Exercise list
Lecture 5
Lecture 6
Exercise 5.1
Given the metric g_{\mu\nu}(X), show that the Euler-Lagrange equation (16) (we drop the “s”), to minimize the action along a trajectory in space-time,
\frac{d}{dt} \frac{\partial \mathcal L}{\partial \dot X^m} = \frac{\partial \mathcal L}{\partial X^m}
where the Lagrangian \mathcal L is
\mathcal L = -m \sqrt{- g_{\mu\nu}(X) \frac{dX^\mu}{dt} \frac{dX^\nu}{dt}}
is equivalent to the definition of a geodesic given by equation (6), which says that the tangent vector to the trajectory in space-time stays constant:
\frac{d^2 X^\mu}{d\tau^2} = - {\Gamma^\mu}_{\sigma\rho}\frac{\mathrm dX^\sigma}{\mathrm d\tau}\frac{dX^\rho}{d\tau}
The Lagrangian is given (we simplify writing g_{\mu\nu}(X) as g_{\mu\nu}):
\mathcal L = -m \sqrt{- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}} \equiv -m \sqrt{F}
The partial derivative with respect to velocity \dot{X}^m is:
\begin{aligned} \frac{\partial \mathcal L}{\partial \dot X^m} &=\frac{\partial}{\partial \dot X^m}\left(-m\sqrt F \right) = \left(\frac{-m}{2 \sqrt F}\right)\frac{\partial F}{\partial \dot X^m} \\ & = -\frac{m}{2 \sqrt F}\frac{\partial }{\partial \dot X^m} \left(- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}\right) \\ & = -\frac{m}{2 \sqrt F}\frac{\partial }{\partial \dot X^m} \left(- g_{\mu\nu} \dot X^\mu \dot X^\nu\right) \\ & = -\frac{m}{2 \sqrt F} (- g_{\mu\nu}) \left(\frac{\partial \dot X^\mu}{\partial \dot X^m} \dot X^\nu + \frac{\partial \dot X^\nu}{\partial \dot X^m} \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}g_{\mu\nu} \left({\delta^\mu}_m \dot X^\nu + {\delta^\nu}_m \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}g_{\mu\nu} \left({\delta^\mu}_m \dot X^\nu + {\delta^\nu}_m \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}\left(g_{m\nu} \dot X^\nu + g_{\mu m} \dot X^\mu \right) \\ & = \frac{m}{2 \sqrt F}\left(2 g_{m\nu} \dot X^\nu \right) \\ & = \frac{m g_{m\nu}}{\sqrt F} \frac{\mathrm dX^\nu}{\mathrm dt} \end{aligned}
From:
\mathrm d\tau^2 = - g_{\mu\nu} \mathrm dX^\mu \mathrm dX^\nu
We have:
\frac{\mathrm d\tau}{\mathrm dt} = \sqrt {\frac{\mathrm d\tau^2}{\mathrm dt^2}} = \sqrt{- g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm dt}} = \sqrt F
Substituting:
\frac{\partial \mathcal L}{\partial \dot X^m} = \frac{m g_{m\nu}}{\sqrt F} \frac{\mathrm dX^\nu}{\mathrm dt} = m g_{m\nu}\frac{\frac{\mathrm dX^\nu}{\mathrm dt}}{\frac{\mathrm d\tau}{\mathrm dt}} = m g_{m\nu}\frac{\mathrm dX^\nu}{\mathrm d\tau}
Now we can compute the full left-hand side of the Euler-Lagrange equation by taking the total derivative with respect to t:
\begin{aligned} \frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot X^m} \right) &= m \frac{\mathrm d}{\mathrm dt} \left( g_{m\nu} \frac{\mathrm dX^\nu}{\mathrm d\tau} \right) \\ &= m \left[ \left( \frac{\mathrm d g_{m\nu}}{\mathrm dt} \right) \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d}{\mathrm dt} \left( \frac{\mathrm dX^\nu}{\mathrm d\tau} \right) \right] \end{aligned}
We can expand these total derivatives using the chain rule, noting that g_{m\nu} depends on position X^\sigma, which depends on t:
\frac{\mathrm d}{\mathrm dt} \left( \frac{\partial \mathcal L}{\partial \dot X^m} \right) = m \left(\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} \frac{\mathrm d\tau}{\mathrm dt} \right)
Next, we compute the right-hand side, \frac{\partial \mathcal L}{\partial X^m}. The dependence on X^m is entirely within the metric tensor:
\begin{aligned} \frac{\partial \mathcal L}{\partial X^m} &= \frac{\partial}{\partial X^m}\left(-m\sqrt F \right) = \left(\frac{-m}{2 F}\right)\frac{\partial \sqrt F}{\partial X^m}\\ & = \left(\frac{m}{2 \sqrt F}\right)\partial_m g_{\mu\nu} \dot{X}^\mu \dot{X}^\nu \end{aligned}
