A Highwayman Astray

At the divergence there had been a sign indicating the abandoned road, yet I took it. A famous passage had streamed through my mind then, Abandon hope all ye who enter here, and I had thought it a warning of sorts, but perished the thought quickly. That was some hours ago, before sunset. The nearly full moon now hangs low over the horizon illuminating the way. What way?

A breeze had picked up a short while ago and the constant rustling of leaves, both in the trees and on the ground, for it is that time of year, have a sedative effect; not even the occasional hooting of an owl seems to rouse my senses. Longing for some sign of civilization, I start imagining rounding each bend in the road and finding myself looking down on some well-lit town in a valley below. I play out scenes in my mind, stopping in at the main tavern for a pint, having lively conversations with one or more of the other patrons, and eventually falling into a comfortable bed at a boarding house. Nearly lost to that simple fiction, the rider atop a horse some distance ahead and blocking the way is seen but not registered immediately.

I take several more strides before stopping in my tracks. My breath catches in my throat and I think to quickly conceal myself using the cover of a nearby tree, but by the time my feet are willing to respond, the horse has begun to slowly bring its master down the road toward me. Feeling no hope of escape, I remain there in the middle of the road, perplexed by the highwayman’s presence.

“You’ll deliver what you have about ye, or I’ll take yer life along with it.”

The pack slung over my shoulder I lower to the ground. The contents of my pockets I place atop it.

“That’s good.”

“If I may be so bold, I’d like to ask a question,” I say.

“Bold indeed. What is it?”

“Is there a town ahead? I’ve no provisions for spending a night in the woods.”

The rider dismounts, eyes me warily, and collects the belongings I’ve surrendered. He answers while securing them in his saddle bags. “Ye’d be better off going back. You’ll only lose more if you don’t quit this road.”

“What more do I have to lose?”

“I don’t know ye, but I’m guessin’ ye don’t really want to know the answer to that question.”

He mounts his horse again and leaves at a gallop in the direction from which I came. Suddenly I’m aware of the stillness all around me. The breeze has stopped and the trees are motionless. What few clouds there were in the sky a short while before have all disappeared.

An owl breaks the silence, and I continue my journey for what seemingly little it is worth.

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Free To Be...

"They're going to take away our freedom,"

says the guy who spends ten hours a day at a job he dislikes bookended by two awful one-hour commutes.

"They're going to take away our freedom,"

says the gal who gave up a career in medicine to marry her college sweetheart and have two kids by the age of twenty-six.

"They're going to take away our freedom,"

says the morbidly obese guy who avoids going upstairs these days because stairs.

"They're going to take away our freedom,"

says the guy who spends three hours each evening arguing online about gun rights.


You keep going on about freedom when all you've ever done is voluntarily give yours away.

Equation of Geodesic From Lagranian Equations of Motion

From Schutz (A First Course in General Relativity, Second Edition) we have:
Eq. 6.7 and Eq. 6.8

$l=\int_{\lambda_{0}}^{\lambda _{1}}\left | g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda}\frac{\mathrm{d} x^{\beta }}{\mathrm{d}\lambda }\right |^{1/2}d\lambda $

$l=\int_{\lambda_{0}}^{\lambda _{1}}\left | \vec{V}\cdot \vec{V}\right |^{1/2}d\lambda $

We'll also need Eq. 6.32 from Schutz:

$\Gamma ^{\alpha }_{\mu \beta }=\frac{1}{2}g^{\alpha \beta }\left ( g_{\beta \mu ,\upsilon } +g_{\beta \upsilon ,\mu }-g_{\mu \upsilon ,\beta }\right )$

We need to end up with Schutz, Eq. 6.51 as our result (geodesic equation):

$\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda } \right )+\Gamma ^{\alpha }_{\mu \beta }\frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }=0$

We'll start with the Euler-Lagrange equations:

$\frac{\partial L}{\partial q_{i}}-\frac{\mathrm{d} }{\mathrm{d} t}\left ( \frac{\partial L}{\partial \dot{q}_{i}} \right )=0$,

where L is the integrand fuction of Eq. 6.7, ${q}_{i}$ are the generalized coordinates of L, and t =$\lambda$

$L=\left | g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }\right |^{1/2}$, and let
$F= g_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }$

switching from Leibniz notation to Newton dot notation... $\frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda }=\dot{x}^{\mu }$ and calculating derivatives:

$\frac{\partial L}{\partial x^{\mu }}=\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu }\dot{x}^{\alpha }\dot{x}^{\beta }$

$\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\partial L}{\partial \dot{x}^{\mu }} \right )=\frac{\mathrm{d} }{\mathrm{d} \lambda }\left [\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta }\left ( \frac{\partial \dot{x}^{\alpha }}{\partial \dot{x}^{\mu }} \dot{x}^{\beta }+\frac{\partial \dot{x}^{\beta }}{\partial \dot{x}^{\mu }}\dot{x}^{\alpha }\right )\right ]$

using $\frac{\partial \dot{x}^{\alpha }}{\partial \dot{x}^{\mu }}=\delta ^{\alpha }_{\mu }$ and making the appropriate substitutions into the Euler-Lagrange equation gives:

$\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu}\dot{x}^{\alpha }\dot{x}^{\beta }-\frac{\mathrm{d} }{\mathrm{d} \lambda }\left [\frac{F}{2\left | F \right |^{3/2}}\left ( g_{\mu \beta } \dot{x}^{\beta }+g_{\alpha \mu }\dot{x}^{\alpha }\right )  \right ]=0$

the derivative on the L.H.S. includes two subterms with a factor of $\frac{\mathrm{d} F}{\mathrm{d} \lambda }$, but since $F=\vec{V}\cdot \vec{V}$, then $\frac{\mathrm{d} F}{\mathrm{d} \lambda }=0$ and we have:

$\frac{F}{2\left | F \right |^{3/2}}*g_{\alpha \beta ,\mu }\dot{x}^{\alpha }\dot{x}^{\beta }-\frac{F}{2\left | F \right |^{3/2}}\left ( g_{\mu \beta ,\alpha } \dot{x}^{\alpha }\dot{x}^{\beta }+g_{\alpha \mu ,\beta }\dot{x}^{\alpha }\dot{x}^{\beta }+g_{\mu \beta }\ddot{x}^{\beta }+g_{\alpha \mu }\ddot{x}^{\alpha }\right )=0$

Almost there... canceling like factors in each term of the L.H.S., using the symmetry of $g_{\alpha \beta }$, and relabeling dummy indices, we end up with:

$g_{\mu \beta ,\alpha }\dot{x}^{\alpha }\dot{x}^{\beta }+g_{\alpha \mu .\beta  }\dot{x}^{\alpha }\dot{x}^{\beta }-g_{\alpha \beta ,\mu  }\dot{x}^{\alpha }\dot{x}^{\beta }+2g_{\alpha \beta }\ddot{x}^{\mu }=0$

Making a substitution into this equation using Schutz Eq. 6.32 and going back to Leibniz notation exclusively for the derivatives gives:

$2g_{\alpha \beta }\Gamma ^{\mu }_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }+2g_{\alpha \beta }\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda } \right )=0$

which obviously becomes:

$\Gamma ^{\mu }_{\alpha \beta }\frac{\mathrm{d} x^{\alpha }}{\mathrm{d} \lambda }\frac{\mathrm{d} x^{\beta }}{\mathrm{d} \lambda }+\frac{\mathrm{d} }{\mathrm{d} \lambda }\left ( \frac{\mathrm{d} x^{\mu }}{\mathrm{d} \lambda } \right )=0$    Q.E.D.