Do vector calculus identities prove symmetries in Euler-Lagrange equations?

Question

Let $\mathrm{S}$ be a differentiable manifold, and $T\mathrm{S}$ its tangent bundle. Let $\mathcal{L}:\mathbb{R}\times T\mathrm{S}\rightarrow \mathbb{R}$ be a Lagrangian functional $\mathcal{L}=\mathcal{L}(t,\mathbf{q},\dot{\mathbf{q}})$. Then, the gradient of $\mathcal{L}$ is: $$\nabla\mathcal{L}=(\frac{\partial\mathcal{L}}{\partial t},\frac{\partial\mathcal{L}}{\partial \mathbf{q}},\frac{\partial\mathcal{L}}{\partial \dot{\mathbf{q}}})$$ By the identity $\nabla\times(\nabla\phi)=\mathbf{0}$ we can assume: $$\nabla\times(\nabla\mathcal{L})=\mathbf{0}$$ So, this gives the system of equations: $$\frac{\partial^2\mathcal{L}}{\partial \mathbf{q}\partial\dot{\mathbf{q}}}-\frac{\partial^2\mathcal{L}}{\partial \dot{\mathbf{q}}\partial\mathbf{q}}=0$$ $$\frac{\partial}{\partial \dot{\mathbf{q}}}\frac{\partial\mathcal{L}}{\partial t}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial \dot{\mathbf{q}}}=0$$ $$\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial \mathbf{q}}-\frac{\partial}{\partial \mathbf{q}}\frac{\partial\mathcal{L}}{\partial t}=0$$ The first equation is no surprise, and its a consequence of Schwarz's theorem on symmetries of partial derivatives, but, if we regard the functions $\mathbf{q}=\mathbf{q}(t)$ and $\dot{\mathbf{q}}=\dot{\mathbf{q}}(t)$ as funtions of the parameter $t$, then $\frac{\partial\mathcal{L}}{\partial t}=\frac{\mathrm{d}\mathcal{L}}{\mathrm{d} t}$, so we have, for the second equation: $$\frac{\partial}{\partial \dot{\mathbf{q}}}\frac{\mathrm{d}\mathcal{L}}{\mathrm{d} t}-\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial\mathcal{L}}{\partial \dot{\mathbf{q}}}=0$$ Or $$\frac{\partial}{\partial \dot{\mathbf{q}}}\frac{\mathrm{d}\mathcal{L}}{\mathrm{d} t}=\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial\mathcal{L}}{\partial \dot{\mathbf{q}}}$$ This already seems contradictory... Also, substituing the LHS into Euler-Lagrange equations gives: $$\frac{\partial\mathcal{L}}{\partial \mathbf{q}}-\frac{\partial}{\partial \dot{\mathbf{q}}}\frac{\mathrm{d}\mathcal{L}}{\mathrm{d} t}=0$$ So... there appears to be some sort of symmetry in E-L equations, after all? Is this a valid proof?

Answer 1

Too long for a comment.

The correct Euler-Lagrange equation is $$ \frac{\partial\mathcal{L}}{\partial \mathbf{q}}-\frac{\mathrm{d}}{\mathrm{d} t}\frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}}=0\,. $$ You derived another equation by claiming that $$ \frac{\partial \mathcal{L}}{\partial t}=\frac{\mathrm{d}\mathcal{L}}{\mathrm{d} t} $$ which is wrong. The LHS is the partial derivative of $\mathcal{L}(t,\mathbf{q},\dot{\mathbf{q}})$ w.r.t $t$ only while the RHS is the total derivative of $$ \mathcal{L}(t,\mathbf{q}(t),\dot{\mathbf{q}}(t))\,. $$ Exercise: express that total derivative as a sum of partial derivatives. Hint: chain rule.