Again, we substitute \sqrt F = \frac{\mathrm d\tau}{\mathrm dt} for the denominator and also re-express the velocities \dot{X}^\mu in terms of proper time (\dot{X}^\mu = \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}):
\begin{aligned} \frac{\partial \mathcal L}{\partial X^m} &= \frac{m}{2 \frac{\mathrm d\tau}{\mathrm dt}} \partial_m g_{\mu\nu} \left( \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \right) \left( \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \right) \\ &= \frac{m}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt} \end{aligned}
Now we equate the left and right sides of the Euler-Lagrange equation and cancel the common factor of m:
\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm dt} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} \frac{\mathrm d\tau}{\mathrm dt} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}
To simplify, we multiply the entire equation by \frac{\mathrm dt}{\mathrm d\tau} and use \frac{\mathrm dX^\sigma}{\mathrm dt} = \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm d\tau}{\mathrm dt}.
\partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} + g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}
Isolating the second derivative term gives:
g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}
We can manipulate the indices:
\begin{aligned} g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} & = \frac{1}{2} \partial_m g_{\mu\nu} \frac{\mathrm dX^\mu}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau}\\ & = \frac{1}{2} \partial_m g_{\sigma\nu} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\nu}{\mathrm d\tau} - \partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}\\ & = \frac{1}{2} \partial_m g_{\sigma\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} - \partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}\\ & = \left( \frac{1}{2} \partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \\ &= \frac{1}{2} \left( \partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} - \partial_\sigma g_{m\rho} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \end{aligned}
We have -\partial_\sigma g_{m\rho} = \frac{1}{2}\left(- \partial_\sigma g_{m\rho} - \partial_\sigma g_{m\rho} \right) that can be written as (just changing the indices):
\partial_\sigma g_{m\rho} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} = \partial_a g_{m b} \frac{\mathrm dX^a}{\mathrm d\tau} \frac{\mathrm dX^b}{\mathrm d\tau} = \partial_\rho g_{m \sigma} \frac{\mathrm dX^\rho}{\mathrm d\tau} \frac{\mathrm dX^\sigma}{\mathrm d\tau} = \partial_\rho g_{m \sigma} \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}
So:
g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{1}{2} \left(\partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} -\partial_\rho g_{m\sigma} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau}
Finally, we multiply by the inverse metric g^{\mu m}:
\begin{aligned} & g^{\mu m} g_{m\nu} \frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = {\delta^\mu}_\nu\frac{\mathrm d^2 X^\nu}{\mathrm d\tau^2} = \frac{\mathrm d^2 X^\mu}{\mathrm d\tau^2} \\ & = \frac{1}{2} g^{\mu m} \left(\partial_m g_{\sigma\rho} - \partial_\sigma g_{m\rho} -\partial_\rho g_{m\sigma} \right) \frac{\mathrm dX^\sigma}{\mathrm d\tau} \frac{\mathrm dX^\rho}{\mathrm d\tau} \end{aligned}
The term on the right involving the metric derivatives is precisely the definition of the Christoffel symbol {\Gamma^\mu}_{\sigma\rho}, with a negative sign:
{\Gamma^\mu}_{\sigma\rho} = \frac{1}{2} g^{\mu m} \left(\partial_\sigma g_{m\rho} + \partial_\rho g_{m\sigma} - \partial_m g_{\sigma\rho} \right)
This leaves us with the final expression:
\frac{\mathrm d^2 X^\mu}{\mathrm d\tau^2} = - {\Gamma^\mu}_{\sigma\rho}\frac{\mathrm dX^\sigma}{\mathrm d\tau}\frac{\mathrm dX^\rho}{\mathrm d\tau}
This result confirms that the path of least action for a free particle in a curved spacetime is the geodesic.
