I'm reading some lecture notes on differential geometry with focus on Newtonian mechanics and applications to fluid mechanics. One theorem claims that the total energy of a system is decreasing. However, if I do the calculations myself, I always end up with the total energy increasing instead, so I am certain that I or the lecture notes are missing a minus sign somewhere, and I hope you can tell me where that missing minus sign is supposed to be.
Definitions (from the lecture notes)
- A configuration space is a Riemannian manifold $(M,\langle\cdot,\cdot\rangle_M)$ with Levi-Civita connection $\nabla$.
- A force field is a covector field $\phi\in\Gamma T^*M=\Omega^1M$. We only consider force fields of the form $\phi=\omega+F$ where $\omega\in\Omega^1M$ is a position-dependent force field and $F\in\Gamma\hom(TM,T^*M)$ is a velocity dependent "force field". In this case the force field $\phi^\gamma$ acting on a curve $\gamma$ in $M$ is defined as $\phi^\gamma:=\omega+F\dot\gamma\in\gamma^*(\Omega^1M)$.
- We can decompose the velocity-dependent component $F$ into a sum $F=B+D$, where $B:=\frac 12(F-F^*)\in\Gamma\hom(TM;T^*M)$ is skew-adjoint under the dual pairing, and therefore induces a 2-form $\beta(X,Y):=(BX)(Y)$ on $M$ called magnetic field, and $D:=\frac12(F+F^*)\in\Gamma\hom(TM;T^*M)$ is self-adjoint under the dual pairing $\langle\,\cdot\mid\cdot\,\rangle\in\Gamma\hom(T^*M,TM;\mathbb R)$, and therefore induces a smooth section $\langle D(\cdot)\mid\cdot\,\rangle\in\Gamma\hom(TM,TM;\mathbb R)$ of symmetric bilinear forms called dissipation tensor. We call $\phi$ physically admissible, if
- $\omega$ is exact, i.e. $\omega=-dU$ for some potential $U\in C^\infty M$,
- $\beta$ is closed, i.e. $d\beta=0$, and
- $D$ is positive semi-definite, i.e. $\langle DX\mid X\rangle=(DX)(X)\geq0$ for all $X\in TM$.
- A smooth curve $\gamma$ in $M$ through a force field $\phi=\omega+F$ is Newtonian, if it satisfies $$\Bigl\langle\frac\nabla{dt}\dot\gamma,\,\cdot\,\Bigr\rangle_M=\phi^\gamma=\omega+F\dot\gamma.$$
- The total energy of a smooth curve $\gamma$ in $M$ through a physically admissible force field $\phi=-dU+F$ is defined as $$E:=\frac12\lvert\dot\gamma\rvert^2+U\circ\gamma.$$
With the definitions in place, here is the (mostly unmodified) theorem and proof from the lecture notes
Theorem and proof
Theorem (First law of thermodynamics). Let $(M,\langle\cdot,\cdot\rangle_M,\nabla)$ be a configuration manifold, let $(dU,B+D)$ be a physically admissible force field, and let $\gamma$ be a Newtonian curve in $M$. Then we have $$\dot E=-\langle D\dot\gamma\mid\dot\gamma\rangle\leq0.$$ Proof. Follows immediately from $$ \begin{align*} \dot E &=\Bigl\langle\frac\nabla{dt}\dot\gamma,\dot\gamma\Bigr\rangle_M+\frac d{dt}(U\circ\gamma) \\[2mm] &=\langle-dU-(B+D)\dot\gamma\mid\dot\gamma\rangle+\langle dU|\dot\gamma\rangle \\[2mm] &=-\langle B\dot\gamma\mid\dot\gamma\rangle-\langle D\dot\gamma\mid\dot\gamma\rangle \\[2mm] &=-\langle D\dot\gamma\mid\dot\gamma\rangle \leq0. \end{align*} $$
My attempt at a proof
We have $$ \begin{align*} \dot E &=\frac12\frac d{dt}\langle\dot\gamma,\dot\gamma\rangle_M+\frac d{dt}(U\circ\gamma) =\langle\nabla_{\dot\gamma}\dot\gamma,\dot\gamma\rangle_M+dU(\dot\gamma) \\[2mm] &=\Bigl\langle\frac\nabla{dt}\dot\gamma,\dot\gamma\Bigr\rangle_M+dU(\dot\gamma), \end{align*} $$ where $$ \Bigl\langle\frac\nabla{dt}\dot\gamma,\,\cdot\,\Bigr\rangle_M =\phi^\gamma =\omega+F\dot\gamma =-dU+B\dot\gamma+D\dot\gamma $$ due to $\gamma$ being Newtonian, so $$ \begin{align*} \dot E &=\langle-dU+B\dot\gamma+D\dot\gamma\mid\dot\gamma\rangle+\langle dU\mid\dot\gamma\rangle =\underbrace{\beta(\dot\gamma,\dot\gamma)}_{=0}+\langle D\dot\gamma\mid\dot\gamma\rangle \geq0. \end{align*} $$
My question
Clearly, my proof is missing a minus sign in front of $F=B+D$. I checked my proof multiple times, and I still don't see why I am missing this crucial minus sign, and where the proof in the lecture notes got this minus sign from. The only possibility I see to get the minus sign into my proof is to change the definition of a force field from $\phi=\omega+F$ to $\phi=\omega-F$. Those two definitions are equivalent for arbitrary $F$, but not in the case of physically admissible force fields, because if $\omega-F$ is physically admissible, then $\omega+F=\omega-(-F)$ has a negative semi-definite dissipation tensor and is therefore not physically admissible (unless $F=0$). Also, the definition $\phi=\omega-F=\omega-B-D$ makes more sense to me as I think of the dissipation tensor $D$ as something that takes away usable energy from the system.
Hopefully, someone with more knowledge in this field can tell me whether I made a mistake in my proof or the definitions are missing a minus sign somewhere to model reality.