Graduate The Connection Between Geodesics and the Lagrangian | Explained in Textbook

[Pentaquark5]

I've recently read in a textbook that a geodesic can be defined as the stationary point of the action

\begin{align}
I(\gamma)=\frac{1}{2}\int_a^b \underbrace{g(\dot{\gamma},\dot{\gamma})(s)}_{=:\mathcal{L}(\gamma,\dot{\gamma})} \mathrm{d}s \text{,}
\end{align}

where $\gamma:[a,b]\rightarrow M$ is a differentiable curve. Thus,

\begin{align}
\mathcal{L}(x^\mu ,\dot{x}^\nu)=\frac{1}{2}g_{\alpha \beta}(x^\mu)\dot{x}^\alpha \dot{x}^\beta\text{.}
\end{align}

How exactly does the Lagrangian in $(2)$ follow from $(1)$?

 

[Geometry_dude]

You read it off?

 

[Pentaquark5]

I guess the issue is that I don't know how to calculate $g(\dot{\gamma},\dot{\gamma})(s)$. From what I know $g(X,Y)=g_{ij}\, dx^i(X) \otimes dx^j(Y)$ for $X,Y$ arbitrary vector fields. How does this definition lead to the Lagrangian above?

 

[Geometry_dude]

Now, let me first say that the term under the integral including the one half is precisely the global expression of the Lagrangian in (2).

If you want to be very precise, which is rare among physicists, then the term under the integral should be
$$\left( \gamma^* g \right) _{s} \left( \frac{\partial}{\partial s}, \frac{\partial}{\partial s} \right)
= g _{\gamma \left( s \right) } \left( \gamma_* \left( \frac{\partial}{\partial s} \right) , \gamma_* \left(\frac{\partial}{\partial s} \right) \right)
$$
Here the $*$ as a superscript denotes the pullback and as a subscript it denotes the pushforward. $\frac{\partial}{\partial s}$ is the standard tangent vector (field) on the domain of the curve $\gamma$. By definition one has
$$\dot \gamma _s= \gamma_* \left( \left( \frac{\partial}{\partial s} \right)_s \right) \, \, $$
for all s in the domain of $\gamma$.
If you now choose local coordinates $x$ with $\gamma$ lying in the domain, then you may write
$$\dot \gamma_s = \dot{x}^i\left(s \right) \, \, \left( \frac{\partial}{\partial x^i} \right)_{\gamma \left(s \right)} \, ,$$
where I used Einstein summation. Plugging this into the expression above yields what you were looking for.

 

[Geometry_dude]

Did that help or did it just make things more confusing?

 

[stevendaryl]

Did that help or did it just make things more confusing?

Well, it's hard for me to imagine what type of person, in first learning relativity, would already know what a push-forward is.

 

[martinbn]

Well, it's hard for me to imagine what type of person, in first learning relativity, would already know what a push-forward is.

A mathematician who has not studied general relativity.

 

[strangerep]

I guess the issue is that I don't know how to calculate $g(\dot{\gamma},\dot{\gamma})(s)$. From what I know $g(X,Y)=g_{ij}\, dx^i(X) \otimes dx^j(Y)$ for $X,Y$ arbitrary vector fields. How does this definition lead to the Lagrangian above?

Think of the $dx^i$ as infinitesimal vectors along a curve "$\gamma$" (where $\gamma$ is really a mapping $\gamma : {\mathbb R} \to {\mathbb R}^4$, i.e., a mapping from an arbitrary parameter $s \in {\mathbb R}$ to a point on the manifold, coordinatized in ${\mathbb R}^4$). So, in component language, we denote the curve as $\gamma^i(s)$.

Then we abuse our notation, and express arbitrary curves as $x^i(s)$, and often drop the $(s)$. [Sigh.]

The tangent vector (aka generalized velocity) at any point on the curve is $d\gamma^i(s)/ds$, or $dx^i(s)/ds$ with the moderately-abused notation, or even just $dx^i/ds$ in disgracefully-abused notation. The Lagrangian is then just the square of the magnitude of the velocity vector, parameterized by $s$ along the curve.

The action $I(\gamma)$ is just a line integral of the Lagrangian along a section of the curve $\gamma$.

 

[Pentaquark5]

Thank you all for your help, I believe I understand now!