Exercise page 183
Finally, when we use this Lagrangian, the Euler-Lagrange equations will of course simply produce Newton’s equation for a particle in a gravitational field U(X), just as it did when we carried out exactly the same calculations in volume 1.
We arrive at
m \ddot X = -m \frac{\partial U}{\partial X}
The approximated action is:
A = \int \left(mc^2 - mU(X) + \frac{m}{2}\dot X^2\right) \mathrm dt
We can recognize the Lagrangian in the form kinetic energy minus potential energy (with a constant term mc^2 which is constant and therefore is not entering in the equation):
\mathcal L = T - U
We can apply the Euler-Lagrange equation:
\frac{\mathrm d}{\mathrm dt} \left(\frac{\partial \mathcal L}{\partial \dot X^m}\right) = \frac{\partial \mathcal L}{\partial X^m}
For the left hand side, there is only one term that depends from \dot X:
\frac{\mathrm d}{\mathrm dt} \left(\frac{\partial \mathcal L}{\partial \dot X^m}\right) = \frac{\mathrm d}{\mathrm dt} \left(m \dot X\right) = m \ddot X
For the right hand side, there is only one term that depends from X:
\frac{\partial \mathcal L}{\partial X^m} = -m \frac{\partial U(X)}{\partial X}
Putting all together we derive the equation of motions:
m \ddot X = -m \frac{\partial U(X)}{\partial X}
This is the standard Newton equation for a particle in a gravitational field U(X).
Since there is a mass on both side, it cancels out:
\ddot X = -\frac{\partial U(X)}{\partial X}
so the motion does not depends from the mass of the object.
Exercise page 192
In our case it becomes simply \mathcal H = p \dot r - \mathcal L. The calculation are left to the reader. The result is
\mathcal H = \frac{m (1 - 2MG/r)}{\sqrt{(1 - 2MG/r) -\dot r/(1- 2MG/r)}} \tag{36}
[…]
Then equation (36), giving the energy, enable us to express \dot r as a function of that energy E. With some algebra we get
\dot r^2 = \left(1 - \frac{2MG}{r}\right)^2 - \frac{\left(1 - \frac{2MG}{r}\right)^3}{E^2} \tag{37}
Note: there is a typo in equation (36). \dot r should be \dot r^2.
The Lagrangian \mathcal L for this system is:
\mathcal L = -m \sqrt{\left(1 - \frac{2MG}{r}\right) - \left(\frac{1}{1 - \frac{2MG}{r}}\right) \dot r^2} \tag{35}
For easy of writing we set:
h(r) \equiv 1 - \frac{2MG}{r}
So the Lagrangian is:
\mathcal L = -m\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}
The Hamiltonian \mathcal H is defined as function of the generalized coordinates q and \dot q and the Lagrangian as:
\mathcal H = \sum_i p_i \dot q_i - \mathcal L
In our case the generalized coordinates are only r and \dot r so we have:
\mathcal H = p \dot r - \mathcal L
To compute it, the first step is to calculate the generalized conjugate momentum of r:
p = \frac{\partial \mathcal L}{\partial \dot r}
Using the definition of \mathcal L:
\begin{aligned} p & = \frac{\partial \mathcal L}{\partial \dot r} \\ & = \frac{\partial }{\partial \dot r} \left[-m \sqrt{h(r) - \frac{1}{h(r)} \dot r^2} \right] \\ & = -m\frac{1}{2}\frac{ -\frac{1}{h(r)}2\dot r}{\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \\ & = \frac{m\dot r}{h(r)\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \end{aligned}
We can now compute the Hamiltonian:
\begin{aligned} \mathcal H & = \frac{m\dot r^2}{h(r)\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} + m\sqrt{h(r) - \frac{1}{h(r)} \dot r^2} \\ & = \frac{m\dot r^2 + m h(r) \left(h(r) - \frac{1}{h(r)} \dot r^2\right)}{h(r)\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \\ & = \frac{m\dot r^2 + m h(r)^2 - m\dot r^2}{h(r)\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \\ & = \frac{m h(r)^2}{h(r)\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \\ & = \frac{m h(r)}{\sqrt{h(r) - \frac{1}{h(r)} \dot r^2}} \\ & = \frac{m \left(1 - \frac{2MG}{r}\right)}{\sqrt{\left(1 - \frac{2MG}{r}\right) - \frac{1}{\left(1 - \frac{2MG}{r}\right)} \dot r^2}} \end{aligned}
The Hamiltonian \mathcal H represents the total conserved energy of the particle. In the physics of gravitation, it is standard to work with quantities expressed per unit mass. This simplifies the equations by removing the test mass m from the dynamics. We define the specific energy E as the Hamiltonian per unit mass.
