{\tau }^{}}+\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\lambda }}^{}}\frac{{\mathrm{d}}^2{x}^{\lambda }}{\mathrm{d}{\tau }^2}$$

So

$$\frac{{\partial }^{}{x}^{\mu }}{\partial {{\xi }^{\alpha }}^{}}\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\lambda }}^{}}\frac{{\mathrm{d}}^2{x}^{\lambda }}{\mathrm{d}{\tau }^2}+\frac{{\partial }^{}{x}^{\mu }}{\partial {{\xi }^{\alpha }}^{}}\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\rho }\partial x^{\sigma }}\frac{{\mathrm{d}}^{}{x}^{\rho }}{\mathrm{d}{\tau }^{}}\frac{{\mathrm{d}}^{}{x}^{\sigma }}{\mathrm{d}{\tau }^{}}=0$$

Given that

$$\frac{{\partial }^{}{x}^{\mu }}{\partial {{\xi }^{\alpha }}^{}}\frac{{\partial }^{}{\xi }^{\alpha }}{\partial {x^{\lambda }}^{}}={\delta }_{\lambda }^{\mu }$$

Therefore

$$\frac{{\mathrm{d}}^2{x}^{\mu }}{\mathrm{d}{\tau }^2}+{\mathrm{\Gamma }}^{\mu }{}_{\rho \sigma }\frac{{\mathrm{d}}^{}{x}^{\rho }}{\mathrm{d}{\tau }^{}}\frac{{\mathrm{d}}^{}{x}^{\sigma }}{\mathrm{d}{\tau }^{}}=0$$

where ${\mathrm{\Gamma }}^{\mu }{}_{\rho \sigma }=\frac{\partial x^{\mu }}{\partial {\xi }^{\alpha }}\frac{{\partial }^2{\xi }^{\alpha }}{\partial x^{\rho }\partial x^{\sigma }}$.