Consider the Lagrangian of the form \begin{equation} \mathcal{L} = \frac{1}{2}g_{\beta\gamma}\dot{q}^\beta\dot{q}^\gamma+A_\beta\dot{q}^\beta-V \end{equation} where $g_{\beta\gamma} = g_{\gamma\beta}$ and $A_\beta,V$ are functions of $q^\alpha$ and time only.

I am asked to show that the "metric" $g_{\beta\gamma}$ is invertible and that the RHS of the Euler-Lagrange equation may be written as \begin{equation} -\frac{\delta \mathcal{L}}{\delta q^\alpha} = g_{\alpha\beta}\ddot{q}^\beta+\Gamma_{\alpha\beta\gamma}\dot{q}^\beta\dot{q}^\gamma-F_{\alpha\beta}\dot{q}^\beta +\frac{\partial V}{\partial q^\alpha} + \frac{\partial}{\partial t}\left(g_{\alpha\beta}\dot{q}^\beta+A_\alpha\right) \end{equation} where the (and I'm just copying here) the "Christoffel symbols of the first kind" $\Gamma_{\alpha\beta\gamma}$and the "field strenghts $F_{\alpha\beta}$ are \begin{equation} \Gamma_{\alpha\beta\gamma} = \frac{1}{2}\left(\frac{g_\alpha\beta}{\partial q^\gamma}+\frac{\partial g_{\alpha\gamma}}{\partial q^\beta}+\frac{\partial g_{\beta\gamma}}{\partial q^\alpha}\right), F_{\alpha\beta} = \frac{\partial A_\beta}{\partial q^\alpha}-\frac{\partial A_\alpha}{\partial q^\beta} \end{equation}

The first question is actually really easy: in my course, the study is limited to non-singular Hessian matrices, in the sense that the determinant of $\frac{\partial^2\mathcal{L}}{\partial\dot{q}^\alpha\dot{q}^\beta}$ in none-zero. However, one notices that $g_{\alpha\beta} = \frac{\partial^2\mathcal{L}}{\partial\dot{q}^\alpha\dot{q}^\beta}$, which solves the question. This result is coherent with the solution given by the professor.

Now I am struggling with the second question. Here is how I tackled the challenge.

The Euler-Lagrange equation tells us that $\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}^\alpha}=\frac{\partial\mathcal{L}}{\partial q^\alpha}$. I therefore start by computing each terms in the equation.

RHS is the easiest, so I'll start there.

\begin{equation} \frac{\partial\mathcal{L}}{\partial q^\alpha} = \frac{1}{2}\frac{\partial g_{\beta\gamma}}{\partial q^\alpha}\dot{q}^\beta\dot{q}^\gamma + \frac{A_\beta}{\partial q^\alpha}\dot{q}^\beta-\frac{\partial V}{\partial q^\alpha} \end{equation} The LHS gives us \begin{equation} \frac{\partial\mathcal{L}}{\partial \dot{q}^\alpha} = g_{\beta\gamma}\dot{q}^\beta+A_\beta \quad\Rightarrow\quad\frac{d}{dt}\frac{\partial\mathcal{L}}{\partial\dot{q}^\alpha} = \dot{g}_{\beta\gamma}\dot{q}^\beta+g_{\beta\gamma}\ddot{q}^\beta+\dot{A}_\beta \end{equation}

Using the Euler-Lagrange equation from there won't give me the saught solution: I don't have enough terms. My guess is that I am missing things in the time derivative, but I don't know what. Could someone point me in the right direction?

Also, on a related note: how is this $\mathcal{L}$agrangian "general"?


1 Answer 1


The "$d/dt$" derivative that appears in the Euler-Lagrange equations is a total derivative, not a partial derivative. For example, in addition to its direct dependence on $t$, $A_\beta$ also depends on $q^\mu$, which in turn depends on $t$. This means that the derivative $d A_\beta/dt$ is $$ \frac{d}{dt} A_\beta(q^\mu(t), t) = \frac{\partial A_\beta}{\partial q^\mu} \frac{d q^\mu}{dt} + \frac{\partial A_\beta}{\partial t}. $$ Similar techniques have to be applied to the $g_{\mu \nu} \dot{q}^\nu$ term.

  • $\begingroup$ Obviously! I am going to put that on the account of my lack of sleep lol. Thank you! $\endgroup$
    – Juian
    Feb 15, 2022 at 22:00

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