E = \frac{\mathcal H}{m}
We can now isolate \dot r^2:
\begin{aligned} & E^2 = \frac{h(r)^2}{h(r) - \frac{1}{h(r)} \dot r^2} \\ & h(r) - \frac{1}{h(r)} \dot r^2 = \frac{h(r)^2}{E^2} \\ & \frac{1}{h(r)} \dot r^2 = h(r) - \frac{h(r)^2}{E^2} \\ & \dot r^2 = h(r)^2 - \frac{h(r)^3}{E^2} \\ & \dot r^2 = \left(1 - \frac{2MG}{r}\right)^2 - \frac{\left(1 - \frac{2MG}{r}\right)^3}{E^2} \end{aligned}
Exercise 5.2
Show that from equation (36) for the energy, and equation (37) for \dot r^2, it follows that
\dot r \approx \sqrt{\frac{r - 2MG}{2MG}} \quad \text{as } r \to 2MG \tag{38}
Note: there is a typo in equation (38). There shouldn’t be the square root \dot r \approx \frac{r - 2MG}{2MG}.
We use the expansion of (1-\frac{A}{x}) about x = A using a Taylor expansion. Let x = A + \varepsilon with \varepsilon \to 0:
1 - \frac{A}{x} = 1 - \frac{A}{A + \varepsilon} = 1 - \frac{A}{A (1 + \frac{\varepsilon}{A})} = 1 - \frac{1}{1 + \frac{\varepsilon}{A}}
We expand 1 / (1 + z) for small z:
\frac{1}{1 + z} \approx 1 - z + z^2 - \cdots
So,
1 - \frac{1}{1 + \frac{\varepsilon}{A}} \approx 1 - [1 - \frac{\varepsilon}{A} + \left(\frac{\varepsilon}{A}\right)^2 - \cdots] = \frac{\varepsilon}{A} - \left(\frac{\varepsilon}{A}\right)^2 + \cdots
Since x = A + \varepsilon,
\varepsilon = x - A
We have:
1 - \frac{A}{x} = \frac{x - A}{A} + \mathcal O((x-A)^2)
In our case then:
1 - \frac{2MG}{r} = \frac{r-2MG}{2MG} + \mathcal O(r - 2MG)^2
Substituting:
\begin{aligned} & \dot r^2 = \left(1 - \frac{2MG}{r}\right)^2 - \frac{\left(1 - \frac{2MG}{r}\right)^3}{E^2} \\ & \dot r^2 = \left(\frac{r-2MG}{2MG}\right)^2 - \frac{\left(\frac{r-2MG}{2MG}\right)^3}{E^2} + \mathcal O(r - 2MG)^4 \\ & \dot r^2 = \left(\frac{r-2MG}{2MG}\right)^2 + \mathcal O(r - 2MG)^3 \\ & \dot r \approx \left(\frac{r-2MG}{2MG}\right)^2 \\ & \dot r \approx \frac{r-2MG}{2MG} \end{aligned}
So we can plot it in the \dot r - r plane.
The radial velocity of the object appear to approach zero linearly as it nears the event horizon.
Exercise 6.1
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