Riemann surface
matlab

Exercise 1
Consider a binary system of gravitating objects of masses $M$ and $m$.

First consider the case in which $m \ll M$ and where the small-mass object is in quasi-circular orbit around the more massive object. Draw the trajectory in two-space and the worldline in a $1+1$ – and in a $2+1$-dimensional spacetime [Hint: use a co-ordinate system centred in $M$ ].

Now let $m=M$ and the binary be in circular orbit around the Newtonian centre of mass of the system. Draw the trajectory in two-space and the worldline in a $1+1$ – and in a $2+1$-dimensional spacetime [Hint: use a co-ordinate system centred in the Newtonian centre of mass].

Figure 1: Trajectories in two-space for the cases $m \ll M$ (left) and $m=M$ (right).

Exercise 2
Consider a two-dimensional space and cover it with two co-ordinate maps: a Cartesian map where $\left{x^\mu\right}=(x, y)$ and a polar map where $\left{x^{\mu^{\prime}}\right}=(r, \theta)$.

Find the co-ordinate transformation $f: x^\mu \rightarrow x^{\mu^{\prime}}$

Find the inverse co-ordinate transformation $f^{-1}: x^{\mu^{\prime}} \rightarrow x^\mu$

Find the components of the transformation matrix $\Lambda_\mu^{\mu^{\prime}}$ and its determinant $J^{\prime}:=\left|\partial x^{\mu^{\prime}} / \partial x^\mu\right|$

Find the components of the inverse transformation matrix $\Lambda_{\mu^{\prime}}^\mu$ and its determinant $J:=\left|\partial x^\mu / \partial x^{\mu^{\prime}}\right|$

Show that $\Lambda_\mu^{\mu^{\prime}} \Lambda_{\nu^{\prime}}^\mu=\delta_{\nu^{\prime}}^{\mu^{\prime}}$ and that $J J^{\prime}=1$

Solution 2
The co-ordinate transformation is given by:
$$f:\left{\begin{array}{l} r=\left(x^2+y^2\right)^{1 / 2} \ \theta=\arctan (y / x) \end{array}\right.$$
The inverse co-ordinate transformation is given by:
$$f^{-1}:\left{\begin{array}{l} x=r \cos \theta \ y=r \sin \theta \end{array}\right.$$
The transformation matrix is given by:
\begin{aligned} \Lambda_\mu^{\mu^{\prime}} & =\frac{\partial x^{\mu^{\prime}}}{\partial x^\mu} \ & =\left(\begin{array}{ll} \partial r / \partial x & \partial r / \partial y \ \partial \theta / \partial x & \partial \theta / \partial y \end{array}\right) \ & =\left(\begin{array}{cc} x\left(x^2+y^2\right)^{-1 / 2} & y\left(x^2+y^2\right)^{-1 / 2} \ -y\left(x^2+y^2\right)^{-1} & x\left(x^2+y^2\right)^{-1} \end{array}\right) \ & \equiv\left(\begin{array}{cc} \cos \theta & \sin \theta \ -\frac{1}{r} \sin \theta & \frac{1}{r} \cos \theta \end{array}\right), \end{aligned}
and its determinant is given by:
\begin{aligned} J^{\prime} & =\left|\partial x^{\mu^{\prime}} / \partial x^\mu\right| \ & =\frac{x^2}{\left(x^2+y^2\right)^{3 / 2}}+\frac{y^2}{\left(x^2+y^2\right)^{3 / 2}} \ & =\left(x^2+y^2\right)^{-1 / 2}, \end{aligned}
3

or alternatively, using equation (4), is given by:
\begin{aligned} J^{\prime} & =\frac{\cos ^2 \theta}{r}+\frac{\sin ^2 \theta}{r} \ & =\frac{1}{r} . \end{aligned}
It is trivial to confirm that both expressions for $J^{\prime}$ are equivalent.
The inverse transformation matrix is given by:
\begin{aligned} \Lambda_{\mu^{\prime}}^\mu & =\frac{\partial x^\mu}{\partial x^{\mu^{\prime}}} \ & =\left(\begin{array}{ll} \partial x / \partial r & \partial x / \partial \theta \ \partial y / \partial r & \partial y / \partial \theta \end{array}\right) \ & =\left(\begin{array}{ll} \cos \theta & -r \sin \theta \ \sin \theta & r \cos \theta \end{array}\right) \ & \equiv\left(\begin{array}{cc} x\left(x^2+y^2\right)^{-1 / 2} & -y \ y\left(x^2+y^2\right)^{-1} & x \end{array}\right) \end{aligned}
and its determinant is given by:
\begin{aligned} J & =\left|\partial x^\mu / \partial x^{\mu^{\prime}}\right| \ & =r \cos ^2 \theta+r \sin ^2 \theta \ & =r \end{aligned}
or alternatively, using equation (8), is given by
\begin{aligned} J & =\frac{x^2}{\left(x^2+y^2\right)^{1 / 2}}+\frac{y^2}{\left(x^2+y^2\right)^{1 / 2}} \ & =\left(x^2+y^2\right)^{1 / 2} \end{aligned}
It is again trivial to confirm that both expressions for $J$ are equivalent.
Matrix multiplication of equations (3) and (8) or equations (4) and (7) yields the identity matrix, confirming the result $\Lambda_\mu^{\mu^{\prime}} \Lambda_{\nu^{\prime}}^\mu=\delta_{\nu^{\prime}}^{\mu^{\prime}}$. It is also straightforward to confirm that $J J^{\prime}=1$ in both co-ordinate systems.

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